Students often refer to Ganita Prakash Class 7 Solutions and Class 7 Maths Chapter 4 Expressions using Letter Numbers NCERT Solutions Question Answer to verify their answers.
NCERT Class 7 Maths Chapter 4 Expressions using Letter Numbers Solutions Question Answer
Ganita Prakash Class 7 Chapter 4 Solutions Expressions using Letter Numbers
NCERT Class 7 Maths Ganita Prakash Chapter 4 Expressions using Letter Numbers Solutions Question Answer
4.1 The Notion of Letter-Numbers
Figure It Out (Pages 84 – 85)
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 = 3s, where ‘s’ is the length of each side of the triangle.
(b) Perimeter = 5t, where ‘t’ is the length of each side of the pentagon.
(c) Perimeter = 6r, where ‘r’ is the length of each side of the hexagon.
Question 2.
Munirathna has a 20-meter-long pipe. However, he wants a longer watering pipe 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:
Initial length of pipe = 20 m
and length of pipe joined = k m
∴ Expression = (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:
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’ kilograms 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:
(d) Given, time to start the machine = 10 s
Time to grind the machine = 8 s
Quantity of grain = y kg
Total time = 10 + 8 × y = (10 + 8y) s
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:
(a) x + 5
(b) y – 4
(c) 13m – 2
(d) 2s – 13
Question 6.
Describe situations corresponding to the following algebraic expressions:
(a) 8 × x + 3 × y
(b) 15 × j – 2 × k
Solution:
(a) If the cost of a table is ₹ x and a chair is ₹ y, then the total cost of 8 tables and 3 chairs is ₹ (8x + 3y).
(b) If an apple weighs j grams and an orange weighs k grams, then the difference in the weights of 15 apples and 2 oranges is (15 × j – 2 × k) grams.
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 the date ‘w’.
Solution:
4.2 Revisiting Arithmetic Expressions, 4.3 Omission of the Multiplication Symbol in Algebraic Expressions, 4.4 Simplification of Algebraic Expressions
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 of different ways and see that you get the same thing.
Solution:
(a) Adding row-wise gives:
(5y – 6 + x) + (x + 2 + 5y) = 2x + 10y – 4
Adding like terms together gives:
(2 × x) + (2 × 5y) + (2 – 6) = 2x + 10y – 4
Adding column-wise gives:
(5y + x) + (-6 + 2) + (x + 5y) = 2x + 10y – 4
(b) Adding row-wise gives:
(2p + 3q – 2 + 3) + (3q + 2p + 3 – 2) + (2p + 3q) + (3q + 2p) = 8p + 12q + 2
Adding like terms together gives:
(4 × 2p) + (4 × 3q) + [2 × (-2)] + (2 × 3) = 8p + 12q + 2
Adding column-wise gives:
(2p + 3q) + (3q + 2p) + (-2 + 3 + 2p + 3q) + (3 – 2 + 3q + 2p) = 8p + 12q + 2
(c) Adding row-wise gives:
(-5g + 5k + 5k – 5g) + (5k + 5k + 5k + 5k) + (5k + 5k + 5k + 5k) + (-5g + 5k + 5k – 5g) = 60k – 20g
Adding like terms together gives:
(12 × 5k) + [4 × (-5g)] = 60k – 20g
Adding column-wise gives:
(-5g + 5k + 5k – 5g) + (5k + 5k + 5k + 5k) + (5k + 5k + 5k + 5k) + (-5g + 5k + 5k – 5g) = 60k – 20g
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
(c) 0, -2q
(d) -d, -c
(e) c, 2d – c
(f) d – 2c
4.5 Pick Patterns and Reveal Relationships
Figure It Out (Page 102)
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 roti costs ₹ 30, and one plate of Pulao costs ₹ 20. If x plates of Jowar roti 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:
(a) 30x + 20y
Here, the cost of one plate of Jowar roti = ₹ 30
Cost of one plate of Pulao = ₹ 20
Total earnings for x plates of Jowar roti = ₹ 30x
Total earnings for y plates of Pulao = ₹ 20y
Total earnings for the day = ₹ 30x + ₹ 20y
Thus, the correct expression is (a) 30x + 20y.
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:
(b) p + q + 2 r
Here, Customers who only bought champak = p
Customers who only bought marigold = q
Customers who bought both = r
Since each customer receives one flag, the total number of flags given away is simply the total number of customers:
Total flags = p + q + r
Thus, the correct answer is (a) p + q + r.
Question 3.
A snail is trying to climb up 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) 10u – 10d, or 10(u – d)
Expression for the snail’s position after 10 days and 10 nights:
Each day, the snail climbs u cm.
Each night, it slips down by d cm.
So, in one full day-night cycle, the net movement of the snail is:
Net movement per day = u-d
After 10 days and 10 nights, the total distance the snail has moved from its starting position is:
Total distance = 10 × (u-d)
Thus, the required expression is 10(u-d)
(b) What happens if (d > u)?
If the snail slips down more than it climbs each day (d > u), then [u-d < 0]
This means the snail is losing ground every day instead of making progress.
Over time, it will move further downward instead of climbing up.
Essentially, the snail will never escape the well and will keep slipping lower and lower.
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:
Radha’s cycling pattern follows a structured increase in distance:
- Week 1: She cycles 5 km per day.
- Week 2: She increases the daily distance by z km, so she cycles (5 + z) km per day.
- Week 3: She again increases the daily distance by z km, so she cycles (5 + 2z) km per day.
Each week has 7 days, so the total distance cycled in 3 weeks is:
Total distance = 7(5) + 7(5 + z) + 7(5 + 2z)
= 35 + 35 + 7z + 35 + 14z
= 105 + 21z
Thus, after 3 weeks, Radha would have cycled (105 + 21z) km.
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) If 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 in travelling = 4t
Time taken during stoppages = 3 × 2 = 6 min
Time taken to travel from Yahapur to Vahapur is 4t + 6 = (4 × 4 + 6) minutes = 22 minutes
(b) Algebraic expression is 4t + 6
Question 7.
Simplify the following expressions:
(a) 3a + 9b – 6 + 8a – 4b – 7a + 16
(b) 3(3a – 3b) – 8a – 4b – 16
(c) 2(2x – 3) + 8x + 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 = 4a + 5b + 10
(b) 3(3a – 3b) – 8a – 4b – 16
= 9a – 9b – 8a – 4b – 16
= a – 13b – 16
(c) 2(2x – 3) + 8x + 12
= 4x – 6 + 8x + 12
= 12x + 6
(d) 8x – (2x – 3) + 12
= 8x – 2x + 3 + 12
= 6x + 15
(e) 8h – (5 + 7h) + 9
= 8h – 5 – 7h + 9
= h + 4
(f) 23 + 4(6m – 3n) – 8n – 3m – 18
= 23 + 24m – 12n – 8n – 3m – 18
= 5 + 21m – 20n
Question 8.
Add the expressions given below:
(а) 4d – 7c + 9 and 8c – 11 + 9d
(b) -6f + 19 – 8s and -23 + 13f + 12s
(c) 8d – 14c + 9 and 16c – (11 + 9d)
(d) 6f – 20 + 8s and 23 – 13f – 12s
(e) 13m – 12n and 12n – 13m
(f) -26m + 24n and 26m – 24n
Solution:
(a) (4d – 7c + 9) + ( 8c – 11 + 9d)
= 4d – 7c + 9 + 8c – 11 + 9d
= 13d + c – 2
(b) (-6f + 19 – 8s) + (-23 + 13f + 12s)
= -6f + 19 – 8s – 23 + 13f + 12s
= 7f + 4s – 4
(c) (8d – 14c + 9) + [16c – (11 + 9d)]
= 8d – 14c + 9 + 16c – 11 – 9d
= -d + 2c – 2
(d) (6f – 20 + 8s) + (23 – 13f – 12s)
= 6f – 20 + 8s + 23 – 13f – 12s
= -7f – 4s + 3
(e) (13m – 12n) + (12n – 13m)
= 13m – 12n + 12n – 13m
= 0
(f) (-26m + 24n) + (26m – 24n)
= -26m + 24n + 26m – 24n
= 0
Question 9.
Subtract the expressions given below:
(а) 9a – 6b + 14 from 6a + 9b – 18
(b) -15x + 13 – 9y from 7y – 10 + 3x
(c) 17g + 9 – 7h from 11 – 10g + 3h
(d) 9a – 6b + 14 from 6a – (9b + 18)
(e) 10x + 2 + 10y from -3y + 8 – 3x
(f) 8g + 4h – 10 from 7h – 8g + 20
Solution:
(a) (6a + 9b – 18) – (9a – 6b + 14)
= 6a + 9b – 18 – 9a + 6b – 14
= -3a + 15b – 32
(b) (7y – 10 + 3x) – (-15x + 13 – 9y)
= 7y – 10 + 3x + 15x – 13 + 9y
= 16y + 18x – 23
(c) (11 – 10g + 3h) – (17g + 9 – 7h)
= 11 – 10g + 3h – 17g – 9 + 7h
= -27g + 10h + 2
(d) [6a – (9b + 18)] – (9a – 6b + 14)
= 6a – 9b – 18 – 9a + 6b – 14
= -3a – 3b – 32
(e) (-3y + 8 – 3x) – (10x + 2 + 10y)
= -3y + 8 – 3x – 10x – 2 – 10y
= -13x – 13y + 6
(f) (7h – 8g + 20) – (8g + 4h – 10)
= 7h – 8g + 20 – 8g – 4h + 10
= 3h – 16g + 30
Question 10.
Describe situations corresponding to the following algebraic expressions:
(a) 8x + 3y
(b) 15x – 2x
Solution:
(a) Meera bought ‘x’ packs of cards that contain 8 cards each. Her friend Sarita bought y packs of cards that contain 3 cards each. Both together bought (8x + 3y) cards.
(b) Pooja bought 15 packs of cards that contain ‘x’ cards each. She gave 2 packs to her friend Sarita. Pooja is now left with (15x-2x) cards.
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:
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:
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:
Colour at 90th position __________ yellow;
Colour at 190th position __________ yellow;
Colour at 343rd position __________ green
Expression for describing the position of red colour __________ 4x – 3
Expression for describing the position of yellow colour __________ 2x
Expression for describing the position of green colour __________ 4x – 1.
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:
General formula __________ x + 1
The formula to count the number of vertices of all the squares __________ 4(4x + 1) = 16x + 4
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) The expression to generate all the numbers in column 1 is 4x – 3.
The expression to generate all the numbers in column 2 is 4x – 2.
The expression to generate all the numbers in column 3 is 4x – 1.
The expression to generate all the numbers in column 4 is 4x.
If r is the row number and c is be column number, then the general formula is 4(r + 1) + c
(b) The number 124 will appear in the 31st row and 4th column.
The number 147 will appear in the 37th row and 3rd column.
The number 201 will appear in the 51st row and 1st column.
(c) 4(r – 1) + c
(d)
