Add the following fractions using Brahmagupta’s method: a. 2 &frasl; 7 + 5 &frasl; 7 + 6 &frasl; 7 b. 3 &frasl; 4 + 1 &frasl; 3 c. 2 &frasl; 3 + 5 &frasl; 6 d. 2 &frasl; 3 + 2 &frasl; 7 e. 3 &frasl; 4 + 1 &frasl; 3 + 1 &frasl; 5 f. 2 &frasl; 3 + 4 &frasl; 5 g. 4 &frasl; 5 + 2 &frasl; 3 h. 3 &frasl;

Brahmagupta’s three steps every time: (1) make the fractional units the same, (2) add the numerators, (3) write the answer in lowest terms. Question With a common fractional unit Answer a. 2/7 + 5/7 + 6/7 (2 + 5 + 6)/7 13/7 = 1 6/7 b. 3/4 + 1/3 9/12 + 4/12 13/12 = 1 1/12 c. 2/3 + 5/6 4/6 + 5/6 = 9/6 3/2 = 1 1/2 d. 2/3 + 2/7 14/21 + 6/21 20/21 e. 3/4 + 1/3 + 1/5 45/60 + 20/60 + 12/60 77/60 = 1 17/60 f. 2/3 + 4/5 10/15 + 12/15 22/15 = 1 7/15 g. 4/5 + 2/3 12/15 + 10/15 22/15 = 1 7/15 h. 3/5 + 5/8 24/40 + 25/40 49/40 = 1 9/40 i. 9/2 + 5/4 18/4 + 5/4 23/4 = 5 3/4 j. 8/3 + 2/7 56/21 + 6/21 62/21 = 2 20/21 k. 3/4 + 1/3 + 1/5 45/60 + 20/60 + 12/60 77/60 = 1 17/60 l. 2/3 + 4/5 + 3/7 70/105 + 84/105 + 45/105 199/105 = 1 94/105 m. 9/2 + 5/4 + 7/6 54/12 + 15/12 + 14/12 83/12 = 6 11/12 Sample working for (c): The smallest common multiple of 3 and 6 is 6 2/3 = (2 × 2)/(3 × 2) = 4/6, 5/6 stays 5/6 4/6 + 5/6 = 9/6 = (9 ÷ 3)/(6 ÷ 3) = 3/2 = 1 1/2 Did you notice? (f) and (g) give the same answer, and so do (e) and (k) — the order of the fractions does not change the sum.

Rahim mixes 2 &frasl; 3 litres of yellow paint with 3 &frasl; 4 litres of blue paint to make green paint. What is the volume of green paint he has made?

Volume = 2/3 + 3/4 Common denominator = 3 × 4 = 12 2/3 = 8/12, 3/4 = 9/12 8/12 + 9/12 = 17/12 = 1 5/12 litres Answer: Rahim has made 17 &frasl; 12 litres, that is 1 5 &frasl; 12 litres of green paint. Check it yourself: 17 &frasl; 12 is a little more than 1 litre — sensible, since 2 &frasl; 3 and 3 &frasl; 4 are each a bit less than a litre.

NCERT Solutions for Class 6th Maths Chapter 7 Brahmagupta’s method for adding fractions — Figure it Out

Book page 179 Updated on2026-09-19

Q1.

Add the following fractions using Brahmagupta’s method: a. ²⁄₇ + ⁵⁄₇ + ⁶⁄₇ b. ³⁄₄ + ¹⁄₃ c. ²⁄₃ + ⁵⁄₆ d. ²⁄₃ + ²⁄₇ e. ³⁄₄ + ¹⁄₃ + ¹⁄₅ f. ²⁄₃ + ⁴⁄₅ g. ⁴⁄₅ + ²⁄₃ h. ³⁄₅ + ⁵⁄₈ i. ⁹⁄₂ + ⁵⁄₄ j. ⁸⁄₃ + ²⁄₇ k. ³⁄₄ + ¹⁄₃ + ¹⁄₅ l. ²⁄₃ + ⁴⁄₅ + ³⁄₇ m. ⁹⁄₂ + ⁵⁄₄ + ⁷⁄₆

Answer

Brahmagupta’s three steps every time: (1) make the fractional units the same, (2) add the numerators, (3) write the answer in lowest terms.

Question	With a common fractional unit	Answer
a. 2/7 + 5/7 + 6/7	(2 + 5 + 6)/7	13/7 = 1 6/7
b. 3/4 + 1/3	9/12 + 4/12	13/12 = 1 1/12
c. 2/3 + 5/6	4/6 + 5/6 = 9/6	3/2 = 1 1/2
d. 2/3 + 2/7	14/21 + 6/21	20/21
e. 3/4 + 1/3 + 1/5	45/60 + 20/60 + 12/60	77/60 = 1 17/60
f. 2/3 + 4/5	10/15 + 12/15	22/15 = 1 7/15
g. 4/5 + 2/3	12/15 + 10/15	22/15 = 1 7/15
h. 3/5 + 5/8	24/40 + 25/40	49/40 = 1 9/40
i. 9/2 + 5/4	18/4 + 5/4	23/4 = 5 3/4
j. 8/3 + 2/7	56/21 + 6/21	62/21 = 2 20/21
k. 3/4 + 1/3 + 1/5	45/60 + 20/60 + 12/60	77/60 = 1 17/60
l. 2/3 + 4/5 + 3/7	70/105 + 84/105 + 45/105	199/105 = 1 94/105
m. 9/2 + 5/4 + 7/6	54/12 + 15/12 + 14/12	83/12 = 6 11/12

Sample working for (c):

The smallest common multiple of 3 and 6 is 6
2/3 = (2 × 2)/(3 × 2) = 4/6, 5/6 stays 5/6
4/6 + 5/6 = 9/6 = (9 ÷ 3)/(6 ÷ 3) = 3/2 = 1 1/2

Did you notice? (f) and (g) give the same answer, and so do (e) and (k) — the order of the fractions does not change the sum.

Q2.

Rahim mixes ²⁄₃ litres of yellow paint with ³⁄₄ litres of blue paint to make green paint. What is the volume of green paint he has made?

Answer

Volume = 2/3 + 3/4
Common denominator = 3 × 4 = 12
2/3 = 8/12, 3/4 = 9/12
8/12 + 9/12 = 17/12 = 1 5/12 litres

Answer: Rahim has made ¹⁷⁄₁₂ litres, that is 1⁵⁄₁₂ litres of green paint.

Check it yourself: ¹⁷⁄₁₂ is a little more than 1 litre — sensible, since ²⁄₃ and ³⁄₄ are each a bit less than a litre.

Q3.

Geeta bought ²⁄₅ meter of lace and Shamim bought ³⁄₄ meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?

Answer

Total lace = 2/5 + 3/4
Common denominator = 5 × 4 = 20
2/5 = 8/20, 3/4 = 15/20
8/20 + 15/20 = 23/20 m = 1 3/20 m

The border needs 1 m and they have ²³⁄₂₀ m, which is more than 1 m (since ²⁰⁄₂₀ = 1).

Answer: total lace = 1³⁄₂₀ m. Yes, it is sufficient — ³⁄₂₀ m of lace is left over.

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