Class 10 Ratio and Proportion Mock Test (ICSE)

Topics in Ratio and Proportion

• Duplicate, triplicate, sub-duplicate and sub-triplicate ratios
• Proportion and continued proportion
• Mean proportional
• Componendo and dividendo
• Properties of proportion

Ratio and Proportion practice questions with answers

1. Q1Multiple choice·1 mark
   If $\frac{a}{b} = \frac{4}{3}$, then the value of $\frac{3a + 2b}{3a - 2b}$ is:

   1. (a)$\frac{3}{7}$
   2. (b)$\frac{1}{3}$
   3. (c)$7$
   4. (d)$3$
   Show answer and step marks

   1. Correct option: (d) $3$. Given $\frac{a}{b} = \frac{4}{3}$, multiplying by $\frac{3}{2}$ gives $\frac{3a}{2b} = \frac{12}{6} = 2$. Applying componendo and dividendo yields $\frac{3a + 2b}{3a - 2b} = \frac{2 + 1}{2 - 1} = 3$.+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   If $5, 10, x$ are in continued proportion, then the value of $x$ is:

   1. (a)$15$
   2. (b)$25$
   3. (c)$20$
   4. (d)$50$
   Show answer and step marks

   1. Correct option: (c) $20$. Since $5, 10, x$ are in continued proportion, $\frac{5}{10} = \frac{10}{x}$, which gives $5x = 100$ and $x = 20$.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   If $\frac{a}{b} = \frac{c}{d}$, which of the following relations is not necessarily true?

   1. (a)$\frac{a-b}{b} = \frac{c-d}{d}$
   2. (b)$\frac{a+b}{b} = \frac{c+d}{d}$
   3. (c)$\frac{a+b}{c} = \frac{b+d}{d}$
   4. (d)$\frac{a}{c} = \frac{b}{d}$
   Show answer and step marks

   1. Correct option: (c) $\frac{a+b}{c} = \frac{b+d}{d}$. Checking the properties of proportion: alternendo, componendo, and dividendo are valid, whereas $\frac{a+b}{c} = \frac{b+d}{d}$ simplifies to $bd = cd$, which requires $b=c$ and is not universally true.+1

   Answer: (c)
4. Q4Multiple choice·1 mark
   The mean proportional between $4$ and $25$ is:

   1. (a)$100$
   2. (b)$14.5$
   3. (c)$10$
   4. (d)$20$
   Show answer and step marks

   1. Correct option: (c) $10$. The mean proportional between $4$ and $25$ is calculated as $\sqrt{4 \times 25} = \sqrt{100} = 10$.+1

   Answer: (c)
5. Q5Long answer·4 marks
   If $a, b, c$ are in continued proportion, prove that $\frac{a^2 + b^2}{b^2 + c^2} = \frac{a}{c}$. Using this result or otherwise, show that $\frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{a}{c}$.
   Show answer and step marks

   1. Since $a, b, c$ are in continued proportion, let $\frac{a}{b} = \frac{b}{c} = k$. Then $b = ck$ and $a = bk = ck^2$.+1
   2. First LHS: $\frac{a^2 + b^2}{b^2 + c^2} = \frac{c^2k^4 + c^2k^2}{c^2k^2 + c^2} = \frac{c^2k^2(k^2 + 1)}{c^2(k^2 + 1)} = k^2$.+1
   3. RHS: $\frac{a}{c} = \frac{ck^2}{c} = k^2$, so $\frac{a^2 + b^2}{b^2 + c^2} = \frac{a}{c}$.+1
   4. Second: $\frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} = \frac{c^2k^2(k^2 + k + 1)}{c^2(k^2 + k + 1)} = k^2 = \frac{a}{c}$, hence proved.+1

   Answer: With $\frac{a}{b} = \frac{b}{c} = k$ (so $b = ck$, $a = ck^2$), both $\frac{a^2 + b^2}{b^2 + c^2}$ and $\frac{a^2 + ab + b^2}{b^2 + bc + c^2}$ simplify to $k^2 = \frac{a}{c}$; hence both results are proved.

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