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: Dec 5, 2025
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Correct Answer: 15
Solution and Explanation
Step 1: Note the given values.
\[
P_A = 120,\quad P_B = 90,\quad P_C = 150.
\]
\[
N_A = 10,\quad N_B + N_C = 20.
\]
Total portfolio value:
\[
3300.
\]
Step 2: Use the share-count condition. \[ N_C = 20 - N_B. \]
Step 3: Write the total value equation. \[ 3300 = (10)(120) + (N_B)(90) + (20 - N_B)(150). \]
Step 4: Simplify. \[ 3300 = 1200 + 90N_B + 3000 - 150N_B. \] Combine constants and coefficients: \[ 3300 = 4200 - 60N_B. \] Rearrange: \[ 60N_B = 4200 - 3300 = 900. \] \[ N_B = \frac{900}{60} = 15. \] Thus, the trader holds: \[ \boxed{15} \] shares of stock B.
Step 2: Use the share-count condition. \[ N_C = 20 - N_B. \]
Step 3: Write the total value equation. \[ 3300 = (10)(120) + (N_B)(90) + (20 - N_B)(150). \]
Step 4: Simplify. \[ 3300 = 1200 + 90N_B + 3000 - 150N_B. \] Combine constants and coefficients: \[ 3300 = 4200 - 60N_B. \] Rearrange: \[ 60N_B = 4200 - 3300 = 900. \] \[ N_B = \frac{900}{60} = 15. \] Thus, the trader holds: \[ \boxed{15} \] shares of stock B.
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