What is an algebraic expression as defined in NCERT Class 7 Maths Chapter 4?

An algebraic expression is a mathematical statement made up of numbers, variables (letters representing unknown values), and arithmetic operations such as addition, subtraction, multiplication, and division. For example, x+5 or 3y−7 are algebraic expressions.

NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letters

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)

30 x + 20 y

(b)

(30 + 20) \times (x + y)

(c)

20 x + 30 y

(d)

(30 + 20) \times x + y

(e)

30 x - 20 y

Sol. Cost of one plate of Jowar roti

= ₹30

∴

Cost of x plate of Jowar roti

= 30 x

Cost of one plate of Pulao

= ₹20

∴

Cost of y plate of Pulao

= 20 y

So, the expression for the total amount earned that day

= 30 x + 20 y

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 + 2 r

(c)

2 \times (p + q + r)

(d)

p + q + r + 2

(e)

p + q + r + 1

(f)

2 \times (p + q)

Sol. Number of customers who bought only champak

= p

Number of customers who bought only marigold

= q

Number of customers who bought both

= r

As Pushpita gave away a tiny national flag to every customer. So, the number of flags she gives away that day

= p + q + r

.

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) During the day, the snail climbs up 'u' cm. During the night snail slips down d cm . So, the net distance covered in one day is u-d. So, in 10 days and 10 nights, the net distance covered by the snail

= 10 (u - d)

. Hence, the expression describing how far away the snail is from it starting position is

10 (u - d) cm

. (b) If

d > u

, the snail slips down more than it climbs. It means the snail will never reach the top.

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. In the first week, Radha cycled 5 km every day. So, she cycled

5 \times 7 = 35 km

in the first week. In the second week, Radha cycles (

5 + z

) km every day. So, she cycled

(5 + z) \times 7 = (35 + 7 z) km

in second week. In the third week, she cycles (

5 + z + z = 5 + 2 z

) km every day. So, she cycled

(5 + 2 z) \times 7 = (35 + 14 z) km

in third week. So, number of kilometres Radha cycled in 3 weeks

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

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

= 105 + 21 z km

In the following figure, observe how the expression

w + 2

becomes

4 w + 20

along one path. Fill in the missing blank on the remaining paths. The ovals contain expressions, and the boxed contain operations.

Sol.

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] Sol. (a) The train from Yahapur to Vahapur stops at 3 stations, and stops for 2 minutes at every station. Time taken in travelling

= 4 t

At

t = 4

, time taken in travelling

= 4 \times 4 = 16

minutes Time taken during stoppages

= 3 \times 2 = 6

minutes So, the time taken to travel from Yahapur to Vahapur

= 16 + 6 = 22

minutes. (b)

Let the time taken to travel from one station to another station

= t

So, time taken to travel from Yahanpur to Vahapur

= 4 t

As there are three stoppages between these two stations, and the train stops for 2 minutes at each stoppage, Therefore, total time taken during stoppages

= 2 \times 3 = 6

minutes So, the algebraic expression for total time taken is

4 t + 6

.

Simplify the following expressions: (a)

3 a + 9 b - 6 + 8 a - 4 b - 7 a + 16

(b)

3 (3 a - 3 b) - 8 a - 4 b - 16

(c)

2 (2 x - 3) + 8 x + 12

(d)

8 x - (2 x - 3) + 12

(e)

8 h - (5 + 7 h) + 9

(f)

23 + 4 (6 m - 3 n) - 8 n - 3 m - 18

Sol. (a)

3 a + 9 b - 6 + 8 a - 4 b - 7 a + 16

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

= 4 a + 5 b + 10

(b)

3 (3 a - 3 b) - 8 a - 4 b - 16

= 9 a - 9 b - 8 a - 4 b - 16

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

= a - 13 b - 16

(c)

2 (2 x - 3) + 8 x + 12

= 4 x - 6 + 8 x + 12

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

= 12 x + 6

(d)

8 x - (2 x - 3) + 12

= 8 x - 2 x + 3 + 12

= 6 x + 15

(e)

8 h - (5 + 7 h) + 9

= 8 b - 5 - 7 b + 9

= (8 b - 7 b) + (- 5 + 9)

= b + 4

(f)

23 + 4 (6 m - 3 n) - 8 n - 3 m - 18

= 23 + 24 m - 12 n - 8 n - 3 m - 18

= (23 - 18) + (24 m - 3 m) + (- 12 n - 8 n)

= 5 + 21 m - 20 n

Add the expressions given below: (a)

4 d - 7 c + 9

and

8 c - 11 + 9 d

(b)

- 6 f + 19 - 8 s

and

- 23 + 13 f + 12 s

(c)

8 d - 14 c + 9

and

16 c - (11 + 9 d)

(d)

6 f - 20 + 8 s

and

23 - 13 f - 12 s

(e)

13 m - 12 n

and

12 n - 13 m

(f)

- 26 m + 24 n

and

26 m - 24 n

Sol. (a)

4 d - 7 c + 9

and

8 c - 11 + 9 d

= 4 d - 7 c + 9 + 8 c - 11 + 9 d

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

= 13 d + c - 2

(b)

- 6 f + 19 - 8 s

and

- 23 + 13 f + 12 s

= - 6 f + 19 - 8 s + - 23 + 13 f + 12 s

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

= 7 f + 4 s - 4

(c)

8 d - 14 c + 9

and

16 c - (11 + 9 d)

= 8 d - 14 c + 9 + 16 c - 11 - 9 d

= 8 d - 9 d - 14 c + 16 c + 9 - 11

= - d + 2 c - 2

= 2 c - d - 2

(d)

6 f - 20 + 8 s

and

23 - 13 f - 12 s

= 6 f - 20 + 85 + 23 - 13 f - 125

= (6 f - 13 f) + (8 s - 12 s) + (- 20 + 23)

= - 7 f - 4 s + 3

(e)

13 m - 12 n

and

12 n - 13 m

= 13 m - 12 n + 12 n - 13 m

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

= 0

(f)

- 26 m + 24 n

and

26 m - 24 n

= - 26 m + 24 n + 26 m - 24 n

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

= 0

Subtract the expressions given below: (a)

9 a - 6 b + 14

from

6 a + 9 b - 18

(b)

- 15 x + 13 - 9 y

from

7 y - 10 + 3 x

(c)

17 g + 9 - 7 h

from

11 - 10 g + 3 h

(d)

9 a - 6 b + 14

from

6 a - (9 b + 18)

(e)

10 x + 2 + 10 y

from

- 3 y + 8 - 3 x

(f)

8 g + 4 h - 10

from

7 h - 8 g + 20

Sol. (a)

(6 a + 9 b - 18) - (9 a - 6 b + 14)

= 6 a + 9 b - 18 - 9 a + 6 b - 14

= (6 a - 9 a) + (9 b + 6 b) + (- 18 - 14)

= - 3 a + 15 b - 32

(b)

(7 y - 10 + 3 x) - (- 15 x + 13 - 9 y)

= 7 y - 10 + 3 x + 15 x - 13 + 9 y

= (7 y + 9 y) + (3 x + 15 x) + (- 10 - 13)

= 16 y + 18 x - 23

(c)

(11 - 10 g + 3 h) - (17 g + 9 - 7 h)

= 11 - 10 g + 3 h - 17 g - 9 + 7 h

= (11 - 9) + (- 10 g - 17 g) + (3 h + 7 h)

= 2 - 27 g + 10 h

= 10 h - 27 g + 2

(d)

6 a - (9 b + 18) - (9 a - 6 b + 14)

= 6 a - 9 b - 18 - 9 a + 6 b - 14

= (6 a - 9 a) + (- 9 b + 6 b) + (- 18 - 14)

= - 3 a - 3 b - 32

= - (3 a + 3 b + 32)

(e)

(- 3 y + 8 - 3 x) - (10 x + 2 + 10 y)

= - 3 y + 8 - 3 x - 10 x - 2 - 10 y

= (- 3 y - 10 y) + (- 3 x - 10 x) + (8 - 2)

= - 13 y - 13 x + 6

(f)

(7 h - 8 g + 20) - (8 g + 4 h - 10)

= 7 h - 8 g + 20 - 8 g - 4 h + 10

= (7 h - 4 h) + (- 8 g - 8 g) + (20 + 10)

= 3 h - 16 g + 30

Describe situations corresponding to the following algebraic expressions: (a)

8 x + 3 y

(b)

15 x - 2 x

Sol. (a) A notebook costs ₹ 8 and a pen costs ₹ 3 . If you buy x notebooks and y pens. Then the total cost bear by you is ₹ (

8 x + 3 y

). (b) A fruit seller has 15 boxes of apples, each box containing

x

apples. Before selling them, he found that 2 boxes of apples were rotten. The number of fresh apples left is

15 x - 2 x

.

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.

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) \times 3

= 4 + 3 (w - 1)

= (4 + 3 w - 3)

= 3 w + 1

matchsticks. To make 10 squares, substitute 10 for w :

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

matchsticks

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, \dots .

. In general,

4 n - 3

positions The sequence of green light:

3, 7, 11, \dots .

. In general;

4 n - 1

positions The sequence of yellow light:

2, 4, 6, \dots .

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

343 \div 4 = 85

quotient +3 remainder. So, it matches a

4 n - 1

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

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 \times 4 = 13

So, number of squares in step

4 = 5 + 4 + 4 + 4 = 5 + 3 \times 4 = 17

So, number of squares in step

10 = 5 + 9 \times 4 = 41

And, number of squares in step

50 = 5 + 49 \times 4 = 201

So, the general formula

= 5 + (n - 1) \times 4 = 5 + 4 (n - 1) = 5 + 4 n - 4 = 4 n + 1

. Since 1 square has 4 vertices, the number of vertices (

4 n + 1

) squares have

4 (4 n + 1) = 16 n + 4

.

Numbers are written in a particular sequence in this endless 4 -column grid.

No. of $₹ 100$ notes	No. of $₹ 20$ notes	No. of $₹ 5$ notes	Expression and total amount
$3$	5	6
			$6 \times 100 + 4 \times 20 + 3 \times 5 = 695$
$8$	4	z
$x$	y	z

No. of ₹100 notes	No. of ₹20 notes	No. of ₹5 notes	Expression and total amount
$3$	5	6	$3 \times 100 + 5 \times 20 + 6 \times 5 = 430$
$6$	4	3	$6 \times 100 + 4 \times 20 + 3 \times 5 = 695$
$8$	4	z	$8 \times 100 + 4 \times 20 + z \times 5 = 880 + 5 z$
$x$	y	z	$x \times 100 + y \times 20 + z \times 5 = 100 x + 20 y + 5 z$

$W - 8$	$W - 7$	$W - 6$
$W - 1$	W	$W + 1$

No. of $₹ 100$ notes	No. of $₹ 20$ notes	No. of $₹ 5$ notes	Expression and total amount
$3$	5	6
			$6 \times 100 + 4 \times 20 + 3 \times 5 = 695$
$8$	4	z
$x$	y	z

No. of ₹100 notes	No. of ₹20 notes	No. of ₹5 notes	Expression and total amount
$3$	5	6	$3 \times 100 + 5 \times 20 + 6 \times 5 = 430$
$6$	4	3	$6 \times 100 + 4 \times 20 + 3 \times 5 = 695$
$8$	4	z	$8 \times 100 + 4 \times 20 + z \times 5 = 880 + 5 z$
$x$	y	z	$x \times 100 + y \times 20 + z \times 5 = 100 x + 20 y + 5 z$

Mon	Tue	Wed	Thu	Fri	Sat	Sun
					1	2
$3$	4	5	6	7	8	9
$10$	11	12	13	14	15	16
$17$	18	19	20	21	22	23
$24$	25	26	27	28	29	30

Frequently Asked Questions

What is an algebraic expression as defined in NCERT Class 7 Maths Chapter 4?

What do NCERT Solutions for Class 7 Maths Chapter 4 – Expressions Using Letter-Numbers cover?

How do NCERT Solutions for Chapter 4 help in understanding algebraic expressions?

What types of problems are included in NCERT Solutions for Class 7 Maths Chapter 4?

How are algebraic expressions simplified in NCERT Solutions for Chapter 4 – Expressions Using Letter-Numbers?

How do NCERT Solutions for Class 7 Maths Chapter 4 improve conceptual clarity?

How are real-life problems solved in NCERT Solutions for Chapter 4?

What makes NCERT Solutions for Class 7 Maths Chapter 4 useful for regular practice?

How are patterns represented in NCERT Solutions for Chapter 4?

Join ALLEN!

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

1.0NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter-Numbers – Download PDF

2.0Key Concepts in Chapter 4 Expressions Using Letter-Numbers

3.0NCERT Solutions for Chapter 4 Expressions Using Letter-Numbers : All Exercises

Exercise 4.1

Exercise 4.2

Exercise 4.3

4.0NCERT Class 7 Maths Chapter 4 Expressions Using Letter-Numbers: Detailed Solutions

FIGURE IT OUT-02

FIGURE IT OUT-03

5.0Key Features of NCERT Solutions Class 7 Maths Chapter 4 : Expressions Using Letter-Numbers

Frequently Asked Questions

What is an algebraic expression as defined in NCERT Class 7 Maths Chapter 4?

What do NCERT Solutions for Class 7 Maths Chapter 4 – Expressions Using Letter-Numbers cover?

How do NCERT Solutions for Chapter 4 help in understanding algebraic expressions?

What types of problems are included in NCERT Solutions for Class 7 Maths Chapter 4?

How are algebraic expressions simplified in NCERT Solutions for Chapter 4 – Expressions Using Letter-Numbers?

How do NCERT Solutions for Class 7 Maths Chapter 4 improve conceptual clarity?

How are real-life problems solved in NCERT Solutions for Chapter 4?

What makes NCERT Solutions for Class 7 Maths Chapter 4 useful for regular practice?

How are patterns represented in NCERT Solutions for Chapter 4?

Join ALLEN!

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

1.0NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter-Numbers – Download PDF

2.0Key Concepts in Chapter 4 Expressions Using Letter-Numbers

3.0NCERT Solutions for Chapter 4 Expressions Using Letter-Numbers : All Exercises

Exercise 4.1

Exercise 4.2

Exercise 4.3

4.0NCERT Class 7 Maths Chapter 4 Expressions Using Letter-Numbers: Detailed Solutions

FIGURE IT OUT-02

FIGURE IT OUT-03

5.0Key Features of NCERT Solutions Class 7 Maths Chapter 4 : Expressions Using Letter-Numbers