Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Chapter 2 Arithmetic Expressions Solutions Question Answer helps develop logical thinking and accuracy.
RBSE Class 7 Maths Chapter 2 Arithmetic Expressions Solutions
Ganita Prakash Class 7 Chapter 2 Solutions
Class 7 Maths Ganita Prakash Part 1 Chapter 2 Solutions
In-Text Questions
Page 26
Question 1.
Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? E × plain your thinking in each case.
(a) 245 + 289 ______ 246 + 285
(b) 273 – 145 ______ 272 – 144
(c) 364 + 587 ______ 363 + 589
(d) 124 + 245 _____ 129 + 245
(e) 213 – 77 ______ 214 – 76
Solution:
(a)
∴ 245 + 289 > 246 + 285
(b) 273 = 272 + 1
145 = 144 + 1
∴ 273 – 145 = 272 – 144
(c) 364 = 363 + 1
587 = 589 – 2
∴ 364 + 587 < 363 + 589
(d) 124 = 129 – 5
∴ 124 + 245 < 129 + 245
(e) 214 = 213 + 1
76 = 77 – 1
∴ 213 – 77 < 214 – 76
Page 28
Question 1.
Check if replacing subtraction by addition in this way does not change the value of expression, by taking different examples.
Solution:
E × ample 1:
23 – 14 = 9
23 + (-14) = 9
∴ 23 – 14 = 23 + (-14)
E × ample 2:
-23 – 14 = – 37
-23 + (-14) = -37
∴ -23 – 14 = -23 + (-14)
Question 2.
Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Solution:
E × ample: Let us subtract : (-7) – (-5)
Solve: (-7) – (-5) is the same as (-7) + (+5).
(-7) – (-5) = -7 + (+5)
-7 + 5 = -7 + 5 .
⇒ -2 = -2
Page 29
Question 1.
Does changing the order in which the terms are added give different values?
Solution:
No! Changing the order in which the terms are added does not give different values.
Question 2.
Will this also hold when there are terms having negative numbers as well? Take some more expressions and check
Solution:
Yes! This will also hold when there are terms having negative numbers as well. For e × ample:
(-6) + (-4) = (-10)
(-4) + (-6) = (-10)
Page 31
Question 1.
Manasa is adding a long list of numbers. it took her five minutes to add them all and she got the answer 1342 11749. Then she realised that she had forgotten to include the fourth 8611 number 9055. Does she have to 9055 start all over again?
Solution:
No! She will not have to start all over again. The required sum
= (11749) + (9055)
= 11749 + 9055
= 20804
Page 32
Question 1.
If the total number of friends goes up to 7 and the tip remains the same, how much will they have to pay? Write an expression for this situation and identify its terms.
Solution:
Total amount with the tip that they will have to pay
= 7 × 23 + 5
= (7 × 23) + 5
= 161 + 5 = 166
Thus, the total cost is ₹ 166.
The terms are 161 and 5.
Question 2.
Think and discuss why she wrote this. The expression in terms (6 × 5) + 3 is ____________
Solution:
Since the teacher called out ‘5’, therefore the students are supposed to arrange in groups of 5. That is why she wrote this.
Note that the total number of students is 3 more than 6 × 5.
Page 33
Question 1.
For each of the cases below, write the expression and identify its terms:
If the teacher had called out ‘4’, Ruby would write …………….
If the teacher had called out ‘7’, Ruby would write …………….
Solution:
If the teacher had called out ‘4’ Ruby would write 8 × 4 + 1, terms = 8 × 4 and 1
If the teacher had cal!ed out ‘7’, Ruby would write 4 × 7+ 5, terms = 4 × 7 and 5.
Page 37
Question 1.
Solution:
Page 41
Question 1.
Use this method to find the following products:
(a) 95 × 8
(b) 104 × 15
(c) 49 × 50
Is this quicker than the multiplication procedure you use generally?
Solution:
(a) 95 × 8 = (100 – 5) × 8
= 100 × 8 = 5 × 8
= 800 – 40 = 760
(b) 104 × 15 = (100 + 4) × 15
= 100 × 15 + 4 × 15
= 1500 + 60 = 1560
(c) 49 × 50 = (50 – 1) × 50
= 50 × 50 – 1 × 50
= 2500 – 50 = 2450
Yes This is quicker than the procedure we use generally.
Question 2.
Which other products might be quicker to find like the ones above?
Solution:
(a) Other products:
99 × 7 = (100 – 1) × 7
= 100 × 7 – 1 × 7
= 700 – 7 = 693
(b) 109 × 17 = (100 + 9) × 17
= 100 × 17 + 9 × 17
= 1700 + 153 = 1853
(c) 69 × 80 = (70 – 1) × 80
= 70 × 80 – 1 × 80
= 5600 – 80 = 5520
Class 7 Maths Ganita Prakash Chapter 2 Solutions
Figure it out (Page 25)
Question 1.
Fill in the blanks to make the expressions equal on both sides of the = sign:
(a) 13 + 4 = ……… + 6
(b) 22 + ………. = 6 × 5
(c) 8 × …………….. = 64 ÷ 2
(d) 34 – ………. – 25
Solution:
(a) 13 + 4 = 11+6
(b) 22 + 8 = 6 × 5
(c) 8 × 4=64-2
(d) 34 – 9 = 25
Question 2.
Arrange the following expressions in ascending (increasing) order of their values:
(a) 67 – 19
(b) 67 – 20
(c) 35 + 25
(d) 5 × 11
(e) 120 ÷ 3
Solution:
120 ÷ 3, 67 – 20, 67 – 19, 5 × 11, 35 +25
Figure it out (Page 34-35)
Question 1.
Find the values of the following expressions by writing the terms in each case.
(a) 28 – 7 + 8
Solution:
28 – 7 + 8
= 28 + (-7) + 8
= 21 + 8 = 29
Terms: 28, -7, 8
(b) 39 – 2 × 6 + 11
Solution:
39 – 2 × 6 + 11 = 39 + (-2 × 6) + 11
= 39 + (- 12) + 11
=39 +(- 1) = 38
Terms: 39, -12, 11
(c) 40 – 10 + 10 + 10
Solution:
40 – 10 + 10 + 10 = 40 +(- 10) + 10 + 10
= 40 + 0 + 10
= 40 + 10 = 50
Terms : 40, – 10, 10, 10
(d) 48 – 10 × 2 + 16 + 2
Solution:
48 – 10 × 2 + 16 ÷ 2 = 48 +(-10 × 2)+ (16 ÷ 2)
= 48 + (-20) + 8
= 48 +(- 12) = 36
Terms: 48, -10 × 2, 16 ÷ 2
(e) 6 × 3 – 4 × 8 × 5
Solution:
6 × 3 – 4 × 8 × 5 = 6 × 3 + (-4 × 8 × 5)
= 18 + (- 160) = -142
Terms: 6 × 3, -4 × 8 × 5
Question 2.
Write a story/situation for each of the following expressions and find their values.
(a) 89 + 21 – 10
Solution:
89 + 21 – 10 = 89 + 21 +(- 10)
= 89 + 11 = 100
Story : The mother of Anna gave her ₹ 89. The father of Anna gave ₹ 21. Anna gave ₹ 10 to her younger brother. Now, what amount is left with her?
(b) 5 × 12 – 6
Solution:
5 × 12 – 6 = 5 × 12 +(-6)
= 60 + (-6) = 54
Story: The cost of a pen is ₹ 5. Apala purchased 12 pens from a book seller. The book seller allowed her a rebate of ₹ 6 on the purchase these pens. What amount has to be paid to the book seller by Apala for the purchase of these pens?
(c) 4 × 9 + 2 × 6
Solution:
4 × 9 + 2 × 6 = (4 × 9)+(2 × 6)
= (36) + (12) = 48
Story : Shalini purchased four note books costing ₹ 9 each and two scales costing ₹ 6 each. Find the total amount to be paid by Shalini to the shopkeeper for the purchase of note books and scales.
Question 3.
For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(a) Queen Alla gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
Solution:
Expression describing how many gold coins Princess Elsa and Princess Anna together have
= 100 × 2 + 100 ÷ 2
= 200 + 50 = 250
Terms: 100 × 2, 100 ÷ 2
(b) A metro train ticket between two stations is ₹ 40 for an adult and ₹ 20 for a child. What is the total cost of tickets:
(i) for four adults and three children?
(ii) for two groups having three adults each?
Solution:
(i) Total cost of tickets for four adults and three children
= 40 × 4 + 20 × 3 = 160 + 60 = 220
Term : 40 × 4, 20 × 3
(ii) Total cost of tickets for two groups having three adults each
= 40 × 3 + 40 × 3 = 120 + 120 = 240
Terms : 40 × 3, 40 × 3
(c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.
Solution:
Total height of the window
= 3 + 5 + 2 + 5 + 2 + 5 + 2 + 5 + 2 + 5 + 2 + 5 + 2 + 5 + 3
= 2 × 6 + 5 × 7 + 3 × 2
= 12 + 35 + 6 = 12 + 41 = 53cm
Terms : 2 × 7, 5 × 7, 3 × 2
Width of border = 3 cm
Width of grill = 2 cm
Width of gap = 5 cm
Figure it Out (Pages 37-38)
Question 1.
Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.
(a) 24 + (6 – 4) = 24 + 6 ☐ ____
(b) 38 + (___ ☐ ______) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 ☐ 6 – 4
(d) 24 – 6 – 4 = 24 ☐ 6 – 4
(e) 27 – (8 + 3) = 27 8 3
(f) 27 – ( ____ ☐ ____ ) = 27 – 8 + 3
Solution:
(a) 24 + (6 – 4) = 24 + 6 – 4
(b) 38 + (9 – 4) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 – 6 – 4
(d) 24 – 6 – 4 = 24 – 6 – 4
(e) 27 – (8 + 3) = 27 – 8 – 3
(f) 27 -( 8 – 13 ) = 27 – 8 + 3
Question 2.
Remove the brackets and write the expression having the same value.
(a) 14 + (12 + 10)
(b) 14 – (12 + 10)
(c) 14 + (12 – 10)
(d) 14 – (12 – 10)
(e) -14 + 12 – 10
(f) 14 – (-12 – 10)
Solution:
(a) 14 + 12 + 1o
(b) 14 – 12 – 10
(e) 14 + 12 – 10
(d) 14 – 12 + 10
(e) -14 + 12 – 10
(f) 14 + 12 + 10
Question 3.
Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(a) (6 + 10) – 2 and 6 + (10 – 2)
Solution:
(6 + 10) – 2 = 16 – 2 = 14
6 + (10 – 2) = 6 + 8 = 14
∴ (6 + 10) – 2 = 6 + (10 – 2)
(b) 16 – (8 – 3) and (16 – 8) – 3
Solution:
16 – (8 – 3) = 16 – 8 + 3 = 11
(16 – 8) – 3 = 16 – 8 – 3 = 5
∴ 16 – (8 – 3) ≠ (16 – 8) – 3
(c) 27 – (18 + 4) and 27 + (-18 – 4)
Solution:
27 – (18 + 4) = 27 – 18 – 4 = 5
27 + (-18 – 4) = 27 – 18 – 4 = 5
∴ 27 – (18 + 4) = 27 + (-18 – 4)
Question 4.
In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms.
(a) 319 + 537, 319 – 537, – 537 + 319, 537 – 319
(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109
Solution:
(a) 319 – 537 – 537 + 319
(b) 87 – 46 + 109 (87 – 46) + 109;
87 + 46 – 109 = 87 + 46 – 109
= 87 + 46 – 109
Question 5.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(a) 34 – 9 + 12 = 13
(b) 56 – 14 – 8 = 34
(c) – 22 – 12 + 10 + 22 = – 22
Solution:
(a) 34 – (9 + 12) = 13
(b) 56 + (-14 – 8) = 34
(c) (-22 + 22) – (12 + 10) – 22
Question 6.
Using only reasoning of how terms change their values fill the blanks to make the expressions on either side of the equality (=) equal.
(a) 423 + …….. = 419 + …………….
(b) 207 – 68 = 210 – ……………..
Solution:
(a) 423 + 419 = 419 + 423
(b) 207 – 68 = 210 – 71
Question 7.
Using the numbers 2, 3 and 5, and the operators, ‘+‘ and ‘-‘, and brackets, as necessary, generate expressions to give as many different values as possible. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0.
Solution:
2 + 3 + 5 = 10;
2 + (3 – 5) = 0;
(-2 + 3) + 5 = 6;
3 – (5 + 2) = -4;
(-2 – 3) + 5 = 0;
-2 – 3 – 5 = -10;
(2 – 3) – 5 = -6
Question 8.
Whenever Jasoda bas to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1.
(a) Do you think she always gets the correct answer? Why?
(b) Can you think of other similar strategies? Give some examples.
Solution:
(a) Yes! She always gets the correct answer because -10 + 1 = -9
(b) Yes, other similar strategies:
• To subtract 11, subtract 10 then subtract 1.
Example: 36 – 11 = 26 – 1 = 25
• To subtract 19, subtract 20 then add 1.
Example: 36 – 19 = 16 + 1 = 17
Question 9.
Consider the two expressions: (a) 73 – 14 + 1, (b) 73 – 14 – 1. For each of these expressions, identify the expressions from the following collection that are equal to it.
(a) 73 – (14 + 1)
(b) 73 – (14 – 1)
(c) 73 + (-14 + 1)
(d) 73 + (- 14 – 1)
Solution:
Given,
(a) 73 – 14 + 1 = 74 – 14 = 60
(b) 73 – 14 – 1 = 73 – 15 = 58
Now, (a) 73 – (14 + 1) = 73 – 15 = 58
(b) 73 – (14 – 1) = 73 – 13 = 60
(c) 73 + (-14 + 1) = 73 – 13 = 60
(d) 73 + (-14 – 1) = 73 – 15 = 58
Hence (a) and (d) are equal to 73 – 14- 1
(b) and(c) are equal to 73 – 14 + 1
Figure it Out (Pages 41-42)
Question 1.
Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.
(a) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 ☐ 3 × ____
(d) (9 + 2) × 4 = 9 × 4 ☐ 2 ×____
(e) 3 × (____ + 4) = 3 ____+____
(f) (____+ 6) × 4 = 13 × 4 + ____
(g) 3 × (____+____) = 3 × 5 + 3 × 2
(h) (____+____)×____= 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × ____
(j) (5 – 2) × 7 = 5 × 7 – 2 × ____
(k) 5 × (8 – 3) = 5 × 8 ☐ 5 × ____
(l) (8 – 3) × 7 = 8 × 7 ☐ 3 × 7
(m) 5 × (12 – ____) =____ ☐ 5 ×____
(n) (15 – ____) × 7 =____ ☐ 6 × 7
(o) 5 × (____ – ____) = 5 × 9 – 5 × 4
(p) (____ – ____) × ____= 17 × 7 – 9 × 7
Solution:
(a) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) > (48 × 4 + 3 × 4
(c) × 5 + 8) = 3 × 5 + 3 × 8
(d) (9 + 2) × 4 = 9 × 4 + 2 × 4
(e) 3 × (7 + 4) = 30 + 3
(f) (13 + 6) × 4 = 13 × 4 + 24
(g) 3 × (5 + 2) = 3 × 5 + 3 × 2
(h) (2 + 3) × 4 = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × 2
(j) (5 – 2) × 7 = 5 × 7 – 2 × 7
(k) 5 × (8 – 3) = 5 × 8 – 5 × 3
(l) (8 – 3) × 7 = 8 × 7 – 3 × 7
(m) 5 × (12 – 4) = 60 – 5 × 4
(n) (15 – 6) × 7 = 105 – 6 × 7
(o) 5 × (9 – 4) = 5 × 9 – 5 × 4
(p) (17 – 9) × 7 = 17 × 7 – 9 × 7
Question 2.
In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.
(a) (8 – 3) × 29 ☐ (3 – 8) × 29
(b) 15 + 9 × 18 ☐ (15 + 9) × 18
(c) 23 × (17 – 9) ☐ 23 × 17 + 23 × 9
(d) (34 – 28) × 42 ☐ 34 × 42 – 28 × 42
Solution:
(a) (8 – 3) × 29 > (3 – 8) × 29
(b) 15 + 9 × 18 < (15 + 9) × 18
(c) 23 × (17 – 9) < 23 × 17 + 23 × 9
(d) (34 – 28) × 42 = 34 × 42 – 28 × 42
Question 3.
Here is one way to make 14: 2 × (1 + 6 ) = 14. Are there other ways of getting 14? Fill them out below:
(a) _____× (_____+_____) = 14
(b) _____× (_____+_____) = 14
(c) _____× (_____+_____) = 14
(d) _____× (_____+_____) = 14
Solution:
(a) 2 × (2 + 5) = 14
(b) 2 × (3 + 4) = 14
(c) 2 × (6 + 1) = 14
(d) 2 × (1 + 1) = 14
Question 4.
Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.
Solution:
First Picture
Sum of the numbers 4 × 5 + 8 × 4
= 20 + 32 = 52
OR
Sum of the numbers = (4 + 8) × 4 + 4
= 48 + 4 = 52
Second Picture
Sum of the numbers = 5 × 8 + 6 × 8
= 40 + 48 = 88
OR
Sum of the numbers 4 × (5 × 2 + 6 × 2)
= 4 × (10 + 12)
= 4 × 22 = 88
Figure it out (Pages 42-43)
Question 1.
Read the situations given below. Write appropriate expressions for each of them and find their values.
(a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
Solution:
Amount of mangoes supplied by Rahim and Shyam in a week to the local district market
= (9 × 7 + 11 × 7) kg
(63 + 77) kg 140 kg
(b) Binu earns ₹ 20,000 per month. She spends ₹ 5,000 on rent, ₹ 5,000 on food, and ₹ 2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
Solution:
The amount Binu will save by the end of a year
= 12 × [₹ 20,000 – (₹ 5,000 + ₹ 5,000 + ₹ 2,000)]
= 12 × [₹ 20,000 – ₹ 12,000]
= 12 × ₹ 8,000 = ₹ 96,000
(c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. in how many days will the snail get the treat?
Solution:
Height climbed up by a snail in 1 day
= 3 cm – 2 cm = 1 cm
Height of post = 10 cm
In 7days = 7 × 1 = 7cm
On 8th day = 7 + 3 = 10 cm
∴ The snail will get the treat in 8 days.
Question 2.
Melvin reads a two-page story everyday except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario?
(a) 5 × 2 × 8
(b) (7 – 2) × 8
(c) 8 × 7
(d) 7 × 2 × 8
(e) 7 × 5 – 2
(f) (7 + 2) × 8
(g) 7 × 8 – 2 × 8
(h) (7 – 5) × 8
Solution:
Number of stories, whose reading is completed, in a week
= 7 – 2 = 5
•. There are 7 days in a week and he does not read story on Tuesdays and Saturdays.
:. Number of stories, whose reading is completed, in 8 weeks
= 5 × 8
= (7 – 2) × 8 = 7 × 8 – 2 × 8
Hence, the expression (b) and (g) describes the scenario.
Question 3.
Find different ways of evaluating the following expressions:
(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
Solution:
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)
= (-1) + (- 1) + (-1) + (-1) + (- 1)
= 5 × (-1) = -5 .
OR
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= (1 +3 + 5 + 7 + 9) – (2 + 4 + 6 + 8 + 10)
= 25 – 30
= -5
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Solution:
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)
= 0 + 0 + 0 +0 + 0
= 0 × 5 – 0
OR
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 + 1 + 1 + 1 + 1) + (-1 – 1 – 1 – 1 – 1)
= 5 + (-5)
= 5 – 5 = 0
Question 4.
Compare the following pairs of expressions using ‘<‘, ‘>‘ or ‘=’. or by reasoning.
(a) 49 – 7 + 8 ________ 49 – 7 + 8
(b) 83 × 42 – 18 ________ 83 × 40 – 18
(c) 145 – 17 × 8 ________ 145 – 17 × 6
(d) 23 × 48 – 35 ________ 23 × (48 – 35)
(e) (16 – 11) × 12 ________ –11 × 12 + 16 × 12
(f) (76 – 53) × 88 ________ 88 × (53 – 76)
(g) 25 × (42 + 16) ________ 25 × (43 + 15)
(h) 36 × (28 – 16) ________ 35 × (27 – 15)
Solution:
(a) 49 – 7 + 8 = 49 – 7 + 8
(b) 83 × 42 – 18 > 83 × 40 – 18
(c) 145 – 17 × 8 < 145 – 17 × 6 (d) 23 × 48 – 35 > 23 × (48 – 35)
(e) (16 – 11) × 12 = -11 × 12 + 16 × 12
(f) (76 – 53) × 88 > 88 × (53 – 76)
(g) 25 (42 + 16) = 25 × (43 + 15)
(h) 36 × (28 – 16) > 35 × (27 – 15)
Question 5.
Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
(a) 83 – 37 – 12
(1) 84-38-12
(ii) 84 – (37 + 12)
(iii) 83 – 38 – 13
(iv)- 37 + 83 – 12
(b) 93 + 37 × 44 + 76
(i) 37 + 93 × 44 + 76
(ii) 93 + 37 × 76 + 44
(iii) (93 + 37) × (44 + 76)
(iv) 37 × 44 + 93 + 76
Solution:
(a) 83 – 37 – 12 = (i) 84 – 38 – 12
= (iv) -37 + 83 – 12
(b) 93 + 37 × 44 + 76 = (iv) 37 × 44 + 93 + 76
Question 6.
Choose a number and create ten different expressions having that value.
Solution:
let us choose a number 2. Then ten different expressions having the value 2 are:
(i) 99 – 97
(ii) 100 – 2 × 49
(iii) 50 – 24 × 2
(iv) 1 + 1
(v) 6 × 2 – 2
(vi) (10 – 9) × 2
(vii) (25 – 24) × 2
(viii) 3 × 5 × 4 – 2 × 29
(ix) 7 × 8 – 27 × 2
(x) 8 × 5 – 2 × 19
Expression Engineer! (Page 4)
Question 1.
Using three 3’s along with the four operations (addition, subtraction, multiplication, and division) and brackets as needed we can create several expressions. For example, (3 + 3)/3 = 2, 3 + 3 – 3 = 3, 3 × 3 + 3 = 12, and so on.
(i) Using four 4’s, create expressions to get all values from 1 to 20.
(ii) Using the numbers 1, 2, 3, 4 and 5 exactly once in any order get as many values as possible between -10 and +10.
(iii) Using the numbers 0 to 9 exactly once in any order, make an expression with a value 100.
Solution:
(i) \(\frac {4+4}{4+4}\) = 1
\(\frac {4+4}{4}\) = 2
\(\frac {4+4+4}{4}\) = 3, etc.
(ii) 1 + 2 + 3 + 4 – 5= 5
1 – 2 + 3 – 4 + 5 = 3
1 × 2 + 3 + 4 + 5 = 14,etc.
(iii) (9 + 8 + 7 + 6 ) × 3 + 5 + 4+ 2 – 1 = 100, etc.
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