Question
David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
Answer: Option A
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Answer : Option A
Explanation :
Let x , y and x be his investments in A, B and C respectively. Then
Then, Interest on x at 10% for 1 year
+ Interest on y at 12% for 1 year
+ Interest on z at 15% for 1 year
= 3200
$MF#%\dfrac{x \times 10 \times 1}{100} + \dfrac{y \times 12 \times 1}{100} + \dfrac{z \times 15 \times 1}{100} = 3200\\\\ \Rightarrow 10x + 12y + 15z = 320000 \quad \color{RED}{---(1)}$MF#%
Amount invested in Scheme C was 240% of the amount invested in Scheme B
$MF#%=> z = \dfrac{240y}{100} = \dfrac{60y}{25} = \dfrac{12y}{5} \color{RED}{---(2)}$MF#%
Amount invested in Scheme C was 150% of the amount invested in Scheme A
$MF#%=> z = \dfrac{150x}{100} = \dfrac{3x}{2}\\\\ => x = \dfrac{2z}{3} = \dfrac{2}{3} \times \dfrac{12y}{5} = \dfrac{8y}{5} \color{RED}{---(3)}$MF#%
From(1),(2) and (3),
10x + 12y + 15z = 320000
$MF#%10\left(\dfrac{8y}{5}\right) + 12y + 15\left(\dfrac{12y}{5}\right) = 320000\\\\ 16y + 12y + 36y = 320000\\\\ 64y = 320000\\\\ y=\dfrac{320000}{64}= \dfrac{10000}{2}=5000$MF#%
i.e.,Amount invested in Scheme B = Rs.5000
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