Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Chapter 4 Expressions using Letter Numbers Solutions Question Answer helps develop logical thinking and accuracy.
RBSE Class 7 Maths Chapter 4 Expressions using Letter Numbers Solutions
Ganita Prakash Class 7 Chapter 4 Solutions
Class 7 Maths Ganita Prakash Part 1 Chapter 4 Solutions
In-text Questions
Page 85
Question 1.
Find the values of the following expressions:
1. 23 – 10 × 2
2. 83 + 28 – 13 + 32
3. 34 – 14 + 20
4. 42 + 15 – (8 – 7)
5. 68 – (18 + 13)
6. 7 × 4 + 9 × 6
7. 20 + 8 × (16 – 6)
Solution:
1. 23 – 10 × 2 = 23 + (-10 × 2)
= 23 + (-20)
= 23 – 20 = 3
2. 83 + 28 – 13 + 32
3. 34 – 14 + 20 = (34 – 14) + 20
= 20 + 20 = 40
4. 42 + 15 -(8 – 7) = 42 + 15 – 1
= 42 + (15 – 1)
= 42 + 14 = 56
5. 68 – (18 + 13)
6. 7 × 4 + 9 × 6 = 28 + 54 = 82
7. 20 + 8 × (16 – 6) = 20 + 8 × 10
= 20 + 80= 100
Page 87
Question 1.
Observe each of them and identify if there is a mistake.
Question 2.
If you think there is a mistake, try to explain what might have gone wrong.
Question 3.
Then, correct it and give the value of the expression.
Solution:
1. 10 – a = 10 – (-4)
= 10 + 4= 14
2. 3d = 3 × 6 = 18
3. 3s – 2 = 3 × (7) – 2
= 21 – 2 = 19
4. 2r + 1 = 2 × 8 + 1
= 16 + 1
= 17
5. 2j = 2 × 5 = 10
6. 3(m + 1) = 3(-6 + 1)
= 3(-5)
= – 15
7. 2f – 2g = 2 × 3 – 2 × 1
= 6 – 2 = 4
8. 2t + b = 2 × 4 + 3
= 8 + 3 = 11
9. h – (3 – n) = 5 – (3 – 6)
= 5 – (-3)
= 5 + 3 = 8
Page 94
Question 1.
Some simplifications of algebraic expressions are done below. The expression on the right-hand side should be in its simplest form.
- Observe each of them and see if there is a mistake.
- If you think there is a mistake, try to explain what might have gone wrong.
- Then, simplify it correctly.
| Expression | Simplest Form | Correct Simplest Form |
| 1. 3a + 2b | 5 | |
| 2. 3b – 2b – b | 0 | |
| 3. 6(p + 2) | 6p + 8 | |
| 4. (4x + 3y) – (3x + 4y) | x + y | |
| 5. 5 – (2 – 6z) | 3 – 6z | |
| 6. 2 + (x + 3) | 2x – 6 | |
| 7. 2y + (3y – 6) | -y + 6 | |
| 8. 7p – p + 5q – 2q | 7p + 3q | |
| 9. 5 (2w + 3x + 4w) | 10w + 15x + 20w | |
| 10. 3j + 6k + 9h + 12 | 3(j + 2k + 3h + 4) | |
| 11. 4(2r + 3s + 5) | -20 – 8r – 12s |
Solution:
Correct simplest form
1. 3a + 2b
2. 0
3. 6p + 12
4. x – y
5. 3 + 6z
6. x + 5
7. 5y – 6
8. 6p + 3q
9. 10w + 15x + 20w
10. 3(j + 2k + 3h + 4)
11. 8r + 12s + 20
Page 95
Question 1.
Find out the formula of this number machine.
Solution:
The formula for the number machine given above is “two times the first number minus the second number”
Expression = 2a – b
= 2 × 6 – 4
= 8
Page 96
Question 1.
Find the formulas of the number machines below and write the expression for each set of inputs.
Solution:
(i) Adding the second number to the first number and subtracting 2.
Expression = a + b – 2
(ii) First number multiplied by second number then adding 1.
Expression = ab + 1
= 10 × 3 + 1 = 31
Class 7 Maths Ganita Prakash Chapter 4 Solutions
Figure it out (Page 84)
Question 1.
Write formulas for the perimeter of:
(a) triangle with all sides equal.
(b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all sidelengths and angle measures are equal)
(c) a regular hexagon
Solution:
(a) Perimeter of triangle = 3 × a
Where a denotes the length of each of the equal sides.
(b) Perimeter of a regular pentagon = 5 × a
Where a denotes the length of each of the equal sides.
(c) Perimeter of a regular hexagon = 6 × a
Where a stands for the length of each of the equal sides.
Question 2.
Munirathna has a 20 m long pipe. However, he wants a longer watering for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.
Solution:
Combined length of the pipe = 20 m + length of the other pipe
= 20 m + k m
= (20 + k) m
Question 3.
What is the total amount Krithika has, if she has the following numbers of notes of ₹ 100, ₹ 20 and ₹ 5? Complete the following table:
Solution:
| No. of ₹100 notes | No. of ₹ 20 notes | No. of ₹ 5 notes | Expression and total amount |
| 3 | 5 | 6 | 3 × 100 + 5 × 20 + 6 × 5 = 300 + 100 + 30 = ₹ 430 |
| 6 | 4 | 3 | 6 × 100 + 4 × 20 + 3 × 5 = ₹ 695 |
| 8 | 4 | z | 8 × 100 + 4 × 20 + z × 5 = 800 + 80 + 5z = ₹(880 + 5z) |
| x | y | z | x × 100 + y × 20 + z × 5 = ₹(100x + 20y + 5z) |
Question 4.
Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially?
(a) 10 + 8 + y
(b) (10 + 8) × y
(c) 10 × 8 × y
(d) 10 + 8 × y
(e) 10 × y + 8
Solution:
Time taken by the roller mill to start running = 10 seconds
∵ Time taken to grind 1 kg of grain into powder = 8 seconds
∴ Time taken to grind y kg of grain into powder = 8 × y seconds
∴ Time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially
= 10 + 8 × y
Hence, the expression (d) 10 + 8 × y is the correct expression.
Question 5.
Write algebraic expressions using letters of your choice.
(a) 5 more than a number
(b) 4 less than a number
(c) 2 less than 13 times a number
(d) 13 less than 2 times a number
Solution:
Let the letter n denotes the given number. Then, the algebraic expression for:
(a) 5 more than a number = x + 5
(b) 4 less than a number = x – 4
(c) 2 less than 13 times a number = 13x – 2
(d) 13 less than 2 times a number = 2x – 13
Question 6.
Describe situations corresponding to the following algebraic expressions :
(a) 8 × x + 3 × y
(b) 15 × j – 2 × k
Solution:
(a) A pen costs f 8 and a pencil cost ₹ 3. Find an expression for the total cost of x pens and y pencils.
(b) The cost of a mango is ₹ 15 and the cost of a banana is ₹ 2. How much is the cost of j mangoes greater than the cost of k bananas, given that j > k.
Question 7.
In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.
Solution:
| w – 8 | w – 7 | w – 6 |
| w – 1 | w | w + 1 |
Figure it Out (Pages 93-94)
Question 1.
Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing.
Solution:
(1) Adding row-wise gives:
5y + (- 6) + x + x + 2 + 5y
= (- 6) + 2 + x + x + 5y + 5y
Adding like terms
= – 4 + 2x + 10y
Adding column-wise gives:
5y + x + (- 6) + 2 + x + 5y
= (- 6) + 2 + x + x + 5y + 5y
Adding like terms
= -4 + 2x+10y
Adding the like terms together gives:
(- 6) + 2 + x + x + 5y + 5y
= – 4 + 2x + 10y
We see that we get the same thing.
(2) Adding row-wise gives:
2p + 3q + (-2) + 3 + 3q + 2p + 3 + (-2) + 2p + 3q + 3q + 2p
= (-2) + 3 + 3 + (-2) + 2p + 2p + 2p + 2p + 3q + 3q + 3q + 3 q
Adding like terms
= 2 + 8p + 12 q
Adding column-wise gives:
2p + 3q + 3q + 2p + (-2) + 3 + 2p + 3q + 3 + (-2) + 3q + 2p
= (-2) + 3 + 3 + (-2) + 2p + 2p + 2p + 2p + 3q + 3q + 3q + 3q
Adding like terms
= 2 + 8p + 12q
Adding like terms together gives:
(-2) + 3 + 3 + (-2) + 2p + 2p + 2p + 2p + 3q + 3q + 3q + 3q
= 2 + 8p + 12 q
We see that we get the same thing.
(3) Adding row-wise gives:
(-5g) + 5k + 5k + (-5g) + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + (-5g) + 5k + 5k + (-5g)
= (-5 g) + (-5 g) + (-5 g) + (-5 g) + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k
Adding like terms
= -20g + 60k
Adding column-wise gives:
(-5g) + 5k + 5k + (-5 g) + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + (-5g) + 5k + 5k + (-5g)
= (-5g) + (-5g) + (-5g) + (-5g) + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k + 5k
= -20g + 60 k.
Adding the upper half and doubling gives:
= 2[6 × (5k) + 2 × (-5g)]
= 2[30k – 10g] = 60k – 20g
We see that we get the same thing.
Question 2.
Simplify each of the following expressions :
(a) p + p + p + p, p + p + p + q, p + q + p – q
(b) p – q + p – q, p + q – p + q
(c) p + q – (p + q), p – q – p – q
(d) 2d – d -d – d, 2d – d – d – c
(e) 2d – d – (d – c), 2d – (d – d) – c
(f) 2d – d – c – c
Solution:
(a) 4p, 3p + q, 2p
(b) 2p – 2q, + 2q
(c) 0, -2q
(d) -d, -c
(e) c, 2d – c
(f) d – 2c
Figure it Out (Pages 102-105)
For the problems asking you to find suitable expression(s), first try to understand the relationship between the different quantities in the situation described. If required, assume some values for the unknowns and try to find the relationship.
Question 1.
One plate of Jowar rod costs ₹ 30 and one plate of pulao costs ₹ 20. If x plates of Jowar rod and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?
(a) 30x + 20y
(b) (30 + 20) × (x + y)
(c) 20x + 30y
(d) (30 + 20) × x + y
(e) 30x – 20y
Solution:
∵ Cost of one plate of Jowar roti = ₹ 30
∴ Cost of x plates of Jowar rod = ₹ 30 x
∵ Cost of one plate of pulao = ₹ 20
∴ Cost of y plates of pulao = ₹ 20y
∴ The expression (a) 30x + 20y describes the total amount of rupees earned that day.
Option (a) is correct.
Question 2.
Pushpita sells two types of flowers on Independence day: champak and marigold.’p’ customers only bought champak,’q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day?
(a) p + q + r
(b) p + q + 2r
(c) 2 × (p + q + r)
(d) p + q + r + 2
(e) p + q + r + 1
(f) 2 × (p + q)
Solution:
Number of customers who bought only champak flowers =p
Number of customers who bought only marigold flowers=q
∴ No. of customers who bought both flowers = r
∴ Number of flags that she gave away that day = p + q + r
∴ Option (a) is correct.
Question 3.
A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.
(a) Write an expression describing how far away the snail is from its starting position.
(b) What can we say about the snail’s movement if d > u?
Solution:
(a) Distance climbed up in 1 day = u cm
Distance slipped down in 1 day = d cm
∴ Distance climbed up in 1 day and 1 night = (u – d) cm If u > d
∴ Distance climbed up in 10 days and 10 nights = 10 (u – d) cm
This is the required expression describing how far away the snail in from its starting position,
(b) The snail will not climb at all if d > u, i.e, the snail will remain at its starting position.
Question 4.
Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?
Solution:
In the first week Y
∵ Distance cycled everyday = 5 km
∴ Distance cycled in the first week
= 5 × 7 km = 35 km
In the second week
∵ Distance cycled everyday = (5 + z) km
∴ Distance cycled in the second week
= (5 + z) 7 km
In the third week
∵ Distance cycled everyday = (5 + z + z) km
= (5 + 2z) km
∴ Distance cycled in the third week
= ( 5 + 2z) 7 km
∴ Total distance cycled in 3 weeks
= 35 km + (5 + z) 7 km + (5 + 2z) 7 km
= (35 + 35 + 7z + 35 + 14z) km
= (35 + 35 + 35 + 7z + 14z) km
Grouping like terms
= (105 + 21z) km
Hence, Radha has cycled (105 + 21z) km after 3 weeks.
Question 5.
In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.
Solution:
Question 6.
A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.
(a) It t = 4, what is the time taken to travel from Yahapur to Vahapur?
(b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation]
Solution:
(a) Time taken from station A to station C = t minutes
Time of stoppage at station C = 2 minutes Time taken from station C to station D = t minutes
Time of stoppage at station D = 2 minutes Time taken from station D to station E = t minutes
Time of stoppage at station E = 2 minutes
Time taken from station E to station B = t minutes
∴ Total time taken to travel from Yahapur to Vahapur = (t + 2 + t + 2 + t + 2 + t) minutes
= (t + t + t + t + 2 + 2 + 2) minutes
Grouping like terms
= (4t + 6) minutes
= (4 × 4 + 6) If t = 4
= 22 minutes
(b) Algebraic expression for the time taken to travel from Yahapur to Vahapur = (4t + 6) minutes
Question 7.
Simplify the following expressions:
(a) 3a + 9b – 6 + &a – 4b – 7a + 16
(b) 3 (3a – 3b) – 8a – 46 – 16
(c) 2 (2x – 3) + + 12
(d) 8x – (2x – 3) + 12
(e) 8h – (5 + 7h) + 9
(f) 23 + 4(6m – 3n) – 8n – 3m – 18
Solution:
(a) 3a + 9b – 6 + 8a – 4b – 7a + 16
= 3a + 8a – 7a + 9b – 4b – 6 + 16
Grouping like terms
= 4a + 56+10
(b) 3 (3a – 3b) – 8a – 4b – 16
= 9a – 9b – 8a – 4b – 16
= 9a – 8a – 9b – 4b – 16
Grouping like terms
= a – 13b – 16
(c) 2 (2x – 3) + 8x + 12
= 4x – 6 + 8x+ 12
= 4x + 8x – 6 + 12
Grouping like terms
= 12x + 6
(d) 8x – (2x – 3) + 12
= 8x – 2x + 3 + 12
= 6x + 15
(e) 8h – (5 + 7h) + 9
= 8h – 5 – 7h + 9
= 86 – 7h – 5 + 9
Grouping like terms
= h + 4
(f) 23 + 4(6m – 3n) – 8n – 3m – 18
= 23 + 24m – 12n – 8n – 3m – 18
= 24m – 3m – 12n – 8n + 23 – 18
Grouping like terms
= 21m – 20n + 5
Question 8.
Add the expressions given below:
(a) 4d – 1c + 9 and 8c – 11 + 9d
Solution:
(4d – 7c + 9) + (8c – 11 + 9d)
= (4d + 9d) + (-7c + 8c) + 9 – 11
Grouping like terms
= 13d + c – 2
(b) -6f + 19 – 8s and – 23 + 13f + 12s
Solution:
(-6f + 19 – 8s) + (-23 + 13f + 12s)
= -6f + 13f – 8s + 12s + 19 – 23
Grouping like terms
= 7f + 4s – 4
(c) 8d – 14c + 9 and 16c – (11 + 9d)
Solution:
(8d – 14c + 9) + {16c – (11 + 9d)}
= 8d – 14c + 9 + 16c – 11 – 9d
= 9 – 11 + 8d – 9d – 14c + 16c
Grouping like terms
= -2 – d + 2c
(d) 6f – 20 + 8s and 23 – 13f – 12s
Solution:
(6f – 20 + 8s) + (23 – 13f – 12s)
= 6f – 13f + 8s – 12s – 20 + 23
Grouping like terms
= -7f – 4s + 3
(e) 13m – 12n and 12n – 13m
Solution:
(13m – 12n) + (12n – 13m)
= (13m – 13m) + (-12n + 12 n)
Grouping like terms
= 0 + 0 = 0
(f) -26m + 24n and 26m – 24n
Solution:
(-26 m + 24n) + (26 m – 24 n)
= (-26m + 26m) + (24n – 24n)
Grouping like terms
= 0 + 0 = 0
Question 9.
Substract the expressions given below :
(a) 9a – 6b + 14 from 6a + 9b- 18
Solution:
(6a + 9b – 18) – (9a – 6b + 14)
= 6a + 9b – 18 – 9a + 6b – 14
= 6a – 9a + 9b + 6b – 18 – 14
Grouping like terms
= -3a + 15b – 32
(b) -15x + 13 – 9y from 7y – 10 + 3x
Solution:
(7y – 10 + 3x) – (-15x + 13 – 9y)
= 7y – 10 + 3x + 15x – 13 + 9y
= 3x + 15x + 7y + 9y – 10 – 13
Grouping like terms
= 18x+ 16y – 23
(c) 17g + 9 – 7h from 11 – 10g + 3h
Solution:
(11 – 10g + 3h) – (17g + 9 – 7h)
= 11 – 10g + 3h – 17g – 9 + 7h
= 3h + 7h – 10g – 11g + 11 – 9
Grouping like terms
= 10h – 27g + 2
(d) 9a – 6b + 14 from 6a – (9b + 18)
Solution:
[6a – (9b + 18)] – (9a – 6b + 14)
= (6a – 9b – 18) – (9a – 6b + 14)
= 6a – 9b – 18 – 9a + 6b – 14
= 6a – 9a – 9b + 6b – 18 – 14
Grouping like terms
= -3a – 3b – 32
(e) 10x + 2 + 10y from -3y + 8 – 3x
Solution:
(-3y + 8 – 3x) – (10x + 2 + 10y)
= -3y + 8 – 3x – 10x – 2 – 10y
= -3x – 10x – 3y – 10y + 8 – 2
= -13x – 13y + 6
(f) 8g + 4h – 10 from 7h – 8g + 20
Solution:
(7h – 8g + 20) – (8g + 4h – 10)
= 7h – 8g + 20 – 8g – 4h + 10
= 7h – 4h – 8g – 8g + 20 + 10
Grouping like terms
= 3h – 16g + 30
Question 10.
Describe situations corresponding to the following algebraic expressions:
(a) 8x + 3y
(b) 15x – 2x
Solution:
(a) The cost of a pen is ₹ 8 and the cost of a pencil is ₹ 3. Meenu purchased 8 pens and 3 pencils. Write an algebraic expression for the total amount Meenu will have to pay to the shopkeeper.
(b) Arun sold 15 handkerchiefs costing ₹ x each from Varun. Later on, Varun returned 2 handkerchiefs of these to Arun. Write an algebraic expression for the amount to be paid by Varun to Arun to purchase the remaining handkerchiefs.
Question 11.
Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?
Solution:
Number of pieces if the rope is cut once = 2
Number of pieces if the rope is folded once and then cut = 3 = 2 + 1
Number of pieces if the rope is folded 2 times and then cut = 4 = 2 + 2
Number of pieces if the rope is folded 10 times and then cut = 2 + 10 = 12
Expression for the number of pieces when the rope is folded r times and then cut = 2 + r
Question 12.
Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?
Solution:
Number of matchsticks required to make 1 square = 4
Number of matchsticks required to make 2 squares = 4 + 3
Number of matchsticks required to make 3 squares = 4 + 2 × 3
∴ Number of matchsticks required to make w squares = 4 + (w – 1) × 3
Question 13.
Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below.
Find the colour at positions 90,190 and 343. Write expressions to describe the positions for each colour.
Solution:
Expression to describe the position of red colour (1, 5, 9,……..)
= 4n – 3
Expression to describe the position of yellow colour (2,4, 6,…….)
= 27n
Expression to describe the position of green colour (3, 7, 11, …………)
= 4n – 1
Position 90
4n – 3 = 90 does not give a natural number value of n.
4n – 1 = 90 does not give a natural number value of n.
2n = 90 ⇒ n = 45 which is a natural number. Colour at position 90 = Yellow.
Position 190
2n – 3 = 190 does not give a natural number value of n.
2n – 1 = 190 does not give a natural number value of w.
2n = 190 ⇒ n = 95 which is a natural number.
∴ Colour at position 190 = Yellow.
Position 343
2n – 3 = 343 does not give a natural value of n. 2n = 343 does not give a natural value of n.
2n – 1 = 343 ⇒ 2n = 344 ⇒ n = 86 which is a natural number.
∴ Colour at position 343 = Green.
Question 14.
Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?
Solution:
Number of squares in step 1 = 5
Number of squares in step 2 = 9 = 5+ 4 = 5 + 1(4)
Number of squares in step 3 = 13 = 5 + 2(4) .
Number of squares in step n
= 5 + (n – 1)4
= 5 + 4n – 4 = 4n + 1
This gives the general formula.
Number of vertices of squares in step 1 = 20
Number of vertices of squares in step 2 = 36 = 20+16 = 20 + 1(16)
Number of vertices of squares in step 3 = 52 = 20 + 32 = 20 + 2(16)
Number of vertices of squares in step n = 20 + (n – 1) (16)
= 20 + 16n – 16
= 16n + 4
Aliter — Formula for the number of vertices of all the squares in step n
= 4(4n + 1) = 16n + 4
∵ A square has 4 vertices
Question 15.
Numbers are written in a particular sequence in this endless 4-column grid.
(a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4).
(b) In which row and column will the following numbers appear:
(i) 124, (ii) 147, (iii) 201
(c) What number appears in row r and column c?
(d) Observe the positions of multiples of 3.
Do you see any pattern in it? List other patterns that you see.
Solution:
(a) Expression to generate all the number given in column 1,i.e., 1, 5, 9, 13, ,i.e., 1,
1 + 4, 1 + 4 + 4, 1 + 4 + 4 + 4, ………….,
i. e., 1 + 0(4), 1 + 1(4), 1 + 2(4), 1 + 3(4), …………. = 1 + (n – 1)4
= 4n – 3
The numbers in column 2 are 2, 6, 10, 14,
i. e., 2, 2 + 4, 2 + 4 + 4, 2 + 4 + 4 + 4,……….
i.e., 2, 2 + 4, 2 + 2(4), 2 + 3(4), ……………
i.e., 2 + 0(4), 2 + 1(4), 2 + 2(4), 2 + 3(4), ………….
∴ Expression to generate all the numbers in column 2
= 2 + 4 (n – 1) = 4n – 2
The numbers in column 3 are:
3, 7, 11, 15,…………
i.e., 3 + 0(4) , 3 + 1(4), 3 + 2(4), 3 + 3(4),
∴ Expression to generate all the numbers in column 3
= 3 + (n – 1)4
= 3 + 4n – 4 = 4n – 1
The numbers in column 4 are 4, 8, 12, 16, ………..
= 4(1), 4(2), 4(3), 4(4),……
∴ Expression to generate all the numbers in column 4 = 4 n
(b) (i) Only 4n = 124 gives the natural number value of n, i.e., n = 31
∴ 124 will appear in 4th column at 31st position.
∴ 124 will appear in 31st row and 4th column.
(ii) Only 4n – 1 = 147 gives the natural number value of n, i.e., n = 37
∴ 147 will appear in third column at 37th position.
∴ 147 will appear in 37th row and 3rd column.
(iii) Only 4n – 3 = 201 will give the natural number value of n, i.e., n = 51
∴ 201 will appear in first column at 51st position.
∴ 201 will appear 51st row and first column.
(c) First number in row r is 1 + 4(r – 1). So number in column c = c + 4(r – 1).
(d) List the first few multiples of 3 :
-3 (row 1, col. 3)
6 (row 2, col. 2)
9 (row 3, col. 1)
12 (row 3, col. 4)
15 (row 4, col. 3)
18 (row 5, col. 2) and so on.
We can see that columns for multiples of 3 move
3 → 2 → 1 → 4 → 3 → 2 → 1 → 4 …………….,
(decreases by 1 each time; wraps from 1 to 4).
Other Patterns: Each column is an arithmetic sequence with common difference 4 and each row contains consecutive numbers.
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