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Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Part 2 Chapter 7 Finding the Unknown Solutions Question Answer helps develop logical thinking and accuracy.

Ganita Prakash Class 7 Part 2 Chapter 7 Solutions

Class 7 Ganita Prakash Part 2 Chapter 7 Solutions

RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 7 Solutions

In-text Questions
Pages 164-166

Question 1.
Find the unknown weights in the following cases :


Solution:

Question 2.
Find the unknown weight of the sack in the following cases. In Fig. 7.10, all the sacks have the same weight.



Solution:
In Fig. 7.9, sack is of 10 kg.
In Fig. 7.10, sack is of 14 kg.
In Fig. 7.11, sack is of 7 kg.
In Fig. 7.12, sack is of 15 kg.

Page 168

Question 1.
The solution to the equation 2n + 1 = 99 is n = 49. Can this equation have any other solution?
Solution:
No, there is an unique solution to any given linear equation.

Question 2.
Try solving 5x – 4 = 7 using trial and error.
Solution:
We can try x = 2,
then LHS = 5 × 2 – 4 = 10-4 = 6 ≠ RHS
Put x = 3,
then LHS = 5 × 3 – 4 = 15-4 = 11 ≠ RHS
Put x = 11/5,
then LHS = 5 × (11/5) – 4 = 11 – 4 = 7

Page 169

Question 1.
It is known that 14593 – 1459 + 145 – 14 + 88 = 13353. To find the value, do we need to evaluate 14593 – 1459 + 145 – 14? No, we can get it by subtracting 88 from 13353.
Why can we do this?
Solution:
We can do this because addition and subtraction are inverse operations. So, the value of the required expression can be found by subtracting 88 from both sides (LHS and RHS), which removes the term 88 and leaves only the expression to be evaluated on the LHS.
14593 – 1459 + 145 – 14 + 88 – 88 = 13353 – 88

Question 2.
It is known that 23 × 41 × 11 × 8 × 7 = 5,80,888. What is the value of the expression 23 × 41 × 11 × 8? Is this the same as dividing both sides by 7, which removes the factor 7 and leaves only the expression to be evaluated on the LHS?
Solution:
Yes, that reasoning is correct.

Page 170

Question 1.
Let us use these ideas to solve the equation 5x – 4 = 7.
Solution:
To retain only 5x on the LHS, we nee’d to remove the term – 4. This can be done by adding 4 to both sides.
Thus, 5x – 4 + 4 = 7 +4.
Hence, 5x = 11.
To retain only the unknown x on the LHS, we need to remove the factor 5. This can be done by dividing both sides by 5.
Thus, \(\frac{5 x}{5}=\frac{11}{5}\)
x = \(\frac {11}{5}\)

Page 173

Question 1.
What happens in cases like \(\frac {u}{15}\) = 6?
Solution:
Multiplying both sides by 15 leaves
u = 6 × 15
u = 90

Page 179

Question 1.
Write equations whose solution is y = 5. Share the equations you made with each other and discuss the methods used.
Solution:
Example 1: y + 3 = 8
Subtract 3 on both sides,
y + 3 – 3 = 8 – 3
y = 5
Example 2: 2y = 10
Divide by 2 on both sides,
2y/2 = 10/2
y = 5

Page 180

Question 1.
Can you form a chain going from the bottom equation to the top?

Solution:
Yes

Question 2.
Without calculating, can you find the value of the unknown in each equation in the chains above?
Solution:
Yes, we can do so.

Pages 181-182

Question 1.
The following are some equations along with the steps used to solve them to find the value of the letter-number. Go through each solution and decide whether the steps are correct If there is a mistake, describe the mistake, correct it and solve the equation.
1. 4x + 6 = 10
4x = 10 + 6
4x = 16
x = 4

2. 7 – 8z = 5
8z = 7 – 5
8z = 2
z = 4

3. 2v – 4 = 6
v – 4 = 6 – 2
v – 4 = 4
v = 8

4. 5z + 2 = 3z – 4
5z + 3z = – 4 + 2
8z = -2
z = – \(\frac {2}{8}\)

5. 15w – 4w = 26
15w = 26 + 4w
15w = 30
w = 2

6. 3 x + 1 = – 12
x + 1 = – \(\frac {12}{3}\)
x + 1 = – 4
x = – 5

7. 4(4q + 2) = 50
4(4q) = 50 – 2
16q = 48
q = 3

8. -2 (3 – 4x) = 14
-6v – 8x = 14
– 8x = 14 + 6
-8x = 20
x = –\(\frac {20}{8}\)

9. 3(7y + 4) = 9 + 5y
7y + 4 = \(\frac {9}{3}\) + 5y
7y + 4 = 3 + 5y
7y – 5y + 4 = 3
2y = 4 – 3
y = \(\frac {1}{2}\)
Solution:
1. Second step is incorrect.
Correct solution is :
4x + 6 = 10
4x = 10 – 6
4x = 4
x = 1

2. Third step is incorrect.
7 – 8z = 5
8z = 7 – 5
8z = 2
z = \(\frac {2}{8}\) = \(\frac {1}{4}\)

3. Second step is incorrect.
2v – 4 = 6
2v = 6 + 4
2v = 10
v = 5

4. Second step is incorrect.
5z + 2 = 3z – 4
5z – 3z = – 4 – 2
2z = – 6
z = \(\frac {-6}{2}\) = -3

5. Third step is incorrect.
15w – 4w = 26
11w = 26
w = \(\frac {26}{11}\)

6. Second step is incorrect.
3x + 1 = – 12
3x = -12 – 1
3x = -13
x = \(\frac {-13}{3}\)

7. Second step is incorrect.
4 (4q + 2) = 50
16q + 8 = 50
16q = 50 – 8
16q = 42
q = \(\frac {42}{16}\) = \(\frac {21}{8}\)

8. Second step is incorrect.
– 2(3 – 4x) = 14
-6 + 8x = 14
8x = 14 + 6
8x = 20
x = \(\frac {20}{8}\) = \(\frac {5}{2}\)

9. Second step is incorrect.
3(7y + 4) = 9 + 5y
21y + 12 = 9 + 5y
21y – 5y = 9 – 12
16y = – 3
y = –\(\frac {3}{16}\)

Page 184

Question 1.
Using this formula can you solve this equation 2x + 3 = 4x + 5?
Solution:
x = \(\frac{5-3}{2-4}=\frac{2}{-2}\) = -1

Class 7 Maths Finding the Unknown Solutions

Figure it Out (Page 172)

Question 1.
Solve these equations and check the solutions.
(a) 3x – 10 = 35
(b) 5s = 3s
(c) 3u – 7 = 2u + 3
(d) 4(m + 6) – 8 = 2m – 4
(e) \(\frac {u}{15}\) = 6
Solution:
(a) 3x – 10 = 35
Adding 10 on both sides, we get
3x – 10 + 10 = 35 + 10
3x = 45
Dividing by 3 on both sides, we get
3x ÷ 3 = 45 ÷ 3
x = 15
Checking : Put x = 15 in LHS, we get
3 × 15 – 10 = 45 – 10 = 35 = RHS
Hence verified.

(b) 5s = 3s
5s – 3s = 0
2s = 0
s = 0
Checking : Put s = 0 in LHS, 5 × 0 = 0
Similarly, put s = 0 in RHS, 3 × 0 = 0
LHS = RHS
Hence verified.

(c) 3u – 7 = 2u + 3
Add 7 on both sides, we get
3u – 7 + 7 = 2u+ 3 + 7
3u = 2u + 10
3u – 2u = 10
u = 10
Checking : LHS : 3 × 10 – 7 = 30 – 7 = 23
RHS : 2 × 10 + 3 = 20 + 3 = 23
Hence verified.

(d) 4 (m + 6) – 8 = 2m – 4
4m + 24 – 8 = 2m – 4
4m + 16 = 2m – 4
Subtracting 16 on both sides, we get
4m + 16 – 16 = 2m – 4 – 16
4m = 2m – 20
4m – 2m = -20
2m = -20
Dividing both sides by 2, we get
m = -10
Checking : Put m = -10 in LHS and RHS,
LHS = 4 (-10 + 6) – 8 = 4 × (-4) – 8 = -16 – 8 = -24
RHS = 2 × (-10) – 4 = -20 – 4 = -24
LHS = RHS
Hence verified.

(e) \(\frac {u}{15}\) = 6
Multiply both sides by 15, we get
\(\frac {u}{15}\) × 5 = 6 × 15
u = 90
Put u = 90 in LHS,
\(\frac {90}{15}\) = 6 = RHS
Hence verified.

Question 2.
Frame an equation that has no solution.
Solution:
An equation has no solution if, after simplifying, it leads to a false statement like 5 = 3
Example : 2x + 3 = 2x + 7

Figure it Out (Page 181)

Question 1.
Write 5 equations whose solution is x = -2.
Solution:
Five equations can be :
(a) x + 2 = 0
(b) 3x + 6 = 0
(c) 2x – 4 = -8
(d) 5x + 10 = 0
(e) -x – 4 = -2

Question 2.
Find the value of each unknown :
(a) 2y = 60
(b) – 8 = 5x – 3
(c) – 53w = -15
(d) 13 – z = 8
(e) k + 8 = 12 – k
(f) 7m = m – 3
(g) 3n = 10 + n
Solution:
(a) 2y = 60
Divide both sides by 2
\(\frac{2 y}{2}=\frac{60}{2}\)
y = 30

(b) – 8 = 5x – 3
-5x = – 3 + 8
– 5x = 5
x = -1

(c) – 53w = -15
w = \(\frac{-15}{-53}=\frac{15}{53}\)

(d) 13 – z = 8
-z = 8 – 13
-z = -5
z = 5

(e) k + 8 = 12 – k
k + k = 12 – 8
2k = 4
k = 2

(f) 7m = m – 3
7m – m = -3
6 m = -3
m = -3/6 = -1/2

(g) 3n = 10 + n
3n – n = 10
2n = 10
n = 10/2 = 5

Question 3.
I am a 3-digit number. My hundred’s digit is 3 less than my ten’s digit. My ten’s digit is 3 less than my unit’s digit. The sum of all the three digits is 15. Who am I?
Solution:
Let unit’s digit = x
Ten’s digit = x – 3
Hundred’s digit = (x – 3) – 3 = x – 6
Sum : x + (x – 3) + (x – 6) = 15
3x – 9 = 15
3x = 24
⇒ x = S
Number : 258 (hundred’s digit = 2, ten’s digit = 5, unit’s digit = 8)

Question 4.
The weight of a brick is 1 kg more than half its weight. What is the weight of the brick?
Solution:
Let weight be w kg.
According to the question,
w = w/2 + 1
2w = w + 2 w = 2 kg

Question 5.
One quarter of a number increased by 9 gives the same number. What is the number?
Solution:
Let number be n.
\(\frac{n}{4}\) + 9 = n
\(\frac{n}{4}\) – n = 9
–\(\frac{3n}{4}\) = -9
n = 12

Question 6.
Given 4k + 1 = 13, find the values of:
(a) 8k+ 2
(b) 4k
(c) k
(d) 4k – 1
(e) – k – 2
Solution:
4k + 1 = 13
4k = 13 – 1
4k = 12
k = 12/4 = 3

(a) 8k + 2
Put k = 3, 8k + 2 = 8 ×, 3 + 2 = 26
(b) 4k
Put k = 3, 4 ×, 3 = 12
(c) k = 3
(d) 4k – 1
Put k = 3
4k – 1 = 4 × 3 – 1 = 12 – 1 = 11
(e) -k – 2
Put k = 3
-k – 2 = -3 – 2 = -5

Figure it Out (Pages 185-189)

Question 1.
Fill in the blanks with integers.
(a) 5 × ___ – 8 = 37
(b) 37 – (33 – ____ ) = 35
(c) – 3 × (-11 + ____ ) = 45
Solution:
(a) Let the number in blank be x
5 × x – 8 = 37
5x – 8 = 37
5x = 37 + 8
5x = 45
x = 9
So, put 9 in the blank.

(b) Let the number in blank be x.
37 – (33 – x) = 35
37 – 33 + x = 35
4 + x = 35
x = 35 – 4 = 31
So, put 31 in the blank.

(c) Let the number in blank be x.
-3 × (-11 + x) = 45
33 – 3x = 45
-3x = 45 – 33
-3x = 12
x = – 4
So, put – 4 in the blank.

Question 2.
Ranju is a daily wage labourer. She earns ₹ 750 a day. Her employer pays her in 50 and 100 rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have?
Solution:
Let number of ₹ 50 notes = x, number of ₹ 100 notes = x
50x + 100x = 750
150x = 750
x = \(\frac {750}{150}\) = 5
∴ She has 5 notes of ₹ 50 and 5 notes of ₹ 100.

Question 3.
In the given picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? Write an equation to describe this problem.

Solution:
Let each blob cover x dots.
Let’s say there are n blobs and some visible dots v
Required equation : nx + v = 25

Question 4.
Here are machines that take an input, perform an operation on it and send out the result as an output.
(a)

Find the inputs in the following cases :

(b)

Find the inputs in the following cases :

Solution:
(a) For first image,
Let input be x.
[(x + 3) × 4] – 5 = 43
(x + 3) × 4 = 48
x + 3 = 12
x =9
For second image,
Let input = x
[(x + 3) × 4] – 5 = 75
(x + 3) × 4 = 80
x + 3 = 20
x = 17

(b) For the first image, let the input number be x.
Then, (x × 3) – (x + 3 ) = 63
3x – x – 3 = 63
2x – 3 = 63
2x = 66
x = 33
Thus, input is 33.
For the second image, let the input number be x.
Then, (x × 3) – ( x + 3) = 227
3x – x – 3 = 227
2x – 3 = 227
2x = 230
x = 115
Thus, input is 115.

Question 5.
What are the inputs to these machines?

Solution:
The required number is : 5 × 3 = 15
15 × 3 =45
In second figure, -11 + 4 = -7
Then, -7 + 4 = -3.

Question 6.
A taxi driver charges a fixed fee of ₹ 800 per day plus ₹ 20 for each kilometer travelled. If the total cost for a taxi ride is ₹ 2200, determine the number of kilometres travelled.
Solution:
Let distance = d km .
800 + 20d = 2200
20d = 2200 – 800
20 d = 1400
d = 70 km

Question 7.
The sum of two numbers is 76. One number is three times the other number. What are the numbers?
Solution:
Let smaller number = x
Larger number – 3x
x + 3x = 76
4x = 76
x = 19
∴ Numbers are 19 and 57.

Question 8.
The figure shows the diagram for a window with a grill. What is the gap between two rods in the grill?

Solution:
Total width = 34 cm, Rod width = 2 cm, End margins = 3 cm each.
Let gap be x cm.
Need to count number of rods and gaps from the diagram.
Assuming 5 rods and 4 gaps :
2(3) + 5(2) + 4g = 34
6 + 10 + 4g = 34
4g = 18
g = 4.5 cm

Question 9.
In a restaurant, a fruit juice costs ₹ 15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹ 600, find the cost of the fruit juice and milkshake.
Solution:
Let milkshake cost be ₹ m.
Fruit juice cost = ₹(m – 15)
4(m – 15) + 7m = 600
4m – 60 + 7m = 600
11m = 660
m = 60
Milkshake = ₹ 60, Fruit juice = ₹ 45

Question 10.
Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.
Solution:
From 28p – 36 = 98 :
28p = 98 + 36 = 134
14p – 19 = \(\frac{28 p}{2}\) – 19
= \(\frac{134}{2}\) – 19
= 67 – 19 =48
28p – 38 = 134 – 38 = 96

Question 11.
The steps to solve three equations are shown below. Identify and correct any mistakes.
(a) 6x + 9 = 66
x + 9 = 11
x = 11 – 9
x = 2
Solution:
The first step in this solution is incorrect.
The correct solution is : 6x + 9 = 66
6x = 66 – 9
6x = 57
x = \(\frac{57}{6}=\frac{19}{2}\)

(b) 14y + 24 = 36
7y + 12 = 18
7y = 6
y = \(\frac{6}{7}\)
Solution:
The solution is correct.

(c) 4x – 5 = 9x + 8
4x = 9x + 8 – 5
4x = 9x + 3
4x – 9x = 3
-5x = 3
x = \(\frac{-5}{3}\)
Solution:
The second step is incorrect, in which when 5 moved to RHS, it does not change its sign.
4x – 5 = 9x + 8
4x = 9x + 8 + 5
4x = 9x + 13
4x – 9x = 13
-5x = 13
x = \(\frac{-13}{5}\)

Question 12.
Find the measures of the angles of these triangles.
Solution:
(a) Sum of all angles in a triangle is 180°.
y+y+ 15+_y + 15 = 180° (opposite sides have equal angle opposite to it)
3y + 30°= 180°
3y = 180° – 30°
3y = 150°
y = 50°
Then, angles of the triangle are 50°, 65° and 65°.

(b) Sum of all angles in a triangle is 180°.
x + x – 10° + x + 10° = 180°
3x = 180°
x = 60°
Then, angles of the triangle are 50°, 60° and 70°.

Question 13.
Write 4 equations whose solution is u = 6.
Solution:
u – 2 = 4
2 u = 12
u + 5 = 11
3u – 10 = 8

Question 14.
The Bakhshali Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second; and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?
Solution:
Let first person get be x.
Second person get be 2x.
Third person = 3(2x) = 6x
Fourth person = 4(6x) = 24x
Total : x + 2x + 6x + 24x = 132
33x = 132
x =4
So, the amount given to the first person is 4.

Question 15.
The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?
Solution:
Let height = h
h = \(\frac {h}{2}\) + 2.5
2h = h + 5
h = 5 meters

Question 16.
Two separate figures are given below. Each figure shows the first few positions in a sequence of arrangements made with sticks. Identify the pattern and answer the following questions for each figure :
(a) How many squares are in position number 11 of the sequence?’
(b) How many sticks are needed to make the arrangement in position number 11 of the sequence?
(c) Can an arrangement in this sequence be made using exactly 85 sticks? If yes, which position number will it correspond to?
(d) Can an arrangement in this sequence be made using exactly 150 sticks? If yes, which position number will it correspond to?

Solution:
(a) In first pattern, there is 1 square increasing in each step. So, at 11^th position there will be 11 squares.
In second pattern, there are 4 squares in 1^st figure, 7 squares in 2^nd figure, 10 squares in 3^rd figure, 13 squares in 4^th figure. They are increasing by 3.
So, at position there will be 34 squares.

(b) In first pattern, number of sticks used are: 6, 9, 12…
So, number of sticks in 11th position will be 36.
In second pattern, number of sticks used are : 13, 22, 31, 40..
So, number of sticks in 11^th position will be 103.

(c) 85 sticks sequence will not occur in pattern 1.
But it will occur in pattern 2 at 9^th term.

(d) In first pattern, using 150 digits the figure will be 49^th.
In second pattern, using 150 digits no figure will be formed.

Question 17.
A number increased by 36 is equal to ten times itself. What is the number?
Solution:
Let number be x.
x + 36 = 10x
36 = 10x – x
36 = 9x
x = 4

Question 18.
Solve these equations :
(a) 5(r + 2) = 10
(b) – 3(u + 2) = 2(u – 1)
(c) 2(7 – 2«) = – 6
(d) 2(x – 4) = – 16
(e) 6(x – 1) = 2(x – 1) – 4
(f) 3 – 7s = 7 – 3s
(g) 2x + 1 = 6 – (2x – 3)
(h) 10 – 5x = 3(x – 4) – 2(x – 7)
Solution:
(a) 5(r + 2) = 10
5r + 10 = 10
5r = 10 – 10
5r = 0
r = 0

(b) -3 = (u + 2) = 2(u – 1)
-3u – 6 = 2u – 2
-3u – 2u = -2 + 6
-5u = 4
u = \(\frac {-4}{5}\)

(c) 2(7 – 2n) = – 6
14 – 4n = -6
-4n = -6 – 14
-4n = -20
n = 5

(d) 2(x – 4) = – 16
2x – 8= -16
2x= -16 + 8
2x = -8
x = \(\frac {-8}{2}\) = -4

(e) 6(x – 1) = 2(x – 1) – 4
6x – 6 = 2x – 2 – 4
6x – 2x = -6 + 6
4x = 0
x = 0

(f) 3 – 7s = 7 – 3s
-7s + 3s = 7 – 3
-4s = 4
s = -1

(g) 2x + 1 = 6 – (2x – 3)
2x + 1 = 6 – 2x + 3
2x + 2x = 9 – 1
4x = 8
x = \(\frac {8}{4}\) = 2

(h) 10 – 5x = 3(x – 4) – 2(x – 7)
10 – 5x = 3x – 12 – 2x + 14
10 – 5x = x + 2
-5x – x = 2 – 10
-6x = – 8
x = \(\frac{-8}{-6}=\frac{4}{3}\)

Question 19.
Solve the equations to find a path from Start to the End. Show your work in the given boxes provided and colour your path as you proceed.

Solution:
Do it yourself

Question 20.
There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?
Solution:
Let children = c, donkeys = d
Equation 1 (heads) : c + d = 28
Equation 2 (feet) : 2c + 4d = 80 (children have 2 feet, donkeys have 4 feet)
From equation 1 : c = 28 – d
Substitute in equation 2 : 2(28 – d) + 4d = 80
56 – 2d + 4d = 80
2d = 24
d = 12
c = 28 – d
c = 28 – 12 = 16
So, there are 12 donkeys and 16 children on the beach.

It’s Puzzle Time
A Magic Trick

Think of any number.
Now multiply it by 2.
Add 10.
Divide by 2.
Now subtract the original number you thought of.
Finally, add 3.

I predict that you now have 8. Am I correct?
Try the trick on your friends and family!
Can you explain why the trick works?
[Hint: Denote the first number thought of by x.]
Can you make your own such tricks?
Solution:
Yes, you are correct! The answer will always be 8, no matter which number is chosen.
Step 1: Translate the steps into algebra
Let’s follow the instructions using the variable x for the original number:

1. Think a number : x
2. Multiply by 2 : 2x
3. Add 10 : 2x + 10
4. Divide by 2 : \(\frac {2x+10}{2}\) = x + 5
5. Subtract the original number : (x + 5) – x = 5
6. Add 3 : 5 + 3 = 8

The trick works because the original number x is cancelled out during the subtraction step. By doubling the number and then dividing by two, you return to the original value, but the constant you added (10) also gets halved (to 5). Since you subtract the original x, you are always left with that halved constant. Adding 3 at the end simply moves that predictable result to 8.
Create your own trick : Do it yourself

Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Part 2 Chapter 6 Constructions and Tilings Solutions Question Answer helps develop logical thinking and accuracy.

Ganita Prakash Class 7 Part 2 Chapter 6 Solutions

Class 7 Ganita Prakash Part 2 Chapter 6 Solutions

RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 6 Solutions

In-text Questions
Page 137

Question 1.
Will the line joining the two points at which the arcs meet, above and below XY always be the perpendicular bisector of XY, i.e., when XY is of any length, and the arcs are drawn using a radius of any length?

Solution:
Yes, it is always true for any line segment and the arcs drawn using a radius of any length.
This can be proved using congruence of triangles.

Question 2.
Which two triangles should be congruent for AB to be the perpendicular bisector of XY (that is, O is the midpoint of XY and AB is perpendicular to XY)?

Solution:
If we show that ∆AOX ≅ ∆AOY, then OX = OY, and ∠AOX = ∠AOY because they are corresponding parts of congruent triangles.
Since, ∠AOX and ∠AOY together form a straight angle, we have ∠AOX + ∠AOY = 180°.
Thus, ∠AOX = ∠AOY = 90°. This establishes that O is the midpoint of XY and AB is perpendicular to XY.

In ∆AOX and ∆AOY, we already know that AX = AY, and AO is common to both triangles. If we can show that ∠XAO = ∠YAO then, by the SAS congruence condition, we can conclude that the triangles are congruent.
To show this, we observe that ∆ABX ≅ ∆ABY. This is so because AX = AY, BX = BY, and AB is common to both the triangles. Thus, we have ∠XAB – ∠YAB, or ∠XAO = ∠YAO because they are corresponding parts of congruent triangles.
Hence, AB is the perpendicular bisector of XY

Pages 138-139

Question 1.
We can have eyes of different shapes.

How do we get these different shapes?
Solution:
We can get the eyes of different shapes by drawing the arcs of different lengths as shown in the figure below.

In this the points C and D are such that CX = CY = DX = DY. An eye of a different shape can be drawn using these points.

Question 2.
Will C and D lie on the perpendicular bisector AB (See figure in above question)?
Solution:
The points C and D are at the same distance from both X and Y. We have just seen that joining any two such points givesthe perpendicular bisector of XY. Since XY has only one perpendicular bisector, which is the line AB, the points C and D must lie on the line AB.

Question 3.
Justify the following statement using the facts that we have established.
“Any point that has the same distance from X and Y lies on the perpendicular bisector of XY.”
Solution:
The perpendicular bisector of a line segment is the set (locus) of all points that are equidistant from the two endpoints of the segment.
Since point P has the same distance from X’ and Y (i.e., PX = PY), it satisfies this property. Therefore, point P must lie on the perpendicular bisector of XY.

Page 141

Question 1.
Can we extend the method of constructing the perpendicular bisector to construct a 90° angle at any point on a line?
Solution:
Yes, we can do so. Draw a line and mark a point O on it. Construct a 90° angle at point O.

• Extend the line on either side of O.
• Using a compass, mark two points X and Y at equal distance from O, so that O is the midpoint of XY.
• The perpendicular bisector of XY will pass through O and is perpendicular to the given line.

Question 2.
In this case, do we need to draw two pairs of intersecting arcs to get the perpendicular bisector of XY?
Solution:
No, we don’t. We already have one point, O, lying on the perpendicular bisector.
We only need to construct one pair of arcs (either above or below the line) from X and Y to get another point on the perpendicular bisector.

Page 143

Question 1.
How do we construct this figure?

Solution:
The supporting lines for this figure will look like a star with 8 equally spaced lines radiating from a center. It is constructed by taking the help of this figure.

Question 2.
What is the angle between two adjacent lines (in the above figure)?
Solution:
We need the angle between every pair of adjacent lines to be equal. Since, 360° is equally divided into 8 parts, every angle is 45°.

Question 3.
How do we construct a 45° angle using only a ruler and a compass?
Solution:
We know how to construct a 90° angle. If we can divide it into two equal parts, or bisect it, then we get a 45° angle.

Page 144

Question 1.
How do we construct these congruent triangles, given the angle?
Solution:
If A and B are marked such that OA = OB, and if C is chosen such that BC = AC, then by the SSS congruence condition, ∆OBC ≅ ∆OAC.
STEPS FOR ANGLE BISECTION

• Mark points A and B such that OA = OB.
  
• Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C
• OC bisects ∠AOB.
  

Page 149

Question 1.
Have you seen this kind of beautiful arch?
Solution:
Examples: Diwan-i-Aam at Red Fort (Delhi), Central Park (New York City).

Question 2.
How did they make these arches?
Solution:
The first step is to be able to draw them on a plane surface such as paper or stone.

Question 3.
Construct this arch shape on a piece of paper.
Solution:
Let us think about the support lines this figure will need.
For symmetry, we should have AB = CD, and
∠BAD = ∠CDA.

Construct equal angles at A and D. Mark B and C such that AB = CD.

Question 4.
How would you construct these support lines?
Solution:
Steps :

1. Draw a horizontal line AD.
2. Construct equal angles at A and D (using angle copying or angle construction methods).
3. Mark B and C such that AB = CD.
4. Use these support lines to construct an arch by drawing arcs from appropriate centres.
5. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

Page 150

Question 1.
What supporting lines will you use to draw this arch?
Solution:
The supporting lines are just two line segments of equal length.

Question 2.
If their midpoints are marked, will you be able to construct a pointed arch?

Solution:

• Yes, mark the midpoints of two equal line segments placed side by side.
• Draw arcs from these midpoints to create the pointed arch shape.

Page 151

Question 1.
How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)?
Solution:
To begin with, try to construct a pentagon and hexagon with equal sidelengths.

Question 2.
Can we break a regular hexagon into smaller pieces that can be constructed?
Solution:
Yes, a regular hexagon can be broken into six congruent equilateral triangles.

• Join the centre of the hexagon to all its vertices.
• This divides the hexagon into six equal triangles.
• Each triangle has all sides equal and can be easily constructed using a compass and ruler.

Page 152

Question 1.
Consider this figure. Will the 70° angle fit into the gap? What is the gap angle ∠AOI?

Solution:
We have : 40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°.
Gap angle = 360° – (40° + 60° + 50° + 30° + 40° + 90°) = 360° – 310° = 50°.
So the 70° angle will not fit the gap (since the gap is only 50°).

Question 2.
In Fig. can you explain why AOD, BOE and COF are straight lines?

Solution:
Since the six equilateral triangles are arranged symmetrically around point O, the opposite vertices form straight lines through O. Each pair of opposite triangles forms a 180° angle at O, making AOD, BOE, and COF straight lines.

Question 3.
Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass.
Solution:
We can construct a regular hexagon more directly if we can construct a 120° angle using a ruler and a compass.

Page 153

Question 1.
Construct a regular hexagon of sidelength 5 cm.
Solution:

1. Draw a line segment AB = 5 cm.
2. At A, construct a 60° angle using the method above.
3. Mark AF = 5 cm on the new line.
4. At B, construct a 60° angle on the other side.
5. Mark BC = 5 cm.
6. Continue this process to construct all six sides, each of 5 cm, with 60° angles between adjacent sides.
7. The six vertices form a regular hexagon.
   

Page 154

Question 1.
How will you construct 30° and 15° angles?
Solution:

• Construct a 60° angle, then bisect it to get 30°.
• Bisect the 30° angle to get 15°.

Question 2.
Construct the following 6-pointed star. Note that it has a rotational symmetry.

Solution:
The steps followed are :

• Construct a regular hexagon.
• Extend alternate sides of the hexagon to form equilateral triangles pointing outward.
• This creates a 6-pointed star.
  

Question 3.
Are the six triangles forming the 6 points of the star — ∆AGH, ∆BHI, ∆CIJ, ∆DJK, ∆ELK, ∆FLG – equilateral? Why?
Solution:
Yes, they are equilateral.

Page 157

Question 1.
Can a 4 × 7 grid be tiled using 2 × 1 tiles?
Solution:
A 4 × 7 grid has 4 × 7 = 28 unit squares.
Each 2 × 1 tile covers 2 unit squares.
Number of tiles needed = 28 ÷ 2 = 14 tiles.
Since 28 is even, a 4 × 7 grid can be tiled using 2 × 1 tiles.

Question 2.
What about a 5 × 7 grid? Complete the justification.
Solution:
A 5 × 7 grid has 5 × 7 = 35 unit squares.
To tile this grid using 2 × 1 tiles, we need each tile to cover exactly 2 unit squares.
Since 35 is odd, it is impossible to tile a 5 × 7 grid using 2 × 1 tiles (because 35 cannot be evenly divided by 2).

Page 158

Question 1.
Is m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the other is odd? If yes, come up with a general strategy to tile it.
Solution:
Yes, if one of m or n is even, then m × n is even.
General strategy : If m is even, use vertical tiles to cover all columns. If n is even, use horizontal tiles to cover all rows.

Question 2.
Is an m × n grid tileable with 2×1 tiles, if both m and n are odd? Give reasons.
Solution:
No, if both m and n are odd, then m × n is odd.
Since each tile covers 2 unit squares, we need an even number of unit squares to tile the grid.
Therefore, a grid with an odd number of unit squares cannot be tiled using 2 × 1 tiles.

Question 3.
Here is a 5 × 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles?

Solution:
A 5 × 3 grid has 15 unit squares. Removing 1 square leaves 14 unit squares (which is even).
However, even though the number of squares is even, the grid may not be tileable depending on which square is removed.
But the figure shown above is tileable by 2 × 1 grid.

Question 4.
Is the following region tileable with 2 × 1 tiles?

Solution:
Yes, it is tileable, as it is shown in the figure below.

Page 159

Question 1.
What about this one?

Solution:
No, it is not tileable with 2 × 1 grid as shown in the figure below.

Question 2.
Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a 5 × 3 grid, makes it non- tileable?
Solution:
No, the above arrangement of tiles cannot be tiled by 2 × 1 grid. We can try out various arrangement to be sure that we cannot tile it.
If we remove a left most unit square from 5 × 3 grid, it will be tileable.

Question 3.
If the plain grid is tileable, is the black-and-white-grid tileable?
Solution:
Yes, if the plain grid can be tiled, the corresponding black-and-white grid can also be tiled.

Question 4.
If the black-and-white grid is tileable, is the plain grid tileable?
Solution:
Yes, if the black-and-white grid can be tiled, the plain grid can also be tiled.

Page 160

Question 1.
Is the black-and-white region in Fig. tileable?

Solution:
Any region tiled with black-and- white-tiles must have an equal number of black tiles and white tiles.
Since the black-and-white region in Fig. has 8 white squares and 6 black squares, it can never be tiled with these tiles.

Question 2.
Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?
Solution:
In a 5 × 3 grid with a chessboard colouring, there are 8 squares of one colour and 7 squares of the other colour.
If we remove a square of the colour that has 8 squares, we get 7 squares of each colour, and the grid can be tiled.
If we remove a square of the colour that has 7 squares, we get 8 squares of one colour and 6 squares of the other, and the grid cannot be tiled.

Class 7 Maths Constructions and Tilings Solutions

Figure it Out (Page 140)

Question 1.
When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer.
[Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector.
[Hint 2 : We can draw the whole line if any two of its points are known ]
Solution:
No, it is not necessary to have the same radius for arcs above and below XY. As long as the arcs from X and Y above XY intersect at a point A, and the arcs from X and Y below XY intersect at a point B, both A and B will be equidistant from X and Y, and thus lie on the perpendicular bisector.
Do construction yourself.

Question 2.
Is it necessary to construct the pairs of arcs above and below XY?
Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.
Solution:
It is not necessary to draw arcs on both sides of XY.
We can draw two pairs of intersecting arcs on the same side of XY using the same radius from X and Y.
Each intersection point is equidistant from X and Y, so both points lie on the perpendicular bisector. Since, a straight line can be drawn using any two points, joining these two points gives the perpendicular bisector of XY.

Question 3.
While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them? Explore this through construction, and then justify your answer.
Solution:
Yes, it is necessary to use the same radii for both arcs (from X and Y) to ensure that the point of intersection is equidistant from X and Y.

Question 4.
Recreate this design using only a ruler and compass-

Solution:
Do construction by yourself.
Use the construction methods you have learned—such as perpendicular bisectors, angle bisectors, and arcs—to recreate the design. After completing it, trace the outer boundary with a ruler or compass using a coloured pencil to highlight the design and distinguish it from the construction lines.

Figure it Out (Page 142)

Question 1.
Justify why AB in the following figure is the perpendicular bisector.

Solution:
When the rope is folded in half and stretched with loops at X and Y, the midpoint of the rope is equidistant from X and Y.
Both points A and B (above and below) are equidistant from X and Y.
Therefore, joining A and B gives the perpendicular bisector of XY.

Question 2.
Can you think of different methods to construct a 90° angle at a given point on a line using a rope?
Solution:
Use the same rope method as above to find the perpendicular bisector, which will form a 90° angle at the given point.

Figure it Out (Pages 144-145 )

Question 1.
Construct at least 4 different angles. Draw their bisectors.
Solution:
One image is shown below. Similar can be constructed.

Question 2.
Construct the 8-petalled figure shown in the following figure.

Solution :

• Start by constructing a 90° angle at a point.
• Bisect it to get 45° angles.
• Construct another 90° angle perpendicular * to the first, then bisect each to create 8 equal angles of 45° each.
• Draw arcs from the center to create the 8-petalled design.
  Do construction by yourself.

Question 3.
In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.

Solution:
Yes, the line OC will still be an angle bisector.
The point C will still be equidistant from A and B, ensuring that ∆OAC ≅ ∆OBC, so OC bisects the angle.
So, 180° – equal angle will also give equal angles.

Question 4.
What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle?
Solution:
Using angle bisection, we can construct angles like 45°, 22.5°, 11.25°, etc. (by successive bisections of 90°).
We can construct 60° (using an equilateral triangle), then bisect to get 30°, 15°, 7.5°, etc.
To construct 65.5°, we would need to construct 60°, + 5.5°, but 5.5° cannot be constructed exactly using ruler and compass alone, so 65.5° cannot be constructed.

Question 5.
Come up with a method to construct the angle bisector using a rope.
Solution:

• Mark two points A and B at equal distances from the vertex O on the two arms of the angle.
• Use a rope with a fixed length to find a point C that is equidistant from A and B (similar to perpendicular bisector method).
• Join O and C to get the angle bisector.

Question 6.
Construct the following figure.

How do we construct the petals so that they are of the maximum possible size within a given square?
Solution:

• Draw a square.
• Draw arcs from each comer of the square with radius equal to half the side of the square.
• The arcs will form petals that touch the midpoints of the sides, creating the maximum possible size.

Figure it Out (Page 147)

Question 1.
Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.

Solution:
Do it yourself

Question 2.
Construct the following figure.

Solution:
Do it yourself

Figure it Out (Page 148)

Question 1.
Construct 4 pairs of parallel lines in different orientations.
Solution:
Do it yourself.

Question 2.
Construct the following figure.

Solution:
Use the construction of parallel lines and copying angles to recreate the given geometric figure.

Figure it Out (Page 151)

Question 1.
Use support lines in the following figure to construct a pointed arch. Make different arches, by changing the radius of the arcs.

Solution:
Change the radius and draw different arcs, do it yourself.

Question 2.
Make your own arch designs.
Solution:
Create your own arch designs using geometric constructions taught above.

Figure it Out (Pages 154-155)

Question 1.
Construct the following figures :

Solution:
All the figures shown above can be made by taking the help of a ruler and a compass.

Question 2.
Optical Illusion : Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

Solution:
Optical illusions often use patterns, angles and symmetry to create visual effects. We can see a triangle inside a triangle in this figure.
Recreate the figure by carefully following the construction steps and observing the patterns.

Question 3.
Construct this figure.
[Hint : Find the angles in this figure ]

Solution:
Analyze the angles in the figure (likely multiples of 60°, 90°, or 45°).
Use angle construction and bisection methods to recreate the figure.

Question 4.
Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P.
[Hint : Find a line segment on / whose perpendicular bisector passes through P ]
Solution:

• From point P, draw two arcs with the same radius that intersect line l at two points, say A and B.
• P is equidistant from A and B.
• Find another point Q (on the opposite side of l from P) that is also equidistant from A and B.
• Join P and Q. The line PQ is the perpendicular bisector of AB and is perpendicular to l.
  

Figure it Out (Page 156)

Question 1.
How can the tangram pieces be rearranged to form each of the following figures?

Solution:
Various shapes can be made through tangram pieces.

No. 1 and 2 : Large right triangles
No. 3 : Parallelogram
No. 4 and 5 : Small right triangles
No. 6 : Square
No. 7 : Medium right triangle
Experiment with different arrangements of the 7 tangram pieces to form the given shapes.

Figure it Out (Page 160)

Question 1.
Are the following tilings possible?

Solution:
Yes, it can be tiled as shown.

Question 2.

Solution:
No it cannot be tiled.

Analyze the specific configuration and use logical reasoning or the black-and-white method to determine if tiling is possible.

Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Part 2 Chapter 5 Connecting the Dots Solutions Question Answer helps develop logical thinking and accuracy.

Ganita Prakash Class 7 Part 2 Chapter 5 Solutions

Class 7 Ganita Prakash Part 2 Chapter 5 Solutions

RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 5 Solutions

In-text Questions
Page 98

Question 1.
Which of the following are statistical questions?
(a) What is the price of a tennis ball in India?
(b) How old are the dogs that live on this street?
(c) What fraction of the students in your class like walking up a hill?
(d) Do you like reading?
(e) Approximately how many bricks are in this wall?
(f) Who was the best bowler in the match yesterday?
(g) What was the rainfall pattern in Barmer last year?
Solution:
(a) Not a statistical question because it expects a single fixed answer rather than data collection and analysis.
(b) Yes, a statistical question because different dogs have different ages and data needs to be collected.
(c) Yes, a statistical question because we need to collect data from students to find the fraction.
(d) Not a statistical question because it expects a simple yes/no answer from one person.
(e) Can be a statistical question if we estimate by measuring and analyzing patterns.
(f) Can be a statistical question if we use data like wickets taken, runs conceded, economy rate, etc.
(g) Yes, a statistical question because it requires collecting and analyzing rainfall data ova- months.

Pages 98-99

Question 1.
The runs scored by Shubman and Yashasvi in a cricket series are given in the table below. Who do you think performed better?

	Match 1	Match 2	Match 3	Match 4
Shubman	0	17	21	90
Yashasvi	67	55	18	35

Solution:
Shreyas says, “Both their perfor¬mances are similar since Yashasvi scored more in the first and second matches, whereas Shubman scored more in the third and fourth”. Vaishnavi says, “I think Shubman performed better because he scored the highest number of run in a match-90!”.
Shreyas says, “No! Yashasvi batted better since the total number of runs he made is 175, while Shubman made only 128”.
Vaishnavi says, “Oh! Also, Yashasvi’s batting is more consistent-the difference between his maximum score and minimum score is lower”.

Question 2.
The table below shows the runs scored by these two players in another series. Who do you think performed better in this series?

	Match 1	Match 2	Match 3	Match 4	Match 5
Shubman	23	07	10	52	18
Yashasvi	26	53	02	–	15

Vaishnavi says, “Here, Shubman performed better since his total is 110 runs, while Yashasvi’s total is 96 runs”. What do you think of Vaishnavi’s statement?
Solution:
Shreyas says, “But Yashasvi made 96 runs in 4 matches and Shubman made 110 runs in 5 matches”.
So, how do we say who performed better? It is often not simple to compare two groups of numbers and clearly say that one is better than the other.

Question 3.
Can a single number act as a representative of a group of numbers?
Solution:
Yes, a single number can act as a representative of a group of numbers. Such a number is called a measure of central tendency.

Page 100

Question 1.
Shreyas and 4 of his friends have collected the following numbers of guavas: 3, 8,10,5, and 4. Parag and 5 of his friends have collected the following numbers of guavas: 5, 4,6,3, 4, and 8. Each group will share their guavas equally amongst themselves. In which group will each member get a bigger share of guavas?
Solution:
To find this out, we first find out how many guavas each group has collected. Then we divide this total by the number of people in the group to get each member’s share. Shreyas’s group has collected 3 + 8 + 10 + 5 + 4 = 30 guavas.
Each member of Shreyas’s group gets 30 ÷ 5 = 6 guavas.
Parag’s group has collected 5 + 4 + 6 + 3 + 4 + 8 = 30 guavas.
Each member of Parag’s group gets 30 ÷ 6 = 5 guavas.
So, the members of Shreyas’s group get 1 more guava each than the members of Parag’s group.

Question 2.
Vaishnavi tracks the number of Hibiscus flowers blooming in her garden each day. The data for the last few days’ is 2, 7, 9, 4, 3. What is the average number of Hibiscus flowers blooming per day in Vaishnavi’s garden?
Solution:
The average
= \(\begin{aligned}
& =\frac{\text { The total number of Hibiscus flowers bloomed }}{\text { Number of days }} \\
& =\frac{2+7+9+4+3}{5}=5 .
\end{aligned}\)

Page 104

Question 1.
Find the average price of onions at Yahapur and Wahapur.
Solution:
Average price in Yahapur
= 458 + 12 = 38.17 rupees per kg
Average price in Wahapur
= 450 – 12 = 37.50 rupees per kg

Page 105

Question 1.
The heights of the family members of Yaangba and Poovizhi are as follows : Yaangba’s family : 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm.
Poovizhi’s family : 170 cm, 173 cm, 165 cm, 118 cm, 175 cm.
Find the average height of each family. Can we say that Yaangba’s family is taller than Poovizhi’s family?
Solution:
Average height of Yaangba’s family
= (169 + 173 + 155 + 165 + 160 + 164) ÷ 6
= 986 ÷ 6 = 164.3 cm
Average height of Poovizhi’s family
= (170 + 173 + 165 + 118 + 175) ÷ 5
= 801 ÷ 5 = 160.2 cm
Although most members in Poovizhi’s family are taller, their family’s average height is less because one child is much younger and not as tall as the rest of the family.
Their average height, 160.2 cm, is less than the heights of 4 out of 5 members.
Here, the average doesn’t seem to represent the data very well.

Page 107

Question 1.
Find the mean and median in Poovizhi’s data without the outlier value 118. What change do you notice?
Solution:
Without 118: Heights are 170, 173, 165, 175
Mean = (170 + 173 + 165 + 175) ÷ 4
= 683 ÷ 4 = 170.75 cm
Median = (170 + 173) ÷ 2 = 171.5 cm
Now the mean and median are close to each other and better represent the data.

Page 115

Question 1.
What is the scale used in this graph?
Solution:
The scale used in this graph is Rs. 10 per unit on y – axis.

Question 2.
Is it now easier to compare month- wise prices in both places?
Solution:
Yes, it is easier, as visual display makes it easier to compare the data carefully and easily.

Page 122

Question 1.

The horizontal line lists the overs starting from 1, and the vertical line indicates the runs scored in each over. The graph shows the number of runs scored per over as a double bar graph- each bar corresponding to a team. Let us call
them the blue team (denoted by blue) and the red team (denoted by red). The scale used for the runs per over is 1 unit = 5 runs. The circles shown on top of the bars indicate that a wicket fell in that over.
Answer the following questions based on the graph :
1. Can we tell who batted first? Who won the match?
2. How many runs did the blue team score in over 12?
3. In which over did the red team score the least number of runs?
4. Is it easy to tell the target set by the team batting first?
Solution:
1. I think blue team batted first. By getting the total of the runs scored by both the teams in each over, we can find out who wins.
2. 15 runs
3. In over 4
4. Not easy, as it is not directly visualised from the graph.

Pages 125-126

Question 1.
Following are the dot plots of heights of boys (in blue) and girls (in orange) of Grades 6, 7 and 8 (in that order) of two different schools. What do you notice? Share your observations.

Solution:
School A :

• Grade 6 : Mean (boys) = 134:8 cm, Mean (girls) = 137.78 cm
• Grade 7 : Mean (boys) = 141.8 cm, Mean (girls) = 141.83 cm
• Grade 8 : Mean (boys) = 149.35 cm, Mean (girls) = 147.81 cm

School B:

• Grade 6 : Mean (boys) = 149.84 cm, Mean (girls) = 150.2 cm
• Grade 7 : Mean (boys) = 156.14 cm,  4Mean (girls) = 155.41 cm
• Grade 8 : Mean (boys) = 156.14 cm, Mean (girls) = 156.83 cm

Observations :

• In School A, girls are taller than boys in Grades 6 and 7, but boys become taller in Grade 8.
• In School B, girls are taller than boys in Grade 6 and Grade 8 but boys are taller than girls in Grade 7.
• School B students are significantly taller than School A students across all grades.

Class 7 Maths Connecting the Dots Solutions

Figure it Out (Pages 101)

Question 1.
Shreyas is playing with a bat and a ball — but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.
Solution:
Total bounces = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40
Number of attempts = 8
Average = 40 = 8
= 5 bounces

Question 2.
Try the activity above on your own. Collect data for 7 or more attempts and find the average.
Solution:
Do it yourself.

Question 3.
Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?
Solution:
Do it yourself.

Question 4.
Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil : 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?
Solution:
Nikhil’s total time = 17 + 18 + 17 + 16 + 19 + 17 + 18 = 122 seconds
Nikhil’s average = 122 ÷ 7 = 17.43 seconds
Sunil’s total time = 20 + 18 + 18 + 17 + 16 + 16 + 17 = 122 seconds
Sunil’s average = 122 ÷ 7 = 17.43 seconds
Both ran at the same average speed.

Question 5.
The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.
Solution:
Total enrolment = 1555 + 1670 + 1750 + 2013 + 2040 + 2126 = 11154
Mean enrolment = 11154 ÷ 6 = 1859 students

Figure it Out (Pages 112-113)

Question 1.
Find the median of onion prices in Yahapur and Wahapur.
Solution:
Yahapur prices : 25, 24, 26, 28, 30, 35, 39, 43, 49, 56, 59, 44
Data in ascending order : 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59
Median = (35 + 39) ÷ 2 = 37 rupees per kg
Wahapur prices : 19, 17, 23, 30, 38, 35, 42, 39, 53, 60, 52, 42
Data in ascending order : 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60
Median = (38 + 39) ÷ 2 = 38.5 rupees per kg

Question 2.
Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, —, 2, 4. Find the mean and median. How would you describe this data?
Solution:
Values (excluding —): 0, 1, 0,4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4
Total = 0 + 1 + 0 + 4 + 8 + 0 + 0 + 2 + 1 + 1 + 5 + 3 + 4 + 0 + 0 + 10 + 25 + 2 + 2 + 4 = 72
Number of values = 20
Mean = 72 ÷ 20 = 3.6 animals
Sorted: 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25
Median = (2 + 2) ÷ 2 = 2 animals
Most students have 0 to 4 animals. The value 3.6 is an outlier. The mean is affected by this outlier, making the median a better representative.

Question 3.
Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height?

Solution:
Total = 50 + 45 + 43 + 52 + 61 + 63 + 46 + 55 + 60 + 55 + 59 + 56 + 56 + 49 + 54 + 65 + 66 + 51 + 44 + 58 + 60 + 54 + 52 + 57 + 61 + 62 + 60 + 60 + 67 = 1611 feet
Number of trees = 29
Mean = 1621 ÷ 29 = 55.89 feet
Data in ascending order : 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55, 56, 56, 57, 58, 59, 60, 60, 60, 60, 61, 61, 62, 63, 65, 66, 67 Median = 15th value = 56 feet
Heights are fairly evenly distributed between
43 and 67 feet, with most trees between 50 and 62 feet.
Trees shorter than average height (55.55): 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55 = 13 trees
Mark the data by yourself on the plot.

Question 4.
The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4.
(a) Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median.
(b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?
Solution:
(a) No, because all values are less than 21. The mean is the average of all values, so it must be less than the maximum value (20.5). The median is the middle value when sorted, so it also cannot be greater than the maximum value.
(b) No, the mean and median always lie within the range of the data (between minimum and maximum).

Question 5.
The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data.

Boys	3.5	4.1	2.6	3.2	3.4	3.8
Girls	4.0	3.1	3.4	3.7	2.5	3.4

Solution:

The dots show the boys and girls is shown the cross.
We can analyse the data, by finding their mean and median.
Boys : Mean = (3.5 +4.1 + 2.6 + 3.2 + 3.4 + 3.8) + 6 = 20.6 – 6 = 3.43 kg
Sorted : 2.6, 3.2, 3.4, 3.5, 3.8, 4.1; Median = (3.4 + 3.5) ÷ 2 = 3.45 kg
Girls : Mean = (4.0 + 3.1 + 3.4 + 3.7 + 2.5 + 3.4) ÷ 6 = 20.1 – 6 =3.35 kg
Sorted : 2.5, 3.1, 3.4, 3.4, 3.7, 4.0; Median = (3.4 + 3.4) ÷ 2 =3.4 kg
Both groups have similar mean and median, indicating newborn boys and girls have comparable weights.

Question 6.
The dot plots of heights of another section of Grade 5 students of the same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures?

Solution:
Whole class :
Mean = 141.21 cm Median = 142.5 cm Boys :
Mean = 142.05 cm Median = 143 cm
Girls :
Mean = 140.14 cm Median = 140 cm
Observations :
In this section, boys are taller than girls on average (opposite of the first section). Heights are more evenly distributed compared to the first section.

Question 7.
The weights of some sumo wrestlers and ballet dancers are : Sumo wrestlers : 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg. Ballet dancers : 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. Approximate^ how many times heavier is a sumo wrestler compared to a ballet dancer?
Solution:
Average weight of sumo wrestlers
\(\begin{aligned}
& =\frac{295.2+250.7+234.1+221.0+200.9}{5} \\
& =\frac{1201.9}{5}=240.38 \mathrm{~kg}
\end{aligned}\)
Average weight of ballet dancers
\(\begin{aligned}
& =\frac{40.3+37.6+38.8+45.5+44.1+48.2}{6} \\
& =\frac{254.5}{6}=42.42 \mathrm{~kg}
\end{aligned}\)
Times heavier = \(\frac{240.38}{42.42}\)
= Approximately 5.7 times

Figure it Out (Pages 122-125)

Question 1.
The following infographic shows the speeds of a few animals in air, on land, and in water. Can we call this graph a bar graph?
(a) What is the scale used in this graph?
(b) What did you find interesting in this infographic? What do you want to explore further?
(c) Identify a pair of creatures where one’s speed is about twice that of the other.
(d) Can we say that a sailfish is about 4 times faster than a humpback whale? Can we say that a sailfish is the fastest aquatic animal in the world?

Solution:
Yes, we can call this a bar graph.
(a) Scale used is 16 km/h per unit.
(b) Do it yourself
(c) Examples — (i) Peregrine falcon and spine tailed swift, (ii) Flying fish and humpback whale, (d) The sailfish travels at about 109 km/h, while the humpback whale swims at around 26 km/h, making the sailfish approximately 4.2 times faster. Based on the given graph, the sailfish appears to be the fastest aquatic animal shown; however, we cannot conclude that it is the fastest in the world without additional data.

Question 2.
Preyashi asked her students ‘If you were to get a super power to become aquatic (water-borne), aerial (air-borne), or spaceborne which one would you choose?’. The responses are shown below. Some chose none. Draw a double-bar graph comparing how both grades chose each option. Choose an appropriate scale.

Grade 5	w, a, a, a, w, n, s, a. n, w, a, a, a, a, a, w, w, s, a, a, n, w, a, a, n
Grade 9	n, w, s, a, s, w, s. s, a, a, w, s, s, a, s, a, n, w, s, s, a, w, a, w, a

Solution:
Count for Grade 5: Air (a) = 13, Water (w) = 6, Space (s) = 2, None (n) = 4
Count for Grade 9 : Air (a) = 8, Water (w) = 6, Space (s) = 9, None (n) = 2

Question 3.
The temperature variation over two days in different months in Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use the scale 1 unit = 4°C. Can you guess which two months these days might belong to?

	12 am	3 am	6 am	9 am	12 pm	3 pm	6 pm	9 pm
Day 1	20°C	18°C	16°C	20°C	26°C	34°C	30°C	24°C
Day 2	37°C	34 °C	30°C	33°C	37°C	43°C	42°C	39°C

Solution:

Day 1 appears to be from a cooler month (like January or February).
Day 2 appears to be from a hotter month (like May or June).

Question 4.
The following clustered-bar graph shows the number of electric vehicles registered in some states every year from 2022 to 2024.

(a) The data (rounded-off to thousands) for the states of Gujarat and Delhi are given in the table below. Mark the corresponding bars on the bar graph. (It is enough if you place the top of the bars between the two appropriate vertical guidelines.)

	2022	2023	2024
Gujarat	69000	89000	78000
Delhi	62000	74000	81000

(b) Notice how the graph is organised, what scale is used, and what patterns the data shows.
(c) How would you describe the change for various states between 2022 and 2024?
(d) Approximately how many more registrations did Assam get in 2023 compared to 2022?
(e) How many times more did the registrations in West Bengal increase from 2022 to 2024?
(f) Is this statement correct-4There were very few new registrations in Uttarakhand in 2023 and 2024, as the increase in the bar lengths is minimal’?
Solution:
(a) Do it yourself.
(b) Scale : 1 cm = 25,000 vehicles registered. It shows that use of electric vehicles is increasing every year.
(c) The trend in increasing in all the states.

(d) In 2022 it is 40,000 (Approx.)
In 2023 it is 60,000 (Approx.)
More registration = 60,000 – 40,000 = 20,000

(e) In 2022 = 11,000 (Approx.)
In 2024 = 44,000 (Approx.) becomes 4 times.

(f) Yes, in the complete bar graph we can observe that Uttarakhand has minimum number of registrations signifying its minimal bar length.

Figure it Out Pages 129-134

Question 1.
The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls

Based on the dot plots, which of the following statements are true?
(a) The data varies more for the boys than for the girls.
(b) The median number of pockets for the boys is more than that for the girls.
(c) The mean number of pockets for the girls is more than that for the boys.
(d) The maximum number of pockets for boys is greater than that for the girls.
Solution:
(a) No, the data varies more for girls than boys.
(b) Yes, the median number of pockets for the boys is more than that for the girls.
(c) Mean for boys is greater than for girls.
(d) Maximum number of pockets for both girls and boys are same.

Question 2.
The following table shows the points scored by each player in four games :

Player	Game 1	Game 2	Game 3	Game 4
A	14	16	10	10
B	0	8	6	4
C	8	11	Did not play	13

Now answer the following questions :
(a) Find the average number of points scored per game by A.
(b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B?
(c) Who is the best performer?
Solution:
(a) Average score of A
\(=\frac{14+16+10+10}{4}=\frac{50}{4}\)
= 12.5

(b) Player C
C played only 3 games (missed Game 3).
Total points by C = 8+ 11 + 13 = 32
To find C’s average, we divide by 3, not 4, because the mean must be calculated using the number of games actually played.
Mean for C = 32 ÷ 3 ≈ 10.67 points per game

Player B
B played all 4 games, even though one score is zero.
Total points by B = 0 + 8 + 6 + 4 = 18
Mean for B = 18 ÷ 4 = 4.5 points per game

(c) Best performer Comparing averages :
A = 12.5 points per game
C ≈ 10.67 points per game
B = 4.5 points per game
Player A is the best performer, having the highest average score per game.

Question 3.
The marks (out of 100) obtained by a group of students in a General Knowledge quiz are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Another group’s scores in the same quiz are 68, 59, 73, 86, 47, 79, 90, 93 and 86. Compare and describe both the groups performance using, mean and median.
Solution:
Given Data
Group 1 – Scores :
85, 76, 90, 85, 39, 48, 56, 95, 81, 75
Group 2 – Scores :
68, 59, 73, 86, 47, 79, 90, 93, 86
Find Mean Group 1
Sum = 85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75
= 730
Number of students = 10
Mean = 730 – 10 = 73
Group 2
Sum = 68 + 59 + 73 + 86 + 47 + 79 + 90 + 93 + 86
= 681
Number of students = 9
Mean = 681 ÷ 9 ≈ 75.67
Find Median
Group 1 – Arrange scores in ascending order:
39, 48, 56, 75, 76, 81, 85, 85, 90, 95
Since there are 10 observations (even number):
Median = average of 5th and 6th values
= (76 + 81) ÷ 2 = 78.5
Group 2 – Arrange scores in ascending order:
47, 59, 68, 73, 79, 86, 86, 90, 93
Since there are 9 observations (odd number):
Median = 5th value = 79
Both the mean and median of Group 2 are higher than those of Group 1. This shows that Group 2 performed better overall in the quiz.

Question 4.
Consider this data collected from a survey of a colony.

Favourite Sports	Cricket	Basket Ball	Swimming	Hockey	Athletics
Watching	1240	470	510	430	250
Participating	620	320	320	250	105

Choose an appropriate scale and draw a double-bar graph. Write down your observations. Solution:
Solution:

Question 5.
Consider a group of 17 students with the following heights (in cm) :
106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101.
The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with height less than a particular height and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the ‘Telling Tall Tales’ section?
Solution:
The 17 students could not be divided into two equal groups because the groups can be formed of 8 or 9, not equal. For doing so, median is the best measure.
Sorted heights (cm):
101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125
One group can have first 6 students and second group can have next 8 students.
So a reasonable guess is that these students are around 5-7 years old (most likely about 6 years).

Question 6.
Describe the mean and median of heights of your class. You can visualise the heights on a dot plot.
Solution:
Do it yourself.

Question 7.
There are two 7th grade sections at a school. Each section has 15 boys and 15 girls. In one section, the mean height of students is 154.2 cm.
From this information, what must be true about the mean height of students in the other section?
(a) The mean height of students in the other section is 154.2 cm.
(b) The mean height of students in the other section is less than 154.2 cm.
(c) The mean height of students in the other section is more than 154.2 cm.
(d) The mean height of students in the other section cannot be determined.
Solution:
(d) The mean height of students in the other section cannot be determined.
No information is given about :

• The heights of boys and girls individually, or
• Whether the students in the two sections are similar in height.

Because the actual heights of students in the second section may be :

• equal to those in the first section,
• generally shorter, or
• generally taller,
• the mean height of the second section could be equal to, less than, or greater than 154.2 cm.

So, from the given information, it is not possible to determine the mean height of the other section

Question 8.
Standing tall in the storm.

(a) Write estimated values for the number of skyscrapers in New York, Tokyo, and London.
(b) Are the following statements valid?
(i) Only 12 cities have more skyscrapers than Mumbai.
(ii) Only 7 cities have fewer skyscrapers than Mumbai.
(iii) The tallest building in the world is in Hong Kong.
Solution:
(a) New York : 309, Tokyo : 168, London : 25 (b) (i) True
(ii) True
(iii) Can’t be said.

Question 9.
Estimate and then measure the objects listed in the following table. Draw a double bar graph based on the data. How accurate were your estimates? Find the average difference between the estimated and measured values.

Object	Estimate (in cm)	Measure (in cm)	Positive Difference
Length of a pen
Length of an eraser
Length of your palm
Length of your geometry box
Length of your math notebook

Solution:
Do it yourself.

Question 10.
Aditi likes solving puzzles. She recently started attempting the ‘Easy’ level Sudoku puzzles. The time she took (in seconds) to solve these puzzles are— 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, 280, 270, 230, 220,
240. The first nine values correspond to Week 1 and the rest to Week 2.
(a) Construct a dot plot below showing the data for both weeks.
(b) Describe the mean, median, and any observations you may have about the data.

Solution:
Given Data
Week 1 (first 9 values, in seconds) :
410, 400, 370, 340, 360, 400, 320, 330, 310
Week 2 (next 8 values, in seconds) :
320, 290, 380, 280, 270, 230, 220, 240
(a) Do it yourself.
(b) Mean, Median, and Observations
Week 1 – Calculations
Sorted data :
310, 320, 330, 340, 360, 370, 400, 400, 410
Mean
Total = 3240
Mean 3240 ÷ 9 = 360 s
Median
Middle value (5th of 9) = 360 s
Week 2 – Calculations
Sorted data :
220, 230, 240, 270, 280, 290, 320, 380 Mean
Total = 2230
Mean – 2230 ÷ 8 ≈ 278.8 s
Median
Average of 4th and 5th values = (270 + 280) ÷ 2 = 275 s
Aditi showed significant improvement from Week 1 to Week 2, as reflected by lower average and median solving times.

Question 11.
Individual Project: Pick at least one of the following :
(a) How Long is a Sentence? Pick any two textbooks from different subjects. Choose any page with a lot of text from each book.
(i) Use a dot plot to describe how many words the sentences have on each page.
(ii) Compare the data of both the pages using mean and median.
(b) What is in a Name? Write down the names of all of your classmates. The following are some interesting things you can do with this data!
(i) Find the mean and median name length (number of letters in a name).
(ii) Visualise the data and describe its variability and central tendency.
(iii) Which starting letters are more popular? Which are less popular?
(iv) What is the median starting letter? What does this say about the number of names starting with the letters A – M and N – Z?
(v) Plot a double-bar graph showing the number of boys’ names and girls’ names that:

• start and end with vowels,
• start with vowels and end with consonants,
• start with consonants and end with vowels,
• start and end with consonants.

Solution:
Do it yourself.

Question 12.
Individual project (long term): This requires collecting data over 2 weeks or more.
In and Out : Track how many times you step out of your house in a day. Do this for a month.
(i) Describe the variability and central tendency of this data. Make a dot plot.
(ii) Do you find anything interesting about this data? Share your observations.
(iii) You can ask any of your family members or friends to do this as well.
Solution:
Do it yourself.

Question 13.
Small-group project: Pick at least one of the following. Make groups of 8 to 10. Collect data individually as needed. Put together everyone’s data and do the appropriate analysis and visualisation.
(a) Our heights vs. our family’s heights: Collect the heights of your family members.
(i) Make a dot plot showing heights of just your family members. Describe its variability and central tendency.
(ii) Make a double-bar graph showing each student’s height next to their family’s mean height.
(iii) Look at everyone’s data and share your observations.

(b) Estimating time: Check the time and close your eyes. Open them when you think 1 minute has passed (no counting). Note down after how many seconds you opened your eyes. Collect this data for yourself and for your family members. Repeat this activity to estimate 3 minutes.
(i) Make two dot plots (for 1 minute and 3 minutes) showing estimates of just your family members.
(ii) Mark these on the respective dot plots. Describe its variability and central tendency.
(iii) Make a double bar graph showing each family’s mean 1 minute estimate and mean 3 minute estimate.
(iv) Look at everyone’s data and share your observations.
Solution:
Do it yourself.

It’s Puzzle Time
Connect the Dots

A number lock has a 3-digit code. Find the code using the hints below.

Solution:
The correct digits can be 1, 4 and 5.
5 is rightly placed at ones place.
So, possible combination can be : 415.

Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Part 2 Chapter 4 Another Peek Beyond the Point Solutions Question Answer helps develop logical thinking and accuracy.

Ganita Prakash Class 7 Part 2 Chapter 4 Solutions

Class 7 Ganita Prakash Part 2 Chapter 4 Solutions

RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 4 Solutions

In-text Questions
Page 67

Question 1.
Jonali and Pallabi play a game. Jonali says a fraction and Pallabi gives the equivalent decimal. Write Pallabi’s answer in the blank spaces.

Solution:

Fractions	Decimals
3/10	0.3
4/100	0.04
67/1000	0.067
457/100	4.57
71/100	0.71
43/100	0.43
9/100	0.09

Question 2.
Jonali goes to the market to buy spices. She purchases 50 g of Cinnamon, 100 g of Cumin seeds, 25 g of Cardamom and 250 g of Pepper. Express each of the quantities in kilograms by writing them in terms of fractions as well as decimals. The fractions Jonali gave Pallabi have denominators 10, 100, 1000, and so on.
Solution:

Quahtitity	Fraction	Decimal
50 g	50/1000 = 5/100 Kg.	0.05 Kg,
100 g	100/1000 = 1/10 Kg.	0.1 Kg.
25 g	25/1000 Kg.	0.025 Kg.
250 g	250/1000 = 25/100 Kg.	0.25 Kg.

Page 68

Question 1.
Write the following fractions as a sum of fractions and also as decimals:

Fraction	Expanding the Numerator	Sum of one-tenths, one-hundredths, one-thousands,….	Decimals
\(\frac {254}{1000}\)	\(\begin{aligned} & \frac{200}{1000}+\frac{50}{1000}+\frac{4}{1000} \\ & =\frac{2}{10}+\frac{5}{100}+\frac{4}{1000} \end{aligned}\)	0.2 + 0.05 + 0.004	0.254
\(\frac {847}{10000}\)
\(\frac {173}{100}\)
\(\frac {23}{1000}\)

Solution:

Fraction	Expanding the Numerator	Sum of one-tenths, one-hundredths, one-thousands,….	Decimals
\(\frac {254}{1000}\)	\(\begin{aligned} & \frac{200}{1000}+\frac{50}{1000}+\frac{4}{1000} \\ & =\frac{2}{10}+\frac{5}{100}+\frac{4}{1000} \end{aligned}\)	0.2 + 0.05 + 0.004	0.254
\(\frac {847}{10000}\)	800/10000+ 40/10000+7/10000= 8/100+4/1000+ 7/ 10000	0.08 + 0.004 + 0.0007	0.0847
\(\frac {173}{100}\)	100/100 + 70/100 + 3/100 = 1 + 7/10 + 3/100	1 + 0.7 + 0.03	1.73
\(\frac {23}{1000}\)	20/1000 + 3/1000 = 2/100 + 3/1000	0.02 + 0.003	0.023

Page 70

Question 1.
Can the product of two decimals be a natural number?
Solution:
Yes, e g., 2.5 × 0.4 = 1

Question 2.
Can the product of a decimal and a natural number be a natural number?
Solution:
Yes, e g., 0.5 × 2 = 1.

Page 71

Question 1.
Suppose we know that 596 × 248 = 147808, can you immediately write down the product of 5.96 × 24.8?
Solution:
5.96 has 2 decimal places and 24.8 has 1 decimal place.
Total decimal places = 2 + 1
= 3 decimal places.
596 × 248 = 147808
Therefore, 5.96 × 24.8 = 147.808

Page 75

Question 1.
What is 0.039 m in centimetres and millimetres?
Solution:
0.039 m = 3.9 cm = 39 mm
Since, 1 m = 100 cm and 1 cm = 10 mm

Question 2.
Complete the following table.

Decimals	÷ 10	÷ 100	÷ 1000	÷ 10000
18.7	1.87	0.187	0.0187	0.00187
21.1
0.13
		2.146
				0.0058

Solution:

Decimals	÷ 10	÷ 100	÷ 1000	÷ 10000
18.7	1.87	0.187	0.0187	0.00187
21.1	2.11	0.211	0.0211	0.00211
0.13	0.013	0.0013	0.00013	0.000013
214.6	21.46	2.146	0.2146	0.02146
58	5.8	0.58	0.058	0.0058

Question 3.
How do we convert \(\frac {1}{2}\) into a decimal?
Solution:
It is easy to express a fraction as a decimal if the denominator is 1, 10, 100, 1000, etc.
\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\)(multiplying the numerator and denominator by 5).
We know that the fraction 5/10 can be represented as a decimal 0.5.

Page 76

Question 1.
Now, what if the ribbon of length 29 m was shared between four friends instead of 2?
Solution:
So, each will get 29 ÷ 4 m, that is, \(\frac {29}{4}\) m
Now, the denominator of the fraction is 4. To convert a fraction to a decimal, it helps if the denominator is of the form 1, 10, 100, 1000, and so on.
So we can get an equivalent fraction of \(\frac {29}{4}\) by multiplying the numerator and denominator by 25.
\(\frac{29 \times 25}{4 \times 25}=\frac{725}{100}=7.25\)
So, each of the 4 friends will get 7.25 m of ribbon.

Page 85

Question 1.
Can you find the quotients of 10 ÷ 9, and 100 ÷ 11?
Solution:
10 ÷ 9 = 1.111… (never ending)
100 ÷ 11 = 9.0909… (never ending)

Question 2.
Now divide 1 by 7 (1 ÷ 7). Will this end?
Solution:

Note all the remainders we get. It starts with 1, then 3, then 2, then 6, and so on. Let us represent this as a chain.

Not only do the remainders repeat in a cycle, the digits of the quotient also repeat in a cycle! 0.142857 142857 14…
This division never ends.

Pages 85-86

Question 1.
Let us consider the number 142857 that arose when dividing 1 by 7. Multiply 142857 by numbers from 1 to 6.
What are the products? What do you notice?
Solution:
142857 × 1 = 142857
142857 × 2 = 285714
142857 × 3 = 428571
142857 × 4 = 571428
142857 × 5 = 714285
142857 × 6 = 857142
Observation : You get the same number back, but with the digits cycled around!

Question 2.
Multiply 142857 by 7. What do you observe?
Solution:
142857 × 7 = 999999. We get the repeating blocks.

Question 3.
Are there other such numbers?
Solution:
Yes, to find one such number, you can find 1 ÷ 17 in decimal, and use the repeating block of digits.
1 ÷ 17 = 0.0588235294117647… (repeating block)
The repeating block is 0588235294117647.

Page 87

Question 1.

What pattern do you observe? Why are 2 and 5 related in this way?
Solution:
\(\frac{1}{2 \times 2 \times 2 \times 2 \times 2}=0.03125\)
and \(\frac{1}{5 \times 5 \times 5 \times 5 \times 5}=0.00032\)

• To find the first value we calculate 2 × 2 × 2 × 2 × 2 = 32, then 1 ÷ 32 = 0.03125
• To find the second value we calculate 5 × 5 × 5 × 5 × 5 = 3125 then 1 ÷ 3125 = 0.00032

We can see the pattern that number of decimal places in the result is equal to the exponent of the denominator.
When a fraction’s denominator contains only powers of 2 or 5, it can be easily written as a terminating decimal. This is because 2 and 5 are the prime factors of 10 and the decimal number system is based on powers of 10.

Page 90

Question 1.
Can you write an expression for the number of days in 100 calendar years with this new adjustment?
Solution:
There are 25 years divisible by 4 in 100 years.
100 is also divisible by 4, but we have to exclude it. So only 24 years have 366 days, and the rest (76 years) have 365 days.
Expression : (24 × 366) + (76 × 365) = 8784 + 27740 = 36,524 days.

Page 93

Question 1.
Do you wonder how people figured out that the Earth completes one revolution around the Sun in exactly 364.2422 days?
Solution:

People discovered how long the Earth takes to go around the Sun through centuries of careful observation and measurement of the sky.
Ancient astronomers watched the repeated pattern of seasons, the position of the Sun among the stars, and events like solstices and equinoxes. By recording the time between two identical events (such as one spring equinox to the next), they could estimate the length of a year. Over many years, averaging these measurements gave a very accurate value. Modem scientists now use precise instruments, telescopes, atomic clocks, and space satellites to measure Earth’s motion, which is how we know the length of a year so accurately (about 364.2422 days).

Class 7 Maths Another Peek Beyond the Point Solutions

Figure it Out (Pages 73-74)

Question 1.
Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products in tenths, hundredths and so on :
(a) 6 × 4 tenths = 24 tenths
(b) 7 × 0.3
(c) 9 × 5 hundredths
Solution:
(a) 6 × 4 tenths = 24 tenths = 2.4
(b) 0.3 = 3 tenths = 7 × 3 tenths = 21 tenths = 2.1
(c) 9 × 5 hundredths = 45 hundredths = 0.45

Question 2.
Find the products :
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23
Solution:
(a) 2734 × 6 = 16404, with 2 decimal places = 164.04
(b) 423 × 37 = 15651, with 2+ 1 = 3 decimal places = 15.651
(c) 432 × 23 = 9936, with 3 + 2 = 5 decimal places = 0.09936

Question 3.
Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?
Solution:
Cloth needed for 1 shirt = 1.65 m
3 shirts need to be made.
Cloth needed = 1.65 × 3 -4.95 m

Question 4.
Meenu bought 4 notebooks and 3 erasers. The cost of each book was ? 15.50 and each eraser was ? 2.75. How much did she spend in all?
Solution:
Cost of 4 notebooks = 4 × 15.50
= ₹ 62.00
Cost of 3 erasers = 3 × 2.75
= ₹ 8.25
Total spent = 62.00+ 8.25
= ₹ 70.25

Question 5.
The thickness of a rupee coin is 1.45 mm. What is the total height of the cylinder formed by placing 36 rupee coins one over the other? Write the answer in centimeters.
Solution:
Total height = 36 × 1.45
= 52.2 mm
1 cm = 10 mm
Therefore, 52.2 mm = 5.22 cm

Question 6.
The price of 1 kg of oranges is ₹ 56.50. What is the price of 2.250 kg. of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why?
Solution:
Price = 56.50 × 2.250 = ₹ 127.125
Yes, we can write 56.50 as 56.5 and 2.250 as 2.25 because trailing zeroes after the decimal don’t change the value.
The product we get will also be the same : 56.5 × 2.25 = 127.125.

Question 7.
Dwarakanath purchases notebooks at a wholesale price of ₹ 23.6 per piece and sells each notebook at ₹ 30/-. How much profit does he make if he sells 50 books in a week?
Solution:
Cost price of 50 notebooks = 50 × 23.6 = ₹ 1180
Selling price of 50 notebooks
= 50 × 30 = ₹ 1500
Profit earned = 1500 – 1180 = ₹ 320 .

Question 8.
Given that 18 × 12 = 216, find the products :
(a) 18 × 1.2
(b) 18 × 0.12
(c) 1.8 × 1.2
(d) 0.18 × 0.12
(e) 0.018 × 0.012
(f) 1.8 × 12
In which of the cases above is the product less than 1?
Solution:
(a) 18 × 1.2 = 21.6 (1 decimal place)
(b) 18 × 0.12 = 2.16 (2 decimal places)
(c) 1.8 × 1.2 = 2.16 (1 + 1 = 2 decimal places)
(d) 0.18 × 0.12 = 0.0216 (2 + 2 = 4 decimal places)
(e) 0.018 × 0.012 =0.000216 (3 + 3 =6 decimal places)
(f) 1.8 × 12 = 21.6 (1 decimal place)
Cases (d) and (e) have products less than 1.

Question 9.
In which of the following multiplications is the product less than 1? Can you find the answer without actually doing the multiplications?
(a) 7 × 0.6
(b) 0.7 × 0.6
(c) 0.7 × 6
(d) 0.07 × 0.06
Solution:
(a) 7 × 0.6 = 4.2, not less than 1.
(b) 0.7 × 0.6 = 0.42, less than 1.
(c) 0.7 × 6 = 4.2, not less than 1.
(d) 0.07 × 0.06 = 0.0042, less than 1.

Question 10.
Multiplying the following numbers by 10,100 and 1000 to complete the table.

	× 10	× 100	× 1000
5.7
23.02
0.92
0.306
24.67

Solution:

	× 10	× 100	× 1000
5.7	57	570	5700
23.02	230.2	2302	23020
0.92	9.2	92	920
0.306	3.06	30.6	306
24.67	246.7	2467	24670

Figure it Out (Page 83)

Question 1.
Find the quotient by converting the denominator into 1, 10, 100 or 1000 and verify the solution by the long division method (division by place value).
(a) \(\frac {18}{5}\)
(b) \(\frac {415}{4}\)
(c) \(\frac {1217}{2}\)
(d) \(\frac {4827}{8}\)
Solution:
(a) Make denominator 10:
\(18 \div 5=\frac{18 \times 2}{5 \times 2}=\frac{36}{10}=3.6\)
Verification: 18 ÷ 5
Dividing I ten and 8 ones into 5 equal parts
1 < 5.
It means we need to regroup 1 ten as 10
ones.
i.e. 10 + 8 = 18 ones.
18 ones ÷ 5, 3 ones remains. To divide 3 ones into 5 equal parts, regroup the 3 ones as 30 tenths (place a decimal while regrouping ones into tenths)
30 tenths ÷ 5 = 6
therefore 18 ÷ 5 = 3.6

Hence verified

(b) Make denominator 100:
\(415 \div 4=\frac{415 \times 25}{4 \times 25}=\frac{10375}{100}=103.75\)
Verifications:
By following the steps:

∴ 415 ÷ 4 = 103.75
Hence verified

(c) Make denominator 10:
\(1217 \div 2=\frac{12.17 \times 5}{2 \times 5}=\frac{6085}{10}=608.5\)
By following the steps:

∴ 1217 ÷ 2 = 608.5
Hence verified

(d) Make denominator 1000:
\(4827 \div 8=\frac{4827 \times 125}{8 \times 125}\)
\(=\frac{603375}{1000}=603.375\)
By following the steps:

∴ 4827 ÷ 8 = 603.375
Hence verified

Question 2.
Choose the correct answer:
(a) \(\frac {1526}{4}\) =
(i) 38.15
(ii) 380.15
(iii) 381.5
(iv) 381.05
Solution:
(iii) 381.5
Explanation : Multiply numerator and denominator by 25.
\(\frac{1526}{4}=\frac{1526 \times 25}{4 \times 25}=\frac{38150}{100}\)
= 381.50

(b) \(\frac {3567}{8}\) =
(i) 4458.75
(ii) 44.5875
(iii) 445.875
(iv) 4458.75
Solution:
(iii) 445.875
Explanation : Multiply numerator and denominator by 125.
\(\frac{3567}{8}=\frac{3567 \times 125}{8 \times 125}=\frac{445875}{1000}\)
= 445.875

Question 3.
What ¡s the quotient?
(a) 132 ÷ 4 =
(b)13.2 ÷ 4 =
(c) 1.32 ÷ 4 =
(d) 0.132 ÷ 4=
Solution:
We solve each by making the. denominator 100(convert ÷ 4 into division by 100 by multiplying numerator and denominator by 25).
(a) 132 ÷ 4 = \(\frac{132}{4}=\frac{132 \times 25}{100}=\frac{3300}{100}=33\)

(b)13.2 ÷ 4 = \(\frac{13.2}{4}=\frac{13.2 \times 25}{100}=\frac{330}{100}=3.3\)

(c) 1.32 ÷ 4 = \(\frac{1.32}{4}=\frac{1.32 \times 25}{100}=\frac{33}{100}=0.33\)

(d) 0.132 ÷ 4 = \(\frac{0.132}{4}=\frac{0.132 \times 25}{100}=\frac{3.3}{100}=0.033\)

Question 4.
What is the quotient?
(a) 126 ÷ 8 =
(b) 12.6 ÷ 8 =
(c) 1.26 ÷ 8 =
(d) 0.126 ÷ 8 =
Solution:
Since 8 × 125 = 1000
(a) \(\frac{126}{8}=\frac{126 \times 125}{8 \times 125}=\frac{15750}{1000}=15.75\)
Dividing by 10 moves the decimal one place
(b) 12.6 ÷ 8 = 1.575
(c) 1.26 — 8 = 0.1575
(d) 0.126 ÷ 8 = 0.01575
(e) 0.0126 ÷ 8 = 0.001575

Figure it Out (Pages 86-87)

Question 1.
Express the following fractions in decimal form :
(a) \(\frac {2}{5}\)
(b) \(\frac {13}{4}\)
(c) \(\frac {4}{50}\)
(d) \(\frac {5}{8}\)
Solution:
(a) Make denominator 10:
\(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}=0.4\)

(b) Make denominator 100:
\(\frac{13}{4}=\frac{13 \times 25}{4 \times 25}=\frac{325}{100}=3.25\)

(c) Make denominator 100:
\(\frac{4}{50}=\frac{4 \times 2}{50 \times 2}=\frac{8}{100}=0.08\)

(d) Make denominator 1000:
\(\frac{5}{8}=\frac{5 \times 125}{8 \times 125}=\frac{625}{1000}=0.625\)

Question 2.
Find the quotients:
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52
Solution:
(a) \(\frac{24.86 \times 10}{1.2 \times 10}=\frac{248.6}{12}=20.716 \ldots \ldots\)
(b) \(\frac{24.86 \times 10}{1.2 \times 10}=\frac{248.6}{12}=20.716 \ldots \ldots\)

Question 3.
Evaluate the following using the information 156 × 12 = 1872.
(a) 15.6 × 1.2 = __________
(b) 187.2 ÷ 1.2 = __________
(c) 18.72 ÷ 15.6 = __________
(d) 0.156 × 0.12 = __________
Solution:
(a) Given : 156 × 12 = 1872.
1 + 1 = 2 decimal places, so 18.72

(b) \(\frac{187.2 \times 10}{1.2 \times 10}=\frac{1872}{12}=156\)

(c) \(\frac{18.72 \times 100}{15.6 \times 100}=\frac{1872}{1560}=1.2\)

(d) 3 + 2 = 5 decimal places, so 156 × 12 = 1872, answer = 0.01872

Question 4.
Evaluate the following:
(a) 25 ÷ ______ = 0.025
(b)25 ÷ ____ = 250
(c)25 ÷ _____ = 2.5
(d) 25 ÷ 10 = 25 × _____
(e) 25 ÷ 0.10 = 25 × ______
(f) 25 ÷ 0.01 = 25 × ______
Solution:
(a) 25 ÷ 1000 = 0.025, so answer 1000
(b) 25 ÷ 0,1 = 250, so answer = 0.1
(c) 25 ÷ 1o = 2.5, so answer = 1o
(d) 25 ÷ 10 = 25 × 0.1, so answer 0.1
(e)25 ÷ 0.10 = 25 × 10, so answer 10
(f) 25 ÷ 0.01 = 25 × 100, so answer = 100

Question 5.
Find the quotient:
(a) 2.46 ÷ 1.5 =
(b) 2.46 ÷ 0.15 =
(c) 2.46 ÷ 0.015 =
Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient obtained in 2.46 ÷ 0.15?
Solution:
(a) 2.46 ÷ 1.5 = \(\frac{2.46 \times 10}{1.5 \times 10}\)
= \(\frac{24.6}{15}\)
= 1.64

(b) 2.45 ÷ 0.15 = \(\frac{2.46 \times 100}{0.15 \times 100}\)
= \(\frac{246}{15}\)
= 16.4

(c) 2.46 ÷ 0.015 = \(\frac{2.46 \times 1000}{0.015 \times 1000}\)
= \(\frac{2460}{15}\)
= 164
24.6 ÷ 1.5 = 16.4 and 2.46 ÷ 0.15 = 16.4. Yes, they are the same

Question 6.
A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the length of each piece?
Solution:
Length of each piece = 4 ÷ 5 = \(\frac{4}{5}\)
\(=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}=0.8 \mathrm{~m}\)

Question 7.
If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side?
Solution:
In regular polygon all sides are equal.
Length of each side = 208.8 ÷ 12 = 17.4 cm

Question 8.
3 litres of watermelon juice is shared among 8 friends equally. How much watermelon juice will each get? Express the quantity of juice in millilitres.
Solution:
Juice per person = 3 ÷ 8 = \(\frac{3}{8}\)
\(=\frac{3 \times 125}{8 \times 125}=\frac{375}{1000}\)
= 0.375 litres
= 375 millilitres

Question 9.
A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?
Solution:
Distance per litre = 234.45 ÷ 12.6
= \(\frac{234.45 \times 10}{12.6 \times 10}\)
= \(\frac{234.45 \times 10}{12.6 \times 10}\)
= 18.6km/litre

Question 10.
13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?
Solution:
Flour per student = 13.5 ÷ 15
= 0.9 kg

Figure it Out (Pages 93-95]

Question 1.
A 210 gram packet of peanut chikki costs ₹ 70.5, while a 110 gram packet of potato chips costs ₹ 33.25. Which is cheaper?
Solution:
Solution : Price per gram of peanut chikki
= 70.5 ÷ 210 = 0.3357 per gram
Price per gram of potato chips
= 33.25 ÷ 110 = 0.3023 per gram
∴ Potato chips are cheaper per gram.

Question 2.
Write the decimal number at the arrow mark:

Solution:
Arrow between 3.1 and 3.2, so answer could be 3.16 (or any value between 3.1 and 3.2).

Arrow between 2.15 and 2.17, so answer could be 2.162 (or any value between 2.15 and 2.17).

Question 3.
Shyamala bought 3 kg bananas at ₹ 30/- per kg. She counted 35 bananas in all. She sells each banana for ₹ 5/-. How much profit does she make selling all the bananas?

Solution:
Cost price = 3 × 30 = ₹ 90
Selling price = 35 ÷ 5 = ₹ 175
Profit = 175 – 90 = ₹ 85

Question 4.
A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The teacher wanted to place 80 textbooks on the shelf. The bookshelf is 160 cm long. How many books could be placed on the shelf? Was there any space left? If yes, how much?
Solution:
Space needed for 80 books
= 80 × 2.5 = 200 cm
Bookshelf length = 160 cm
Number of books that can fit
= 160 ÷ 2.5 = 64 books
Space left = 160 – (64 × 2.5)
= 160 – 160 = 0 cm (no space left)

Question 5.
Fill in the following blanks appropriately : 1 cm = 10 mm 1 m = 100 cm 1 km = 1000 m

1 cm = 10 mm 1 m = 100 cm 1 km = 1000 m

1 kg = 1000g 1 g = 1000 mg

1 l = 1000 ml

5.5 km = ____ m	35 cm =__ m	14.5 cm = ____ mm
68 g = ___ kg	9.02 m = _____ mm	125.5 ml = ___ l

Solution:

5.5 km = 5500 m	35 cm = 0.35 m	14.5 cm = 145 mm
68 g = 0.068 kg	9.02 m = 902o mm	125.5 ml = [0.1255] l

Question 6.
The following problem was set by Sridharacharya in his book, Patiganita. “6\(\frac {1}{4}\) is divided by 2\(\frac {1}{2}\), and 60\(\frac {1}{4}\) is divided by 3\(\frac {1}{2}\). Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?
Solution:
6\(\frac {1}{4}\) = 6 + 0.25 = 6.25 4
and 2\(\frac {1}{2}\) = 2 + 0.5 = 2.5
∴ 6.25 ÷ 2.5 =2.5 (Quotient)
and 60\(\frac {1}{4}\) = 60 + 0.25 = 60.25
3\(\frac {1}{2}\) = 3 + 0.5 = 3.5
∴ 60.25 ÷ 3.5 = 17.21 (Quotient)

Question 7.
Fill the boxes in at least 2 different ways :
(a) ☐ × ☐ = 2.4
(b) ☐ × ☐ = 14.5
Solution:
(a) 1.2 × 2 = 2.4, or 0.6 × 4 = 2.4, or 2.4 × 1 = 2.4
(b) 29 × 0.5 = 14.5, or 5 × 2.9 = 14.5

Question 8.
Find the following quotients given that 756 ÷ 36 = 21 :
(a) 75.6 ÷ 3.6
(b) 7.56 ÷ 0.36
(c) 756 ÷ 0.36
(d) 75.6 ÷ 360
(e) 7560 ÷ 3.6
(f) 7.56 ÷ 0.36
Solution:
(a) Both have 1 decimal place, so quotient = 21
(b) Both have 2 decimal places, so quotient = 21
(c) (756 × 100)/(0.36 × 100) = 75600/36 = 2100
(d) 75.6 ÷ 360 = 0.21
(e) (7560 × 10)/(3.6 × 10) = 75600/36 = 2100
(f) (7.56 × 100)/(0.36 × 100) = 756/36 = 21

Question 9.
Find the missing cells if each represents a ÷ b:

b↓ a→	1517	151.7	15.17	1.517	15170
37	41
3.7			4.1
0.37
0.037		4100
370

Solution:

b↓ a→	1517	151.7	15.17	1.517	15170
37	41	4.1	0.41	0.041	410
3.7	410	41	4.1	0.41	4100
0.37	4100	410	41	4.1	41000
0.037	41000	4100	410	41	4,10,000
370	4.1	0.41	0.041	0.0041	41

Question 10.
Using the digits 2, 4, 5, 8, and 0 fill the boxes get the : ☐☐ . ☐ × ☐ . ☐ to get the:
(a) maximum product
(b) minimum product
(c) product greater than 150
(d) product nearest to 100
(e) product nearest to 5
Solution:
(a) 42.0 × 8.5 = 357
(b) 45.8 × 0.2 = 9.16
(c) 85.4 × 2.0 = 170.8
(d) 20.5 × 4.8 = 98.4
(e) 45.8 × 0.2 = 9.16

Question 11.
Sort the following expressions in increasing order:
(a) 245.05 × 0.942368
(b) 245.05 × 7.9682
(c) 245.05 ÷ 7.9682
(d) 245.05 ÷ 0.942368
(e) 245.05
(f) 7.9682
Solution:
(a) 245.05 × 0.942368
Value ≈ 230.93
(b) 245.05 × 7.9682
Value ≈ 1952.32
(c) 245.05 ÷ 7.9682
Value ≈ 30.75
(d) 245.05 ÷ 0.942368
Value ≈ 260.02
(e) 245.05
Value = 245.05
(f) 7.9682
Value = 7.9682
Increasing oder: (f) < (c) < (a) < (e) < (d) < (b)

It’s Puzzle Time
Hidato

Puzzle

Solution:

In Hidato, a grid of cells is given. It is usually square-shaped, like Sudoku or Kakuro, but it can also include hexagons or any shape that forms a tessellation. It can have inner holes (like a disc), but it is made of only one piece. Usually, in every Hidato puzzle the lowest and the highest numbers are given on the grid. Your task is to fill the grid such that there is a continuous path of consecutive numbers from the lowest to the highest number. The next number must be in any one of the adjacent cells, including diagonally adjacent cells.

The grid comes pre-filled with some numbers (with values between the smallest and the highest) to ensure that these puzzles have a single solution.
Try solving the following Hidato puzzles.

Solution:

Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Part 2 Chapter 3 Finding Common Ground Solutions Question Answer helps develop logical thinking and accuracy.

Ganita Prakash Class 7 Part 2 Chapter 3 Solutions

Class 7 Ganita Prakash Part 2 Chapter 3 Solutions

RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 3 Solutions

In-text Questions
Page 48

Question 1.
Do you remember the ‘Jump Jackpot’ game from Grade 6 (see the chapter ‘Prime Time’)? Grumpy places a treasure on a number and Jumpy chooses a jump size and tries to collect the treasure. In each case below, the two numbers upon which treasures are kept are given. Find the longest jump size (starting from 0) using which Jumpy can land on both the numbers having the treasure.
(a) 14 and 30
(b) 7 and 11
(c) 30 and 50
(d) 28 and 42
Solution:
(a) 14 and 30
Factors of 14 : 1, 2, 7, 14
Factors of 30 : 1, 2, 3, 5, 6, 10, 15, 30
Common factors : 1, 2
HCF = 2

(b) 7 and 11
Both are prime numbers.
Common factor = 1
HCF = 1

(c) 30 and 50
Factors of 30 : 1, 2, 3, 5, 6, 10, 15, 30
Factors of 50 : 1,2, 5, 10, 25, 50
Common factors: 1, 2, 5, 10
HCF = 10

(d) 28 and 42
Factors of 28 : 1, 2, 4, 7, 14, 28
Factors of 42 : 1, 2, 3, 6, 7, 14, 21, 42
Common factors : 1, 2, 7, 14
HCF = 14

Question 2.
Is the longest jump size for the numbers the same as their HCF? Explain why it is so.
Solution:
Yes, the longest jump size is the same as the HCF of the two numbers. Reason : To collect both the treasures, using the jump of longest size.

Page 51

Question 1.
Similarly, is 2 × 7 = 14 a factor of 840? Why or why not? Is 2 × 2 × 2 a factor of 840? Why or why not? Is 3 × 3 × 3 a factor of 840? Why or why not? Can we use this idea to list down all the possible factors of a number using just its prime factors?
Solution:
Yes,
Both 2 and 7 appear in the prime factorisation of 840.
Since 14 can be formed using the primes of 840, it divides 840 exactly.
Yes,
Three 2s occur in the prime factorisation, so 8 (which uses all three 2s) is also a factor of 840.
No,
The number 840 has only one factor of 3. Since 27 needs three 3 s, it cannot be formed from the prime factors of 840 and therefore does not divide 840.
Yes, we can use this idea to list down all the possible factors of a number using just its prime factors.

Page 55

Question 1.
What about the largest common multiple? Does such a number exist?
Solution:
For any common multiple of two numbers, we can always create a larger one by multiplying it by 2, 3, or any other whole number. This process can continue forever. For example, if 24 is a common multiple of two numbers, then :
48, 72, 96, 120, … are also common multiples.
Since this list never ends, there is no greatest (largest) common multiple – common multiples are unlimited.
Only the least common multiple (LCM) exists, not a largest one.

Page 56

Question 1.
Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find the first number for which ‘idli-vada’ will be called out:
(a) 4 and 6
(b) 7 and 11
(c) 14 and 30
(d) 15 and 55
Is the answer always the LCM of the two numbers? Explain.
Solution:
In the Idli-Vada game, ‘idli-vada’ is called when a number is a common multiple of both given numbers.
The first such number is the LCM (Least Common Multiple) of the two numbers.
(a) 4 and 6
LCM of 4 and 6 = 12 First ‘idli-vada’ at 12

(b) 7 and 11
Since both are primes,
LCM = 7 × 11 = 77 First ‘idli-vada’ at 77

(c) 14 and 30
14 = 2 × 7
30 = 2 × 3 × 5
LCM = 2 × 7 × 3 × 5 = 210
First ‘idli-vada’ at 210

(d) 15 and 55
15 = 3 × 5
55 = 5 × 11
LCM = 3 × 5 × 11 = 165
First ‘idli-vada’ at 165
Yes, the answer is always the LCM.
Reason:
Players call ‘idli-vada’ at numbers that are multiples of both chosen numbers (common multiples).
The least such number is exactly the LCM, so it is always the first number where ‘idli-vada’ is called.

Page 59

Question 1.
For number pairs satisfying this property (i.e., one of the numbers is the HCF),
(a) if m is a number, what could be the other number?
(b) if 7k is a number, what could be the other number?
Solution:
(a) If m is a number. The other number can be any multiple of m, i.e., m and km.
Here, k is any positive integer. m = 6⇒ pairs : (6, 12), (6, 18), (6, 24), …
In all cases, HCF = m.

(b) The other number could be any factor of 7k such that the smaller number divides the larger one.
So a suitable pair is : 7k, k.
If 7k = 21 (so k = 3), the pair is (21, 3) and HCF = 3.

Page 59

Question 1.
What happens to the HCF of two numbers if both numbers are doubled?
Take some pairs of numbers and explore. Are you able to see why the HCF will also double?
Solution:
If both numbers are doubled, then both numbers get an extra factor of 2 in their prime factors. This 2 will be included as a factor in the largest common subpart, and so the HCF will double. For example, consider the numbers 270 and 50.

HCF = 2 × 5 = 10
Let us double these numbers to get 540 and 100.

HCF = 2 × 2 × 5 = 20.

Page 60

Question 1.
Consider the following two multiples of 14 – 14 × 6, 14 × 9. What is their HCF?
Solution:
Clearly, 14 is a common factor. Let us calculate the prime factorisations.

HCF = 14 × 3 = 42.

Question 2.
Here are some more numbers where both numbers are multiples of the same number. Find their HCF:
(a) 18 × 10, 18 × 15
(b) 10 × 38, 10 × 21
(c) 5 × 13, 5 × 20
(d) 12 × 16, 12 × 20
Solution:
(a) HCF = 18 × HCF (10, 15)
HCF (10, 15) = 5
HCF = 18 × 5 = 90

(b) HCF = 10 × HCF (38, 21)
HCF (38, 21) = 1
HCF = 10

(c) HCF = 5 × HCF (13, 20)
HCF (13, 20) = 1
HCF = 5

(d) HCF = 12 × HCF (16, 20)
HCF (16, 20) = 4
HCF – 12 × 4 = 48

Question 3.
In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is 10? Explore a few more examples of this type to understand when this happens.
Solution:
Part (c) in above question also have the same thing.
Other examples are :
7 × 4 and 7 × 9
HCF (4, 9) = 1
HCF = 7
12 × 5 and 12 × 11
HCF (5, 11) = 1
HCF = 12

Page 62

Question 1.
You can try this method (division method) for these pairs of numbers.
(a) 90 and 150
Solution:
(a)

LCM = 2 × 3 × 3 × 5 × 1 × 5 = 450

(b)

LCM = 2 × 2 × 3 × 7 × 11 = 924

Page 62

Question 1.
Which is greater-the LCM of two numbers or their product?
Solution:
The product of two numbers is always greater than or equal to their LCM

Question 2.
You could analyse the above statement using examples. Then try to reason or prove, why the LCM is never greater than the product of the numbers.
Solution:
Example 1 (numbers with a common factor):
Take 12 and 18.
Prime factorisations : 12 = 2² × 3, 18 = 2 × 3².
LCM = 2² × 3² = 36. Product = 12 × 18 = 216. So LCM (36) is much less than the product (216).

Example 2 (coprime numbers) :
Take 8 and 15.
They have no common prime factors, so LCM = 8 × 15 = 120. Here LCM equals the product because the numbers are coprime.
The product a × b contains all prime factors of both a and b, but with multiplicities multiplied. The LCM only needs each prime to appear with the highest power that appears in either number, not multiplied. So the LCM can’t be larger than the raw product – at worst (when there are no common primes) it equals the product.
The LCM of two positive integers is always less than or equal to their product; equality holds precisely when the two numbers are coprime.

Question 3.
Consider the numbers 105 and 95. Find their LCM.
Solution:
Factorising them into their primes :
105 = 3 × 5 × 7
95 = 5 × 19
LCM = 3 × 5 × 7 × 19.
Let us consider the product in the factorised form:
105 × 95 = 3 × 5 × 5 × 7 × 19

Question 4.
Is the LCM a factor of the product? If yes, what should it be multiplied with to get the product?
Solution:
Yes, the LCM is a factor of the product.
To get the product from the LCM, multiply by the remaining factor 5 :
LCM × 5 = 105 × 95
So, the LCM must be multiplied by 5 to obtain the product.

Question 5.
Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers :
(a) 45, 105
(b) 275, 352
(c) 222, 370
Solution:
(a) Prime factorisation
45 = 3² × 5
105 = 3 × 5 × 7
LCM = 32 × 5 × 7 = 315
45 × 105 = 4725
So, 4725 ÷ 315 = 15
HCF = 3 × 5 = 15
So, LCM need to be multiplied by HCF to get the product of numbers.
Same pattern is observed in all.

Page 63

Question 1.
Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF?
Solution:
Yes.
Our observations seem to suggest the following:
HCF × LCM = Product of the two numbers.

Question 2.
Why does this happen? Can you give an explanation or proof?
Solution:
This happens because of how the prime factors of the two numbers are shared between the HCF and the LCM.
The common prime factors of both numbers make up the HCF.
The highest powers of all prime factors (common and non-common) make up the LCM.
When we multiply HCF and LCM, every prime factor from both numbers is included exactly as it appears in the product of the two numbers.
So, the prime factors of the two numbers are correctly “divided” between the HCF and the LCM, and together they recreate the full product of the numbers :
HCF × LCM = Product of the two numbers

Question 3.
Explore whether this property holds when 3 numbers are considered.
Solution:
No, this property does not hold for three numbers.
Example : 2, 3, 4
HCF (2, 3, 4) = 1
LCM (2, 3, 4) = 12
Product = 2 × 3 × 4 = 24
HCF × LCM = 1 × 12 = 12 × 24

Class 7 Maths Finding Common Ground Solutions

Figure it Out (Page 51)

Question 1.
List all the factors of the following numbers :
(a) 90
(b) 105
(c) 132
(d) 360 (this number has 24 factors)
(e) 840 (this number has 32 factors)
Solution:
(a) Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.
(b) Factors of 105 are 1, 3, 5, 7, 15, 21, 35, 105.
(c) Factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
(d) Factors of 360 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
(e) Factors of 840 are 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

Figure it Out (Page 53)

Question 1.
Find the common factors and the HCF of the following numbers :
(a) 50, 60
(b) 140, 275
(c) 77, 725
(d) 370, 592
(e) 81, 243
How do we directly find the HCF without listing all the factors?
Solution:
(a) Factors of 50 : 1,2, 5, 10, 25,
Factors of 60 : 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factors : 1, 2, 5, 10 HCF = 10

(b) Factors of 140 : 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140
Factors of 275 : 1, 5, 11, 25, 55, 275 Common factors : 1, 5 HCF = 5

(c) Factors of 77 : 1, 7, 11, 77
Factors of 725 : 1, 5, 25, 29, 145, 725
Common factor: 1
HCF = 1

(d) Factors of 370 : 1, 2, 5, 10, 37, 74, 185, 370
Factors of 592 : 1, 2, 4, 8, 16, 37, 74, 148, 296, 592
Common factors : 1, 2, 37, 74
HCF = 74

(e) Factors of 81 : 1, 3, 9, 27, 81
Factors of 243 : 1, 3, 9, 27, 81, 243
Common factors : 1, 3, 9, 27, 81
HCF = 81
We can directly find the HCF without listing prime factors by finding the prime factorisation of the given numbers simultaneously and identifying the common prime factors by which both numbers can be divided. The product of these common prime factors is their HCF.

Figure it Out (Page 54)

Question 1.
Find the HCF of the following numbers :
(a) 24, 180
(b) 42, 75, 24
(c) 240, 378
(d) 400, 2500
(e) 300, 800
Solution:
(a) Prime factorisation :
24 = 2 × 2 × 2 × 3 = 2³ × 3
180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5
Common prime factors : 2² × 3
HCF = 4 × 3 = 12

(b) Prime factorisation:
42 = 2 × 3 × 7
75 = 3 × 5 × 5 24 = 2 × 2 × 2 × 3 = 2³ × 3
Common prime factor to all : 3
HCF = 3

(c) Prime factorisation:
240 = 2⁴ × 3 × 5
378 = 2 × 3³ × 7
Common prime factors : 2 × 3
HCF = 6

(d) Prime factorisation:
400 = 2⁴ × 5²
2500 = 2² × 5⁴
Common prime factors : 2² × 5²
HCF = 4 × 25 = 100

(e) Prime factorisation:
300 = 2² × 3 × 5²
800 = 2⁵ × 5²
Common prime factors : 2² × 5²
HCF = 4 × 25 = 100

Question 2.
Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as : 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?
Solution:
No, we cannot say that 72 and 144 have no common factor other than 1.
Even though the composite factors shown (6, 12, 8, 18) do not look the same, this does not mean the numbers have no common factors. Composite factors can be further broken into prime factors.
72 = 2³ × 3²
144 = 2⁴ × 3²
They clearly share common prime factors 2 and 3, so they have several common factors such as 2, 3, 4, 6, 9, 12, 18, 36 and more.
Their HCF is 72.
Thus, we must always use prime factorisation to correctly find common factors, not just compare composite factors.

Figure it Out (Page 58)

Question 1.
Find the LCM of the following numbers :
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222, 370
Solution:
(a) Prime factorisation :
30 = 2 × 3 × 5
72 = 2³ × 3²
Take highest powers of all primes :
LCM = 2³ × 3² × 5
LCM = 360

(b) Prime factorisation:
36 = 2² × 3²
54 = 2 × 3³
Highest powers:
LCM = 2² × 3³
LCM = 108

(c) Prime factorisation:
105 = 3 × 5 × 7
195 = 3 × 5 × 13
65 = 5 × 13
Highest powers :
LCM = 3 × 5 × 7 × 13
LCM = 1365

(d) Prime factorisation :
222 = 2 × 3 × 37
370 = 2 × 5 × 37
Highest powers :
LCM = 2 × 3 × 5 × 37
LCM = 1110

Figure it Out (Page 59)

Question 1.
Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold.
(a) Two consecutive even numbers
(b) Two consecutive odd numbers
(c) Two even numbers
(d) Two consecutive numbers
(e) Two co-prime numbers
Share your observations with the class.
Solution:
(a) They always have 2 as a common factor, so their HCF is 2.
(b) They have no common factor other than 1, so their HCF is 1.
(c) They are always divisible by 2, so their HCF is at least 2 (it may be greater depending on the numbers).
(d) They always have no common factor except 1.
(e) They have no common factor other than 1.

Question 2.
The LCM of 3 and 24 is 24 (it is one of the two given numbers).
(a) Find more such number pairs where the LCM is one of the two numbers.
(b) Make a general statement about such numbers. Describe such number pairs using algebra.
Solution:
(a) The LCM of two numbers is one of the numbers when one number is a multiple of the other.
Examples :

• LCM (4, 20) = 20
• LCM (6, 18) = 18
• LCM (5, 25) = 25
• LCM (7, 35) = 35
• LCM (8, 32) = 32

(b) Let the two numbers be n and kn, where A: is a positive integer.
Since kn is already a multiple of n, the smallest number divisible by both is kn.
LCM (n, kn) = kn

Question 3.
Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold.
(a) Two multiples of 3
(b) Two consecutive even numbers
(c) Two consecutive numbers
(d) Two co-prime numbers
Solution:
(a) If a = 3m and b = 3n, then both numbers contain the factor 3.
Therefore, LCM (a, b) = 3 × LCM (m, n)
The LCM will always be a multiple of 3.

(b) Let the numbers be 2k and 2 (k + 1).
They share only the common factor 2, so :
LCM 2k, (2k + 2) = \(\frac{(2 k)(2 k+2)}{2}\)
The LCM is always greater than the larger number.

(c) Let the numbers be n and n + 1.
Since consecutive numbers are always co¬prime :
gcd (n, n + 1) = 1
LCM (n, n + 1) = n(n + 1)
The LCM equals the product of the numbers.

(d) Let the numbers be a and b such that :
gcd (a, b) = 1
LCM (a, b) = a × b
The LCM is simply the product of the two numbers.

Figure it Out (Pages 63-64)

Question 1.
In the two rows below, colours repeat as shown. When will the blue stars meet next?

Solution:
In the first row blue star is repeated at the position 4, 10..
So, next blue star will be at 16. (at the interval of 6)
In the second row, blue stars are at the position 4, 8, 12..
So, next blue star will be at 16. (at the interval of 4)

Question 2.
(a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2?
(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?
Solution:
(a) To be a multiple, the first number must contain all prime factors of the second number in equal or greater powers.
First number : 5¹ × 7¹ × 11²
Second number : 2 × 5 × 7² × 11
The first number is missing the factor 2 and has only one 7 instead of two.
No, it is not a multiple.

(b) To be a factor, every prime factor of the smaller number must appear in the larger number with equal or greater power.
The first number needs 11², but the second number has only one 11.
No, it is not a factor.

Question 3.
Find the HCF and LCM of the following (state your answers in the form of prime factorisations) :
(a) 3 3x5x7x7 and 12x7x11
(b) 45 and 36
Solution: (a) Numbers :
1st = 3 × 3 × 5 × 7 × 7 = 3² × 5 × 7²
2nd = 12 × 7 × 11 = (2² × 3) × 7 × 11 = 2² × 3 × 7 × 11
HCF = 3 × 7 = 21
LCM = 2² × 3² × 5 × 7² × 11

(b) Numbers :
45 = 3² × 5
36 = 2² × 3²
HCF : Common prime factors with lowest powers.
Then, HCF = 3²
LCM : All prime factors with highest powers.
Then, LCM = 2² × 3² × 5

Question 4.
Find two numbers whose HCF is 1 and LCM is 66.
Solution:
Since, HCF × LCM = Product of the two numbers
and the HCF = 1, the two numbers must be co-prime and their product must equal the LCM:
1 × 66 = 66
So we need two co-prime factors of 66.
Prime factorisation : 66 = 2 × 3 × 11
Split the primes into two groups with no common factor :
6 = 2 × 3 and 11 These are co-prime.
The two numbers are 6 and 11.
HCF (6, 11) = 1
LCM (6, 11) = 66

Question 5.
A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)
Solution:
We need a number of cows that is divisible by 3, 5, and 7, and is less than 200.
The LCM of 3, 5 and 7 is LCM (3, 5, 7) = 3 × 5 × 7 = 105
Any number that works must be a multiple of 105. The multiples are 105, 210,only 105 is less than 200.
So the cowherd had 105 cows.

Question 6.
The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?
(a) 9 cm
(b) 6 cm
(c) 4 cm
(d) 3 cm
(e) 2 cm
Solution:
To pack cubes into the box without leaving gaps, the edge length of each cube must exactly divide all three dimensions of the box : 12 cm, 18 cm, and 36 cm.
Find the common factors of 12, 18, and 36.
Prime factorisations :
12 = 2² × 3
18 = 2 × 3²
36 = 2² × 3²
Common factors of 12, 18 and 36 are : 1, 2, 3, 6
Out of the given options, the common factor condition is satisfied by 4, 3 and 2.
Cubes that can be packed without gaps :
6 cm, 3 cm, and 2 cm Correct options : (b), (d), and (e)

Question 7.
Among the numbers below, which is the largest number that perfectly divides both 306 and 36?
(a) 36
(b) 612
(c) 18
(d) 3
(e) 2
(f) 360
Solution:
Prime factorisation :
306 = 2 × 3 × 3 × 17 = 2 × 3² × 17
36 = 2² × 3²
HCF = 2 × 3² = 2 × 9 = 18
The largest number that perfectly divides both 306 and 36 is : (c) 18 (their HCF).

Question 8.
Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.
Solution:
Prime factorisation :
3 = 3
4 = 2²
5 = 5
7 = 7
LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
So, the number must be a multiple of420. It should leave a remainder of 10 when divided by 11.
We need : N ≡ 10 (mod 11) ⇒ 420k ≡ 10 (mod 11)
420 mod 11 = ?
11 × 38 = 418 → 420 – 418 = 2
420 ≡ 2 (mod 11)
So : 2k ≡ 10 (mod 11)
The modular inverse of 2 modulo 11 is 6 (since 2 × 6 = 12 ≡ 1 mod 11).
k ≡ 10 × 6 ≡ 60 ≡ 5 (mod 11)
Smallest positive k = 5k = 5k = 5
N = 420 × 5 = 2100

Question 9.
Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out How many children could have been playing initially?
(a) 72
(b) 90
(c) 45
(d) 3
(e) 36
(f) None of these
Solution:
When 6 was called, no one got out -»total number of children is divisible by 6 When 9 was called, no one got out → total number of children is divisible by 9 When 10 was called, some got out → total number of children is not divisible by 10 Let the total number of children be N.
Find the LCM of 6 and 9
Prime factorisation:
6 = 2 × 3
9 = 3²
LCM (6, 9) = 2 × 3 × 3 = 18 So number of children must be multiple of 18.
Here 72, 90 and 36 are multiple of 18 in which only 72 and 36 are not multiple of 10.
Hence, number of children could have been played initially = 72 or 36.
So option (a) and (e) are correct.

Question 10.
Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be :
(a) Less than both numbers
(b) In between the two numbers
(c) Greater than both numbers
(d) Less than m × n
(e) Greater than m × n
Solution:
Two different prime numbers have no common factor other than 1.
So, HCF (m, n) = 1
LCM (m, n) = m × n/HCF (m, n) = m × n
Analyze the options
(a) Less than both numbers → (LCM = m × n > m and > n), so it is wrong statement.
(b) In between the two numbers → so it is wrong statement.
(c) Greater than both numbers → (LCM = m × n > m and > n) so it is a correct statement.
(d) Less than m × n → (LCM = m × n) so it is wrong statement.
(e) Greater than m × n → so it is wrong statement.
So, correct statement is (c).

Question 11.
A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?
Solution:
Rabbit moves 7 feet per leap → total distance = 7n
Dog moves 9 feet per leap → total distance = 9n
The rabbit has a head start of 150 feet, so the dog catches the rabbit when :
9 n = 7n + 150
9n – 7n = 150
2n = 150
n = 75

Question 12.
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?
Solution:

Number	Prime Factorisation
1	–
2	2
3	3
4	22
5	5
6	2 × 3
8	23
9	32
10	2 × 5

Prime Factorisation
Take the highest powers of all primes

• 2 → highest power = 2³ (from 8)
• 3 → highest power = 3² (from 9)
• 5 → highest power = 5¹ (from 5 or 10)

LCM = 2³ × 3² × 5 = 360
The smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10 is 360

Question 13.
Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together \(\frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63} \text { and } \frac{1}{21}\). What do you get? How can we find this sum efficiently?
Solution:
We need to add :
\(\frac{8}{15}+\frac{1}{20}+\frac{7}{36}+\frac{11}{63}+\frac{1}{21}\)
Find the LCM of all denominators
Prime factorisations :
15 = 3 × 5
20 = 2² × 5
36 = 2² × 3²
63 = 3² × 7
21 = 3 × 7
LCM = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 × 1260
Convert each fraction to denominator 1260
\(\begin{aligned}
& \frac{8}{15}=\frac{672}{1260} \\
& \frac{1}{20}=\frac{63}{1260} \\
& \frac{7}{36}=\frac{245}{1260} \\
& \frac{11}{63}=\frac{220}{1260} \\
& \frac{1}{21}=\frac{60}{1260}
\end{aligned}\)
Add the numerators : 672 + 63 + 245 + 220 + 60 = 1260
Then, final answer is : \(\frac {1260}{1260}\) = 1
Using LCM avoids repeated simplifications and gives a clean, fast solution – just as used by ancient mathematicians like Mahaviracharya.

It’s Puzzle Time!
Mystery Colours!

You might have noticed and wondered about these different circle designs around the page numbers on each page!
The picture below shows all the designs for the numbers from 1 to 100.

Try to decode the colour scheme for each number.
There are several interesting patterns here.
Share your observations with your classmates.
Extending this scheme, colour the page numbers from 101-110.
Solution:
The colour scheme is according to number of factors a particular number. The number of colours around the number in a circle is equal to the number of factors that number had.
The colouring work will be done by students themself