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IPMAT Indore 2024 (MCQ) PYQs

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IPMAT Indore 2024

Arithmetic
>
Simple & Compound Interest

Hard

Sagarika divides her savings of 1000010000 rupees to invest across two schemes A and B. Scheme A offers an interest rate of 10%10\% per annum, compounded half-yearly, while scheme B offers a simple interest rate of 12%12\% per annum. If at the end of first year, the value of her investment in scheme B exceeds the value of her investment in scheme A by 23102310 rupees, then the total interest, in rupees, earned by Sagarika during the first year of investment is

✅ Correct Option: 4
A2−A1=2310A1=(10,000−x)(1+102×100)2=(10,000−x)(441440)A2=x+x×12100=112x100\begin{gathered}A_{2}-A_{1}=2310 \\ A_{1}=(10,000-x)\left(1+\frac{10}{2 \times 100}\right)^{2}=(10,000-x)\left(\frac{441}{440}\right) \\ A_{2}=x+\frac{x \times 12}{100}=\frac{112 x}{100}\end{gathered}
2310=112x100−(441004−441x400)2310=448x+441x400−44100413335=889x400⇒x=15×400=6000\begin{aligned} 2310 & =\frac{112 x}{100}-\left(\frac{44100}{4}-\frac{441 x}{400}\right) \\ 2310 & =\frac{448 x+441 x}{400}-\frac{44100}{4} \\ 13335=\frac{889 x}{400} \quad \Rightarrow \quad x & =15 \times 400 \\ & =6000 \end{aligned} \newline SI=6000×12×1100=720CI=P(1+R2×100)2−P=4000[(1−102×100)2−1]=4000×41400=410\begin{aligned} S I=\frac{6000 \times 12 \times 1}{100} & =720 \\ C I=P\left(1+\frac{R}{2 \times 100}\right)^{2} & -P=4000\left[\left(1-\frac{10}{2 \times 100}\right)^{2}-1\right] \\ & =4000 \times \frac{41}{400}=410 \end{aligned}
∴SI+CI=720+410=1130\therefore S I+C I=720+410=1130

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