Q: One quarter of a number increased by 9 gives the same number. What is the number?

Let the number be x. According to the question, $\frac {1}{4}$ × x + 9 = x $\frac {x}{4}$ + 9 = x Subtract $\frac {x}{4}$ from both sides $\frac{x}{4}+9-\frac{x}{4}=\frac{x}{1}-\frac{x}{4}$ ⇒ 9 = $\frac{4 x-x}{4}$ [LCM of 1 and 4 = 4] ⇒ 9 = $\frac {3x}{4}$ Divide both sides by $\frac {3}{4}$ ⇒ $9 \div \frac{3}{4}=\frac{3 x}{4} \div \frac{3}{4}$ ⇒ $9 \times \frac{4}{3}=\frac{3 x}{4} \times \frac{4}{3}$ ⇒ 12 = x The required number is 12.

Q: Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.

28p – 36 = 98 ⇒ 28p = 98 + 36 ⇒ 28p = 134 ⇒ p = $\frac {134}{28}$ ⇒ p = $\frac {67}{14}$ Now to find the value of (i) 14p – 19 = 14 × $\frac {67}{14}$ – 19 = 67 – 19 = 48 (ii) 28p – 38 = 28 × $\frac {67}{14}$ – 38 = 134 – 38 = 96

Question 1

Solve these equations and check the solutions.
(a) 3x – 10 = 35
(b) 5s = 3s
(c) 3u – 7 = 2u + 3
(d) 4(m + 6) – 8 = 2m – 4
(e) $\frac {u}{15}$ = 6

Accepted Answer

(a) 3x – 10 = 35
Add 10 on both sides
3x – 10+ 10 = 35 + 10
⇒ 3x = 45
Divide by 3 on both sides
⇒ $\frac{3 x}{3}=\frac{45}{3}$
⇒ x = 15
Check:
LHS = 3x – 10 for x = 15
= 3 × 15 – 10
= 45 – 10
= 35
LHS = RHS
Hence checked.
(b) 5s = 3s
Bring unknown terms on one side by subtracting 3s
⇒ 5s – 3s = 3s – 3s
⇒ 2s = 0
Divide by 2 on both sides 2
⇒ $\frac{2 s}{2}=\frac{0}{2}$
⇒ s = 0
Check:
LHS = 5s = 5 × 0 = 0
RHS = 3s = 3 × 0 = 0
LHS = RHS
Hence checked.
(c) 3u – 7 = 2u + 3
Bring the unknown terms to one side
By subtracting 2u from both sides
⇒ 3u – 7 – 2u = 2u + 3 – 2u
⇒ u – 7 = 3
Add 7 to both sides
⇒ u – 7 + 7 = 3 + 7
⇒ u = 10
Check: For u = 10
LHS = 3u – 7
= 3 × 10 – 7
= 30 – 7
= 23
RHS = 2u + 3
= 2 × 10 + 3
= 20 + 3
= 23
LHS = RHS
Hence checked.
(d) 4(m + 6) – 8 = 2m – 4
Apply the distributive property
4m + 24 – 8 = 2m – 4
⇒ 4m + 16 = 2m – 4
Subtract 2m from both sides
4m + 16 – 2m = 2m – 4 – 2m
⇒ 2m + 16 = -4
Subtract 16 from both sides
2m + 16 – 16 = -4 – 16
⇒ 2m = -20
Divide by 2 on both sides
2m ÷ 2 = -20 ÷ 2
⇒ m = -10
Check:
LHS = 4(m + 6) – 8
= 4(-10 + 6) – 8
= 4(-4) – 8
= -16 – 8
= -24
RHS = 2m – 4
= 2(-10) – 4
= -20 – 4
= -24
LHS = RHS
Hence checked.
(e) $\frac {u}{15}$ = 6
Multiply both sides by 15
$\frac {u}{15}$ × 15 = 6 × 15
⇒ u = 90
Check:
LHS = $\frac {u}{15}$
$\frac {90}{15}$ = 6
RHS = 6
LHS = RHS
Hence checked.

Question 2

Frame an equation that has no solution.
[Hint: 4 more than a number, and 5 more than a number can never be equal!]

Accepted Answer

Equation u + 6 = u + 2 has no solution
Subtract u from both sides
u + 6 – u = u + 2 – u
⇒ 6 = 2, which is an incorrect statement.
Figure It Out (Page 181)

Source: This question is from Finding the Unknown, Mathematics &mdash; Class 7, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Finding the Unknown: Frame, Equation, Solution, Hint, Number, Equal. These are fundamental topics in Mathematics that students are expected to master as part of the CBSE Class 7 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Mathematics papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberAlways show your working steps clearly.
Verify your answer by substituting values back into the equation.
Practice similar problems from the textbook exercises.
Memorise important formulae and their conditions of applicability.
Related Questions from Finding the UnknownQ1: Solve these equations and check the solutions. (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 = 2u + 3 (d)...
Q10: Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.
Q11: The steps to solve three equations are shown below. Identify and correct any mistakes. (a) 6x + 9 =...
Q12: Find the measures of the angles of these triangles.
Q13: Write 4 equations whose solution is u = 6.
Practice more questions from Finding the Unknown &mdash; Mathematics, Class 7 CBSE

Question 3

I am a 3-digit number. My hundred’s digit is 3 less than my ten’s digit. My ten’s digit is 3 less than my unit’s digit. The sum of all three digits is 15. Who am I?

Accepted Answer

Let the ten’s place digit be x.
Then hundred’s place digit = x – 3
Unit’s digit = x + 3
Sum of digits = 15
According to the question,
(x – 3) + x + (x + 3) = 15
⇒ 3x = 15
Divide both sides by 3
$\frac{3 x}{3}=\frac{15}{3}$
⇒ x = 5
Ten’s digit = 5
Hundred’s digit = 5 – 3 = 2
Unit’s digit = 5 + 3 = 8
Number formed = 258

Question 4

The weight of a brick is 1kg more than half its weight. What is the weight of the brick?

Accepted Answer

Let the weight of a brick be x kg.
According to the question,
x = $\frac {x}{2}$ + 1
Subtract $\frac {x}{2}$ from both sides
$\frac{x}{1}-\frac{x}{2}=\frac{x}{2}+1-\frac{x}{2}$
⇒ $\frac{2 x-x}{2}$ = 1
⇒ $\frac {x}{2}$ = 1
Multiply both sides by 2
$\frac {x}{2}$ × 2 = 1 × 2
⇒ x = 2
∴ Weight of the brick = 2 kg

Question 5

One quarter of a number increased by 9 gives the same number. What is the number?

Accepted Answer

Let the number be x.
According to the question,
$\frac {1}{4}$ × x + 9 = x
$\frac {x}{4}$ + 9 = x
Subtract $\frac {x}{4}$ from both sides
$\frac{x}{4}+9-\frac{x}{4}=\frac{x}{1}-\frac{x}{4}$
⇒ 9 = $\frac{4 x-x}{4}$ [LCM of 1 and 4 = 4]
⇒ 9 = $\frac {3x}{4}$
Divide both sides by $\frac {3}{4}$
⇒ $9 \div \frac{3}{4}=\frac{3 x}{4} \div \frac{3}{4}$
⇒ $9 \times \frac{4}{3}=\frac{3 x}{4} \times \frac{4}{3}$
⇒ 12 = x
The required number is 12.

Question 6

Given 4k + 1 = 13, find the values of:
(a) 8k+ 2
(b) 4k
(c) k
(d) 4k – 1
(e) -k – 2

Accepted Answer

4k + 1 = 13
Subtract 1 from both sides
4k + 1 – 1 = 13 – 1
⇒ 4k = 12
Divide both sides by 4
4k ÷ 4 = 12 ÷ 4
⇒ k = 3
(a) 8k + 2 = 8(3)+ 2
= 24 + 2
= 26
(b) 4k = 4 × 3 = 12
(c) k = 3
(d) 4k – 1 = 4(3) – 1
= 12 – 1
= 11
(e) -k – 2 = -3 – 2 = -5
7.3 Mind the Mistake, Mend the Mistake, 7.4 A Pinch of History
Figure It Out (Pages 185-189)

Question 7

The sum of two numbers is 76. One number is three times the other number. What are the numbers?

Accepted Answer

Let the two numbers be x and 3x.
According to the question
x + 3x = 76
⇒ 4x = 76
⇒ x = 19
The two numbers are 19 and 3 × 19, i.e., 19 and 57.

Source: This question is from Finding the Unknown, Mathematics &mdash; Class 7, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Finding the Unknown: Sum, Numbers, Number, Times, Finding the Unknown, Mathematics. These are fundamental topics in Mathematics that students are expected to master as part of the CBSE Class 7 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Mathematics papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberAlways show your working steps clearly.
Verify your answer by substituting values back into the equation.
Practice similar problems from the textbook exercises.
Memorise important formulae and their conditions of applicability.
Related Questions from Finding the UnknownQ1: Solve these equations and check the solutions. (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 = 2u + 3 (d)...
Q10: Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.
Q11: The steps to solve three equations are shown below. Identify and correct any mistakes. (a) 6x + 9 =...
Q12: Find the measures of the angles of these triangles.
Q13: Write 4 equations whose solution is u = 6.
Practice more questions from Finding the Unknown &mdash; Mathematics, Class 7 CBSE

Question 8

The figure shows the diagram for a window with a grill. What is the gap between the two rods in the grill?

Accepted Answer

Total height of the window, including the frame = 34 cm
Total height of the window excluding the frame = 34 cm – 3 cm – 3 cm = 28 cm
Width of one rod of grill = 2 cm
No. of rods of the grill = 5
Total width of 5 rods = 5 × 2 cm = 10 cm
Let the height of 1 gap be x cm.
No. of gaps = 6
Total height of gaps = 6x
According to the question
10 + 6x = 28
⇒ 6x = 28 – 10
⇒ 6x = 18
⇒ x = 3
Gap between two rods = 3 cm.

Question 9

In a restaurant, a fruit juice costs ₹ 15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹ 600, find the cost of the fruit juice and milkshake.

Accepted Answer

Let the cost of a chocolate milkshake be ₹ x.
Cost of fruit juice = ₹(x – 15)
Cost of 4 fruit juices and 7 chocolate milkshakes = ₹ 600
According to the question
4(x – 15) + 7x = 600
⇒ 4x – 60 + 7x = 600
⇒ 4x + 7x = 600 + 60
⇒ 11x = 660
⇒ x = 60
Cost of 1 chocolate milkshake = ₹ 60
Cost of 1 fruit juice = ₹ 60 – ₹ 15 = ₹ 45

Question 10

Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.

Accepted Answer

28p – 36 = 98
⇒ 28p = 98 + 36
⇒ 28p = 134
⇒ p = $\frac {134}{28}$
⇒ p = $\frac {67}{14}$
Now to find the value of
(i) 14p – 19 = 14 × $\frac {67}{14}$ – 19
= 67 – 19
= 48
(ii) 28p – 38 = 28 × $\frac {67}{14}$ – 38
= 134 – 38
= 96

Question 11

The steps to solve three equations are shown below. Identify and correct any mistakes.
(a) 6x + 9 = 66
x + 9 = 11
x = 11 – 9
x = 2
(b) 14y + 24 = 36
7y + 12 = 18
7y = 6
y = $\frac {6}{7}$
(c) 4x – 5 = 9x + 8
4x = 9x + 8 – 5
4x = 9x + 3
4x – 9x = 3
-5x = 3
x = $\frac {-5}{3}$

Accepted Answer

(a) 6x + 9 = 66 → Error
Variable terms should be on one side and constants on the other side before separating the coefficient from the variable.
Correction
6x + 9 = 66
6x = 66 – 9
6x = 57
x = $\frac {57}{6}$
x = $\frac {19}{2}$
x = 9$\frac {1}{2}$
(b) 14y + 24 = 36
7y + 12 = 18 [Dividing by 2 throughout]
7y = 6
y = $\frac {6}{7}$ → No Error
(c) 4x – 5 = 9x + 8
4x = 9x + 8 – 5 → Error
4x = 9x + 3
4x – 9x = 3
-5x = 3
x = $\frac {-5}{3}$ → Error
Correction
4x – 5 = 9x + 8
4x = 9x + 8 + 5
4x = 9x + 13
4x – 9x = 13
-5x = 13
x = $\frac {-13}{5}$
x = -2$\frac {3}{5}$

Question 12

Find the measures of the angles of these triangles.

Accepted Answer

(i) In ∆ABC,
AB = AC [Given]
∠B = ∠C [Angles opposite to equal sides in a triangle are equal]
So ∠B = y + 15
∠A + ∠B + ∠C = 180° [Angle sum property of ∆]
⇒ y + y + 15 + y + 15 = 180°
⇒ 3y + 30 = 180°
⇒ 3y = 180 – 30
⇒ 3y = 150
⇒ y = 50
∠A = 50°, ∠B = 50° + 15° = 65°, and ∠C = 50° + 15° = 65°
(ii) In ∆ABC,
∠A + ∠B + ∠C = 180° [Sum of angles of a triangle is 180°]
⇒ x + x – 10 + x + 10 = 180°
⇒ 3x = 180°
⇒ x = 60°
∠A = 60°, ∠B = 60° – 10° = 50° and ∠C = 60° + 10° = 70°

Question 13

Write 4 equations whose solution is u = 6.

Accepted Answer

(i) u = 6
Multiply both sides by $\frac {2}{3}$
⇒ $\frac {2}{3}$u = $\frac {2}{3}$ × 6
⇒ $\frac {2}{3}$u = 4
(ii) u = 6
Add 7 to both sides u + 7 = 6 + 7
⇒ u + 1 = 13
(iii) u = 6
Multiply both sides by 2
2u = 12
Add 3 to both sides
⇒ 2u + 3 = 12 + 3
⇒ 2u + 3 = 15
(iv) u = 6
Multiply both sides by 3
3u = 18
Subtract 5 from both sides
3u – 5 = 18 – 5
⇒ 3u – 5 = 13

Question 14

The Bakhshali Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second, and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?

Accepted Answer

Let the amount given to the first person be x.
Amount given to second person = 2x
Amount given to third person = 3(2x) = 6x
Amount given to fourth person= 4(6x) = 24x
Total amount = 132
⇒ x + 2x + 6x + 24x = 132
⇒ 33x = 132
⇒ x = $\frac {132}{4}$
⇒ x = 4
The amount given to the first person is 4.

Question 15

The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?

Accepted Answer

Let the height of a giraffe be x m.
According to the question
∴ Height of the giraffe = 5 m.

Source: This question is from Finding the Unknown, Mathematics &mdash; Class 7, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Finding the Unknown: Height, Giraffe, Half, Metres, Tall, Finding the Unknown. These are fundamental topics in Mathematics that students are expected to master as part of the CBSE Class 7 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Mathematics papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberAlways show your working steps clearly.
Verify your answer by substituting values back into the equation.
Practice similar problems from the textbook exercises.
Memorise important formulae and their conditions of applicability.
Related Questions from Finding the UnknownQ1: Solve these equations and check the solutions. (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 = 2u + 3 (d)...
Q10: Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.
Q11: The steps to solve three equations are shown below. Identify and correct any mistakes. (a) 6x + 9 =...
Q12: Find the measures of the angles of these triangles.
Q13: Write 4 equations whose solution is u = 6.
Practice more questions from Finding the Unknown &mdash; Mathematics, Class 7 CBSE

Question 16

Two separate figures are given below. Each figure shows the first few positions in a sequence of arrangements made with sticks. Identify the pattern and answer the following questions for each figure:
(a) How many squares are in position number 11 of the sequence?
(b) How many sticks are needed to arrange position number 11 of the sequence?
(c) Can an arrangement in this sequence be made using exactly 85 sticks? If yes, which position number will it correspond to?
(d) Can an arrangement in this

Accepted Answer

So, no. of sticks needed in 11th position will be 13 + 9 × 10 = 103
(c) Let the nth arrangement have 85 sticks.
13 + (n – 1) × 9 = 85
⇒ 13 + 9n – 9 = 85
⇒ 4 + 9n = 85
⇒ 9n = 85 – 4
⇒ 9n = 81
⇒ n = 9
Yes, the arrangement will have 85 sticks.
(d) Let the nth arrangement have 150 sticks.
13 + (n – 1) × 9 = 150
⇒ 13 + 9n – 9 = 150
⇒ 4 + 9n = 150
⇒ 9n = 150 – 4
⇒ 9n = 146
⇒ n = $\frac {146}{9}$
⇒ n = 16$\frac {2}{9}$
It is not a whole number.
So, no arrangement can be made with 150 sticks.

Question 17

A number increased by 36 is equal to ten times itself. What is the number?

Accepted Answer

Let the number be x.
According to the question
x + 36 = 10 × x
⇒ x + 36 = 10x
⇒ 36 = 10x – x
⇒ 36 = 9x
⇒ 36 ÷ 9 = x
⇒ 4 = x

Source: This question is from Finding the Unknown, Mathematics &mdash; Class 7, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Finding the Unknown: Number, Increased, Equal, Ten, Times, Itself. These are fundamental topics in Mathematics that students are expected to master as part of the CBSE Class 7 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Mathematics papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberAlways show your working steps clearly.
Verify your answer by substituting values back into the equation.
Practice similar problems from the textbook exercises.
Memorise important formulae and their conditions of applicability.
Related Questions from Finding the UnknownQ1: Solve these equations and check the solutions. (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 = 2u + 3 (d)...
Q10: Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.
Q11: The steps to solve three equations are shown below. Identify and correct any mistakes. (a) 6x + 9 =...
Q12: Find the measures of the angles of these triangles.
Q13: Write 4 equations whose solution is u = 6.
Practice more questions from Finding the Unknown &mdash; Mathematics, Class 7 CBSE

Question 18

Solve these equations:
(a) 5(r + 2) = 10
(b) -3(u + 2) = 2(u – 1)
(c) 2(7 – 2n) = -6
(d) 2(x – 4) = -16
(e) 6(x – 1) = 2(x – 1) – 4
(f) 3 – 7s = 7 – 3s
(g) 2x + 1 = 6 – (2x – 3)
(h) 10 – 5x = 3(x – 4) – 2(x – 7)

Accepted Answer

(a) 5(r + 2) =10
⇒ 5 × (r + 2) = 10
⇒ (r + 2) = 10 ÷ 5
⇒ r + 2 = 2
⇒ r = 2 – 2
⇒ r = 0
(b) -3(u + 2) = 2(u – 1)
⇒ -3u – 6 = 2u – 2
⇒ -3u – 2u = -2 + 6
⇒ -5u = 4
⇒ u = $\frac {-4}{5}$
(c) 2(7 – 2n) = -6
⇒ 2 × (7 – 2n) = -6
⇒ 7 – 2n = -6 ÷ 2
⇒ 7 – 2n = -3
⇒ -2n = -3 – 7
⇒ -2n = -10
⇒ n = 5
(d) 2(x – 4) = -16
⇒ x – 4 = -8
⇒ x = -8 + 4
⇒ x = -4
(e) 6(x – 1) = 2(x – 1) – 4 [Distributive property]
⇒ 6x – 6 = 2x – 2 – 4
⇒ 6x – 6 = 2x – 6
⇒ 6x – 2x = -6 + 6
⇒ 4x = 0
⇒ x = 0
(f) 3 – 7s = 7 – 3s
⇒ 3 – 7 = -3s + 7s
⇒ -4 = 4s
⇒ s = -1
(g) 2x + 1 = 6 – (2x – 3)
⇒ 2x + 1 = 6 – 2x + 3
⇒ 2x + 2x = 6 + 3 – 1
⇒ 4x = 8
⇒ x = 2
(h) 10 – 5x = 3(x – 4) – 2(x – 7)
⇒ 10 – 5x = 3x – 12 – 2x + 14
⇒ 10 – 5x = 3x – 2x – 12 + 14
⇒ 10 – 5x = x + 2
⇒ 10 – 2 = x + 5x
⇒ 8 = 6x
⇒ x = $\frac {4}{3}$

Question 19

There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?

Accepted Answer

Let the number of children be x.
Let the number of donkeys be y.
x + y = 28
y = 28 – x ……(i)
Children have 2 feet, but donkeys have 4 feet.
According to the question
2x + 4y = 80
⇒ 2x + 4(28 – x) = 80 [From (i)]
⇒ 2x + 112 – 4x = 80
⇒ -2x + 112 = 80
⇒ -2x = 80 – 112
⇒ -2x = -32
⇒ x = 16
There are 16 children, and (28 – 16) = 12 donkeys.