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Question

A, B and C divide a certain sum of money among themselves. The average of the amount with them is Rs.4520. Share of A is \(10\frac{2}{3}%\) % more than share of B and  \(33\frac{1}{3}%\) % less than share of C. What is the share of B (in Rs.)?

The correct answer is

3600

Understanding the Money Division Problem

This question involves dividing a sum of money among three individuals, A, B, and C, based on their average share and specific percentage relationships between their shares. We are given the average amount they have and how A's share relates to B's and C's shares. Our goal is to find the exact share of B.

Calculating the Total Sum

We are given that the average amount of money with A, B, and C is Rs. 4520. Since there are three people, the total sum of money is the average amount multiplied by the number of people.

Total Sum = Average Amount \(\times\) Number of People

Total Sum = \(4520 \times 3\)

Total Sum = \(13560\) Rs.

So, the sum of the shares of A, B, and C is Rs. 13560.

\(A + B + C = 13560\)

Analyzing Percentage Relationships

We are given two percentage relationships:

  1. Share of A is \(10\frac{2}{3}\)% more than share of B.
  2. Share of A is \(33\frac{1}{3}\)% less than share of C.

Let's convert these percentages into fractions:

  • \(10\frac{2}{3}\)% = \(\frac{10 \times 3 + 2}{3}\)% = \(\frac{32}{3}\)% = \(\frac{32}{300}\) = \(\frac{8}{75}\)
  • \(33\frac{1}{3}\)% = \(\frac{33 \times 3 + 1}{3}\)% = \(\frac{100}{3}\)% = \(\frac{100}{300}\) = \(\frac{1}{3}\)

Now, we can write the relationships algebraically:

  • A is \(\frac{8}{75}\) more than B: \(A = B + \frac{8}{75}B = \left(1 + \frac{8}{75}\right)B = \frac{75+8}{75}B = \frac{83}{75}B\)
  • A is \(\frac{1}{3}\) less than C: \(A = C - \frac{1}{3}C = \left(1 - \frac{1}{3}\right)C = \frac{3-1}{3}C = \frac{2}{3}C\)

Expressing B and C in terms of A

From the relationships above, we can express B and C in terms of A:

  • From \(A = \frac{83}{75}B\), we get \(B = A \times \frac{75}{83} = \frac{75}{83}A\)
  • From \(A = \frac{2}{3}C\), we get \(C = A \times \frac{3}{2} = \frac{3}{2}A\)

Setting up the Equation and Solving for A

We know that \(A + B + C = 13560\). Substitute the expressions for B and C in terms of A into this equation:

\(A + \frac{75}{83}A + \frac{3}{2}A = 13560\)

To solve for A, find a common denominator for the fractions. The denominators are 1, 83, and 2. The least common multiple of 83 and 2 is \(83 \times 2 = 166\).

\(\frac{166}{166}A + \frac{75 \times 2}{83 \times 2}A + \frac{3 \times 83}{2 \times 83}A = 13560\)

\(\frac{166}{166}A + \frac{150}{166}A + \frac{249}{166}A = 13560\)

Combine the fractions on the left side:

\(\frac{166 + 150 + 249}{166}A = 13560\)

\(\frac{565}{166}A = 13560\)

Now, solve for A:

\(A = 13560 \times \frac{166}{565}\)

Let's simplify the fraction. Both 13560 and 565 are divisible by 5.

\(13560 \div 5 = 2712\)

\(565 \div 5 = 113\)

So, \(A = 2712 \times \frac{166}{113}\)

Now, check if 2712 is divisible by 113. \(2712 \div 113 = 24\).

So, \(A = 24 \times 166\)

\(A = 3984\)

A's share is Rs. 3984.

Calculating the Share of B

We have the relationship \(B = \frac{75}{83}A\). Substitute the value of A we just found:

\(B = \frac{75}{83} \times 3984\)

Check if 3984 is divisible by 83. \(3984 \div 83 = 48\).

So, \(B = 75 \times 48\)

\(B = 3600\)

B's share is Rs. 3600.

Verifying the Solution (Optional but Recommended)

Let's find C's share: \(C = \frac{3}{2}A = \frac{3}{2} \times 3984 = 3 \times 1992 = 5976\).

Total sum: \(A + B + C = 3984 + 3600 + 5976 = 13560\). This matches the total sum calculated from the average.

Percentage checks:

  • A vs B: A is \(3984 - 3600 = 384\) more than B. Percentage more = \( \frac{384}{3600} \times 100 = \frac{384}{36} = \frac{32}{3} = 10\frac{2}{3}\)%. Correct.
  • A vs C: A is \(5976 - 3984 = 1992\) less than C. Percentage less = \( \frac{1992}{5976} \times 100 = \frac{1}{3} \times 100 = 33\frac{1}{3}\)%. Correct.

All conditions are satisfied, confirming that B's share is Rs. 3600.

Individual Share (Rs.)
A 3984
B 3600
C 5976
Total 13560

Share Calculation Conclusion

The share of B is Rs. 3600.

Revision Table: Key Concepts

Concept Description Formula Used
Average Sum divided by the number of items. Average = Total Sum / Number of Items
Total Sum from Average The sum of amounts when the average is known. Total Sum = Average \(\times\) Number of Items
Percentage Increase A quantity is (percentage)% more than another. New Value = Original Value \(\times\) \((1 + \text{percentage}/100)\)
Percentage Decrease A quantity is (percentage)% less than another. New Value = Original Value \(\times\) \((1 - \text{percentage}/100)\)
Solving Linear Equations Finding the value of an unknown variable in an equation. Requires algebraic manipulation.

Additional Information: Handling Percentages in Equations

When dealing with percentage increases or decreases in algebraic equations, it's often helpful to represent the percentages as decimals or fractions. For example:

  • If value A is 20% more than value B, this means A = B + 20% of B = \(B + 0.20B = 1.20B\). Or as fractions: \(A = B + \frac{20}{100}B = B + \frac{1}{5}B = \frac{6}{5}B\).
  • If value A is 25% less than value C, this means A = C - 25% of C = \(C - 0.25C = 0.75C\). Or as fractions: \(A = C - \frac{25}{100}C = C - \frac{1}{4}C = \frac{3}{4}C\).

Using fractions can sometimes make calculations easier, especially when dealing with repeating decimals or fractions like \(10\frac{2}{3}\)% or \(33\frac{1}{3}\)%, as seen in this problem.

In this solution, we used the fractional approach: \(10\frac{2}{3}\)% more means multiplying by \(1 + \frac{8}{75} = \frac{83}{75}\), and \(33\frac{1}{3}\)% less means multiplying by \(1 - \frac{1}{3} = \frac{2}{3}\).

Setting up the initial equation relating A, B, and C using one variable (in this case, A) was crucial for solving the problem effectively.

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Important Questions from Ratio and Proportion

  1. The cost of a diamond is directly proportional to the square of its weight. The cost of a 14 gm diamond is Rs. 2560. This diamond got broken down into two pieces in the ratio of 5 ∶ 9. How much loss percent is incurred due to this breakage ? (Correct to two decimal places)

  2. Atul purchased Bread costing Rs.20 and gave a 100 rupee note to the shopkeeper. The shopkeeper gave the balance money in coins of denomination Rs.2, Rs.5 and Rs.10. If these coins are in the ratio 5 ∶ 4 ∶ 1, then how many Rs.5 coins did the shopkeeper give?

  3. A person divides a certain amount among his three sons in the ratio of 3 ∶ 4 ∶ 5. If he had divided this amount in the ratio of 1/3,1/4,1/5, his son, who had got the lowest share earlier, would get Rs.1,188 more. Find the amount (in Rs).

  4. In a school 3/8 of the number of students are girls and the rest are boys. One-third of the number of boys are below 10 years and 2/3 the number if girls are also below 10 years. If the number of students of age 10 or more years is 260. then the number of boys in the school is:

  5. If a : b : c = \(\frac{1}{4} : \frac{1}{3} : \frac{1}{2}, \)  then  \( \ \frac{a}{b} : \frac{b}{c} : \frac{c}{a} = ?\)

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