NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter-Numbers Exercise 4.3

Exercise 4.3 of Class 7 Maths Chapter 4 : Expressions Using Letter-Numbers brings students a real-life application of algebra. In this exercise the student will develop, and use equations based on examples of word problems. It consolidates the learning of previous exercises dealing with variables and simple equations and applying it to real situations. 

All the questions in this exercise are designed following the latest NCERT syllabus and the CBSE question pattern. These types of questions are important to help students solve mathematical problems encountered at home or school or anywhere else in life.

The NCERT Solutions are presented thoroughly in a step by step manner for a clear understanding. Download the PDFs and start your practice today.

1.0Download NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter-Numbers Exercise 4.3: Free PDF

In exercise 4.3 students learn how to form and solve equations from word problems which helps develop real-world maths skills. The NCERT Solutions for Class 7 Maths Chapter 4 provides a step-by-step explanation that is easy to understand. You can download the free PDF from below:

2.0Key Concepts in Exercise 4.3 of Class 7 Maths Chapter 4

This exercise focuses on applying algebra to everyday problems through simple equations.

• Understanding and identifying the unknown quantity in a word problem
• Converting word statements into algebraic equations
• Solving equations step by step to find the correct answer
• Applying basic operations to isolate and solve for the variable
• Interpreting the result in the context of the original problem

3.0NCERT Class 7 Maths Chapter 4: Other Exercises

4.0NCERT Class 7 Maths Chapter 4 Exercise 4.3: Detailed Solutions

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

Sol.

Cost of x plate of Jowar roti = 30x

Cost of y plate of Pulao = 20y

Total amount earned = 30x + 20y

Answer: (a) 30x + 20y

2. Pushpita sells two types of flowers: 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)

Sol.

Number of flags = Total number of unique customers.

Total customers = (only champak) + (only marigold) + (bought both)

Total customers = p + q + r

Answer: (a) p + q + r

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?

Sol.

(a) Net distance covered in one day and one night = u - d.

In 10 days and 10 nights, net distance covered = 10 * (u - d) cm.

Expression: 10(u - d) cm.

(b) If d > u, the net movement is negative (downward), meaning the snail will continue to move away from the top or remain at the bottom, never reaching the top (if the well is deep enough).

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?

Sol.

Week 1 daily distance: 5 km

Week 1 total distance (7 days): 5 * 7 = 35 km

Week 2 daily distance: (5 + z) km

Week 2 total distance: (5 + z) * 7 = 35 + 7z km

Week 3 daily distance: (5 + z + z) = (5 + 2z) km

Week 3 total distance: (5 + 2z) * 7 = 35 + 14z km

Total kilometers after 3 weeks = Week 1 + Week 2 + Week 3

= 35 + (35 + 7z) + (35 + 14z)

= (35 + 35 + 35) + (7z + 14z)

= 105 + 21z km

5. In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blank on the remaining paths.

Path 1: w + 2 [multiply by 4] -> 4w + 8 [add 12] -> 4w + 20 (This is the implied completed path for 4w+20)

Sol.

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?

Sol.

The journey is divided into 4 travel segments (Y to S1, S1 to S2, S2 to S3, S3 to V).

There are 3 stoppage stations (S1, S2, S3).

(a) Time taken in travelling = 4 * t

At t = 4, travel time = 4 * 4 = 16 minutes.

Time taken during stoppages = 3 stations * 2 minutes/station = 6 minutes.

Total time = 16 + 6 = 22 minutes.

(b) 

Time taken to travel = 4t

Time taken during stoppages = 3 * 2 = 6 minutes.

Algebraic expression for total time = 4t + 6.

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

Sol.

(a) (3a+8a-7a) + (9b-4b) + (-6+16) = 4a + 5b + 10

(b) 9a-9b-8a-4b-16 = (9a-8a) + (-9b-4b) - 16 = a - 13b - 16

(c) 4x-6+8x+12 = (4x+8x) + (-6+12) = 12x + 6

(d) 8x-2x+3+12 = 6x + 15

(e) 8h-5-7h+9 = (8h-7h) + (-5+9) = h + 4

(f) 23+24m-12n-8n-3m-18 = (23-18) + (24m-3m) + (-12n-8n) = 5 + 21m - 20n

8. Add the expressions given below:

(a) 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

Sol.

(a) (4d+9d) + (-7c+8c) + (9-11) = 13d + c - 2

(b) (-6f+13f) + (-8s+12s) + (19-23) = 7f + 4s - 4

(c) 8d-14c+9 + 16c-11-9d = (8d-9d) + (-14c+16c) + (9-11) = -d + 2c - 2

(d) (6f-13f) + (8s-12s) + (-20+23) = -7f - 4s + 3

(e) (13m-13m) + (-12n+12n) = 0

(f) (-26m+26m) + (24n-24n) = 0

9. Subtract the expressions given below:

(a) 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

Sol. (Subtracting Expression A from Expression B means B - A)

(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 = 2 - 27g + 10h

(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 = -13y - 13x + 6

(f) (7h-8g+20) - (8g+4h-10) = 7h-8g+20-8g-4h+10 = 3h - 16g + 30

10. Describe situations corresponding to the following algebraic expressions:

(a) 8x + 3y

(b) 15x - 2x

Sol.

(a) Situation: A fruit basket costs ₹ 8, and a bottle of juice costs ₹ 3. If a customer buys x fruit baskets and y bottles of juice, the total cost is ₹ (8x + 3y).

(b) Situation: A theater has 15 rows of seats, and each row has x seats. If 2 full rows are reserved (and taken out of general sale), the number of seats available for general sale is 15x - 2x.

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?

Sol.

Step 1 ( 0 fold): We get 0+2=2 pieces

Step 2 ( 1 fold): We get 1+2=3 pieces

Step 3 (2 folds): We get 2+2=4 pieces

In the same way, if the rope is folded 10 times and cut, we get 10+2=12 pieces.

In the same way, when the rope is folded r times and cut, we get r+2 pieces.

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?

Sol. Step 1: To make 1 square, we need 4 matchsticks.

Step 2: To make 2 squares, we need 4+3=7 matchsticks

Step 3: To make 3 squares, we need 4+3+3=10 matchsticks.

And to make w squares we need =4+(w−1)×3

=4+3(w−1)

=(4+3w−3)

=3w+1 matchsticks.

To make 10 squares, substitute 10 for w :

3(10)+1=30+1=31 matchsticks

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.

Sol. The sequence of red light: 1,5,9,…..

In general, 4n−3 positions

The sequence of green light: 3,7,11,…..

In general; 4n−1 positions

The sequence of yellow light: 2,4,6,…..

In general, 2 n positions

Since 90 and 190 are even numbers, it will be 2 n positions.

Now, 343÷4=85 quotient +3 remainder.

So, it matches a 4n−1 position.

So, colour at positions 90, 190, and 343 are yellow, yellow, and green, respectively.

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?

Sol. Number of squares in step 1=5

Number of squares in step 2=5+4=9

Number of squares in step 3=5+4+4=5+2×4=13

So, number of squares in step 4=5+4+4+4=5+3×4=17

So, number of squares in step 10=5+9×4=41

And, number of squares in step 50=5+49×4=201

So, the general formula =5+(n−1)×4=5+4(n−1)=5+4n−4=4n+1.

Since 1 square has 4 vertices, the number of vertices ( 4n+1 ) squares have 4(4n+1)=16n+4.

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.

Sol. (a) Expression to generate all the numbers in a given column (1,2,3,4)

Let r be the row number.

Column 1: 1,5,9,13,……. which starts at 1 and adds 4 each row.

So, number in the rth  row of column 1=4×(r−1)+1

Column 2: 4×(r−1)+2

Column 3: 4×(r−1)+3

Column 4: 4×(r−1)+4

If c is the column number, then the general formula to generate all numbers is 4×(r−1)+c.

(b) (i) We divide each number by 4 to find its row and column

124÷4⇒ Quotient =31 and remainder is 0

∴124=4×31+0 or 4×30+4

Comparing it with 4×(r−1)+c, we get

r−1=30,c=4

So, r=31 and c=4

So, row is 31 and column is 4

(ii) 147÷4⇒ Quotient =36 and remainder is 3

∴147=4×36+3

Comparing it with 4×(r−1)+c, we get

r−1=36,c=3

So, 147 will appear at row 36+1=37 and column 3

(iii) 201÷4⇒ Quotient =50 and remainder is 1

∴201=4×50+1

Comparing it with 4×(r−1)+c, we get

r−1=50,c=1

So, 201 will appear at row 51 and column 1 .

(c) The number that appears in row r and column c is 4(r−1)+c.

(d) Every third number is a multiple of 3.

We can observe that even numbers always appear in column 2 and column 4.

Odd numbers always appear in column 1 and column 3.

Every row has 2 odd and 2 even numbers.

The sum of each row increases by 16 .

(e.g., Row 1: 1+2+3+4=10, Row 2: 5+6+7+8=26, Row 3: 9+10+11+12=42 )

5.0Key Features and Benefits of Class 7 Maths Chapter 4 Exercise 4.3

• The exercise is structured according to the recent NCERT syllabus and the current CBSE exam pattern.
• The solutions provided are step-wise. Hence it develops students' response techniques for resolving algebraic problems.
• Regular practice will develop students' logical thinking, better comprehension of equations, and improved analytical techniques.
• These exercises can aid in the preparation for ​school Olympiads​ and other reasoning-based exams.
• Develops confidence in the use of algebra in real-life situations and awareness to advanced maths ​discussions.