Question:
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds is:
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds is:
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When the total number of shares of two assets is fixed, express one in terms of the other and substitute into the value equation. This reduces the problem to a simple linear equation.
Updated On: Jul 4, 2026
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Correct Answer: 15
Approach Solution - 1
Approach: Stock A is fully known, so peel its value off the total first. What remains is a simple two-variable system for B and C, where the share-count constraint lets you eliminate one variable instantly.
Step 1: Remove stock A. Value of A held: \[ 10 \times 120 = 1200. \] Remaining value in B and C: \[ 3300 - 1200 = 2100. \]
Step 2: Set up B and C. Let B-shares \(= x\), C-shares \(= y\). We are told \[ x + y = 20, \qquad 90x + 150y = 2100. \]
Step 3: Eliminate \(y\). Substitute \(y = 20 - x\): \[ 90x + 150(20 - x) = 2100 \implies 90x + 3000 - 150x = 2100 \implies -60x = -900 \implies x = 15. \]
Step 4: Sanity check. Then \(y = 5\): value \(= 90(15) + 150(5) = 1350 + 750 = 2100\), and with A's \(1200\) the total is \(3300\). Correct.
\[ \boxed{x = 15 \text{ shares of B}} \]
Step 1: Remove stock A. Value of A held: \[ 10 \times 120 = 1200. \] Remaining value in B and C: \[ 3300 - 1200 = 2100. \]
Step 2: Set up B and C. Let B-shares \(= x\), C-shares \(= y\). We are told \[ x + y = 20, \qquad 90x + 150y = 2100. \]
Step 3: Eliminate \(y\). Substitute \(y = 20 - x\): \[ 90x + 150(20 - x) = 2100 \implies 90x + 3000 - 150x = 2100 \implies -60x = -900 \implies x = 15. \]
Step 4: Sanity check. Then \(y = 5\): value \(= 90(15) + 150(5) = 1350 + 750 = 2100\), and with A's \(1200\) the total is \(3300\). Correct.
\[ \boxed{x = 15 \text{ shares of B}} \]
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Approach Solution -2
Approach: Instead of substituting variables, use alligation: find the average price of the combined 20 shares of B and C, then split them by weighted average between the two known prices.
Step 1: Value contributed by B and C together. Total portfolio value is 3300, and 10 shares of A contribute \(10\times120=1200\). So B and C's 20 shares contribute \(3300-1200=2100\).
Step 2: Average price of those 20 shares.
\[ \text{average price} = \frac{2100}{20} = 105. \]
Step 3: Alligation between B (Rs 90) and C (Rs 150). The ratio of shares of B to C is inversely proportional to how far the average sits from each price:
\[ \frac{\text{shares of B}}{\text{shares of C}} = \frac{150-105}{105-90} = \frac{45}{15} = \frac{3}{1}. \]
So B and C split the 20 shares in ratio \(3:1\), giving B \(= \frac{3}{4}\times 20 = 15\) shares.
\[ \boxed{\text{Shares of B} = 15} \]
Step 1: Value contributed by B and C together. Total portfolio value is 3300, and 10 shares of A contribute \(10\times120=1200\). So B and C's 20 shares contribute \(3300-1200=2100\).
Step 2: Average price of those 20 shares.
\[ \text{average price} = \frac{2100}{20} = 105. \]
Step 3: Alligation between B (Rs 90) and C (Rs 150). The ratio of shares of B to C is inversely proportional to how far the average sits from each price:
\[ \frac{\text{shares of B}}{\text{shares of C}} = \frac{150-105}{105-90} = \frac{45}{15} = \frac{3}{1}. \]
So B and C split the 20 shares in ratio \(3:1\), giving B \(= \frac{3}{4}\times 20 = 15\) shares.
\[ \boxed{\text{Shares of B} = 15} \]
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