WBBSE Class 10 Maths Algebra Chapter 2 Ratio And Proportion Multiple Choice Questions

WBBSE Class 10 Maths Algebra Chapter 2 Ratio And Proportion Multiple Choice Questions

Example 1. The fourth proportion of 3, 4, and 6 is

1. 8
2. 10
3. 12
4. 24

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Solution: 3 : 4 : : 6 : ? or, \(\frac{3}{4}\) = \(\frac{6}{?}\)

⇒ or, ? = \(\frac{6 \times 4}{3}\) = 8

∴ The correct answer is 1. 8

Example 2. The 3rd proportion of 8 and 12 is

1. 12
2. 16
3. 18
4. 20

Solution: 8 x third proportional = 12²

⇒ or, third proportional = \(\frac{144}{8}\) = 18

∴ The correct answer is 3. 18

Example 3. The mean proportional of 16 and 25 is

1. ± 400
2. ± 100
3. ± 20
4. ± 40

Solution: Mean proportional = ± \(\sqrt{16 \times 25}\) = ± 20

∴ The correct answer is 3. ± 20

Wbbse Class 10 Maths Algebra Notes

Example 4. a is a positive integer and a: \(\frac{27}{64}\) = \(\frac{3}{4}\): a, then the value of a is

1. \(\frac{81}{256}\)
2. 9
3. \(\frac{9}{16}\)
4. \(\frac{16}{9}\)

Solution: \(\frac{a}{\frac{27}{64}}=\frac{3}{\frac{4}{a}}\)

⇒ or, \(a^2=\frac{27}{64} \times \frac{3}{4}\)

⇒ or, a = \(\frac{9}{16}\)

∴ The correct answer is 3. \(\frac{9}{16}\)

Example 5. If 2a = 3b = 4c then a: b: c is

1. 3: 4: 6
2. 4: 3: 6
3. 3: 6: 4
4. 6: 4: 3

Solution: Let 2a = 3b = 4c = k (≠0)

∴ a : b : c = \(\frac{k}{2}: \frac{k}{3}: \frac{k}{4}\) =6:4:3

∴ The correct answer is 4. 6: 4: 3

Example 6. If 2x = 3y = 5z then x : y : z

1. 15: 10: 6
2. 6: 10: 15
3. 15: 6: 10
4. none of these

Solution: Let 2x = 3y = 5z = k (≠0)

∴ x: y: z = \(\frac{k}{2}: \frac{k}{3}: \frac{k}{4}\) = 15 : 10 : 6

∴ The correct answer is 1. 15: 10: 6

Example 7. If x + y = z and 2x – z = y then x: y: z =

1. 1: 2: 3
2. 2: 1: 3
3. 3: 1: 2
4. none of these

Solution: x = z – y = \(\frac{y+z}{2}\)

⇒ or, 2z- 2y = y + z or, z = 3y

⇒ Now, x = z- y = 3y-y = 2y

∴ x: y: z = 2y: y: 3y = 2: 1: 3

∴ The correct answer is 2. 2: 1: 3

Class 10 Maths Algebra Chapter 2 MCQs

Example 8. If \(\frac{m}{n}\) = \(\frac{2}{3}\) then \(\frac{n+m}{n-m}\) is

1. 2
2. 4
3. 5
4. 10

Solution: \(\frac{n}{m}=\frac{3}{2} \quad \text { or, } \quad \frac{n+m}{n-m}=\frac{3+2}{3-2}\) [componendo & dividendo]

⇒ or, \(\frac{n+m}{n-m}=5\)

∴ The correct answer is 3. 5

Example 9. If (a + b) : √ab = 4 : 1 then a : b =

1. (2+√3):(2-√3)
2. (2-√3) : (2+√3)
3. 1:1
4. None of these

Solution: \(\frac{a+b}{\sqrt{a b}}=\frac{4}{1}\)

⇒ or, \(\quad \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{4+2}{4-2}\) [Componendo & dividendo]

⇒ or, \(\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=3\)

⇒ or, \(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\sqrt{3}\), Again by componendo and dividendo,

⇒ We get, \(\frac{2 \sqrt{a}}{2 \sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\),

⇒ \(\frac{a}{b}=\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)^2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\)

∴ The correct answer is 1.

Example 10. If \(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{2 a-3 b+c}{p}\) = \(\frac{2 a-3 b+c}{p}\) then P =

1. 0
2. 1
3. 2
4. 3

Solution: Let \(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}\) = k (≠ 0)

∴ a = 3k, b = 4k, c = 7k.

∴ \(\frac{2 \cdot 3 k-3 \cdot 4 k+7 k}{p}=k\)

⇒ or, \(\frac{k}{p}\)

∴ p = 1

∴ The correct answer is 2. 1