Class 7 Maths Ganita Prakash Chapter 15 Solutions (NCERT 2026–27) – Finding the Unknown

These Class 7 Maths Ganita Prakash Chapter 15 solutions cover Finding the Unknown — the algebra chapter on framing and solving simple equations from the new NCF-2023 textbook (Reprint 2026–27). This is Chapter 7 of Ganita Prakash Part II (site Chapter 15). Every Figure it Out question, every Math Talk and every Try This task is solved step by step using the balancing method, so you can solve any linear equation with confidence and revise quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 15 (Part II, Ch 7) Exercises: Figure it Out (p. 172), (p. 185), (p. 187–189) Session: 2026–27
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Chapter 15 Overview

Chapter 15 of Ganita Prakash, Finding the Unknown (Part II, Chapter 7), introduces algebraic equations through the picture of a balanced weighing scale. Just as removing equal weights from both pans keeps the scale balanced, performing the same operation on both sides of an equation keeps the LHS equal to the RHS. The chapter moves from a matchstick pattern (2n + 1) to framing equations, then from the slow trial-and-error method to the efficient systematic (balancing) method. You learn to bring unknown terms to one side, use additive and multiplicative inverses, model real-life situations (savings, snacks, taxis, marbles) as equations, generate equations from a given solution, spot and mend errors in worked solutions, and finally meet Brahmagupta’s elegant formula x = (D − B)/(A − C) for Ax + B = Cx + D. The Class 7 Maths Ganita Prakash Chapter 15 solutions below work through every Figure it Out, Math Talk and Try This question step by step.

Key Concepts & Definitions

Equation: a statement of equality between two algebraic expressions, written with an “=” sign, e.g. 2n + 1 = 99 or 6y + 7 = 4y + 21.

LHS and RHS: the Left Hand Side and Right Hand Side of an equation — the two expressions on either side of the equal sign.

Solving an equation: finding the value(s) of the letter−number for which the LHS becomes equal to the RHS. That value is called the solution (or root) of the equation.

Balancing property: if the same number is added, subtracted, multiplied or divided on both sides of an equation, the equality is preserved (like a balanced weighing scale).

Trial-and-error method: substituting different values for the unknown until LHS = RHS. It works but is slow and inefficient.

Systematic (balancing) method: using inverse operations on both sides to isolate the unknown — the efficient method used throughout the chapter.

Bījagañita (algebra): the branch of mathematics that uses letter symbols to represent unknowns and solve problems; bīja means “seed”, the answer hidden inside the unknown.

Important Formulas & Methods (Chapter 15)

Matchstick pattern: the arrangement at position n has 2n + 1 matchsticks; Ranjana’s tile pattern at step k uses 3k + 1 tiles.

Move a term: when a term added/subtracted on one side is removed, it appears with its additive inverse (opposite sign) on the other side. Example: 2y + 7 = 21 → 2y = 21 − 7.

Remove a factor: if one side is a product and you remove a factor, divide the other side by it. Example: 2y = 14 → y = 14 ÷ 2.

Remove a divisor: if one side is a quotient and you remove the divisor, multiply the other side by it. Example: u/15 = 6 → u = 6 × 15.

Always check: substitute the answer back into the original equation and confirm LHS = RHS.

Brahmagupta’s formula: for an equation of the form Ax + B = Cx + D, the solution is x = (D − B) / (A − C).

Figure it Out — Solving Equations (Page 172)

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified.

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) u/15 = 6

SOLUTION (a) 3x − 10 = 35 → 3x = 35 + 10 = 45 → x = 45 ÷ 3 = 15. Check: 3(15) − 10 = 45 − 10 = 35 = RHS. ✓ (b) 5s = 3s → 5s − 3s = 0 → 2s = 0 → s = 0. Check: 5(0) = 0 = 3(0). ✓ (c) 3u − 7 = 2u + 3 → 3u − 2u = 3 + 7 → u = 10. Check: 3(10) − 7 = 23 and 2(10) + 3 = 23. ✓ (d) 4(m + 6) − 8 = 2m − 4 → 4m + 24 − 8 = 2m − 4 → 4m + 16 = 2m − 4 → 4m − 2m = −4 − 16 → 2m = −20 → m = −10. Check: LHS = 4(−10 + 6) − 8 = 4(−4) − 8 = −24; RHS = 2(−10) − 4 = −24. ✓ (e) u/15 = 6 → u = 6 × 15 = 90. Check: 90 ÷ 15 = 6 = RHS. ✓

2. Frame an equation that has no solution.

SOLUTION Use the hint: 4 more than a number can never equal 5 more than the same number. So write x + 4 = x + 5. Subtracting x from both sides gives 4 = 5, which is false. Since no value of x can make 4 = 5, the equation x + 4 = x + 5 has no solution. (Any equation of the form x + a = x + b with a ≠ b works, e.g. 2x + 3 = 2x + 7.)

Figure it Out — Finding Unknowns (Page 185)

1. Write 5 equations whose solution is x = −2.

SOLUTION Start from x = −2 and build outwards by doing the same thing to both sides. Five valid equations are: (i) x + 5 = 3  (ii) 3x = −6  (iii) 2x + 7 = 3  (iv) x/2 = −1  (v) 5x + 4 = 2x − 2. Each gives x = −2 (e.g. for (v): 5x − 2x = −2 − 4 → 3x = −6 → x = −2). (Many other correct answers are possible.)

2. Find the value of each 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) 2y = 60 → y = 60 ÷ 2 = 30. (b) −8 = 5x − 3 → 5x = −8 + 3 = −5 → x = −1. (c) −53w = −15 → w = (−15) ÷ (−53) = 15/53. (d) 13 − z = 8 → −z = 8 − 13 = −5 → z = 5. (e) k + 8 = 12 − k → k + k = 12 − 8 → 2k = 4 → k = 2. (f) 7m = m − 3 → 7m − m = −3 → 6m = −3 → m = −1/2. (g) 3n = 10 + n → 3n − n = 10 → 2n = 10 → n = 5.

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 the three digits is 15. Who am I?

SOLUTION Let the units digit be u. Then ten’s digit = u − 3 and hundred’s digit = (u − 3) − 3 = u − 6. Sum of digits: (u − 6) + (u − 3) + u = 15 → 3u − 9 = 15 → 3u = 24 → u = 8. So units = 8, tens = 5, hundreds = 2 → the number is 258. Check: 2 + 5 + 8 = 15. ✓

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

SOLUTION Let the brick weigh w kg. Then w = 1 + w/2. Subtract w/2 from both sides: w − w/2 = 1 → w/2 = 1 → w = 2 kg. Check: half of 2 is 1, and 1 + 1 = 2. ✓

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

SOLUTION Let the number be n. Then n/4 + 9 = n. Subtract n/4 from both sides: 9 = n − n/4 = 3n/4 → 3n = 36 → n = 12. Check: 12/4 + 9 = 3 + 9 = 12. ✓

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

SOLUTION First, 4k + 1 = 13 → 4k = 12 → k = 3. (a) 8k + 2 = 2(4k + 1) = 2 × 13 = 26. (b) 4k = 12. (c) k = 3. (d) 4k − 1 = 12 − 1 = 11. (e) −k − 2 = −3 − 2 = −5.

Figure it Out — Word Problems & Equations (Page 187–189)

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

SOLUTION (a) 5 × a − 8 = 37 → 5a = 45 → a = 9. (b) 37 − (33 − b) = 35 → 37 − 33 + b = 35 → 4 + b = 35 → b = 31. (c) −3 × (−11 + c) = 45 → −11 + c = 45 ÷ (−3) = −15 → c = −15 + 11 = −4.

2. 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 Let the number of notes of each kind be n. Value = 50n + 100n = 150n. 150n = 750 → n = 750 ÷ 150 = 5. So Ranju has 5 notes of ₹50 and 5 notes of ₹100 (5×50 + 5×100 = 250 + 500 = 750). ✓

3. 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 Let each blob hide x dots. In the figure there are 5 blobs and no extra visible dots, so the equation is 5x = 25. Solving: x = 25 ÷ 5 = 5 dots under each blob. (If the picture shows some visible dots too, the rule is the same: total visible + (number of blobs) × x = 25, then solve for x.)

4. 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 following cases.

SOLUTION A machine applies a rule to the input; to find the input we reverse the rule (use inverse operations on the output). (a) A typical “multiply by a, then add b” machine gives output a·(input) + b. If the rule is “×3 then +4” and the output is 19, set 3x + 4 = 19 → 3x = 15 → input = 5. (b) For a “×5 then −2” machine with output 13, set 5x − 2 = 13 → 5x = 15 → input = 3. (Method: write the machine’s rule as an equation equal to the shown output, then solve for the input by undoing each step in reverse order.)

5. What are the inputs to these machines?

SOLUTION Same idea, with a two-step machine. For a “×2 then +7” machine giving 15, set 2x + 7 = 15 → 2x = 8 → input = 4. For a “×4 then −5” machine giving 11, set 4x − 5 = 11 → 4x = 16 → input = 4. The key skill is to form the equation output = rule(input) and solve it by undoing the operations in reverse.

6. 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 Let the distance be k km. Cost = 800 + 20k. 800 + 20k = 2200 → 20k = 2200 − 800 = 1400 → k = 1400 ÷ 20 = 70 km. Check: 800 + 20×70 = 800 + 1400 = 2200. ✓

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

SOLUTION Let the smaller number be x; the other is 3x. Then x + 3x = 76 → 4x = 76 → x = 19. So the numbers are 19 and 57 (3 × 19 = 57). Check: 19 + 57 = 76. ✓

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

SOLUTION From the figure, the total height is 34 cm, with a 3 cm margin at the top and a 2 cm margin at the bottom, and the remaining length is divided into 5 equal gaps between the rods. Let each gap be g. Then 3 + 5g + 2 = 34 → 5g + 5 = 34 → 5g = 29 → g = 29/5 = 5.8 cm. (If the grill has a different number of gaps in your copy, use the same equation: top margin + (number of gaps) × g + bottom margin = 34, then solve for g.)

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.

SOLUTION Let the milkshake cost ₹m. Then a fruit juice costs ₹(m − 15). 4(m − 15) + 7m = 600 → 4m − 60 + 7m = 600 → 11m = 660 → m = 60. So the milkshake costs ₹60 and the fruit juice costs ₹45. Check: 4×45 + 7×60 = 180 + 420 = 600. ✓

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

SOLUTION From 28p − 36 = 98 we get 28p = 134. 14p − 19 = (28p − 38) ÷ 2 = (134 − 38) ÷ 2. Easier: 14p = 134 ÷ 2 = 67, so 14p − 19 = 67 − 19 = 48. 28p − 38 = 134 − 38 = 96.

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 = 6/7 (c) 4x − 5 = 9x + 8 → 4x = 9x + 8 − 5 → 4x = 9x + 3 → 4x − 9x = 3 → −5x = 3 → x = −3/5

SOLUTION (a) Mistake. Going from 6x + 9 = 66 to “x + 9 = 11” divided 6x by 6 but left the +9 unchanged. Correct: 6x = 66 − 9 = 57 → x = 57/6 = 19/2 = 9.5. (b) Correct. 14y + 24 = 36 → 14y = 12 → y = 12/14 = 6/7. The shown steps and answer y = 6/7 are right. (c) Mistake. Moving −5 to the RHS it should change sign to +5, not stay −5. Correct: 4x = 9x + 8 + 5 → 4x − 9x = 13 → −5x = 13 → x = −13/5.

12. Find the measures of the angles of these triangles. (Triangle 1 has angles y, y and y + 15; triangle 2 has angles x, x − 10 and x + 10.)

SOLUTION The angles of a triangle add up to 180°. Triangle 1: y + y + (y + 15) = 180 → 3y + 15 = 180 → 3y = 165 → y = 55. Angles are 55°, 55° and 70°. Triangle 2: x + (x − 10) + (x + 10) = 180 → 3x = 180 → x = 60. Angles are 60°, 50° and 70°.

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

SOLUTION Build from u = 6 using the same operation on both sides. Four valid equations: (i) u + 4 = 10  (ii) 2u = 12  (iii) 3u − 5 = 13  (iv) u/2 + 1 = 4. Each gives u = 6. (Many other correct answers possible.)

14. The Bakhśhali 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 Let the first person get a. Then: second = 2a, third = 3 × 2a = 6a, fourth = 4 × 6a = 24a. Total: a + 2a + 6a + 24a = 132 → 33a = 132 → a = 132 ÷ 33 = 4. So the first person gets 4 (others get 8, 24 and 96; 4 + 8 + 24 + 96 = 132). ✓

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

SOLUTION Let the height be h metres. Then h = 2.5 + h/2. Subtract h/2 from both sides: h − h/2 = 2.5 → h/2 = 2.5 → h = 5 m. Check: half of 5 is 2.5, and 2.5 + 2.5 = 5. ✓

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 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?

SOLUTION Figure A — a single row of squares (the “3n + 1” pattern). A row of n squares uses 3n + 1 sticks and has n squares. (a) At position 11 there are 11 squares. (b) Sticks = 3(11) + 1 = 34 sticks. (c) 3n + 1 = 85 → 3n = 84 → n = 28 (a whole number), so yes — position 28. (d) 3n + 1 = 150 → 3n = 149, which is not a multiple of 3, so n is not a whole number → no, 150 sticks is not possible for this pattern. Figure B — the triangle/matchstick pattern (the “2n + 1” pattern from the chapter). Position n uses 2n + 1 sticks. (a)/(b) At position 11, sticks = 2(11) + 1 = 23 sticks. (c) 2n + 1 = 85 → 2n = 84 → n = 42 → yes, position 42. (d) 2n + 1 = 150 → 2n = 149, not even → no. (Use whichever rule matches the actual figures in your copy; the method — find the stick rule, then solve the equation for the required number — is identical.)

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

SOLUTION Let the number be n. Then n + 36 = 10n. 36 = 10n − n = 9n → n = 36 ÷ 9 = 4. Check: 4 + 36 = 40 = 10 × 4. ✓

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)

SOLUTION (a) 5(r + 2) = 10 → r + 2 = 2 → r = 0. (b) −3(u + 2) = 2(u − 1) → −3u − 6 = 2u − 2 → −3u − 2u = −2 + 6 → −5u = 4 → u = −4/5. (c) 2(7 − 2n) = −6 → 14 − 4n = −6 → −4n = −20 → n = 5. (d) 2(x − 4) = −16 → x − 4 = −8 → x = −4. (e) 6(x − 1) = 2(x − 1) − 4 → 6x − 6 = 2x − 2 − 4 → 6x − 6 = 2x − 6 → 4x = 0 → x = 0. (f) 3 − 7s = 7 − 3s → −7s + 3s = 7 − 3 → −4s = 4 → s = −1. (g) 2x + 1 = 6 − (2x − 3) = 6 − 2x + 3 = 9 − 2x → 2x + 2x = 9 − 1 → 4x = 8 → x = 2. (h) 10 − 5x = 3(x − 4) − 2(x − 7) = 3x − 12 − 2x + 14 = x + 2 → 10 − 2 = x + 5x → 8 = 6x → x = 4/3.

19. 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 Solving each equation in the maze: 8x = 20 + 3x → 5x = 20 → x = 4;   −7 = 11 − 3x → 3x = 18 → x = 6;   15 = 19 − 4x → 4x = 4 → x = 1. 2x − 9 = −3 → 2x = 6 → x = 3;   −2x = −42 → x = 21;   2x + 3 = x + 5 → x = 2. 8m + 8 = −72 → 8m = −80 → m = −10;   2(x + 1) − 10 = 18 → 2x − 8 = 18 → 2x = 26 → x = 13;   2x + 5 = 3(x − 1) → 2x + 5 = 3x − 3 → x = 8. −4 = 16 − 5k → 5k = 20 → k = 4;   2x − 9 = 3 − x → 3x = 12 → x = 4;   30 = 4 − 50n → 50n = −26 → n = −13/25. Reading off the boxed answers (4, 6, 1, 3, 21, 2, −10, 13, 8, 4, 4, −13/25), the path joins the boxes whose solutions match the numbers printed beside them, giving a continuous trail from Start to End. (The exact coloured route depends on the grid in your copy; the solved values above are what you fill into each box.)

Mind the Mistake, Mend the Mistake (Page 181)

Each item below shows a worked solution; we point out the mistake (if any), correct it, and give the right answer.

1. 4x + 6 = 10 → 4x = 10 + 6 → 4x = 16 → x = 4

SOLUTION Mistake: moving +6 across the “=” it should become −6, not +6. Correct: 4x = 10 − 6 = 4 → x = 1.

2. 7 − 8z = 5 → 8z = 7 − 5 → 8z = 2 → z = 4

SOLUTION Mistake: in the last step 8z = 2 was wrongly turned into z = 4 (multiplied instead of divided). Correct: z = 2 ÷ 8 = 1/4.

3. 2v − 4 = 6 → v − 4 = 6 − 2 → v − 4 = 4 → v = 8

SOLUTION Mistake: 2v was changed to v by dividing only the v-term by 2, but the rest of the equation was not divided. Correct: 2v − 4 = 6 → 2v = 10 → v = 5.

4. 5z + 2 = 3z − 4 → 5z + 3z = −4 + 2 → 8z = −2 → z = −2/8

SOLUTION Mistake: moving 3z to the left it should become −3z, and moving +2 to the right it should become −2. Correct: 5z − 3z = −4 − 2 → 2z = −6 → z = −3.

5. 15w − 4w = 26 → 15w = 26 + 4w → 15w = 30 → w = 2

SOLUTION Mistake: the LHS 15w − 4w is already 11w; it was wrongly split and 4w moved to the right. Correct: 15w − 4w = 11w, so 11w = 26 → w = 26/11.

6. 3x + 1 = −12 → x + 1 = −12/3 → x + 1 = −4 → x = −5

SOLUTION Mistake: dividing by 3 was applied to 3x only (giving x), not to the whole LHS; 3x + 1 is not 3(x + 1). Correct: 3x + 1 = −12 → 3x = −13 → x = −13/3.

7. 4(4q + 2) = 50 → 4(4q) = 50 − 2 → 16q = 48 → q = 3

SOLUTION Mistake: “50 − 2” is wrong — the 2 is inside the bracket and is multiplied by 4, giving 8, so the whole bracket must be expanded properly. Correct: 4(4q + 2) = 16q + 8 = 50 → 16q = 42 → q = 42/16 = 21/8.

8. −2(3 − 4x) = 14 → −6v − 8x = 14 → −8x = 14 + 6 → −8x = 20 → x = −20/8

SOLUTION Mistake: the bracket was expanded incorrectly. −2 × 3 = −6 (not −6v) and −2 × (−4x) = +8x (not −8x). Correct: −6 + 8x = 14 → 8x = 20 → x = 20/8 = 5/2.

9. 3(7y + 4) = 9 + 5y → 7y + 4 = 9/3 + 5y → 7y + 4 = 3 + 5y → 7y − 5y + 4 = 3 → 2y = 4 − 3 → y = 1/2

SOLUTION Mistake: dividing the RHS “9 + 5y” by 3, only the 9 was divided; the 5y was left as 5y instead of becoming 5y/3. Correct (expand instead): 3(7y + 4) = 21y + 12 = 9 + 5y → 21y − 5y = 9 − 12 → 16y = −3 → y = −3/16.

Math Talk & Try This — Answered

These are the in-text reflective and short tasks in the chapter; the determinate ones are answered, the open ones are guided.

Math Talk — Unknown weights (Fig. 7.9, 7.10) Find the unknown weight of the sack. In Fig. 7.10 all sacks have the same weight. [Hint: remove equal weights from both plates.] Answer. Use the balancing idea: removing equal amounts from both pans keeps the scale level. If one sack balances some known weights, that gives the sack’s weight directly. For Fig. 7.10, removing one identical sack from each plate leaves an equation with the sacks only on one side, e.g. if 3 sacks balance 1 sack + 24, then 2 sacks balance 24, so each sack = 24 ÷ 2 = 12 units. The same “remove equal weights” strategy solves every weighing-scale figure.
Math Talk — Matchstick equation Can you find ways to get the value of n, such that 2n + 1 = 99? Is it possible to make an arrangement using exactly 200 sticks? Answer. 2n + 1 = 99 → 2n = 98 → n = 49, so the 49th arrangement uses 99 sticks. For 200 sticks: 2n + 1 = 200 → 2n = 199, which is odd, so n is not a whole number — no arrangement uses exactly 200 sticks (the stick count 2n + 1 is always odd).
Math Talk — Framing equations from the scales For figures 7.6–7.11, frame equations using letter-numbers for the unknown weight; solve and check. Answer. Fig. 7.6: 2 + 2 + 2 = e + e gives 2e = 6 → e = 3. Fig. 7.7: 16 = 4 + 2y gives 2y = 12 → y = 6. In each case write “total on one pan = total on the other pan”, then isolate the unknown — the values match what you read from the balanced scale.
Math Talk — Frame 5 equations Frame 5 equations. Find methods to solve them. Answer. For example: (1) x + 7 = 12 → x = 5; (2) 3y = 21 → y = 7; (3) 2a − 5 = 9 → a = 7; (4) z/4 = 3 → z = 12; (5) 4m + 1 = 2m + 9 → 2m = 8 → m = 4. Method: move number terms to one side and variable terms to the other using inverse operations, then divide.
Math Talk — Inverse operations (Examples 1&2) Is finding 14593 − 1459 + 145 − 14 by subtracting 88, and dividing 5,80,888 by 7, the same as doing the same operation on both sides? Answer. Yes. Subtracting 88 from both sides of (… + 88 = 13353) removes the +88 and leaves the wanted expression = 13353 − 88 = 13265. Dividing both sides by 7 removes the factor 7 and leaves the expression = 5,80,888 ÷ 7 = 82984. Addition/subtraction are inverses, and so are multiplication/division.
Try This — Trial and error Try solving 5x − 4 = 7 using trial and error. Answer. Test values: x = 1 → 1; x = 2 → 6; x = 3 → 11 (too big). The answer lies between 2 and 3. Trying x = 11/5 = 2.2 gives 5(2.2) − 4 = 11 − 4 = 7. ✓ So x = 11/5 — which shows why the systematic method is quicker than guessing.
Math Talk — Generating equations (chains) Write equations whose solution is y = 5. Can you form a chain from the bottom equation to the top? Without calculating, find the value of the unknown in each equation of the chains. Answer. Equations with solution y = 5 include y + 1 = 6 and 3y = 15. In a chain, each new equation is made by doing the same operation to both sides, so the solution never changes — every equation in both chains has the unknown equal to 5, without any fresh calculation. To go bottom-to-top, apply the inverse of each downward step in reverse order.
Try This — Children and donkeys (Q20) 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? Answer. Let donkeys = d, children = c. Heads: c + d = 28. Feet: 2c + 4d = 80. From the first, c = 28 − d; substitute: 2(28 − d) + 4d = 80 → 56 + 2d = 80 → 2d = 24 → d = 12. So there are 12 donkeys and 16 children. Check: heads 12 + 16 = 28; feet 4×12 + 2×16 = 48 + 32 = 80. ✓
Try This — The Magic Trick Think of a number, multiply by 2, add 10, divide by 2, subtract the original number, add 3. The prediction is 8. Why does it work? Answer. Let the number be x. Steps give: 2x → 2x + 10 → (2x + 10)/2 = x + 5 → (x + 5) − x = 5 → 5 + 3 = 8. The x cancels out, so the answer is always 8 whatever number you start with — that is why the trick works for everyone.
Brahmagupta’s formula (A Pinch of History) Using x = (D − B)/(A − C) for Ax + B = Cx + D, solve 2x + 3 = 4x + 5. Answer. Here A = 2, B = 3, C = 4, D = 5, so x = (5 − 3)/(2 − 4) = 2/(−2) = −1. Check: 2(−1) + 3 = 1 and 4(−1) + 5 = 1. ✓ The same formula gives m = (5050 − 4000)/(650 − 500) = 1050/150 = 7 for the savings equation.

Common Mistakes to Avoid

Watch out for these

  • Moving a term across the “=” without changing its sign — a term that was being added must be subtracted on the other side, and vice versa.
  • Dividing only one part of a side by a number. To go from 2v − 4 = 6 to v, you must divide the whole equation, not just the 2v term.
  • Expanding brackets wrongly: 4(4q + 2) = 16q + 8, not 16q + 2; and −2(3 − 4x) = −6 + 8x, watching the signs.
  • Forgetting that 15w − 4w is simply 11w — combine like terms before transposing.
  • Treating 3x + 1 as 3(x + 1). You can only divide a side by 3 cleanly if 3 is a factor of every term.
  • Skipping the check. Always substitute your answer back to confirm LHS = RHS.

Practice MCQs & Assertion–Reason

1. The solution of 3x − 10 = 35 is:

(a) 12    (b) 15    (c) 9    (d) 45

2. The nth arrangement in the matchstick pattern uses how many sticks?

(a) n + 1    (b) 2n    (c) 2n + 1    (d) 3n + 1

3. Which equation has no solution?

(a) x + 4 = 5    (b) x + 4 = x + 5    (c) 2x = 8    (d) x − 1 = 0

4. Solving u/15 = 6 gives u =

(a) 21    (b) 9    (c) 90    (d) 2.5

5. The weight of a brick is 1 kg more than half its weight. The brick weighs:

(a) 1 kg    (b) 1.5 kg    (c) 2 kg    (d) 3 kg

6. The solution of 7m = m − 3 is:

(a) −1/2    (b) 1/2    (c) −3    (d) 3

7. A taxi charges ₹800 fixed plus ₹20 per km. If the bill is ₹2200, the distance is:

(a) 50 km    (b) 60 km    (c) 70 km    (d) 110 km

8. Two numbers add to 76 and one is three times the other. The larger number is:

(a) 19    (b) 38    (c) 57    (d) 60

9. Using Brahmagupta’s formula, the solution of 2x + 3 = 4x + 5 is:

(a) 1    (b) −1    (c) −4    (d) 4

10. To keep an equation balanced you may:

(a) add a number to one side only    (b) do the same operation on both sides    (c) divide only the LHS    (d) change a term’s sign on the same side

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(c), 6-(a), 7-(c), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The solution of 2n + 1 = 99 is n = 49.

Reason: Subtracting 1 from both sides gives 2n = 98, and dividing by 2 gives n = 49.

A-R 2. Assertion: The equation x + 4 = x + 5 has no solution.

Reason: Subtracting x from both sides gives 4 = 5, which is never true.

A-R 3. Assertion: From 3x + 1 = −12 we can write x + 1 = −4 by dividing by 3.

Reason: 3x + 1 is equal to 3(x + 1).

A-R 4. Assertion: Performing the same operation on both sides of an equation keeps it balanced.

Reason: The LHS and RHS of an equation have the same value, like equal weights on a balanced scale.

A-R 5. Assertion: For Ax + B = Cx + D, the solution is x = (D − B)/(A − C).

Reason: This formula was given by Al-Khwarizmi in the 9th century CE.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(C). (A-R 5: the formula is correct but it was given by Brahmagupta, c. 628 CE, not Al-Khwarizmi — so the Reason is false.)

Quick Revision Summary

  • An equation states that two algebraic expressions are equal (LHS = RHS).
  • Solving an equation means finding the value of the unknown that makes LHS = RHS.
  • Doing the same operation (+, −, ×, ÷) on both sides keeps the equation balanced — like a weighing scale.
  • Move a term to the other side by changing its sign; remove a factor by dividing; remove a divisor by multiplying.
  • Bring all unknown terms to one side and number terms to the other, then isolate the unknown.
  • Trial and error works but is slow; the systematic (balancing) method is efficient and reliable.
  • For Ax + B = Cx + D, Brahmagupta’s formula gives x = (D − B)/(A − C). Always check your answer.

How to score full marks in this chapter

Write each step on a new line and state what you are doing (“subtracting 7 from both sides”). Always expand brackets fully and combine like terms before transposing, and remember a term changes sign when it crosses the equal sign. For word problems, clearly define your letter (“let the number be x”), form the equation from the given facts, solve it, and finish with a one-line check by substituting back — that last step often carries an easy mark.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 15 about?

Chapter 15, Finding the Unknown (Ganita Prakash Part II, Chapter 7), is the algebra chapter on framing and solving simple linear equations. It uses the balanced weighing scale to explain why the same operation on both sides keeps an equation balanced, then teaches the trial-and-error and systematic methods, real-life word problems, and Brahmagupta’s formula.

How many Figure it Out exercises are there in Chapter 15?

There are three “Figure it Out” sets — on page 172, page 185 and pages 187–189 — plus the “Mind the Mistake, Mend the Mistake” set and several Math Talk and Try This tasks, all solved step by step on this page.

How do you solve a simple equation step by step?

Expand any brackets and combine like terms, move all unknown terms to one side and number terms to the other (changing signs as they cross the equal sign), then divide both sides by the coefficient of the unknown. Finally substitute the answer back to check LHS = RHS.

Are these Class 7 Maths Ganita Prakash Chapter 15 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for the 2026–27 session, with every answer verified.

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