Question

David invested in three schemes A, B, C at $10\%$, $12\%$, $15\%$ p.a. respectively. The total interest in one year was ₹3200. Also, $C$ was $150\%$ of $A$ and $240\%$ of $B$. What was the amount invested in $B$? 

• A. ₹5000
• B. ₹6500
• C. ₹8000
• D. cannot be determined

Hint:

Translate percentage relations into multipliers (e.g., $150\%\!=1.5$). Reduce to one variable, then use the interest equation.

Answer 1

The Correct Option is A

Let amounts be $A=a,\;B=b,\;C=c$. Given $c=1.5a$ and $c=2.4b\Rightarrow b=\frac{c}{2.4}=0.625a$.
Interest eqn: $0.10a+0.12b+0.15c=3200$. Substitute $b,c$: \[ 0.10a+0.12(0.625a)+0.15(1.5a)= (0.10+0.075+0.225)a=0.40a=3200 \Rightarrow a=8000. \] Hence $b = 0.625a = 0.625 \times 8000 = \boxed{\text{₹}\,5000}$. 

Answer 2

Rather than expressing everything in terms of \(A\), we can express both \(A\) and \(C\) in terms of \(B\) and solve directly for the amount invested in \(B\), then check each option.

1. Option A (₹5000): Since \(C=240\%\) of \(B\), we have \(C=2.4B\). Since \(C=150\%\) of \(A\), \(A=\dfrac{C}{1.5}=\dfrac{2.4B}{1.5}=1.6B\). The total interest equation is \[ 0.10A+0.12B+0.15C=3200. \] Substituting \(A=1.6B\) and \(C=2.4B\): \[ 0.10(1.6B)+0.12B+0.15(2.4B)=0.16B+0.12B+0.36B=0.64B=3200\;\Rightarrow\;B=5000. \] This matches option A.
2. Option B (₹6500): Substituting \(B=6500\) gives \(0.64\times6500=4160\ne3200\), so this is incorrect.
3. Option C (₹8000): This is actually the value of \(A\), not \(B\); substituting \(B=8000\) in \(0.64B\) gives \(5120\ne3200\), so it is incorrect for \(B\).
4. Option D (cannot be determined): Since the system above yields a single unique value \(B=5000\), the amount is fully determined, so this option is incorrect.

Expressing \(A\) and \(C\) in terms of \(B\) and solving the interest equation gives a unique value for \(B\).

Hence, the correct answer is option A: ₹5000.