Ratio, Proportion, Indices, LogarithmMCQMTP Dec 22 - Series IQuestion 839 of 302
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A bag contains $25\displaystyle paise, $10 paise, and 5\displaystyle 5 paise in a ratio of $3:2:1$. The total value of $Rs.\displaystyle \text{Rs.} 40$, the number of $5$ paise coins is

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option c — $40$

All Options:

  • A$45$
  • B$48$
  • C$40$
  • D$50$

Detailed Solution & Explanation

Let N25\displaystyle N_{25}, N10\displaystyle N_{10}, and N5\displaystyle N_5 represent the number of $25\displaystyle paise, $10 paise, and 5\displaystyle 5 paise coins, respectively. The problem states that the coins are in a ratio of $3:2:1$. 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 Rs.\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  Rupee=100\displaystyle \text{ Rupee} = 100  paise\displaystyle \text{ paise} So, the total value in paise is: Total Value=Rs.40×100 paise/Rupee=4000 paise\displaystyle \text{Total Value} = \text{Rs.} 40 ×\displaystyle \times 100  paise/Rupee=4000\displaystyle \text{ paise/Rupee} = 4000  paise\displaystyle \text{ paise} Now, we calculate the total value contributed by each type of coin. Value contributed by $25paisecoins\displaystyle paise coins= N_{25} ×\displaystyle \times 25$ paise. Value contributed by $10paisecoins\displaystyle paise coins= N_{10} ×\displaystyle \times 10$ paise. Value contributed by 5\displaystyle 5 paise coins =N5×5\displaystyle = N_5 ×\displaystyle \times 5 paise. The sum of these values must equal the total value of $4000$ paise: (N25×25)+(N10×10)+(N5×5)=4000(N_{25} ×\displaystyle \times 25) + (N_{10} ×\displaystyle \times 10) + (N_5 ×\displaystyle \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 ×\displaystyle \times 25) + (2k ×\displaystyle \times 10) + (k ×\displaystyle \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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