NCERT Solutions for Class 7 maths Chapter 4: Expressions using Letter Numbers

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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

Use this expression to find A’ftab’s age if Shabnam’s age is 20.
Solution:
According to the expression,
Aftab’s age = Shabnam’s Age – 3
∴ Aftab’s age = 20 – 3 = 17 years

How much should she pay if she buys 8 coconuts and 9 kg jaggery?
Solution:
Cost of 8 coconuts = 8 × ₹ 35 = ₹ 280
Cost of 9 kg jaggery = 9 × ₹ 60 = ₹ 540
Total cost = ₹ 280 + ₹ 540 = ₹ 820

Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery.
Solution:
Expression for total amount is c × 35 + j × 60
Replacing c by 7 and j by 4 in the expression, we get
c × 35 + j × 60 = 7 × 35 + 4 × 60
= 245 + 240
= 485
Hence, the total amount to be paid for 7 coconuts and 4 kg of jaggery is ₹ 485.

What is the perimeter of a square with sidelength 7 cm? Use the expression to find out.
Solution:
We have, perimeter of a square of sidelength q = 4 × q
Perimeter of a square of sidelength 7 cm = 4 × 7 cm = 28 cm

Figure it Out (Pages 84-85)

(b) Let sidelength of a regular pentagon = a
So, perimeter of the regular pentagon = 5a

(c) Let sidelength of a regular hexagon = a
So, perimeter of the regular hexagon = 6a

(b) Rohan earns ₹ 15 per glass by selling lemonade in his school fete. To maintain the cleanliness he further announced ₹ 2 discount on each glass, the customer return to him. If he sold j number of glasses and k number of glasses were returned to him, what is the total earnings at the end of the day?

Mind the Mistake, Mend the Mistake

NCERT In-Text Questions (Page 87)

Below are some simplifications, where the letter numbers are replaced by numbers, and the value of the expression is obtained.
1. Observe each of them and identify if there is a mistake.
2. If you think there is a mistake, try to explain what might have gone wrong.
3. Then, correct it and give the value of the expression.

Solution:
1. If a = -4, then 10 – a = 6 is wrong.
As 10 – a = 10 – (-4) = 10 + 4 = 14
So, if a = -4, then the correct expression is 10 – a = 14.

Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Solution:
If the price per eraser is d, then
total money earned by selling erasers = 4 × d + 6 × d + 1 × d
= 4d + 6d + d
= (4 + 6 + 1) × d
= 11 × d
= 11d

Check that both expressions take the same value when c is replaced by different numbers.
Solution:
Let c = 5, then
5c + 3c + 10c = 5 × 5 + 3 × 5 + 10 × 5
= 25 + 15 + 50
= 40 + 50
= 90
And 18c = 18 × 5 = 90
Hence, both expressions take the same value when c is replaced by the number 5.
Similarly, we can verify for other numbers.

NCERT In-Text Questions (Pages 91-93)

Could we have written the initial expression as (40x + 75y) + (-6x – 10y)?
Solution:
Yes, as (40x + 75y) + (-6x – 10y)
= (40x + 75y) + -(6x + 10y)
= (40x + 75y) – (6x + 10y)

Example 8.
Charu has been through three rounds of a quiz. Her scores in the three rounds are 7p – 3q, 8p – 4q, and 6p – 2q. Here, p represents the score for a correct answer and q represents the penalty for an incorrect answer.

What are her scores in the second and third rounds?
Solution:
Her score in the second round = 8p – 4q
= 8 × 4 – 4 × 1 [As p = 4 and q = 1]
= 32 – 4
= 28
Her score in the third round = 6p – 2q
= 6 × 4 – 2 × 1 [As p = 4, q = 1]
= 24 – 2
= 22

What if there is no penalty? What will be the value of q in that situation?
Solution:
If there is no penalty, then the value of q will be 0 in that situation.

Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.
Solution:
Some possible scores of Krishita in three rounds may be 8p – 4q, 9p – 2q, and 6p – q
or 7p – 3q, 10p – 3q, and 6p – q
Here 8p – 4q + 9p – 2q + 6p – q
= (8 + 9 + 6)p – (4 + 2 + 1)q
= 23p – 7q
Also 7p – 3q + 10p – 3q + 6p – q
= (7 + 10 + 6)p – (3 + 3 + 1)q
= 23p – 7q

Can we say who scored more? Can you explain why?
How much more has Krishita scored than Charu? This can be found by finding the difference between the two scores.
23p – 7q – (21p – 9q)

Simplify this expression further.
Solution:
23p – 7q – (21p – 9q)
= 23p – 7q – + 21p – 9q
= 23p – 21p – 7q + 9q
= 23p – 21p + 9q – 7q
= (23 – 21)p + (9 – 7)q
= 2p + 2q
= 2 × (p + q)
= 2(p + q)
The value of 2p + 2q is always positive as long as both p > 0 and q > 0 (since marks and penalties are positive). So, Krishita scored 2(p + q) marks more than Charu.

Fill in the blanks below by replacing the letter-numbers with numbers; an example is shown. Then compare the values that 5u and 5 + u take.

Solution:

We see that the values of 5u and 5 + u are not equal for different values of u. So, the expressions 5u and 5 + u are not equal.

Are the expressions 10y – 3 and 10(y – 3) equal?
10y – 3, short for 10 × y – 3, means 3 less than 10 times y,
10(y – 3), short for 10 × (y – 3), means 10 times (3 less than y).
Let us compare the values that these expressions take for different values of y.

Solution:

After filling in the two diagrams, do you think the two expressions are equal?
Solution:
We see that the values of 10y – 3 and 10(y – 3) are not equal for different values of y. So, the expressions 10y – 3 and 10(y – 3) are not equal.

Figure it Out (Pages 93-94)

(b) p – q + p – q = 2p – 2q
p + q – p + q = 2q

(c) p + q – (p + q) = 0
p – q – p – q = -2q

(d) 2d – d – d – d = -d
2d – d – d – c = -c

(e) 2d – d – (d – c) = c
2d – (d – d) – c = 2d – c

(f) 2d – d – c – c = d – 2c

Mind the Mistake, Mend the Mistake

NCERT In-Text Questions (Pages 94-95)

Some simplifications of algebraic expressions are done below. The expression on the right-hand side should be in its simplest form.

Solution:

Take a look at all the corrected simplest forms (i.e., brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter-numbers these expressions have?
Solution:
Yes

Find the formulas of the number machines below and write the expression for each set of inputs.

Solution:
The formula for the number machines in the first row is “sum of first number and second number minus two,” and the expression is a + b – 2.
The expression for each set of inputs is:
5 + 2 – 2 = 5, 8 + 1 – 2 = 7, 9 + 11 – 2 = 18, 10 + 10 – 2 = 18, and a + b – 2
The formula for the number machines in the second row is “product of first number and second number plus one,” and the expression is a × b + 1.
The expression for each set of inputs is:
4 × 1 + 1 = 5, 6 × 0 + 1 = 1, 3 × 2 + 1 = 7, 10 × 3 + 1 = 31, and a × b + 1 = ab + 1.

Now, make a formula on your own. Write a few number machines as examples using that formula. Challenge your classmates to figure it out!
Solution:
Do it yourself.

Algebraic Expressions to Describe Patterns

NCERT In-Text Questions (Pages 96-97)

Example 12.
Somjit noticed a repeating pattern along the border of a saree.

Use this to find what design appears at positions 99, 122, and 148.
Solution:

Patterns in a Calendar

NCERT In-Text Questions (Page 99)

Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’.
Solution:
Let a = 2, then

Here, the diagonal sums are 2 + 10 = 12 and 9 + 3 = 12
And 2a + 8 = 2 × 2 + 8 = 12
Hence, the diagonal sum is equal to 2a + 8.

Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that form this shape. What do you observe?
Solution:
Sum of all numbers = 8 + 14 + 15 + 16 + 22 = 75
The sum is 5 times the number in the centre.
Now, let the number at the centre: 20, then the shape is

Sum of all the numbers = 13 + 19 + 20 + 21 + 27 = 100 = 20 × 5
Further, let the number at the centre: 12, then the shape is

Sum of all numbers = 5 + 11 + 12 + 13 + 19 = 60 = 12 × 5
Hence, we see that the total sum is always 5 times the number in the centre.

Will this always happen? How do you show this?
Solution:
The general formula for a 3 × 3 square with centre number a is shown here,

Here, the sum is a – 7 + a – 1 + a + a + 1 + a + 7 = a + a + a + a + a – 7 – 1 + 1 + 7
So, the sum of numbers = 5a (5 times the number in the centre).

Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers.
Solution:
Do it yourself.

Matchstick Patterns

Look at the picture below. It is a pattern using matchsticks. Can you identify what the pattern is?

We can see that Step 1 has 1 triangle, Step 2 has 2 triangles, Step 3 has 3 triangles, and so on.

NCERT In-Text Questions (Page 101)

What are these numbers in Step 3 and Step 4?
Solution:
In step 3, there are 3 matchsticks placed horizontally and 4 matchsticks placed diagonally.
In step 4, there are 4 matchsticks placed horizontally and 5 matchsticks placed diagonally.

How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y in each orientation. Do the two expressions add up to 2y + 1?
Solution:
The pattern in the horizontally placed matchsticks is: 1, 2, 3, 4,….
In general, n for the nth step.
The pattern in the diagonally placed matchsticks is: 2, 3, 4, 5,….
In general, n + 1 for the nth step.
∴ The number of matchsticks placed horizontally at step ‘y’ is y.
The number of matchsticks placed diagonally at step ‘y’ is y + 1.
Now, these two expressions add up to y + y + 1 = 2y + 1.

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.

(b) If d > u, the snail slips down more than it climbs.
It means the snail will never reach the top.

(b)

Let the time taken to travel from one station to another station = t
So, time taken to travel from Yahanpur to Vahapur = 4t
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 × 3 = 6 minutes
So, the algebraic expression for total time taken is 4t + 6.

(b) 3(3a – 3b) – 8a – 4b – 16
= 9a – 9b – 8a – 4b – 16
= (9a – 8a) + (-9b – 4b) – 16
= a – 13b – 16

(c) 2(2x – 3) + 8x + 12
= 4x – 6 + 8x + 12
= (4x + 8x) + (-6 + 12)
= 12x + 6

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

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

(f) 23 + 4(6m – 3n) – 8n – 3m – 18
= 23 + 24m – 12n – 8n – 3m – 18
= (23 – 18) + (24m – 3m) + (-12n – 8n)
= 5 + 21m – 20n

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

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

(d) 6f – 20 + 8s and 23 – 13f – 12s
= 6f – 20 + 85 + 23 – 13f – 125
= (6f – 13f) + (8s – 12s) + (-20 + 23)
= -7f – 4s + 3

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

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

(b) (7y – 10 + 3x) – (-15x + 13 – 9y)
= 7y – 10 + 3x + 15x – 13 + 9y
= (7y + 9y) + (3x + 15x) + (-10 – 13)
= 16y + 18x – 23

(c) (11 – 10g + 3h) – (17g + 9 – 7h)
= 11 – 10g + 3h – 17g – 9 + 7h
= (11 – 9) + (-10g – 17g) + (3h + 7h)
= 2 – 27g + 10h
= 10h – 27g + 2

(d) 6a – (9b + 18) – (9a – 6b + 14)
= 6a – 9b – 18 – 9a + 6b – 14
= (6a – 9a) + (-9b + 6b) + (-18 – 14)
= -3a – 3b – 32
= -(3a + 3b + 32)

(e) (-3y + 8 – 3x) – (10x + 2 + 10y)
= -3y + 8 – 3x – 10x – 2 – 10y
= (-3y – 10y) + (-3x – 10x) + (8 – 2)
= -13y – 13x + 6

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

(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 15x – 2x.

(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)