NCERT Solutions for Class 7th Maths Chapter 7 .4 A Pinch of History — Figure it Out

Book page 185 – 1897 Updated on2026-09-19

Q1.

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

Answer

(a) 9 (b) 31 (c) − 4

(a) 5x − 8 = 37
5x = 37 + 8 = 45
x = 45 ÷ 5 = 9
Check: 5 × 9 − 8 = 45 − 8 = 37 ✓

(b) 37 − (33 − x) = 35
37 − 33 + x = 35 (the minus sign changes both terms in the bracket)
4 + x = 35
x = 35 − 4 = 31
Check: 37 − (33 − 31) = 37 − 2 = 35 ✓

(c) − 3 × (− 11 + x) = 45
− 11 + x = 45 ÷ (− 3) = − 15 (divide both sides by − 3)
x = − 15 + 11 = − 4
Check: − 3 × (− 11 − 4) = − 3 × (− 15) = 45 ✓

Why it happens: in (b) the whole bracket is being subtracted, so − (33 − x) becomes − 33 + x. In (c) dividing by a negative number gives a negative result on the right. Both are places where a sign is easily lost — writing the step out fully protects you.

Q2.

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?

Answer

5 notes of ₹50 and 5 notes of ₹100.

Let n be the number of notes of each kind.
Value of the ₹50 notes = 50n
Value of the ₹100 notes = 100n

Total = ₹750, so
50n + 100n = 750
150n = 750
n = 750 ÷ 150 = 5

Check: 5 × 50 + 5 × 100 = 250 + 500 = ₹750 ✓

Why it happens: because the two counts are equal, one ₹50 note and one ₹100 note can be treated as a single bundle worth ₹150. Then the question is simply "how many ₹150 bundles make ₹750?" — and 750 ÷ 150 = 5.

Q3.

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.

Three black blobs and the blue dots that are not covered; altogether there are 25 dots.

Answer

Each blob hides 7 dots. The equation is 3b + 4 = 25.

Let b be the number of dots hidden by one blob.
Dots hidden by 3 blobs = 3b
Dots you can see = 4

3b + 4 = 25
3b = 25 − 4 = 21
b = 21 ÷ 3 = 7

Check: 3 × 7 + 4 = 21 + 4 = 25 ✓

Why it happens: the 4 visible dots are a fixed part of the total, so they can be taken away from both sides at once. That leaves 21 dots shared equally among 3 blobs. It is the same move as removing a fixed weight from both pans of a balance.

Tip: count the blobs and the loose dots carefully before writing the equation. Everything else follows from those two numbers.

Q4.

Here are machines that take an input, perform an operation on it and send out the result as an output. (a) [Machine: input → +3 → ×4 → −5 → output. The worked example is 12 → 15 → 60 → 55.] Find the inputs in the following cases: output 43, and output 75. (b) [Machine: the input is sent along two branches, one ×3 and one +3, and the second result is subtracted from the first. The worked example is 12 → 36 and 15 → 36 − 15 = 21.] Find the inputs in the following cases: output 63, and output 227.

Answer

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

(a) The machine does: x → x + 3 → 4(x + 3) → 4(x + 3) − 5
So output = 4(x + 3) − 5 = 4x + 7
Check with the worked example: 4 × 12 + 7 = 55 ✓

Output 43: 4x + 7 = 43 → 4x = 36 → x = 9
Trace: 9 → 12 → 48 → 43 ✓

Output 75: 4x + 7 = 75 → 4x = 68 → x = 17
Trace: 17 → 20 → 80 → 75 ✓

(b) The machine does: (x × 3) − (x + 3)
So output = 3x − x − 3 = 2x − 3
Check with the worked example: 2 × 12 − 3 = 21 ✓

Output 63: 2x − 3 = 63 → 2x = 66 → x = 33
Trace: 33 × 3 = 99, 33 + 3 = 36, 99 − 36 = 63 ✓

Output 227: 2x − 3 = 227 → 2x = 230 → x = 115
Trace: 115 × 3 = 345, 115 + 3 = 118, 345 − 118 = 227 ✓

Why it happens: instead of undoing the machine step by step, we write one expression for the whole machine — 4x + 7 in (a), 2x − 3 in (b). After that any output can be handled with a single equation. Notice how the numbers 3, 4 and 5 in (a) collapse into just 4x + 7.

Tip: you can also work backwards through the boxes. For output 43 in (a): 43 + 5 = 48, 48 ÷ 4 = 12, 12 − 3 = 9. Same answer, opposite direction.

Q5.

What are the inputs to these machines? [Machine 1: input → ÷3 → ÷3 → output +5. Machine 2: input → −4 → −4 → output −11.]

Answer

Machine 1: input 45. Machine 2: input − 3.

Machine 1 — the input is divided by 3 twice, and the result is 5.
(x ÷ 3) ÷ 3 = 5
x ÷ 9 = 5
x = 5 × 9 = 45
Trace: 45 ÷ 3 = 15, 15 ÷ 3 = 5 ✓

Machine 2 — 4 is subtracted twice, and the result is − 11.
x − 4 − 4 = − 11
x − 8 = − 11
x = − 11 + 8 = − 3
Trace: − 3 − 4 = − 7, − 7 − 4 = − 11 ✓

Why it happens: two identical boxes in a row can be replaced by a single one. Dividing by 3 twice is dividing by 9; subtracting 4 twice is subtracting 8. Combining the boxes first turns a two-step puzzle into a one-step equation.

Q6.

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.

Answer

70 kilometres.

Let k be the number of kilometres travelled.
Cost = 800 + 20k

800 + 20k = 2200
20k = 2200 − 800 = 1400
k = 1400 ÷ 20 = 70 km

Check: 800 + 20 × 70 = 800 + 1400 = ₹2200 ✓

Why it happens: the ₹800 is paid whether the taxi moves or not, so it is removed from the bill first. The remaining ₹1400 was earned at ₹20 a kilometre, and 1400 ÷ 20 = 70. This is the same shape as Madhubanti's party bill — a fixed part plus a rate.

Q7.

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

Answer

The numbers are 19 and 57.

Let the smaller number be x. Then the other is 3x.
x + 3x = 76
4x = 76
x = 76 ÷ 4 = 19
The other number = 3 × 19 = 57

Check: 19 + 57 = 76 ✓ and 57 = 3 × 19 ✓

Why it happens: naming the smaller number keeps the algebra whole — the bigger one is then just 3x. If you had called the bigger number y, you would get y + y/3 = 76, which works but brings a fraction in. Think of 76 as being made of 4 equal parts: 1 part is the smaller number, 3 parts the larger.

Q8.

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

The window grill: it is 34 cm from top to bottom, the band at each end is 3 cm wide and each rod between them is 2 cm wide.

Answer

The gap between two rods is 3 cm.

Read the grill from top to bottom. Between the 3 cm band at the top and the 3 cm band at the bottom there are 5 rods, each 2 cm wide, and 6 equal gaps.

Let the gap be g cm.
Total height = top band + bottom band + 5 rods + 6 gaps
34 = 3 + 3 + 5 × 2 + 6g
34 = 6 + 10 + 6g
34 = 16 + 6g
6g = 34 − 16 = 18
g = 18 ÷ 6 = 3 cm

The fixed parts use 16 cm of the height; the six gaps share the remaining 18 cm.

Check: 3 + 3 + 3 + 2 + 3 + 2 + 3 + 2 + 3 + 2 + 3 + 2 + 3 = 34 cm ✓

Why it happens: the metal itself takes up a fixed height — 3 + 3 for the two end bands and 5 × 2 for the rods, which is 16 cm in all. Whatever height is left over, 18 cm, is shared equally by the six openings, so each opening is 3 cm.

Tip: in problems like this, count the rods and the gaps separately. With rods in a row there is always one more gap than the number of rods between the two end bands.

Q9.

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.

Answer

A fruit juice costs ₹45 and a chocolate milkshake costs ₹60.

Let the milkshake cost ₹m. Then a juice costs ₹(m − 15).

4 juices + 7 milkshakes = ₹600
4(m − 15) + 7m = 600
4m − 60 + 7m = 600
11m − 60 = 600
11m = 600 + 60 = 660
m = 660 ÷ 11 = ₹60 (milkshake)

Juice = 60 − 15 = ₹45

Check: 4 × 45 + 7 × 60 = 180 + 420 = ₹600 ✓ and 60 − 45 = ₹15 ✓

Why it happens: there are two unknown prices, but one is described in terms of the other, so a single letter is enough. Naming the milkshake keeps the juice as m − 15; naming the juice instead would give j and j + 15, and 4j + 7(j + 15) = 600 leads to the same answer, j = 45.

Q10.

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

Answer

14p − 19 = 48 and 28p − 38 = 96. Neither answer needs the value of p.

From 28p − 36 = 98:
28p = 98 + 36 = 134

28p − 38 = 134 − 38 = 96

14p − 19: notice that 14p is half of 28p, so 14p = 134 ÷ 2 = 67
14p − 19 = 67 − 19 = 48

Another way for 14p − 19: it is exactly half of 28p − 38.
So 14p − 19 = 96 ÷ 2 = 48 — the same answer.

Why it happens: this is Example 1 and Example 2 again in a new dress. You are given the value of a whole expression, and every expression asked for can be reached from it by simple arithmetic. Solving for p would give p = 134/28 = 67/14, a fraction — and then the working would be messier for no gain.

Check it yourself: with p = 67/14, 14p = 67 so 14p − 19 = 48 ✓, and 28p = 134 so 28p − 38 = 96 ✓

Q11.

The steps to solve three equations are shown below. Identify and correct any mistakes.

(a)

6x + 9 = 66¹¹
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 = –5/3

Answer

(a) is wrong (x = 19/2), (b) is completely correct, and (c) has two mistakes (x = − 13/5).

(a) The mistake: both sides were divided by 6, but on the LHS only the 6x was divided — the 9 was left as it was.
Correct solution
6x + 9 = 66
6x = 66 − 9 = 57
x = 57 ÷ 6 = 19/2 (that is 9.5)
Check: 6 × 9.5 + 9 = 57 + 9 = 66 ✓

(b) No mistake. Every step is correct.
14y + 24 = 36 → divide both sides by 2 → 7y + 12 = 18
7y = 18 − 12 = 6, so y = 6/7
Check: 14 × 6/7 + 24 = 12 + 24 = 36 ✓

(c) Two mistakes. The term − 5 was removed from the LHS but written as − 5 on the RHS instead of + 5; and at the end, − 5x = 3 was turned into x = − 5/3 instead of x = 3 ÷ (− 5).
Correct solution
4x − 5 = 9x + 8
4x = 9x + 8 + 5 = 9x + 13
4x − 9x = 13
− 5x = 13
x = 13 ÷ (− 5) = − 13/5
Check: LHS = 4(− 13/5) − 5 = − 52/5 − 25/5 = − 77/5. RHS = 9(− 13/5) + 8 = − 117/5 + 40/5 = − 77/5 ✓

Why it happens: in (a) dividing a side means dividing every term of it — that is why (b), where 14, 24 and 36 are all even, is safe while (a) is not. In (c) both slips are sign or order slips: a removed term changes sign, and dividing by − 5 means writing 3 ÷ (− 5), not (− 5) ÷ 3.

Q12.

Find the measures of the angles of these triangles. [First triangle: the two slanting sides are marked equal; the top angle is y and the marked base angle is y + 15. Second triangle: the top angle is x and the two base angles are x – 10 and x + 10.]

Answer

First triangle: 50°, 65°, 65°. Second triangle: 60°, 50°, 70°.

First triangle. The two slanting sides are marked equal, so the two base angles are equal — each is y + 15.
Angle sum of a triangle = 180°
y + (y + 15) + (y + 15) = 180
3y + 30 = 180
3y = 180 − 30 = 150
y = 150 ÷ 3 = 50

Angles: y = 50°, and y + 15 = 65° twice.
Check: 50 + 65 + 65 = 180 ✓

Second triangle.
x + (x − 10) + (x + 10) = 180
3x = 180 (the − 10 and + 10 cancel)
x = 180 ÷ 3 = 60

Angles: x = 60°, x − 10 = 50°, x + 10 = 70°
Check: 60 + 50 + 70 = 180 ✓

Why it happens: the tick marks in the first triangle are doing real work — they say the triangle is isosceles, so the angles opposite the equal sides are equal. Without them the problem would have too little information. In the second triangle the two base angles are 10° below and 10° above the top angle, so their extra and their shortfall cancel, leaving 3x = 180.

Tip: always check that the three angles you get really add to 180° and that each one is positive.

Q13.

Write 4 equations whose solution is u = 6.

Answer

Build them from u = 6 by doing the same thing to both sides.

Equation	How it was built	Check with u = 6
u − 6 = 0	Subtract 6 from both sides	6 − 6 = 0 ✓
2u = 12	Multiply both sides by 2	12 = 12 ✓
u ÷ 3 = 2	Divide both sides by 3	6 ÷ 3 = 2 ✓
5u + 1 = 31	Multiply by 5, then add 1	30 + 1 = 31 ✓
4u − 7 = 3u − 1	Two expressions that both give 17 at u = 6	24 − 7 = 17 and 18 − 1 = 17 ✓

Why it happens: the solution travels with the equation. Because every step was applied to both sides, u = 6 satisfies each new equation just as it satisfied u = 6 itself. Solving any one of them would take you back down to u = 6.

Q14.

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?

Answer

The first person is given 4.

Let the first person get x.
Second = 2 × x = 2x
Third = 3 × 2x = 6x
Fourth = 4 × 6x = 24x

Total = 132, so
x + 2x + 6x + 24x = 132
33x = 132
x = 132 ÷ 33 = 4

Person	Expression	Amount
First	x	4
Second	2x	8
Third	6x	24
Fourth	24x	96
Total	33x	132

Check: 4 + 8 + 24 + 96 = 132 ✓

Why it happens: each amount is built on the one before it, not on the first one. "Thrice as much as the second" is 3 × 2x = 6x, not 3x. Reading the chain carefully is the whole difficulty; after that the equation 33x = 132 is easy.

Did you know? The Bakhshali Manuscript, found near Peshawar, is one of the oldest surviving Indian mathematical texts and contains one of the earliest known uses of a symbol for zero.

Q15.

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

Answer

The giraffe is 5 metres tall.

Let the height be h metres.
h = h/2 + 2.5
h − h/2 = 2.5 (subtract h/2 from both sides)
h/2 = 2.5
h = 2.5 × 2 = 5 m

Check: half of 5 m is 2.5 m, and 2.5 + 2.5 = 5 m ✓

Why it happens: a whole height minus half of it is the other half. So that other half is exactly the 2½ m mentioned, and the full height is twice that. It is the same reasoning as the brick problem on page 181, with 2.5 in place of 1.

Q16.

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?

Figure 1

Figure 2

The first few positions of each stick sequence.

Answer

Figure 1: (a) 11 squares (b) 36 sticks (c) no (d) yes, position 49.
Figure 2: (a) 24 squares (b) 73 sticks (c) yes, position 13 (d) no.

Figure 1 — the rule.
Position 1: 1 square + arrowhead = 4 + 2 = 6 sticks
Position 2: 2 squares + arrowhead = 7 + 2 = 9 sticks
Position 3: 3 squares + arrowhead = 10 + 2 = 12 sticks
Each new square adds 3 sticks, so position n has 3n + 3 sticks and n squares.

(a) Position 11 has 11 squares.
(b) Sticks = 3 × 11 + 3 = 36.
(c) 3n + 3 = 85 → 3n = 82. But 82 is not divisible by 3, so no arrangement uses exactly 85 sticks.
(d) 3n + 3 = 150 → 3n = 147 → n = 49. Yes — position 49.

Figure 2 — the rule.
Position 1: 2 rows of 2 squares = 4 squares, 13 sticks
Position 2: 3 rows of 2 squares = 6 squares, 19 sticks
Position 3: 4 rows of 2 squares = 8 squares, 25 sticks
Each new step adds 2 squares and 6 sticks.
Position n has 2n + 2 squares and 6n + 7 sticks.

(a) Squares = 2 × 11 + 2 = 24.
(b) Sticks = 6 × 11 + 7 = 66 + 7 = 73.
(c) 6n + 7 = 85 → 6n = 78 → n = 13. Yes — position 13.
(d) 6n + 7 = 150 → 6n = 143. But 143 is not divisible by 6, so no arrangement uses exactly 150 sticks.

Why it happens: the counting works because neighbouring squares share sticks. In Figure 1 the first square needs 4 sticks and each square after it needs only 3 more, plus 2 for the arrowhead — giving 3n + 3. In Figure 2, each extra step adds one more row of two squares, and joining it to the staircase costs 6 new sticks, giving a jump of 6 each time: 13, 19, 25, 31, … .

Check it yourself: Figure 1, position 49 → 3(49) + 3 = 150 ✓. Figure 2, position 13 → 6(13) + 7 = 85 ✓. The two figures answer opposite halves of parts (c) and (d) — a neat piece of design by the book.

Q17.

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

Answer

The number is 4.

Let the number be x.
x + 36 = 10x
36 = 10x − x (subtract x from both sides)
36 = 9x
x = 36 ÷ 9 = 4

Check: 4 + 36 = 40 and 10 × 4 = 40 ✓

Why it happens: ten times the number is nine extra copies of it. Those nine extra copies must be worth the 36 that was added, so one copy is 36 ÷ 9 = 4.

Q18.

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)

Answer

(a) r = 0 (b) u = − 4/5 (c) n = 5 (d) x = − 4 (e) x = 0 (f) s = − 1 (g) x = 2 (h) x = 4/3

(a) 5(r + 2) = 10 → r + 2 = 10 ÷ 5 = 2 → r = 2 − 2 = 0
Check: 5(0 + 2) = 10 ✓

(b) − 3(u + 2) = 2(u − 1)
− 3u − 6 = 2u − 2
− 3u − 2u = − 2 + 6 → − 5u = 4 → u = − 4/5
Check: LHS = − 3(− 4/5 + 2) = − 3(6/5) = − 18/5. RHS = 2(− 4/5 − 1) = 2(− 9/5) = − 18/5 ✓

(c) 2(7 − 2n) = − 6 → 7 − 2n = − 3 → − 2n = − 10 → n = 5
Check: 2(7 − 10) = 2(− 3) = − 6 ✓

(d) 2(x − 4) = − 16 → x − 4 = − 8 → x = − 4
Check: 2(− 4 − 4) = 2(− 8) = − 16 ✓

(e) 6(x − 1) = 2(x − 1) − 4
6x − 6 = 2x − 2 − 4 = 2x − 6
6x − 2x = − 6 + 6 → 4x = 0 → x = 0
Check: LHS = 6(− 1) = − 6. RHS = 2(− 1) − 4 = − 6 ✓

(f) 3 − 7s = 7 − 3s
− 7s + 3s = 7 − 3 → − 4s = 4 → s = − 1
Check: 3 + 7 = 10 and 7 + 3 = 10 ✓

(g) 2x + 1 = 6 − (2x − 3)
2x + 1 = 6 − 2x + 3 = 9 − 2x
2x + 2x = 9 − 1 → 4x = 8 → x = 2
Check: LHS = 5. RHS = 6 − (4 − 3) = 6 − 1 = 5 ✓

(h) 10 − 5x = 3(x − 4) − 2(x − 7)
10 − 5x = 3x − 12 − 2x + 14 = x + 2
10 − 2 = x + 5x → 8 = 6x → x = 4/3
Check: LHS = 10 − 20/3 = 10/3. RHS = 4/3 + 2 = 10/3 ✓

Why it happens: in (a), (c) and (d) the bracket is multiplied by a number that divides the other side exactly, so dividing first is quickest. In (e) the same bracket (x − 1) appears on both sides, so you could also subtract 2(x − 1) from both sides to get 4(x − 1) = − 4, then x − 1 = − 1 and x = 0. In (g) and (h) a minus sign stands in front of a bracket, and it changes the sign of every term inside.

Tip: part (e) has the surprising answer x = 0. Zero is a perfectly good solution — always test it rather than assuming a solution must be non-zero.

Q19.

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.

Page 189 — the puzzle grid: twelve boxes, each holding an equation, joined by numbered arrows. The Start box at the top left is worked out for you.

Answer

Solve every box first, then follow the arrow whose label matches your answer.

Box	Equation	Working	Solution
Start	8x = 20 + 3x	8x − 3x = 20; 5x = 20	x = 4
Row 1 middle	− 7 = 11 − 3x	3x = 11 + 7 = 18	x = 6
Row 1 right	15 = 19 − 4x	4x = 19 − 15 = 4	x = 1
Row 2 left	2x − 9 = − 3	2x = − 3 + 9 = 6	x = 3
Row 2 middle	− 2x = − 42	x = (− 42) ÷ (− 2)	x = 21
Row 2 right	2x + 3 = x + 5	2x − x = 5 − 3	x = 2
Row 3 left	8m + 8 = − 72	8m = − 72 − 8 = − 80	m = − 10
Row 3 middle	2(x + 1) − 10 = 18	2x + 2 = 28; 2x = 26	x = 13
Row 3 right	2x + 5 = 3(x − 1)	2x + 5 = 3x − 3; 8 = x	x = 8
Row 4 left	− 4 = 16 − 5k	5k = 16 + 4 = 20	k = 4
Row 4 middle	2x − 9 = 3 − x	3x = 12	x = 4
Row 4 right	30 = 4 − 50n	50n = 4 − 30 = − 26	n = − 13/25

The path from Start to End:

8x = 20 + 3x gives x = 4 → take the arrow marked 4 downwards.
2x − 9 = − 3 gives x = 3 → take the arrow marked 3 to the right.
− 2x = − 42 gives x = 21 → take the slanting arrow marked 21 up to the right.
15 = 19 − 4x gives x = 1 → take the arrow marked 1 downwards.
2x + 3 = x + 5 gives x = 2 → take the arrow marked 2 downwards.
2x + 5 = 3(x − 1) gives x = 8 → take the arrow marked 8 to the left.
2(x + 1) − 10 = 18 gives x = 13 → take the arrow marked 13 to the left.
8m + 8 = − 72 gives m = − 10 → take the arrow marked − 10 downwards.
− 4 = 16 − 5k gives k = 4 → take the arrow marked + 4 to the right.
2x − 9 = 3 − x gives x = 4 → take the arrow marked 4 to the right, and you reach END.

Why it happens: the maze is a checking device. If your answer does not match any arrow leaving your box, you have made a slip — go back and check that box before moving on. The decoy arrows (− 4, − 1, − 44, 5, 6, − 8, − 2, − 5) belong to boxes that are not on the route.

Tip: two different boxes on the path both give 4, and the arrows leaving them are marked "+ 4" and "4". They mean the same number.

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

There are 12 donkeys and 16 children.

Let there be d donkeys. Since every child and every donkey has one head,
number of children = 28 − d

Feet: a child has 2, a donkey has 4.
4d + 2(28 − d) = 80
4d + 56 − 2d = 80
2d + 56 = 80
2d = 80 − 56 = 24
d = 24 ÷ 2 = 12 donkeys

Children = 28 − 12 = 16

Check: heads = 12 + 16 = 28 ✓ feet = 12 × 4 + 16 × 2 = 48 + 32 = 80 ✓

Why it happens: here is a neat way to see it without algebra. If all 28 were children, there would be 28 × 2 = 56 feet. But there are 80 feet — 24 extra. Every donkey we swap in for a child adds 2 extra feet, so there must be 24 ÷ 2 = 12 donkeys. That is exactly the arithmetic the equation carries out.

Try This: change the numbers — 30 heads and 96 feet. All-children would give 60 feet, 36 extra, so 18 donkeys and 12 children. Check: 18 × 4 + 12 × 2 = 72 + 24 = 96 ✓

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