NCERT Solutions Ganita Prakash (Part 2) Chapter 3 Figure it Out — Dividing a Whole in a Given Ratio

Book page 60 Updated on2026-09-05

Q1.

A cricket coach schedules practice sessions that include different activities in a specific ratio — time for warm-up/cool-down : time for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5. If each session is 150 minutes long, how much time is spent on each activity?

Answer

Add the terms, find the value of one part, then multiply back.

Number of parts = 3 + 4 + 3 + 5 = 15
One part = 150 ÷ 15 = 10 minutes
Warm-up/cool-down = 3 × 10 = 30 minutes
Batting = 4 × 10 = 40 minutes
Bowling = 3 × 10 = 30 minutes
Fielding = 5 × 10 = 50 minutes

Check: 30 + 40 + 30 + 50 = 150 minutes. ✓

Why it happens: The ratio does not say how long anything lasts — it says how the 150 minutes are shared. Cutting the session into 15 equal parts of 10 minutes each and handing out 3, 4, 3 and 5 of them is exactly what the formula 150 × 3/15, 150 × 4/15, … is doing.

Tip: Fielding gets the largest slice here — 5 parts out of 15, that is one-third of the whole session.

Q2.

A school library has books in different languages in the following ratio — no. of Odiya books : no. of Hindi books : no. of English books :: 3 : 2 : 1. If the library has 288 Odiya books, how many Hindi and English books does it have?

Answer

Here the whole is not given — one term is. So find the value of one part from that term.

Odiya = 3 parts = 288 books
One part = 288 ÷ 3 = 96 books
Hindi = 2 parts = 2 × 96 = 192 books
English = 1 part = 96 books

Total in the library = 288 + 192 + 96 = 576 books.

Why it happens: This is the mirror image of Q1. There the whole was known and the parts unknown; here one part is known and the whole is unknown. Both are solved by the same single number — the size of one part. Once you have it, every term of the ratio can be filled in by multiplication.

Check it yourself: 288 : 192 : 96, divided throughout by their HCF 96, gives back 3 : 2 : 1.

Q3.

I have 100 coins in the ratio — no. of ₹10 coins : no. of ₹5 coins : no. of ₹2 coins : no. of ₹1 coins :: 4 : 3 : 2 : 1. How much money do I have in coins?

Answer

The ratio counts coins, not rupees — so divide the 100 coins first, and only then work out the money.

Number of parts = 4 + 3 + 2 + 1 = 10
One part = 100 ÷ 10 = 10 coins

Coin	Parts	Number of coins	Value
₹10	4	40	₹400
₹5	3	30	₹150
₹2	2	20	₹40
₹1	1	10	₹10
Total	10	100 coins	₹600

Why it happens: The number of coins is in the ratio 4 : 3 : 2 : 1, but the money is in the ratio 400 : 150 : 40 : 10, which is not 4 : 3 : 2 : 1. Each count has been multiplied by a different value (10, 5, 2, 1), so the proportion is destroyed. A ratio only survives multiplication by the same factor.

Q4.

Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not?

Answer

Choose any convenient unit — say 1 unit = 1 cm — and the sides become 3 cm, 4 cm and 5 cm.

Draw the longest side AB = 5 cm.
With A as centre and radius 4 cm, draw an arc.
With B as centre and radius 3 cm, draw another arc cutting the first at C.
Join AC and BC. Triangle ABC has sides 3 : 4 : 5.

Same ratio, same shape, different size — similar but not congruent.

No, they will not all be congruent.

Why it happens: The ratio fixes only the shape. Taking 1 unit = 1.5 cm gives sides 4.5 cm, 6 cm and 7.5 cm — still in the ratio 3 : 4 : 5, since 4.5 : 6 : 7.5 divided by 1.5 is 3 : 4 : 5. Its three angles are equal to those of the first triangle, but every side is 1.5 times as long, so it cannot be placed exactly on top of the first one. Congruence needs the sides to be equal, not merely proportional. Triangles like these, with equal angles and proportional sides, are called similar.

Did you know? Any triangle with sides in the ratio 3 : 4 : 5 is right-angled, because 3² + 4² = 9 + 16 = 25 = 5². By the Baudhāyana–Pythagoras theorem of the previous chapter, the angle opposite the longest side is 90°. This is why a knotted 3–4–5 rope is still used by masons to set out a right angle on site.

Q5.

Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not?

Answer

No. Such a triangle cannot exist, whatever unit you choose.

Let the sides be 1 unit, 3 units and 5 units.
Sum of the two shorter sides = 1 + 3 = 4 units
Longest side = 5 units
4 < 5 — the two shorter sides together are still shorter than the longest side.

The two shorter sides, laid flat against the longest side, together fall short of it — so their ends can never meet above it.

Why it happens: In any triangle, the sum of any two sides must be greater than the third — the straight route from one end of the base to the other can never be longer than a bent route between the same two points. Draw the base 5 units long and swing the two short sides inwards: their tips fall 1 unit apart and no vertex is formed. Compare this with Q4, where 3 + 4 = 7 > 5, so the arcs do cross.

Try This: The angle ratio 1 : 3 : 5 does work — it gives 20°, 60° and 100° (Example 5). Ratios of angles only have to add to 180°; ratios of sides must also pass the triangle inequality.

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