Ratio, Proportion, Indices, LogarithmMCQMTP Dec 22 - Series IQuestion 839 of 305
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A bag contains 25\displaystyle 25 paise, 10\displaystyle 10 paise, and 5\displaystyle 5 paise in a ratio of 3:2:1\displaystyle 3:2:1. The total value of Rs.40\displaystyle \text{Rs.} 40, the number of 5\displaystyle 5 paise coins is

Options

A45\displaystyle 45
B48\displaystyle 48
C40\displaystyle 40
D50\displaystyle 50
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Correct Answer

✅ Option c — 40\displaystyle 40

All Options:

  • A45\displaystyle 45
  • B48\displaystyle 48
  • C40\displaystyle 40
  • D50\displaystyle 50

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Detailed Solution & Explanation

Let N25\displaystyle N_{25}, N10\displaystyle N_{10}, and N5\displaystyle N_5 represent the number of 25\displaystyle 25 paise, 10\displaystyle 10 paise, and 5\displaystyle 5 paise coins, respectively. The problem states that the coins are in a ratio of 3:2:1\displaystyle 3:2:1. Therefore, we can write: N25:N10:N5=3:2:1N_{25} : N_{10} : N_5 = 3:2:1 This implies that there exists a common multiplier, say k\displaystyle k, such that: N25=3kN_{25} = 3k N10=2kN_{10} = 2k N5=1k=kN_5 = 1k = k Next, we need to consider the total value of these coins. The total value is given as Rs.40\displaystyle \text{Rs.} 40. To work with a consistent unit, we convert the total value into paise, since the coin denominations are in paise. 1 Rupee=100 paise1 \text{ Rupee} = 100 \text{ paise} So, the total value in paise is: Total Value=Rs.40×100 paise/Rupee=4000 paise\text{Total Value} = \text{Rs.} 40 \times 100 \text{ paise/Rupee} = 4000 \text{ paise} Now, we calculate the total value contributed by each type of coin. Value contributed by 25\displaystyle 25 paise coins =N25×25\displaystyle = N_{25} \times 25 paise. Value contributed by 10\displaystyle 10 paise coins =N10×10\displaystyle = N_{10} \times 10 paise. Value contributed by 5\displaystyle 5 paise coins =N5×5\displaystyle = N_5 \times 5 paise. The sum of these values must equal the total value of 4000\displaystyle 4000 paise: (N25×25)+(N10×10)+(N5×5)=4000(N_{25} \times 25) + (N_{10} \times 10) + (N_5 \times 5) = 4000 Substitute the expressions for N25\displaystyle N_{25}, N10\displaystyle N_{10}, and N5\displaystyle N_5 in terms of k\displaystyle k into the equation: (3k×25)+(2k×10)+(k×5)=4000(3k \times 25) + (2k \times 10) + (k \times 5) = 4000 Perform the multiplications: 75k+20k+5k=400075k + 20k + 5k = 4000 Combine the terms involving k\displaystyle k: (75+20+5)k=4000(75 + 20 + 5)k = 4000 100k=4000100k = 4000 Now, solve for k\displaystyle k: $$k = \frac{4000}{

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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