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NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –105Section 4.5 Pick Patterns and Reveal Relationships — Figure it Out

Book page 102 Updated on2026-09-05

Q1.
One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day? (a) 30x + 20y (b) (30 + 20) × (x + y) (c) 20x + 30y (d) (30 + 20) × x + y (e) 30x – 20y
Answer

The correct expression is (a) 30x + 20y.

Money from jowar roti = 30 × x = 30x
Money from pulao = 20 × y = 20y
Total = 30x + 20y
Why it happens: (b) would be right only if every customer took one of each. (c) swaps the two rates. (d) multiplies only x by 50. (e) subtracts the pulao money instead of adding it.
Check it yourself: For x = 3 and y = 2, the true earning is ₹90 + ₹40 = ₹130, and 30 × 3 + 20 × 2 = 130 ✔
Q2.
Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day? (a) p + q + r (b) p + q + 2r (c) 2 × (p + q + r) (d) p + q + r + 2 (e) p + q + r + 1 (f) 2 × (p + q)
Answer

The correct expression is (a) p + q + r.

Total customers = p + q + r
One flag per customer
Flags given away = p + q + r
Why it happens: Each flag goes to a customer, not to a flower. A customer who bought both kinds is still one person and still gets one flag — which is why (b) and (c) are wrong.
Q3.
A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights. (a) Write an expression describing how far away the snail is from its starting position. (b) What can we say about the snail’s movement if d > u?
Answer

(a) In one full day-and-night the snail gains (u – d) cm. This repeats 10 times.

Distance from start = 10 × (u – d)
= 10(u – d) cm = 10u – 10d cm

(b) If d > u, then u – d is negative.

10(u – d) is negative → the snail ends up below its starting point

It slips down more each night than it climbs during the day, so it keeps losing ground and will never reach the top.

Check it yourself: If u = 8 cm and d = 3 cm, the snail is 10 × 5 = 50 cm up. If u = 3 cm and d = 8 cm, it is 10 × (–5) = –50 cm, i.e. 50 cm below where it began.
Q4.
Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?
Answer

Work out each week separately — a week has 7 days.

WeekDaily distanceDistance that week
15 km7 × 5 = 35
2(5 + z) km7(5 + z) = 35 + 7z
3(5 + 2z) km7(5 + 2z) = 35 + 14z
Total = 35 + (35 + 7z) + (35 + 14z)
= (35 + 35 + 35) + (7z + 14z)
= 105 + 21z km
Check it yourself: If z = 1 km, the totals are 35 + 42 + 49 = 126 km, and 105 + 21 × 1 = 126 ✔
Q5.
In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.
Answer

Follow each arrow away from the centre oval w + 2, applying one box at a time.

PathStepsResult
Top right (given)w + 2 → +3 → w + 5 → ×44w + 20
Top leftw + 2 → –5 → w – 3 → ×33w – 9
Bottom rightw + 2 → –4 → w – 2 → ×33w – 6
Bottom leftw + 2 → –8 → w – 6 → –4w – 10
Top left: (w – 3) × 3 = 3w – 9
Bottom right: (w + 2) – 4 = w – 2, and (w – 2) × 3 = 3w – 6 ✔
Bottom left: (w + 2) – 8 = w – 6, and (w – 6) – 4 = w – 10
Why it happens: ×3 multiplies every term, so (w – 3) × 3 = 3w – 9, not 3w – 3. Checking the bottom-right path backwards from 3w – 6 confirms the missing oval must be w – 2.
Q6.
A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations. (a) If t = 4, what is the time taken to travel from Yahapur to Vahapur? (b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation]
Answer

Three stations lie between the two towns, so the journey is cut into 4 stretches, with 3 halts.

Yahapur Vahapur t t t t 2 2 2
Four travel stretches of t minutes each, with a 2-minute halt at each of the three middle stations.
(b) Total time = 4 × t + 3 × 2
= 4t + 6 minutes

(a) When t = 4: 4 × 4 + 6 = 16 + 6 = 22 minutes
Why it happens: Three stops in between create four gaps, not three — the same reason 3 cuts on a rope give 4 pieces.
Q7.
Simplify the following expressions: (a) 3a + 9b – 6 + 8a – 4b – 7a + 16 (b) 3 (3a – 3b) – 8a – 4b – 16 (c) 2 (2x – 3) + 8x + 12 (d) 8x – (2x – 3) + 12 (e) 8h – (5 + 7h) + 9 (f) 23 + 4(6m – 3n) – 8n – 3m – 18
Answer
(a) 3a + 8a – 7a = 4a; 9b – 4b = 5b; –6 + 16 = 10
= 4a + 5b + 10

(b) 3(3a – 3b) = 9a – 9b
9a – 8a = a; –9b – 4b = –13b
= a – 13b – 16

(c) 2(2x – 3) = 4x – 6
4x + 8x = 12x; –6 + 12 = 6
= 12x + 6

(d) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12
= 6x + 15

(e) 8h – (5 + 7h) + 9 = 8h – 5 – 7h + 9
= h + 4

(f) 4(6m – 3n) = 24m – 12n
24m – 3m = 21m; –12n – 8n = –20n; 23 – 18 = 5
= 21m – 20n + 5
Check it yourself: In (d) put x = 2: 16 – (4 – 3) + 12 = 16 – 1 + 12 = 27, and 6 × 2 + 15 = 27 ✔
Q8.
Add the expressions given below: (a) 4d – 7c + 9 and 8c – 11 + 9d (b) – 6f + 19 – 8s and – 23 + 13f + 12s (c) 8d – 14c + 9 and 16c – (11 + 9d) (d) 6f – 20 + 8s and 23 – 13f – 12s (e) 13m – 12n and 12n – 13m (f) – 26m + 24n and 26m – 24n
Answer
(a) (4d + 9d) + (–7c + 8c) + (9 – 11) = 13d + c – 2

(b) (–6f + 13f) + (–8s + 12s) + (19 – 23) = 7f + 4s – 4

(c) 16c – (11 + 9d) = 16c – 11 – 9d
(8d – 9d) + (–14c + 16c) + (9 – 11) = –d + 2c – 2

(d) (6f – 13f) + (8s – 12s) + (–20 + 23) = –7f – 4s + 3

(e) (13m – 13m) + (–12n + 12n) = 0

(f) (–26m + 26m) + (24n – 24n) = 0
Why it happens: In (e) and (f) the second expression is the exact opposite of the first, so every term cancels. Such a pair is said to add up to zero.
Q9.
Subtract the expressions given below: (a) 9a – 6b + 14 from 6a + 9b – 18 (b) – 15x + 13 – 9y from 7y – 10 + 3x (c) 17g + 9 – 7h from 11 – 10g + 3h (d) 9a – 6b + 14 from 6a – (9b + 18) (e) 10x + 2 + 10y from –3y +8 – 3x (f) 8g + 4h – 10 from 7h – 8g + 20
Answer

“Subtract A from B” means work out B – A. Change the sign of every term of A, then add.

(a) (6a + 9b – 18) – (9a – 6b + 14)
= 6a + 9b – 18 – 9a + 6b – 14 = –3a + 15b – 32

(b) (7y – 10 + 3x) – (–15x + 13 – 9y)
= 7y – 10 + 3x + 15x – 13 + 9y = 18x + 16y – 23

(c) (11 – 10g + 3h) – (17g + 9 – 7h)
= 11 – 10g + 3h – 17g – 9 + 7h = –27g + 10h + 2

(d) 6a – (9b + 18) = 6a – 9b – 18
(6a – 9b – 18) – (9a – 6b + 14) = –3a – 3b – 32

(e) (–3y + 8 – 3x) – (10x + 2 + 10y)
= –3y + 8 – 3x – 10x – 2 – 10y = –13x – 13y + 6

(f) (7h – 8g + 20) – (8g + 4h – 10)
= 7h – 8g + 20 – 8g – 4h + 10 = 3h – 16g + 30
Tip: The order matters. ‘Subtract A from B’ is B – A, never A – B. Reversing it flips the sign of the whole answer.
Q10.
Describe situations corresponding to the following algebraic expressions: (a) 8x + 3y (b) 15x – 2x
Answer

(a) 8x + 3y

A fruit seller sells mangoes at ₹x each and bananas at ₹y each. A customer buys 8 mangoes and 3 bananas, so the bill is 8x + 3y rupees.

(b) 15x – 2x

A shopkeeper has a packet of 15 pencils, each costing ₹x. Two of them are damaged and cannot be sold. The money received for the packet is 15x – 2x = 13x rupees.

Why it happens: In (b) both terms carry the same letter-number x, so they are like terms and merge into a single term 13x. In (a) the letters differ and the two terms must stay apart.
Q11.
Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?
Answer

Each extra fold gives one extra piece.

Number of folds0123…10
Pieces after one cut2345…12
Folded 10 times and cut → 10 + 2 = 12 pieces
Folded r times and cut → r + 2 pieces
Why it happens: With r folds the cut passes through r + 1 strands, giving 2(r + 1) little lengths. But the r folded ends each join two of them back together, so the number of separate pieces is 2(r + 1) – r = r + 2.
Q12.
Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?
Answer

The first square needs 4 matchsticks; every square after that shares one side, so it needs only 3 more.

Squares1234…10
Matchsticks471013…31
10 squares = 4 + 3 × 9 = 4 + 27 = 31 matchsticks
w squares = 4 + 3 × (w – 1)
= 4 + 3w – 3
= 3w + 1 matchsticks
Check it yourself: w = 3 gives 3 × 3 + 1 = 10 ✔, and w = 10 gives 31 ✔
Q13.
Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.
Answer

The printed sequence is red, yellow, green, yellow, red, … so the block red – yellow – green – yellow repeats every 4 positions.

Position12345678
ColourRedYellowGreenYellowRedYellowGreenYellow
90 = 4 × 22 + 2 → remainder 2 → Yellow
190 = 4 × 47 + 2 → remainder 2 → Yellow
343 = 4 × 85 + 3 → remainder 3 → Green

Expressions for the positions:

ColourPositionsExpression
Red1, 5, 9, 13, …4n – 3
Green3, 7, 11, 15, …4n – 1
Yellow2, 4, 6, 8, … (every even position)2n
Why it happens: Yellow appears twice in each block of four — once between red and green and once between green and red. That is why yellow fills all the even positions, while red and green take alternate odd positions.
Q14.
Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?
Answer

Each step is an X with one centre square and four arms. Step 1 has arms of length 1, Step 2 arms of length 2, and so on.

StepCentreFour armsSquares
114 × 15
214 × 29
314 × 313
414 × 417
1014 × 1041
5014 × 50201
Number of squares in Step n = 4n + 1
Step 4 → 17, Step 10 → 41, Step 50 → 201

Counting vertices: every square has 4 corners, so

Number of vertices = 4 × (4n + 1)
= 16n + 4
Why it happens: The 4 arms grow by one square each per step; the centre square never changes. That gives the “+ 1” in 4n + 1. Counting 4 vertices for each square simply multiplies the whole formula by 4.
Q15.
Numbers are written in a particular sequence in this endless 4-column grid. (a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4). (b) In which row and column will the following numbers appear: (i) 124 (ii) 147 (iii) 201 (c) What number appears in row r and column c? (d) Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.
Answer

Row 1 holds 1–4, row 2 holds 5–8, row 3 holds 9–12 … Each row uses up 4 numbers, so row r begins at 4(r – 1) + 1.

(a) Column expressions (with r the row number):

ColumnNumbersExpression
11, 5, 9, 13, …4(r – 1) + 1 = 4r – 3
22, 6, 10, 14, …4(r – 1) + 2 = 4r – 2
33, 7, 11, 15, …4(r – 1) + 3 = 4r – 1
44, 8, 12, 16, …4(r – 1) + 4 = 4r

(b) Locating numbers — divide by 4 and read the remainder as the column (remainder 0 means column 4).

(i) 124 = 4 × 31 + 0 → column 4, and 4(r – 1) + 4 = 124 gives r – 1 = 30 → row 31
(ii) 147 = 4 × 36 + 3 → column 3, r – 1 = 36 → row 37
(iii) 201 = 4 × 50 + 1 → column 1, r – 1 = 50 → row 51

(c) The number in row r and column c is

4(r – 1) + c

(d) Multiples of 3. Their columns run 3, 2, 1, 4, 3, 2, 1, 4, … repeating.

Multiple of 33691215182124
Column32143214

Other patterns:

  • Every number in column 4 is a multiple of 4.
  • Even numbers appear only in columns 2 and 4; odd numbers only in columns 1 and 3.
  • Going down any column, the numbers rise by 4 each time.
  • Along a row the numbers rise by 1; along a downward diagonal they rise by 5.
Why it happens: Adding 3 each time while the grid width is 4 means the column shifts back by one every step (3 = 4 – 1), which is exactly the repeating cycle 3, 2, 1, 4.
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