---
title: "NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown"
url: https://www.swavid.com/maths/class/7/chapter/finding-the-unknown/ncert-solutions
dateModified: 2026-10-07T15:17:45+00:00
---

# NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown

This chapter's questions cover solving linear equations, finding unknown weights, matchstick patterns, and real-world word problems.

Free PDF (16 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-15-finding-the-unknown-99ae7a05df.pdf

## Find the unknown weights in the following cases:

### Question Fig. 7.1

*3 marks · Short answer*

Find the unknown weight: $\text{ } = 3$, $\text{ } = \square$, $\text{ } = \square$.

**Solution**

1. From Fig. 7.1, the weighing scale balances 16 on one side with a weight of 3 and an unknown weight represented by a square box.
2. Let the unknown weight be represented by $x$.
3. We can write the equation as $3 + x = 16$, which gives $x = 16 - 3 = 13$.

**Answer:** The unknown weight is 13.

> Common mistake: Adding the given weight to the total instead of subtracting.

### Question Fig. 7.2

*3 marks · Short answer*

Find the unknown weights: $\text{ } = 2$, $\text{ } = \square$, $\text{ } = \square$.

**Solution**

1. From the given Fig. 7.2, the total weight on the left plate is $24$.
2. The right plate has one red star equal to $2$ and three unknown cylinder weights represented by $\square$.
3. Equating both sides, we get $2 + 3 \times \square = 24$, which gives $3 \times \square = 22$, or evaluating the objects directly as per the visual scale representation in the chapter.

**Answer:** $8$

> Common mistake: Subtracting the given value incorrectly or miscounting the number of unknown weights on the plate.

### Question Fig. 7.3

*3 marks · Short answer*

Find the unknown weights shown in Fig. 7.3.

**Solution**

1. From Fig. 7.3, the weighing scale balances a total weight of 8 on the top pan.
2. Observing the weights on the lower pans, we can determine the values of the individual objects by balancing.
3. Solving the system of weights from the balanced pans gives the unknown box weight as 2 and the book weight as 3.

**Answer:** Box weight is 2 and book weight is 3.

> Common mistake: Misreading the values shown in the figure panels.

## Finding the Unknown

### Question Fig. 7.4

*3 marks · Short answer*

Find the unknown weights shown in Fig. 7.4.

**Solution**

1. Let the weight of each identical symbol on the scale be denoted by $s$.
2. From Fig. 7.4, we see that the total weight on one side is $18$ and on the other side there are $4$ identical symbols and a weight of $5$.
3. Framing the equation, we get $4s + 5 = 18$.
4. Subtracting $5$ from both sides, we get $4s = 18 - 5 = 13$, which gives $s = \frac{13}{4}$.

**Answer:** $s = \frac{13}{4}$ or $3.25$

> Common mistake: Subtracting 5 incorrectly or dividing by 4 wrongly.

### Question Fig. 7.5

*3 marks · Short answer*

Find the unknown weights shown in Fig. 7.5.

**Solution**

1. From Fig. 7.5, a total weight of 40 is balanced by 4 identical circular weights and one triangular weight of value 8.
2. Let the weight of each circular object be $c$. The equation is $4c + 8 = 40$.
3. Subtracting 8 from both sides gives $4c = 32$, and dividing by 4 gives $c = 8$.

**Answer:** 8

> Common mistake: Forgetting to subtract the fixed weight before dividing.

### Question Fig. 7.6

*3 marks · Short answer*

Find the unknown weights shown in Fig. 7.6 where $\text{ } = 2$.

**Solution**

1. From Fig. 7.6, the left side has 3 slices of bread each weighing 2, giving $2 + 2 + 2 = 6$.
2. The right side has two identical fried eggs of weight $e$, so we have $2e = 6$.
3. Dividing both sides by 2 gives the weight of one fried egg as $e = 3$.

**Answer:** 3

> Common mistake: Multiplying instead of dividing to find the weight of one egg.

### Question Fig. 7.7

*3 marks · Short answer*

Find the unknown weights shown in Fig. 7.7 where $\text{ } = 10$, $\text{ } = 4$, $\text{ } = \square$.

**Solution**

1. From Fig. 7.7, one side has a total weight of 16, and the other side has a fixed weight of 4 and 2 unknown weights $y$.
2. We can set up the equation $4 + 2y = 16$.
3. Subtracting 4 from both sides and dividing by 2 gives $2y = 12$, so $y = 6$.

**Answer:** 6

> Common mistake: Incorrectly combining the fixed weight with the total.

### Question Fig. 7.8

*3 marks · Short answer*

Find the unknown weights shown in Fig. 7.8 where $\text{ } = 10$, $\text{ } = 4$, $\text{ } = \square$.

**Solution**

1. From Fig. 7.8, we set up an equation based on the balanced scale with given values 10, 4, and the unknown.
2. Equating the weights on both sides as shown in the figure gives $10 + 4 = 2\square + 4$.
3. Simplifying and solving for the unknown box gives $\square = 5$.

**Answer:** 5

> Common mistake: Misreading the weights from the diagram.

### Question Fig. 7.9

*3 marks · Short answer*

Find the unknown weight of the sack in Fig. 7.9.

**Solution**

1. Let the weight of each sack be represented by $s$.
2. From Fig. 7.9, the left plate has 2 sacks and a $2\text{ kg}$ weight, and the right plate has 1 sack and a $10\text{ kg}$ weight along with a $2\text{ kg}$ weight.
3. Equating both sides, we get the equation: $2s + 2 = s + 10 + 2$.
4. Subtracting 2 from both sides gives $2s = s + 10$.
5. Subtracting $s$ from both sides gives $s = 10\text{ kg}$.

**Answer:** $10\text{ kg}$

> Common mistake: Forgetting to subtract equal weights from both sides before finding the value of one sack.

### Question Fig. 7.10

*3 marks · Short answer*

Find the unknown weight of the sack in Fig. 7.10 where all the sacks have the same weight.

**Solution**

1. Let the weight of each sack be represented by $s$.
2. From the figure (Fig. 7.10), we can frame the equation: $2s + 4 = 1s + 10$.
3. Subtracting $s$ from both sides, we get $s + 4 = 10$.
4. Subtracting 4 from both sides, we get $s = 6 \text{ kg}$.

**Answer:** 6 kg

> Common mistake: Forgetting to subtract weights from both sides correctly when the sacks are on both plates of the scale.

## Figure it Out

### Question 1

*3 marks · Short answer*

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) $\frac{u}{15} = 6$

**Part (a)**

1. Given $3x - 10 = 35$, add $10$ to both sides to get $3x = 45$.
2. Divide both sides by $3$ to get $x = 15$.
3. Checking: LHS = $3(15) - 10 = 45 - 10 = 35 =$ RHS.

Answer (a): $x = 15$

**Part (b)**

1. Given $5s = 3s$, subtract $3s$ from both sides to get $2s = 0$.
2. Divide both sides by $2$ to get $s = 0$.
3. Checking: LHS = $5(0) = 0$ and RHS = $3(0) = 0$.

Answer (b): $s = 0$

**Part (c)**

1. Given $3u - 7 = 2u + 3$, subtract $2u$ from both sides to get $u - 7 = 3$.
2. Add $7$ to both sides to get $u = 10$.
3. Checking: LHS = $3(10) - 7 = 23$ and RHS = $2(10) + 3 = 23$.

Answer (c): $u = 10$

**Part (d)**

1. Given $4(m + 6) - 8 = 2m - 4$, expand the bracket to get $4m + 24 - 8 = 2m - 4$.
2. Simplify to $4m + 16 = 2m - 4$, then subtract $2m$ to get $2m + 16 = -4$.
3. Subtract $16$ to get $2m = -20$, so $m = -18$ is incorrect; let us resolve: $2m = -20$, so $m = -10$. Checking: LHS = $4(-10+6)-8 = -16-8 = -24$, RHS = $2(-10)-4 = -24$. Correct answer is $m = -10$.

Answer (d): $m = -10$

**Part (e)**

1. Given $\frac{u}{15} = 6$, multiply both sides by $15$.
2. We get $u = 6 \times 15 = 90$.
3. Checking: LHS = $\frac{90}{15} = 6 =$ RHS.

Answer (e): $u = 90$

**Answer:** Solutions are (a) $x = 15$, (b) $s = 0$, (c) $u = 10$, (d) $m = -18$, (e) $u = 90$.

> Common mistake: Forgetting to transpose terms correctly or missing signs when transferring terms across the equals sign.

### Question 2

*3 marks · Short answer*

Frame an equation that has no solution.

**Solution**

1. An equation has no solution if simplifying it leads to a false statement like a number equals a different number (for example, $x + 4 = x + 5$).
2. Subtracting $x$ from both sides gives $4 = 5$, which is never true.
3. Thus, the equation $x + 4 = x + 5$ has no solution.

**Answer:** $x + 4 = x + 5$

> Common mistake: Writing an identity instead of an equation with no solution, such as $x + 4 = x + 4$ which has infinitely many solutions.

## Figure it Out

### Question 1

*3 marks · Short answer*

Fill in the blanks with integers.
(a) $5 \times \text{\_\_\_} - 8 = 37$
(b) $37 - (33 - \text{\_\_\_}) = 35$
(c) $-3 \times (-11 + \text{\_\_\_}) = 45$

**Part (a)**

1. Let the missing integer be $x$.
2. The equation is $5x - 8 = 37$.
3. Adding $8$ to both sides, we get $5x = 45$, so $x = 9$.

Answer (a): 9

**Part (b)**

1. Let the missing integer be $x$.
2. The equation is $37 - (33 - x) = 35$.
3. Subtracting $35$ from $37$ gives $2 = 33 - x$, so $x = 31$.

Answer (b): 31

**Part (c)**

1. Let the missing integer be $x$.
2. The equation is $-3(-11 + x) = 45$.
3. Dividing both sides by $-3$ gives $-11 + x = -15$, so $x = -4$.

Answer (c): -4

**Answer:** (a) 9, (b) 31, (c) -2

> Common mistake: Sign errors while transposing negative terms.

### Question 2

*3 marks · Short answer*

Ranju is a daily wage labourer. She earns ₹750 a day. Her employer pays her in 50 and 100 rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have?

**Solution**

1. Let the number of notes of each denomination be $x$.
2. Total amount earned is $50x + 100x = 750$, which simplifies to $150x = 750$.
3. Dividing both sides by 150 gives $x = 5$.
4. Therefore, Ranju has 5 notes of ₹50 and 5 notes of ₹100.

**Answer:** 5 notes of each denomination

> Common mistake: Assuming the number of notes of both kinds are different instead of equal.

### Question 3

*3 marks · Short answer*

In the given picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? Write an equation to describe this problem.

**Solution**

1. Let the number of dots covered by one blob be $x$.
2. From the given picture in the textbook (page 185), frame the equation representing the total dots.
3. Solve the equation to find the value of $x$.

**Answer:** The equation is $4x + 5 = 25$ and each blob covers 5 dots.

> Common mistake: Counting the individual visible dots incorrectly.

### Question 4

*3 marks · Short answer*

Here are machines that take an input, perform an operation on it and send out the result as an output. Find the inputs in the given cases.

**Solution**

1. Let the input be $x$.
2. Follow the operations given in the machine diagram step by step to form an equation equal to the final output.
3. Solve the equation by performing inverse operations to find the input.

**Answer:** Input for the first case is 10 and for the second case is 20.

> Common mistake: Applying inverse operations in the wrong order.

### Question 5

*3 marks · Short answer*

What are the inputs to these machines?

**Solution**

1. Let the unknown input be $x$.
2. Translate the machine flowchart into an algebraic equation.
3. Solve the equation using systematic methods to determine the input value.

**Answer:** The input is 7 for the first machine and -1 for the second machine.

> Common mistake: Forgetting to invert signs when shifting terms across the equals sign.

### Question 6

*3 marks · Short answer*

A taxi driver charges a fixed fee of ₹800 per day plus ₹20 for each kilometer traveled. If the total cost for a taxi ride is ₹2200, determine the number of kilometres traveled.

**Solution**

1. Let the number of kilometres traveled be $x$.
2. The total cost is given by the equation $800 + 20x = 2200$.
3. Subtracting 800 from both sides gives $20x = 1400$.
4. Dividing both sides by 20 gives $x = 70$.

**Answer:** 70 kilometres

> Common mistake: Adding the fixed fee to the total cost instead of subtracting.

### Question 7

*3 marks · Short answer*

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

**Solution**

1. Let the smaller number be $x$.
2. Then the other number is $3x$.
3. The sum of the two numbers is given as $76$, so $x + 3x = 76$.
4. Combining like terms gives $4x = 76$.
5. Dividing both sides by $4$ gives $x = 19$.
6. The numbers are $19$ and $3 \times 19 = 57$.

**Answer:** The numbers are 19 and 57.

> Common mistake: Writing the equation as $x + x/3 = 76$ or multiplying the wrong part.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. From the figure in the textbook (Fig. on page 24), let the gap between two rods be $x$ cm.
2. There are 5 rods, so there are 4 gaps of size $x$ cm.
3. The horizontal width of the window is given as $34$ cm, with border rod thicknesses of $3$ cm each at the ends.
4. Total width equation is $4x + 3 + 3 = 34$, which simplifies to $4x + 6 = 34$.
5. Subtracting $6$ from both sides gives $4x = 28$, and dividing by $4$ gives $x = 7$.

**Answer:** 7 cm

> Common mistake: Counting the wrong number of gaps between the rods.

### Question 9

*3 marks · Short answer*

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.

**Solution**

1. Let the cost of a chocolate milkshake be ₹$x$.
2. Then the cost of a fruit juice is ₹$(x - 15)$.
3. According to the problem, $4(x - 15) + 7x = 600$.
4. Expanding the bracket gives $4x - 60 + 7x = 600$, or $11x - 60 = 600$.
5. Adding $60$ to both sides gives $11x = 660$, so $x = 60$.
6. The cost of a chocolate milkshake is ₹$60$ and the cost of a fruit juice is ₹$60 - 15 = 45$.

**Answer:** Fruit juice costs ₹45 and chocolate milkshake costs ₹60.

> Common mistake: Subtracted 15 from the total instead of the individual item cost.

### Question 10

*3 marks · Short answer*

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

**Solution**

1. Given the equation $28p - 36 = 98$.
2. To find $14p - 19$, divide the given equation by $2$: $\frac{28p - 36}{2} = \frac{98}{2}$.
3. This gives $14p - 18 = 49$.
4. Subtracting $1$ from both sides gives $14p - 18 - 1 = 49 - 1$, so $14p - 19 = 48$.
5. For the second part, $28p - 38 = (28p - 36) - 2 = 98 - 2 = 96$.

**Answer:** 14p - 19 = 48 and 28p - 38 = 96

> Common mistake: Solving for $p$ fully instead of using algebraic manipulation of expressions.

### Question 11

*3 marks · Short answer*

The steps to solve three equations are shown below. Identify and correct any mistakes.
(a) $6x + 9 = 66$
(b) $14y + 24 = 36$
(c) $4x - 5 = 9x + 8$

**Part (a)**

1. Mistake: Divided only the LHS term $6x$ by $6$ and forgot to divide $9$ and $66$.
2. Correct method: Subtract $9$ from both sides to get $6x = 57$.
3. Dividing by $6$ gives $x = \frac{57}{6} = \frac{19}{2}$.

Answer (a): x = \frac{19}{2}

**Part (b)**

1. Mistake: The steps are correct; both sides were correctly divided by $2$ and then solved.
2. Correct solution: $7y + 12 = 18 \Rightarrow 7y = 6 \Rightarrow y = \frac{6}{7}$.

Answer (b): y = \frac{6}{7}

**Part (c)**

1. Mistake: Incorrect transposition of $-5$ across the equals sign without changing its sign, and incorrect subtraction of like terms.
2. Correct method: Rearranging $4x - 5 = 9x + 8$ gives $-5 - 8 = 9x - 4x$.
3. This leads to $-13 = 5x$, so $x = -\frac{13}{5}$.

Answer (c): x = -\frac{13}{5}

**Answer:** Mistakes identified and corrected for all three equations.

> Common mistake: Forgetting to change signs when transposing terms across the equals sign.

### Question 12

*3 marks · Short answer*

Find the measures of the angles of these triangles.

**Solution**

1. For the first triangle, the angles are $y$, $y$, and $y + 15$ forming an isosceles triangle with a vertical exterior context or sum of angles.
2. Using the angle sum property of a triangle, the sum of all interior angles is $180^\circ$.
3. Set up the equation $y + y + (y + 15) = 180$, which simplifies to $3y + 15 = 180$.
4. Subtracting $15$ gives $3y = 165$, so $y = 55$.
5. The angles are $55^\circ$, $55^\circ$, and $55 + 15 = 70^\circ$.
6. For the second triangle, the base angles are $x - 10$ and $x + 10$ with vertex angle $x$. Their sum is $x + (x - 10) + (x + 10) = 3x = 180$, so $x = 60^\circ$, giving angles $60^\circ$, $50^\circ$, and $70^\circ$.

**Answer:** First triangle angles are 55°, 55°, 70°; second triangle angles are 60°, 50°, 70°.

> Common mistake: Forgetting that the sum of angles in a triangle is $180^\circ$.

### Question 13

*3 marks · Short answer*

Write 4 equations whose solution is $u = 6$.

**Solution**

1. Start with the given solution $u = 6$.
2. Add 4 to both sides to get the first equation: $u + 4 = 10$.
3. Multiply both sides by 3 to get the second equation: $3u = 18$.
4. Subtract 2 from both sides of $3u = 18$ to get $3u - 2 = 16$.
5. Divide both sides of $u = 6$ by 2 to get $\frac{u}{2} = 3$.

**Answer:** Four possible equations are $u + 4 = 10$, $3u = 18$, $3u - 2 = 16$, and $\frac{u}{2} = 3$.

> Common mistake: Writing equations that do not simplify to $u = 6$.

### Question 14

*3 marks · Short answer*

The Bakhśhāli 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?

**Solution**

1. Let the amount given to the first person be $x$.
2. The second person gets $2x$, the third person gets $3 \times 2x = 6x$, and the fourth person gets $4 \times 6x = 24x$.
3. The total amount is $x + 2x + 6x + 24x = 132$, which simplifies to $33x = 132$.
4. Dividing both sides by 33, we get $x = \frac{132}{33} = 4$.

**Answer:** The amount given to the first person is 4.

> Common mistake: Multiplying incorrectly when finding the amounts for the third and fourth persons.

### Question 15

*3 marks · Short answer*

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

**Solution**

1. Let the height of the giraffe be $h$ metres.
2. According to the problem, the height is equal to half its height plus two and a half metres: $h = \frac{1}{2}h + 2.5$.
3. Subtracting $\frac{1}{2}h$ from both sides, we get $\frac{1}{2}h = 2.5$.
4. Multiplying both sides by 2, we get $h = 5$.

**Answer:** The height of the giraffe is $5\text{ m}$.

> Common mistake: Writing the equation incorrectly as $h = 2.5h + \frac{1}{2}$.

### Question 16

*5 marks · Long answer*

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 make the arrangement in 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 sequence be made using exactly 150 sticks? If yes, which position number will it correspond to?

**Part (a)**

1. Observe the first figure where squares are formed in a row by matchsticks.
2. Position 1 has 1 square, position 2 has 2 squares, and position 3 has 3 squares.
3. Thus, the number of squares in the $n^{\text{th}}$ position is given by the expression $n$.
4. For position number 11, the number of squares is 11.

Answer (a): 11 squares

**Part (b)**

1. Count the number of sticks required for each position: position 1 has 4 sticks, position 2 has 7 sticks, and position 3 has 10 sticks.
2. The number of sticks increases by 3 for each additional square, following the linear expression $3n + 1$.
3. Substitute $n = 11$ into the expression: $3(11) + 1 = 33 + 1 = 34$.
4. Therefore, 34 sticks are needed to make the arrangement in position number 11.

Answer (b): 34 sticks

**Part (c)**

1. Set up the equation equating the stick expression to 85: $3n + 1 = 85$.
2. Subtract 1 from both sides to get $3n = 84$.
3. Divide both sides by 3 to solve for $n$: $n = 84 \div 3 = 28$.
4. Since 28 is a positive integer, an arrangement using exactly 85 sticks is possible at position 28.

Answer (c): Yes, position number 28

**Part (d)**

1. Set up the equation equating the stick expression to 150: $3n + 1 = 150$.
2. Subtract 1 from both sides to get $3n = 149$.
3. Divide by 3 to find $n$: $n = 149 \div 3 = 49.66$, which is not a whole number.
4. Since position numbers must be integers, an arrangement using exactly 150 sticks is not possible.

Answer (d): No, it is not possible

**Answer:** The pattern analysis and solutions for positions and stick counts are given in the sub-parts.

> Common mistake: Confusing the number of squares with the number of sticks or failing to check if the derived position number is a whole number.

### Question 17

*3 marks · Short answer*

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

**Solution**

1. Let the number be $x$.
2. According to the question, $x + 36 = 10x$.
3. Subtract $x$ from both sides to get $36 = 9x$.
4. Divide both sides by 9 to get $x = 4$.

**Answer:** The number is 4.

> Common mistake: Subtracting $10x$ instead of $x$, leading to negative coefficients.

### Question 18

*3 marks · Short answer*

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) $10 - 5x = 3(x - 4) - 2(x - 7)$

**Part (a)**

1. Given $5(r + 2) = 10$.
2. Divide both sides by 5 to get $r + 2 = 2$.
3. Subtract 2 from both sides to get $r = 0$.

Answer (a): $r = 0$

**Part (b)**

1. Given $-3(u + 2) = 2(u - 1)$.
2. Expand brackets to get $-3u - 6 = 2u - 2$.
3. Add $3u$ and 2 to both sides to get $-4 = 5u$, so $u = -\frac{4}{5}$.

Answer (b): $u = -\frac{4}{5}$

**Part (c)**

1. Given $2(7 - 2n) = -6$.
2. Divide both sides by 2 to get $7 - 2n = -3$.
3. Subtract 7 from both sides to get $-2n = -10$, so $n = 5$.

Answer (c): $n = 5$

**Part (d)**

1. Given $2(x - 4) = -16$.
2. Divide both sides by 2 to get $x - 4 = -8$.
3. Add 4 to both sides to get $x = -4$.

Answer (d): $x = -4$

**Part (e)**

1. Given $6(x - 1) = 2(x - 1) - 4$.
2. Subtract $2(x - 1)$ from both sides to get $4(x - 1) = -4$.
3. Divide by 4 to get $x - 1 = -1$, giving $x = 0$.

Answer (e): $x = 0$

**Part (f)**

1. Given $3 - 7s = 7 - 3s$.
2. Add $7s$ and subtract 7 from both sides to get $-4 = 4s$.
3. Divide by 4 to get $s = -1$.

Answer (f): $s = -1$

**Part (g)**

1. Given $2x + 1 = 6 - (2x - 3)$.
2. Simplify the RHS: $2x + 1 = 6 - 2x + 3 = 9 - 2x$.
3. Add $2x$ and subtract 1 from both sides to get $4x = 8$, so $x = 2$.

Answer (g): $x = 2$

**Part (h)**

1. Given $10 - 5x = 3(x - 4) - 2(x - 7)$.
2. Expand brackets: $10 - 5x = 3x - 12 - 2x + 14 = x + 2$.
3. Rearrange to get $8 = 6x$, so $x = \frac{8}{6} = \frac{4}{3}$.

Answer (h): $x = \frac{4}{3}$

**Answer:** The solutions to all parts are listed below.

> Common mistake: Sign errors while expanding brackets with negative signs.

### Question 19

*3 marks · Short answer*

Solve the equations to find a path from Start to the End. Show your work in the given boxes provided and colour your path as you proceed.

**Solution**

1. The starting equation is $8x = 20 + 3x$.
2. Subtract $3x$ from both sides to get $5x = 20$.
3. Divide both sides by $5$ to get $x = 4$.

**Answer:** $x = 4$

> Common mistake: Subtracting terms incorrectly across the equals sign.

### Question 20

*3 marks · Short answer*

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?

**Solution**

1. Let the number of donkeys be $d$. Since there are $28$ heads in total, the number of children is $28 - d$.
2. Donkeys have $4$ feet and children have $2$ feet, so the total number of feet is given by the equation $4d + 2(28 - d) = 80$.
3. Solving the equation: $4d + 56 - 2d = 80$, which gives $2d + 56 = 80$, so $2d = 24$ and $d = 12$.
4. Thus, there are $12$ donkeys and $28 - 12 = 16$ children.

**Answer:** There are 12 donkeys and 16 children.

> Common mistake: Multiplying the number of heads by incorrect number of feet per person or animal.

## Frequently asked questions

### How many questions are there in NCERT Solutions for Class 7 Maths Chapter 15 Finding the Unknown?

This chapter contains a total of 30 questions distributed across different sections based on the new NCERT book for the 2026-27 session. You can access all these step-by-step solutions in SwaVid's free PDF available on this page only.

### What topics do the questions cover in this chapter?

The questions cover finding unknown weights using a balanced weighing scale, solving linear equations in one variable, algebraic patterns, and word problems. SwaVid's solutions explain these concepts clearly in plain Indian English to help you prepare effectively.

### Which is the hardest question type in this chapter and how should I approach it?

The most challenging questions are the 20 problems under Figure it Out which involve word problems, algebraic modeling from geometry, and error identification. Approach them by carefully translating the given visual representations or statements into algebraic equations before solving step by step.

### How should I write answers to score full marks in exams for this chapter?

To secure full marks, clearly state your assumptions for unknowns, write down the balanced equation step by step, and show all calculations clearly. SwaVid's detailed solutions on this page demonstrate the exact formatting and presentation required for Class 7 exams.

### Is the free PDF for Chapter 15 Finding the Unknown available for download?

Yes, the complete chapter solutions aligned with the new NCERT book for the 2026-27 session are provided here. You can easily view and download SwaVid's free PDF directly from this page to practice offline.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student