Chapter 7: Finding the Unknown Class 8th Mathematics (Ganita Prakash) NCERT Solution

July 13, 2026 • Admin ETeam • Class 7, Class 7 NCERT Solution

NCERT Ganita Prakash • Grade 7 • Part II

Chapter 7 — Finding the Unknown

Complete step-by-step solutions for every in-text question, Math Talk, Mind-the-Mistake, and Figure-it-Out exercise — with diagrams and worked-out algebra.

7.1 Find the Unknowns Figure It Out (page 172) 7.2 Worked Examples Figure It Out (page 181) 7.3 Mind the Mistake 7.4 A Pinch of History Main Exercise (Q1–Q20) Puzzle Time — Magic Trick

7.1 Find the Unknowns — Unknown Weights

In-text questions from the “Unknown Weights” activity

Warm-up Verify the two introductory weighing-scale examples

A weighing scale shows the total weight of everything hanging from it.

Answer

Scale 1: two weights of 2 units each are hung together, so the scale reads \(2+2=4\). ✔

Scale 2: weights of 4 units and 3 units are hung together, so the scale reads \(4+3=7\). ✔

Fig. 7.1 Find the unknown weights (total = 16)

Fig. 7.1: total weight = 16

Two unknowns, one equation — many valid answers exist

Since Flower = 3, the diamond and pink flower together must weigh \(16-3=13\). Any pair of whole numbers adding to 13 is valid — this is an open exploration question (that is why the book asks you to “discuss and give reasons”).

One valid answer Diamond = 6 and Pink flower = 7, since \(3+6+7=16\).
(Diamond = 5, Pink = 8 or Diamond = 8, Pink = 5 are equally correct — as long as Diamond + Pink = 13.)

Fig. 7.2 Find the unknown weights (total = 24)

Fig. 7.2: total weight = 24

Two unknowns, one equation — many valid answers exist

Two starfish weigh \(2+2=4\), so Fish + Seahorse \(=24-4=20\). Any whole-number pair adding to 20 works.

One valid answerFish = 9 and Seahorse = 11, since \(2+2+9+11=24\).

Fig. 7.3 Find the unknown weights (total = 8)

Fig. 7.3: total weight = 8

Because the scale is balanced, the left pan equals the right pan: Book = Money + Money = 2 × Money. Also, the sum of everything is 8: Book + Money + Money = 8.

Solution Substituting Book \(=2m\): \(2m + m + m = 8 \Rightarrow 4m = 8 \Rightarrow m = 2\).
So Money = 2 and Book \(=2\times2=4\).

Book = 4, Money = 2

Fig. 7.4 Find the unknown weights (total = 18)

Fig. 7.4: total weight = 18 (1 sun + 5 clouds + 1 lightning)

Two unknowns, one equation — many valid answers exist

Sun = 5, so \(5\times\text{Cloud} + 5 + \text{Lightning} = 18\), i.e. \(5c+l=13\).

One valid answerCloud = 1 and Lightning = 8, since \(5(1)+5+8=18\).

Fig. 7.5 Find the unknown weights (total = 40)

Fig. 7.5: total weight = 40 (3 crowns + 3 diamonds)

Two unknowns, one equation — many valid answers exist

With 3 crowns and 3 diamonds: \(3k+3d=40\), i.e. \(k+d=\tfrac{40}{3}\). No whole-number pair fits exactly here — which is exactly why this activity is meant for open class discussion rather than one forced numeric answer. Any pair satisfying \(3k+3d=40\) is acceptable, e.g. Crown \(=5\), Diamond \(=\tfrac{25}{3}\).

Fig. 7.6 Find the weight of one egg

Fig. 7.6: 3 bread slices = 2 eggs

Left pan: \(2+2+2=6\). Right pan: \(e+e=2e\). Balanced scale ⇒ \(2e = 6\).

Solution\(2e=6 \Rightarrow e = 3\)

Weight of one egg = 3

Fig. 7.7 Find the weight of one circle

Fig. 7.7: 4 stars = 1 star + 2 circles

Left pan (4 stars): \(4\times4=16\). Right pan: \(4+2y\). Balanced scale ⇒ \(4+2y=16\).

Solution\(4+2y=16 \Rightarrow 2y = 12 \Rightarrow y = 6\)

Weight of one circle = 6

Fig. 7.8 Find the weight of one banana

Fig. 7.8: watermelon + orange = 2 bananas

Left pan: \(10+4=14\). Right pan: \(b+b=2b\). Balanced scale ⇒ \(2b=14\).

Solution\(2b=14 \Rightarrow b=7\)

Weight of one banana = 7

Fig. 7.9 Find the weight of one sack

Fig. 7.9: sack + 2 kg = 10 kg + 2 kg

Balanced scale: \(s + 2 = 10 + 2\).

Solution — remove the equal 2 kg weight from both pans\(s = 10\) kg

Weight of one sack = 10 kg

Fig. 7.10 Find the weight of one sack (all sacks equal)

Fig. 7.10: 2 sacks = 10 kg + 1 sack + 4 kg

Balanced scale: \(s+s = 10+s+4\).

Solution — remove one sack from each pan (hint given in the book) Left becomes just \(s\); right becomes \(10+4=14\).
So \(s = 14\) kg.

Weight of one sack = 14 kg

Fig. 7.11 Find the weight of one sack

Fig. 7.11: 4 sacks + 1 kg = 2 sacks + 10 kg + 10 kg

Balanced scale: \(4s+1 = 2s+10+10\).

Solution — remove 2 sacks from each pan (hint given in the book) Left becomes \(2s+1\); right becomes \(20\).
\(2s+1=20 \Rightarrow 2s=19 \Rightarrow s = 9.5\) kg

Weight of one sack = 9.5 kg

Fig. 7.12 Find the weight of one sack

Fig. 7.12: 90 sacks + 50 kg = 60 sacks + 500 kg

Balanced scale: \(90s + 50 = 60s + 500\).

Solution \(90s-60s = 500-50\)
\(30s = 450\)
\(s = 15\) kg

Weight of one sack = 15 kg

In-text Matchstick Pattern — arrangement with 99 sticks

Positions 1, 2, 3 of the matchstick-triangle sequence

Position \(n\) needs \(2n+1\) matchsticks (as shown in the book).

(a) Position that uses exactly 99 sticks Solve \(2n+1=99\):
\(2n = 98\)
\(n = 49\)

The arrangement with 99 sticks is at position 49.

(b) Can an arrangement use exactly 200 sticks? Solve \(2n+1=200\):
\(2n=199\)
\(n = 99.5\)

No — \(n\) is not a whole number. \(2n+1\) is always odd, so an even number like 200 is impossible.

Math Talk Solve \(5x-4=7\) by trial and error, then algebraically

Trial and error: at \(x=2\), LHS \(=6\) (too small); at \(x=3\), LHS \(=11\) (too big). So the answer lies between 2 and 3 — trial and error alone won’t land on it exactly.

Algebraic solution \(5x-4=7\)
\(5x = 11\)
\(x = \dfrac{11}{5} = 2.2\)

x = 11/5 = 2.2

Figure It Out (page 172)

Q1. Solve these equations and check the solutions

(a) \(3x-10=35\)

\(3x = 35+10 = 45\)

\(x = 45 \div 3 = 15\)

x = 15

Check: 3(15) − 10 = 45 − 10 = 35 ✔

(b) \(5s=3s\)

\(5s-3s=0\)

\(2s=0\)

\(s=0\)

s = 0

Check: 5(0) = 0 and 3(0) = 0 ✔

(c) \(3u-7=2u+3\)

\(3u-2u = 3+7\)

\(u = 10\)

u = 10

Check: 3(10) − 7 = 23 and 2(10) + 3 = 23 ✔

(d) \(4(m+6)-8=2m-4\)

\(4m+24-8=2m-4\)

\(4m+16=2m-4\)

\(2m=-20\)

\(m=-10\)

m = −10

Check: 4(−10 + 6) − 8 = 4(−4) − 8 = −24; 2(−10) − 4 = −24 ✔

(e) \(\dfrac{u}{15}=6\)

\(u = 6 \times 15\)

u = 90

Check: 90 ÷ 15 = 6 ✔

Q2 Frame an equation that has no solution

Hint: “4 more than a number” and “5 more than a number” can never be equal.

Example equation with no solution Let the number be \(x\).
\(x + 4 = x + 5\)
Subtracting \(x\) from both sides gives \(4 = 5\), which is never true.

x + 4 = x + 5 has no solution.

7.2 Solving Equations Systematically — Worked Examples

Quick-check answers for the Math-Talk prompts inside the worked examples

Example 7 Tile pattern — can Ranjana make an arrangement with 100 tiles?

The number of tiles needed at step \(k\) is \(3k+1\).

Solve \(3k+1=100\)\(3k = 99\)
\(k = 33\)

Yes — Step 33 uses exactly 100 tiles.

Example 9 — Math Talk Check: Jahnavi & Sunita save equal amounts after 7 months

Check Jahnavi: \(4000+650(7) = 4000+4550 = 8550\)
Sunita: \(5050+500(7) = 5050+3500 = 8550\)
Both equal ₹8550 — confirmed correct.

Example 12 Marbles — find how many marbles each boy has

Equation formed: \(2y+30=60\), where \(y\) = Suresh’s marbles.

Solve\(2y = 60-30 = 30\)
\(y = 15\)

Suresh has 15 marbles; Ramesh has 15 + 30 = 45 marbles. (Total: 15 + 45 = 60 ✔)

Math Talk Write equations whose solution is y = 5, then chain them

Two example equations\(y+1=6\) and \(3y=15\) — both give y = 5.

Reverse chain check Starting at \(y=5\): add 2 → \(y+2=7\); multiply by 3 → \(3y+6=21\); subtract 6 → \(3y=15\). This retraces the original chain in reverse, confirming that performing the same operation on both sides never changes the solution.

Example 13 — Math Talk Give a real-life situation modelled by \(100x + 75 = 250\)

Sample situation A shop sells notebooks at ₹100 each and charges a fixed ₹75 delivery fee. If Riya’s total bill is ₹250, then \(100x + 75 = 250\). In general, this equation models any situation with a per-unit cost of 100, a fixed extra cost of 75, and a total of 250.

A Pinch of History — Math Talk Use Brahmagupta’s formula to solve \(2x + 3 = 4x + 5\)

Brahmagupta’s rule for \(Ax+B=Cx+D\) is \(x = \dfrac{D-B}{A-C}\). Here A = 2, B = 3, C = 4, D = 5.

Solve\(x = \dfrac{5-3}{2-4} = \dfrac{2}{-2} = -1\)

x = −1

Check: LHS = 2(−1) + 3 = 1; RHS = 4(−1) + 5 = 1 ✔

Figure It Out (page 181)

Q1 Write 5 equations whose solution is x = −2

Sample equations (many are possible) \(x+2=0\) • \(2x=-4\) • \(3x+1=-5\) • \(x-3=-5\) • \(5x+10=0\)

Q2. Find the value of each unknown

(a) \(2y=60\) → \(y=30\) y = 30

(b) \(-8=5x-3 \Rightarrow -5=5x \Rightarrow x=-1\) x = −1

(c) \(-53w=-15 \Rightarrow w=\dfrac{-15}{-53}=\dfrac{15}{53}\) w = 15/53

(d) \(13-z=8 \Rightarrow z=13-8=5\) z = 5

(e) \(k+8=12-k \Rightarrow 2k=4 \Rightarrow k=2\) k = 2

(f) \(7m=m-3 \Rightarrow 6m=-3 \Rightarrow m=-\dfrac{1}{2}\) m = −1/2

(g) \(3n=10+n \Rightarrow 2n=10 \Rightarrow n=5\) n = 5

Q3 I am a 3-digit number. Hundred’s digit is 3 less than ten’s digit; ten’s digit is 3 less than unit’s digit. Digit sum = 15. Who am I?

Solution Let unit’s digit \(=u\). Then ten’s digit \(=u-3\), hundred’s digit \(=u-6\).
\((u-6)+(u-3)+u = 15\)
\(3u-9=15\)
\(3u=24 \Rightarrow u=8\)
Ten’s digit \(=5\), hundred’s digit \(=2\).

The number is 258.

Q4 The weight of a brick is 1 kg more than half its weight. Find the weight.

Solution Let weight \(=w\).
\(w = 1+\dfrac{w}{2}\)
\(w-\dfrac{w}{2}=1 \Rightarrow \dfrac{w}{2}=1\)
\(w=2\)

The brick weighs 2 kg.

Q5 One quarter of a number increased by 9 gives the same number. Find the number.

Solution Let the number \(=x\).
\(\dfrac{x}{4}+9 = x\)
\(9 = x-\dfrac{x}{4} = \dfrac{3x}{4}\)
\(x = 12\)

The number is 12.

Q6 Given \(4k + 1 = 13\), find the value of each expression

First find k\(4k=12 \Rightarrow k=3\)

(a) 8k + 2 = 2(4k) + 2 = 2(12) + 2 = 26

(b) 4k = 12

(c) k = 3

(d) 4k − 1 = 12 − 1 = 11

(e) −k − 2 = −3 − 2 = −5

7.3 Mind the Mistake, Mend the Mistake

Each shown solution has an error — the mistake is identified and the equation is solved correctly

PROBLEM 1 — shown (incorrect) working

4x + 6 = 10 4x = 10 + 6 4x = 16 x = 4

Mistake: The +6 was moved to the RHS without changing its sign. It should be subtracted from both sides.

CORRECT WORKING

\(4x = 10-6 = 4 \Rightarrow x = 1\)

x = 1

PROBLEM 2 — shown (incorrect) working

7 − 8z = 5 8z = 7 − 5 8z = 2 z = 4

Mistake: The division in the last step is wrong: 2 ÷ 8 = 1/4, not 4.

CORRECT WORKING

\(8z=2 \Rightarrow z=\dfrac{2}{8}=\dfrac{1}{4}\)

z = 1/4

PROBLEM 3 — shown (incorrect) working

2v − 4 = 6 v − 4 = 6 − 2 v − 4 = 4 v = 8

Mistake: Only the ‘2v’ term was divided by 2 while the right side had 2 subtracted — the same operation was not applied to both sides.

CORRECT WORKING

\(2v=6+4=10 \Rightarrow v=5\)

v = 5

PROBLEM 4 — shown (incorrect) working

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

Mistake: Sign errors while moving terms: 3z should be subtracted (not added), and 2 should be subtracted from −4 (not added).

CORRECT WORKING

\(5z-3z=-4-2 \Rightarrow 2z=-6 \Rightarrow z=-3\)

z = −3

PROBLEM 5 — shown (incorrect) working

15w − 4w = 26 15w = 26 + 4w 15w = 30 w = 2

Mistake: There was no need to move −4w at all — the LHS simplifies directly: 15w − 4w = 11w.

CORRECT WORKING

\(11w=26 \Rightarrow w=\dfrac{26}{11}\)

w = 26/11

PROBLEM 6 — shown (incorrect) working

3x + 1 = −12 x + 1 = −12/3 x + 1 = −4 x = −5

Mistake: Only the 3x term was divided by 3; the +1 was left untouched. The +1 should first be removed by subtraction.

CORRECT WORKING

\(3x=-12-1=-13 \Rightarrow x=-\dfrac{13}{3}\)

x = −13/3

PROBLEM 7 — shown (incorrect) working

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

Mistake: The bracket was not expanded correctly — 4 × 2 = 8 (not 2), since 4(4q + 2) = 16q + 8.

CORRECT WORKING

\(16q+8=50 \Rightarrow 16q=42 \Rightarrow q=\dfrac{21}{8}\)

q = 21/8

PROBLEM 8 — shown (incorrect) working

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

Mistake: Distributing −2 gives −6 + 8x (not −6 − 8x), since −2 × −4x = +8x.

CORRECT WORKING

\(-6+8x=14 \Rightarrow 8x=20 \Rightarrow x=\dfrac{5}{2}\)

x = 5/2

PROBLEM 9 — shown (incorrect) working

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

Mistake: Dividing by 3 was applied only to the 9 on the right, not to the 5y as well. The bracket should be expanded first.

CORRECT WORKING

\(21y+12=9+5y \Rightarrow 16y=-3 \Rightarrow y=-\dfrac{3}{16}\)

y = −3/16

7.4 A Pinch of History — Example

Example 16 Bhāskarāchārya, Bījagaṇita (1150 CE): One man has ₹300 and 6 horses. Another has 10 horses and a debt of ₹100. They are equally rich. Find the price of one horse.

Let the price of one horse \(=x\).

Solution \(300+6x = 10x-100\)
\(300+100 = 10x-6x\)
\(400 = 4x\)
\(x = 100\)

Price of one horse = ₹100

Figure It Out — Main Exercise (Q1–Q20)

Pages 185–189

Q1. Fill in the blanks with integers

(a) \(5\times\square-8=37\) → \(5\times\square=45\) → Blank = 9

(b) \(37-(33-\square)=35\) → \(33-\square=2\) → Blank = 31

(c) \(-3\times(-11+\square)=45\) → \(-11+\square=-15\) → Blank = −4

Q2 Ranju earns ₹750/day, paid in equal numbers of ₹50 and ₹100 notes. How many of each?

Solution Let the number of each note \(=n\).
\(50n+100n=750\)
\(150n=750\)
\(n=5\)

5 notes of ₹50 and 5 notes of ₹100.

Q3 Each black blob hides an equal number of blue dots. Total dots = 25. How many dots does one blob hide?

2 identical blobs + 3 visible dots = 25

Let each blob hide \(x\) dots. Equation: \(2x+3=25\).

Solution\(2x=22\)
\(x=11\)

Each blob hides 11 dots.

Q4 (a) Input machine: +3, ×4, −5. Example: input 12 → output 55. Find inputs for outputs 43 and 75.

Worked example: 12 → +3 → 15 → ×4 → 60 → −5 → 55

Work backwards using inverse operations in reverse order: add 5, then divide by 4, then subtract 3.

Output = 43\(43+5=48\)
\(48\div4=12\)
\(12-3=9\)

Input = 9

Output = 75\(75+5=80\)
\(80\div4=20\)
\(20-3=17\)

Input = 17

Q4 (b) Two-branch machine: (Input × 3) − (Input + 3). Example: input 12 → output 21. Find inputs for outputs 63 and 227.

Two-branch subtract machine

Let the input be \(x\). The rule simplifies: \(3x-(x+3) = 2x-3\).

Output = 63\(2x-3=63\)
\(2x=66 \Rightarrow x=33\)

Input = 33

Output = 227\(2x-3=227\)
\(2x=230 \Rightarrow x=115\)

Input = 115

Q5 Find the inputs to these machines

Machine 1: input → ÷3 → ÷3 → output = 5.

Solution Let input \(=x\). \(\dfrac{x}{3\times3}=5 \Rightarrow \dfrac{x}{9}=5\)
\(x = 45\)

Input = 45

Machine 2: input → −4 → −4 → output = −11.

Solution\(x-4-4=-11\)
\(x-8=-11 \Rightarrow x=-3\)

Input = −3

Q6 A taxi charges ₹800 fixed + ₹20 per km. Total fare = ₹2200. Find the distance travelled.

Solution Let distance \(=k\) km.
\(800+20k=2200\)
\(20k=1400\)
\(k=70\)

Distance travelled = 70 km

Q7 The sum of two numbers is 76. One number is 3 times the other. Find the numbers.

Solution Let the smaller number \(=x\); larger \(=3x\).
\(x+3x=76 \Rightarrow 4x=76 \Rightarrow x=19\)

The numbers are 19 and 57.

Q8 Window grill: top gap 3 cm, next gap 2 cm, then equally spaced rods span the remaining 34 cm. Find the gap between two rods.

Top frame → 3 cm → rod → 2 cm → rod → (equal gaps) → 34 cm → bottom frame

Reading the figure: after the first two gaps (3 cm, then 2 cm), the remaining rods are evenly spaced and together span 34 cm across 4 equal gaps.

Solution Let the equal gap \(=g\).
\(4g = 34\)
\(g = 8.5\) cm

Gap between two equally-spaced rods = 8.5 cm

Q9 Fruit juice costs ₹15 less than a chocolate milkshake. 4 juices + 7 milkshakes cost ₹600. Find each price.

Let milkshake \(=m\); juice \(=m-15\).

Solution \(4(m-15)+7m=600\)
\(4m-60+7m=600\)
\(11m=660\)
\(m=60\)

Milkshake = ₹60, Fruit juice = ₹45

Q10 Given \(28p – 36 = 98\), find \(14p – 19\) and \(28p – 38\)

First find 28p\(28p = 98+36 = 134\)

14p − 19\(14p = \dfrac{134}{2} = 67\)
\(14p-19 = 67-19 = 48\)

28p − 38\(28p-38 = 134-38 = 96\)

14p − 19 = 48

28p − 38 = 96

Q11 Identify and correct the mistakes in these three solutions

(a)

6x + 9 = 66 x + 9 = 11 x = 11 − 9 x = 2

Issue: Dividing by 6 was applied to 6x and 66, but not to the 9 — the whole equation must be divided, or better, subtract 9 first.

CORRECT

\(6x=66-9=57 \Rightarrow x=\dfrac{57}{6}=9.5\)

x = 9.5

(b)

14y + 24 = 36 7y + 12 = 18 7y = 6 y = 6/7

Issue: No mistake here — every step is correct!

CORRECT

\(7y=18-12=6 \Rightarrow y=\dfrac{6}{7}\)

y = 6/7 (already correct)

(c)

4x − 5 = 9x + 8 4x = 9x + 8 − 5 4x = 9x + 3 4x − 9x = 3 −5x = 3 x = −5/3

Issue (two errors): Adding 5 to both sides should give 8 + 5 = 13 on the right, not 8 − 5. Also the last step inverts the division: from −5x = 3 we get x = −3/5, not −5/3.

CORRECT

\(4x=9x+13 \Rightarrow -5x=13 \Rightarrow x=-\dfrac{13}{5}\)

x = −13/5

Q12 Find the measures of the angles of these triangles

Isosceles: two equal marked sides

Scalene triangle

First triangle is isosceles (the tick marks show the two slanted sides are equal), so both base angles equal \(y+15\). Angle sum \(=180°\):

Solve \(y+(y+15)+(y+15)=180\)
\(3y+30=180 \Rightarrow y=50\)

Apex angle = 50°, base angles = 65° and 65°

Second triangle:

Solve \(x+(x-10)+(x+10)=180\)
\(3x=180 \Rightarrow x=60\)

Angles = 60°, 50° and 70°

Q13 Write 4 equations whose solution is u = 6

Sample equations \(u-6=0\) • \(2u=12\) • \(3u+1=19\) • \(\dfrac{u}{2}=3\)

Q14 Bakhśhālī Manuscript (300 CE): 1st person gets x; 2nd gets 2× the 1st; 3rd gets 3× the 2nd; 4th gets 4× the 3rd. Total = 132. Find the 1st person’s share.

1st = \(x\); 2nd = \(2x\); 3rd = \(3(2x)=6x\); 4th = \(4(6x)=24x\).

Solution \(x+2x+6x+24x=132\)
\(33x=132\)
\(x=4\)

The first person receives 4.

Q15 The height of a giraffe is 2.5 m more than half its height. Find the height.

Solution Let height \(=h\).
\(h = 2.5+\dfrac{h}{2}\)
\(\dfrac{h}{2}=2.5 \Rightarrow h=5\)

The giraffe is 5 metres tall.

Q16 Two stick-arrangement sequences — find squares/sticks at position 11, and check 85 & 150 sticks

Pattern A — arrow/pencil shape (squares in a row + triangular tip)

Reading the figure: position \(n\) has \((n+1)\) unit squares in a row plus one triangular tip, using \(3n + 4\) sticks in total.

(a) Squares at position 11\(\text{squares}(n) = n+1 \Rightarrow \text{squares}(11) = 12\)

(b) Sticks at position 11\(\text{sticks}(n) = 3n+4 \Rightarrow \text{sticks}(11) = 3(11)+4 = 37\)

Yes — position 27 uses exactly 85 sticks.

(d) Exactly 150 sticks?\(3n+4=150 \Rightarrow 3n=146 \Rightarrow n \approx 48.67\)

No — \(n\) is not a whole number, so 150 sticks is not possible.

Pattern B — staircase of unit squares

Reading the figure: position \(n\) is a staircase of \(n\) unit squares, each new square sharing one edge with the previous, using \(3n + 1\) sticks.

(a) Squares at position 11\(\text{squares}(n) = n \Rightarrow \text{squares}(11) = 11\)

(b) Sticks at position 11\(\text{sticks}(n) = 3n+1 \Rightarrow \text{sticks}(11) = 3(11)+1 = 34\)

Yes — position 28 uses exactly 85 sticks.

(d) Exactly 150 sticks?\(3n+1=150 \Rightarrow 3n=149 \Rightarrow n \approx 49.67\)

No — \(n\) is not a whole number, so 150 sticks is not possible.

Q17 A number increased by 36 is equal to ten times itself. Find the number.

Solution \(x+36=10x\)
\(36=9x\)
\(x=4\)

The number is 4.

Q18. Solve these equations

(a)

\(5(r+2)=10\)

\(r+2=2 \Rightarrow r=0\)

r = 0

(b)

\(-3(u+2)=2(u-1)\)

\(-3u-6=2u-2 \Rightarrow -4=5u \Rightarrow u=-\dfrac{4}{5}\)

u = −4/5

(c)

\(2(7-2n)=-6\)

\(14-4n=-6 \Rightarrow -4n=-20 \Rightarrow n=5\)

n = 5

(d)

\(2(x-4)=-16\)

\(x-4=-8 \Rightarrow x=-4\)

x = −4

(e)

\(6(x-1)=2(x-1)-4\)

Let \(a=x-1\): \(6a=2a-4 \Rightarrow 4a=-4 \Rightarrow a=-1 \Rightarrow x=0\)

x = 0

(f)

\(3-7s=7-3s\)

\(-7s+3s=7-3 \Rightarrow -4s=4 \Rightarrow s=-1\)

s = −1

(g)

\(2x+1=6-(2x-3)\)

\(2x+1=9-2x \Rightarrow 4x=8 \Rightarrow x=2\)

x = 2

(h)

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

\(10-5x=3x-12-2x+14 = x+2\)

\(10-2=6x \Rightarrow 8=6x \Rightarrow x=\dfrac{4}{3}\)

x = 4/3

Q19. Solve the equations to find a path from Start to End

Each tile in the maze is a small equation. Solving all of them (the values are used to trace the coloured path) gives:

\(8x=20+3x\) → x = 4

\(-7=11-3x\) → x = 6

\(15=19-4x\) → x = 1

\(2x-9=-3\) → x = 3

\(-2x=-42\) → x = 21

\(2x+3=x+5\) → x = 2

\(8m+8=-72\) → m = −10

\(2(x+1)-10=18\) → x = 13

\(2x+5=3(x-1)\) → x = 8

\(-4=16-5k\) → k = 4

\(2x-9=3-x\) → x = 4

\(30=4-50n\) → n = −13/25

Q20 (Try This) 28 heads and 80 feet on a beach — some children, some donkeys. How many of each?

Children have 2 feet, donkeys have 4 feet. Let children \(=c\) and donkeys \(=d\).

Solution \(c+d=28 \Rightarrow c=28-d\)
\(2c+4d=80 \Rightarrow c+2d=40\)
Substituting: \((28-d)+2d=40 \Rightarrow d=12\)
So \(c=28-12=16\)

16 children and 12 donkeys.

Puzzle Time — A Magic Trick

Puzzle Think of a number → ×2 → +10 → ÷2 → subtract the original number → +3. The result is always 8. Why?

Algebraic explanation Let the starting number \(=x\).
\(x \xrightarrow{\times 2} 2x \xrightarrow{+10} 2x+10 \xrightarrow{\div 2} x+5 \xrightarrow{-x} 5 \xrightarrow{+3} 8\)

The \(x\) cancels out at the “subtract the original number” step, so the answer is always 8 — no matter which number is chosen.

Chapter 7: Finding the Unknown Class 8th Mathematics (Ganita Prakash) NCERT Solution

Chapter 7 — Finding the Unknown

7.1 Find the Unknowns — Unknown Weights

Figure It Out (page 172)

7.2 Solving Equations Systematically — Worked Examples

Figure It Out (page 181)

7.3 Mind the Mistake, Mend the Mistake

7.4 A Pinch of History — Example

Figure It Out — Main Exercise (Q1–Q20)

Puzzle Time — A Magic Trick

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