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NCERT Solutions · Class 7 Maths · Ganita Prakash Part 2 · Chapter 7

Chapter 7: Finding the Unknown (Equations)

Step-by-step answers to every "Figure it Out", Mind the Mistake and in-text question of Part 2, Chapter 7, Finding the Unknown (NCERT Class 7 Maths, Ganita Prakash Part 2, 2026-27): balance and mobile puzzles, framing and solving linear equations, word problems, number machines, Brahmagupta's formula, the equation path puzzle and the magic trick. All 42 questions are answered, with the key answer highlighted.

7.1 Find the Unknowns

Find the unknown weights in the hanging mobiles (Figs. 7.1 to 7.8). A balanced mobile has equal weights on its two sides; the number at the top is the total.

Solution

Fig.	Reasoning	Unknown weights
7.1 (total 16)	each side 8; leaf + bird + leaf = 3 + b + 3 = 8; flower = 8	bird 2, flower 8
7.2 (total 24)	each side 12; star + fish + star = 2 + f + 2 = 12; fish + submarine = 12	fish 8, submarine 4
7.3 (total 8)	each side 4; book = 4; the two pencil boxes balance each other: 2 each	book 4, box 2
7.4 (total 18)	each side 9; sun 5 + 4 clouds = 9; 3 lightning = 9	cloud 1, lightning 3
7.5 (total 40)	each side 20; 4 crowns = 20; crown + 5 gems = 20	crown 5, gem 3
7.6	3 breads = 6 = 2 eggs	egg 3
7.7	4 crosses = 16 = donut + cross + donut, so 2 donuts = 12	donut 6
7.8	watermelon 10 = banana + orange + banana, so 2 bananas = 6	banana 3

7.1: 2, 8; 7.2: 8, 4; 7.3: 4, 2; 7.4: 1, 3; 7.5: 5, 3; 7.6: 3; 7.7: 6; 7.8: 3

Find the unknown weight of the sack (Figs. 7.9 to 7.12; all sacks in a figure weigh the same).

Solution

7.9: sack + 2 kg = 10 kg + 2 kg. Remove 2 kg from both plates: sack = 10 kg.
7.10: 2 sacks = 10 kg + 4 kg + 1 sack. Remove one sack from both: sack = 14 kg.
7.11: 5 sacks = 10 + 10 + 1 kg + 2 sacks. Remove 2 sacks: 3 sacks = 21 kg, so sack = 7 kg.
7.12: 90 sacks + 50 kg = 60 sacks + 500 kg. Remove 60 sacks and 50 kg: 30 sacks = 450 kg, so sack = 15 kg.

10 kg, 14 kg, 7 kg, 15 kg

Matchstick Pattern and Equations

Find n such that 2n + 1 = 99. Is it possible to make an arrangement with exactly 200 sticks?

Solution

$2 n + 1 = 99$ , so $2 n = 98$ and n = 49: the 49th arrangement uses 99 sticks.

For 200 sticks: $2 n + 1 = 200$ gives $n = 99.5$ , not a whole number (2n + 1 is always odd). So no arrangement uses exactly 200 sticks.

n = 49; 200 sticks is impossible (the number of sticks is always odd).

Frame equations for Figs. 7.6 to 7.11 and solve them. Then frame 5 equations of your own.

Solution

Fig.	Equation	Solution
7.6	2e = 6	e = 3
7.7	4 + 2y = 16	y = 6
7.8	2b + 4 = 10	b = 3
7.9	s + 2 = 12	s = 10
7.10	2s = s + 14	s = 14
7.11	5s = 2s + 21	s = 7

These match the answers found earlier. Five more equations: $x + 7 = 12$ (x = 5), $3 a = 27$ (a = 9), $2 p - 5 = 11$ (p = 8), $\frac{k}{4} = 6$ (k = 24), $5 m + 2 = 3 m + 10$ (m = 4).

2e = 6, 4 + 2y = 16, 2b + 4 = 10, s + 2 = 12, 2s = s + 14, 5s = 2s + 21; solutions 3, 6, 3, 10, 14, 7.

7.2 Solving Equations Systematically

Can 2n + 1 = 99 have any other solution? Try solving 5x − 4 = 7 by trial and error.

Solution

No: increasing n always increases 2n + 1, so only one value (49) can give exactly 99.

For $5 x - 4 = 7$ : $x = 2$ gives 6 and $x = 3$ gives 11, so the answer is between 2 and 3; trying decimals, $x = 2.2$ gives $11 - 4 = 7$ ✓. Trial and error is slow here because the solution, $x = \frac{11}{5}$ , is not a whole number.

Only n = 49; 5x − 4 = 7 gives x = 11/5 = 2.2 (hard to find by trial and error).

Given 23 × 41 × 11 × 8 × 7 = 5,80,888, find 23 × 41 × 11 × 8. Is this the same as dividing both sides by 7?

Solution

$5, 80, 888 \div 7 =$ 82,984. Yes: dividing both sides by 7 removes the factor 7 from the left side, leaving exactly the expression we want.

82,984 (divide both sides by 7).

Figure it Out (page 172)

Solve and check: (a) 3x − 10 = 35 (b) 5s = 3s (c) 3u − 7 = 2u + 3 (d) 4(m + 6) − 8 = 2m − 4 (e) u/15 = 6

Solution

(a) $3 x = 45$ , x = 15. Check: $45 - 10 = 35$ ✓

(b) $5 s - 3 s = 0$ , so $2 s = 0$ , s = 0. Check: $0 = 0$ ✓

(d) $4 m + 24 - 8 = 2 m - 4$ , so $4 m + 16 = 2 m - 4$ , $2 m = - 20$ , m = −10. Check: $4 (- 4) - 8 = - 24$ and $- 20 - 4 = - 24$ ✓

(e) $u = 6 \times 15$ , u = 90. Check: $90 \div 15 = 6$ ✓

(a) 15 (b) 0 (c) 10 (d) −10 (e) 90

Frame an equation that has no solution.

Solution

$x + 4 = x + 5$ : subtracting x from both sides gives $4 = 5$ , which is never true. So no value of x works. (Another: $2 y + 1 = 2 y - 3$ .)

E.g. x + 4 = x + 5 (it reduces to 4 = 5).

Solving Problems

Ranjana's tile pattern uses 3k + 1 tiles at Step k. Can she make an arrangement with 100 tiles? At which step?

Solution

$3 k + 1 = 100$ , so $3 k = 99$ and k = 33. Yes: Step 33 uses exactly 100 tiles.

Yes, Step 33.

Check the savings answer (Jahnavi ₹4000 + ₹650 a month; Sunita ₹5050 + ₹500 a month; equal after 7 months).

Solution

After 7 months: Jahnavi $4000 + 650 \times 7 = 4000 + 4550 = 8550$ ; Sunita $5050 + 500 \times 7 = 5050 + 3500 = 8550$ . Both have ₹8,550 ✓.

Both have ₹8,550 after 7 months.

Riyaz's trick (subtract 3, multiply by 4, add 8): try different starting numbers. What simple rule gives the starting number from the final answer?

Solution

The final answer is $4 (x - 3) + 8 = 4 x - 4 = 4 (x - 1)$ . E.g. start 10 → 32; start 5 → 16.

Rule: divide the final answer by 4 and add 1. (For 24: $24 \div 4 + 1 = 7$ .)

Final answer = 4(x − 1), so starting number = answer ÷ 4 + 1.

Ramesh and Suresh have 60 marbles; Ramesh has 30 more. Use 2y + 30 = 60 to find both.

Solution

$2 y = 30$ , so $y = 15$ : Suresh has 15 and Ramesh has 15 + 30 = 45 marbles.

Suresh 15, Ramesh 45

Generating Equations

Write equations whose solution is y = 5. Form chains from the bottom equation to the top and compare the operations. Without calculating, what is the unknown in each equation of the chains?

Solution

Examples: $y + 1 = 6$ , $3 y = 15$ , $2 y - 4 = 6$ , $\frac{y}{5} = 1$ , $10 - y = 5$ .

Reverse chains:

$5 = y$ → subtract 6: $- 1 = - 6 + y$ → subtract y: $- y - 1 = - 6$ → multiply by −1: $y + 1 = 6$ .
$y = 5$ → add 2: $y + 2 = 7$ → multiply by 3: $3 y + 6 = 21$ → subtract 6: $3 y = 15$ .

Going upwards uses the inverse operations in the reverse order (add ↔ subtract, multiply ↔ divide; multiplying by −1 undoes itself).

Every equation in these chains has the same solution, y = 5, because doing the same operation to both sides keeps the equality true for y = 5.

E.g. y + 1 = 6, 3y = 15; going back up uses the inverse operations in reverse order; every equation in the chains has the solution y = 5.

Figure it Out (page 181)

Write 5 equations whose solution is x = −2.

Solution

$x + 5 = 3$ ; $3 x = - 6$ ; $2 x + 7 = 3$ ; $10 - x = 12$ ; $4 (x + 3) = 4$

E.g. x + 5 = 3, 3x = −6, 2x + 7 = 3, 10 − x = 12, 4(x + 3) = 4

Find the unknown: (a) 2y = 60 (b) −8 = 5x − 3 (c) −53w = −15 (d) 13 − z = 8 (e) k + 8 = 12 − k (f) 7m = m − 3 (g) 3n = 10 + n

Solution

(a) y = 30 (b) $5 x = - 5$ , x = −1 (c) w = 15/53 (d) z = 5 (e) $2 k = 4$ , k = 2 (f) $6 m = - 3$ , m = −1/2 (g) $2 n = 10$ , n = 5

(a) 30 (b) −1 (c) 15/53 (d) 5 (e) 2 (f) −1/2 (g) 5

I am a 3-digit number. My hundreds digit is 3 less than my tens digit, which is 3 less than my units digit. The sum of my digits is 15. Who am I?

Solution

Let the units digit be u. Then tens $= u - 3$ and hundreds $= u - 6$ .

$(u - 6) + (u - 3) + u = 15$ , so $3 u - 9 = 15$ , $3 u = 24$ , $u = 8$ . Digits: 2, 5, 8.

258

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

Solution

$w = \frac{w}{2} + 1$ , so $\frac{w}{2} = 1$ and w = 2 kg.

2 kg

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

Solution

$\frac{n}{4} + 9 = n$ , so $9 = \frac{3 n}{4}$ and n = 12. (Check: 3 + 9 = 12 ✓)

Given 4k + 1 = 13, find: (a) 8k + 2 (b) 4k (c) k (d) 4k − 1 (e) −k − 2

Solution

(a) $8 k + 2 = 2 (4 k + 1) = 26$ (b) $4 k = 12$ (c) $k = 3$ (d) $4 k - 1 = 11$ (e) $- k - 2 = - 5$

(a) 26 (b) 12 (c) 3 (d) 11 (e) −5

7.3 Mind the Mistake, Mend the Mistake

Check each solution, describe any mistake, and solve correctly: (1) 4x + 6 = 10 → 4x = 10 + 6 (2) 7 − 8z = 5 → 8z = 7 − 5 → z = 4 (3) 2v − 4 = 6 → v − 4 = 6 − 2 (4) 5z + 2 = 3z − 4 → 5z + 3z = −4 + 2 (5) 15w − 4w = 26 → 15w = 26 + 4w → 15w = 30 (6) 3x + 1 = −12 → x + 1 = −12/3 (7) 4(4q + 2) = 50 → 4(4q) = 50 − 2 (8) −2(3 − 4x) = 14 → −6v − 8x = 14 (9) 3(7y + 4) = 9 + 5y → 7y + 4 = 9/3 + 5y

Solution

	Mistake	Correct solution
1	Moving +6 to the other side should give −6, not +6.	4x = 4, x = 1
2	8z = 2 is right (adding 8z and subtracting 5), but then z = 2 ÷ 8, not 2 × 2.	z = 1/4
3	The 2 is a factor of 2v only; it cannot be "subtracted". First add 4.	2v = 10, v = 5
4	Moving 3z and +2 across must change their signs.	2z = −6, z = −3
5	26 + 4w cannot become 30 (unlike terms); simplify the left side first.	11w = 26, w = 26/11
6	Dividing by 3 must divide every term: (3x + 1) ÷ 3 is not x + 1. First subtract 1.	3x = −13, x = −13/3
7	4(4q + 2) is 16q + 8; the 4 also multiplies the 2.	16q = 42, q = 21/8
8	−2 × 3 = −6 and −2 × (−4x) = +8x; the "v" in the working is a slip.	−6 + 8x = 14, 8x = 20, x = 5/2
9	Dividing by 3 must divide 5y as well; better to expand first.	21y + 12 = 9 + 5y, 16y = −3, y = −3/16

Correct answers: (1) x = 1 (2) z = 1/4 (3) v = 5 (4) z = −3 (5) w = 26/11 (6) x = −13/3 (7) q = 21/8 (8) x = 5/2 (9) y = −3/16

7.4 A Pinch of History

Using Brahmagupta's formula

x = \frac{D - B}{A - C}

for

A x + B = C x + D

, solve 5x + 4 = 3x + 8, 3x − 6 = 2x + 4 and 2x + 3 = 4x + 5.

Solution

$5 x + 4 = 3 x + 8$ : $x = \frac{8 - 4}{5 - 3} = \frac{4}{2} =$ 2
$3 x - 6 = 2 x + 4$ : $x = \frac{4 - ( - 6 )}{3 - 2} =$ 10
$2 x + 3 = 4 x + 5$ : $x = \frac{5 - 3}{2 - 4} = \frac{2}{- 2} =$ −1

(Why it works: $A x - C x = D - B$ , i.e. $(A - C) x = D - B$ .)

2, 10 and −1

Figure it Out (page 185)

Fill in the blanks with integers: (a) 5 × __ − 8 = 37 (b) 37 − (33 − __) = 35 (c) −3 × (−11 + __) = 45

Solution

(a) $5 x = 45$ , 9 (b) $37 - 33 + x = 35$ , so $x = 31$ : 31 (c) $- 11 + x = - 15$ , −4

(a) 9 (b) 31 (c) −4

Ranju earns ₹750 a day in ₹50 and ₹100 notes, an equal number of each. How many notes of each?

Solution

$50 n + 100 n = 750$ , so $150 n = 750$ and $n = 5$ : 5 notes of ₹50 and 5 notes of ₹100.

5 of each

Each of the 3 black blobs hides the same number of blue dots; 4 dots are visible and there are 25 dots in all. How many dots does one blob cover? Write the equation.

Solution

$3 n + 4 = 25$ , so $3 n = 21$ and n = 7 dots under each blob.

3n + 4 = 25, n = 7

Number machines: (a) +3, ×4, −5 (example 12 → 55). Find the inputs for outputs 43 and 75. (b) The input goes through ×3 and through +3, and the second result is subtracted from the first (example 12 → 36 − 15 = 21). Find the inputs for outputs 63 and 227.

Solution

(a) $4 (x + 3) - 5 = 43$ , so $4 (x + 3) = 48$ , $x + 3 = 12$ , x = 9. For 75: $4 (x + 3) = 80$ , $x + 3 = 20$ , x = 17.

(b) $3 x - (x + 3) = 2 x - 3$ . For 63: $2 x = 66$ , x = 33. For 227: $2 x = 230$ , x = 115.

(a) 9 and 17 (b) 33 and 115

What are the inputs to these machines? (i) ÷3 then ÷3 gives 5 (ii) −4 then −4 gives −11

Solution

(i) $\frac{x}{9} = 5$ , x = 45 (ii) $x - 8 = - 11$ , x = −3

(i) 45 (ii) −3

A taxi charges ₹800 a day plus ₹20 per km. The total is ₹2200. How many km were travelled?

Solution

$800 + 20 k = 2200$ , so $20 k = 1400$ and k = 70 km.

70 km

The sum of two numbers is 76; one is three times the other. Find them.

Solution

$x + 3 x = 76$ , so $4 x = 76$ and $x = 19$ : the numbers are 19 and 57.

19 and 57

The window is 34 cm high, with a 3 cm border at the top and bottom and 5 rods 2 cm thick making 6 equal gaps. What is the gap between two rods?

Solution

$2 \times 3 + 5 \times 2 + 6 g = 34$ , so $16 + 6 g = 34$ , $6 g = 18$ , and g = 3 cm.

3 cm

A fruit juice costs ₹15 less than a chocolate milkshake. 4 juices and 7 milkshakes cost ₹600. Find each price.

Solution

Let a milkshake cost m rupees; a juice costs $m - 15$ .

$4 (m - 15) + 7 m = 600$ , so $11 m - 60 = 600$ , $11 m = 660$ , $m = 60$ .

Milkshake ₹60, juice ₹45.

Milkshake ₹60, fruit juice ₹45

Given 28p − 36 = 98, find 14p − 19 and 28p − 38.

Solution

$28 p = 134$ , so $14 p = 67$ : $14 p - 19 =$ 48. And $28 p - 38 = (28 p - 36) - 2 = 98 - 2 =$ 96.

48 and 96

Identify and correct the mistakes: (a) 6x + 9 = 66 → x + 9 = 11 → x = 2 (b) 14y + 24 = 36 → 7y + 12 = 18 → 7y = 6 → y = 6/7 (c) 4x − 5 = 9x + 8 → 4x = 9x + 8 − 5 → … → x = −5/3

Solution

(a) Wrong: dividing by 6 must divide 9 too. Correct: $6 x = 57$ , $x = \frac{57}{6} = 9.5$ .

(b) Correct: dividing every term by 2 is fine, and $y = \frac{6}{7}$ .

(c) Two mistakes: moving −5 to the right should give +5, so $4 x = 9 x + 13$ ; and at the end $- 5 x = 3$ would give $x = - \frac{3}{5}$ , not $- \frac{5}{3}$ . Correct: $- 5 x = 13$ , $x = - \frac{13}{5}$ .

(a) x = 9.5 (b) correct, y = 6/7 (c) x = −13/5

Find the angles of the triangles: (i) isosceles with apex angle y and base angle y + 15 (ii) angles x, x − 10, x + 10.

Solution

(i) The two base angles are equal (opposite equal sides): $y + 2 (y + 15) = 180$ , so $3 y = 150$ , $y = 50$ . Angles: 50°, 65°, 65°.

(ii) $x + (x - 10) + (x + 10) = 180$ , so $3 x = 180$ , $x = 60$ . Angles: 60°, 50°, 70°.

(i) 50°, 65°, 65° (ii) 60°, 50°, 70°

Write 4 equations whose solution is u = 6.

Solution

$u + 4 = 10$ ; $2 u = 12$ ; $3 u - 8 = 10$ ; $\frac{u}{2} = 3$

E.g. u + 4 = 10, 2u = 12, 3u − 8 = 10, u/2 = 3

Bakhśhālī Manuscript: the second person gets twice the first, the third thrice the second, the fourth four times the third. The total is 132. Find the first person's amount.

Solution

First $a$ , second $2 a$ , third $6 a$ , fourth $24 a$ . $a + 2 a + 6 a + 24 a = 33 a = 132$ , so a = 4.

4 (the shares are 4, 8, 24, 96)

A giraffe's height is 2.5 m more than half its height. How tall is it?

Solution

$h = \frac{h}{2} + 2.5$ , so $\frac{h}{2} = 2.5$ and h = 5 m.

5 m

For each stick pattern (I: n squares in a row with a triangle at the end; II: a staircase of squares with 4, 7, 10, 13, … squares), find (a) the squares in position 11 (b) the sticks in position 11 (c) whether exactly 85 sticks can be used (d) whether exactly 150 sticks can be used.

Solution

Pattern I: position n has n squares; a row of n squares needs $3 n + 1$ sticks and the triangle adds 2, so $3 n + 3$ sticks.

(a) 11 squares (b) $3 \times 11 + 3 = 36$ sticks (c) $3 n + 3 = 85$ gives $n = \frac{82}{3}$ : no (d) $3 n + 3 = 150$ gives $n = 49$ : yes, position 49.

Pattern II: position n has 3n + 1 squares (4, 7, 10, 13, …). The squares form a chain where each square shares one side with the next, so sticks $= 4 (3 n + 1) - 3 n = 9 n + 4$ (13, 22, 31, 40, …).

(a) $3 \times 11 + 1 = 34$ squares (b) $9 \times 11 + 4 = 103$ sticks (c) $9 n + 4 = 85$ gives $n = 9$ : yes, position 9 (d) $9 n + 4 = 150$ gives $n = \frac{146}{9}$ : no.

I: 11 squares, 36 sticks; 85 no; 150 yes (position 49). II: 34 squares, 103 sticks; 85 yes (position 9); 150 no.

A number increased by 36 equals ten times itself. Find it.

Solution

$x + 36 = 10 x$ , so $9 x = 36$ and x = 4.

Solve: (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)

Solution

(a) $r + 2 = 2$ , r = 0

(b) $- 3 u - 6 = 2 u - 2$ , $- 5 u = 4$ , u = −4/5

(d) $x - 4 = - 8$ , x = −4

(e) $6 x - 6 = 2 x - 6$ , $4 x = 0$ , x = 0

(f) $- 4 s = 4$ , s = −1

(g) $2 x + 1 = 9 - 2 x$ , $4 x = 8$ , x = 2

(h) RHS $= 3 x - 12 - 2 x + 14 = x + 2$ ; $10 - 5 x = x + 2$ , $6 x = 8$ , x = 4/3

(a) 0 (b) −4/5 (c) 5 (d) −4 (e) 0 (f) −1 (g) 2 (h) 4/3

Solve the equations to find a path from Start to End.

Solution

Solutions of the boxes: $8 x = 20 + 3 x$ → 4; $2 x - 9 = - 3$ → 3; $- 2 x = - 42$ → 21; $15 = 19 - 4 x$ → 1; $2 x + 3 = x + 5$ → 2; $2 x + 5 = 3 (x - 1)$ → 8; $2 (x + 1) - 10 = 18$ → 13; $8 m + 8 = - 72$ → −10; $- 4 = 16 - 5 k$ → 4; $2 x - 9 = 3 - x$ → 4; ( $- 7 = 11 - 3 x$ → 6, not on the path).

Path: Start ( $x = 4$ ) → $2 x - 9 = - 3$ (3) → $- 2 x = - 42$ (21) → $15 = 19 - 4 x$ (1) → $2 x + 3 = x + 5$ (2) → $2 x + 5 = 3 (x - 1)$ (8) → $2 (x + 1) - 10 = 18$ (13) → $8 m + 8 = - 72$ (−10) → $- 4 = 16 - 5 k$ (4) → $2 x - 9 = 3 - x$ (4) → $30 = 4 - 50 n$ → End.

(Follow each arrow labelled with the answer of the box you are in. The last box gives $50 n = - 26$ , $n = - \frac{13}{25}$ .)

4 → 3 → 21 → 1 → 2 → 8 → 13 → −10 → 4 → 4 → End

Children and donkeys on a beach have 28 heads and 80 feet. How many of each?

Solution

Let there be d donkeys; then $28 - d$ children. Feet: $4 d + 2 (28 - d) = 80$ , so $2 d + 56 = 80$ , $d = 12$ .

12 donkeys and 16 children. (Check: 48 + 32 = 80 feet ✓)

12 donkeys, 16 children

A Magic Trick

Think of a number, double it, add 10, halve, subtract the original number, add 3. Why do you always get 8? Make your own trick.

Solution

With x: $2 x$ → $2 x + 10$ → $x + 5$ → $x + 5 - x = 5$ → $5 + 3 = 8$ . The x cancels out, so the answer is always 8, whatever number you start with.

Your own trick: think of a number, add 4, multiply by 3, subtract 12, divide by 3, subtract the original number, add 7: $\frac{3 ( x + 4 ) - 12}{3} - x + 7 = x - x + 7 = 7$ , always 7.

(2x + 10) ÷ 2 − x + 3 = 8 for every x, because x cancels out.

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