Smriti invested the same sum of money in two different banks at 11 - p-a for 3 1-2 years and 4 1-2 years respectively- If the difference in interests for two periods is - 412-50- then the sum of money invested in each bank is

The correct option is B 3750Principal amount = P SI 1 = P ×11/100×7/2= 77P/200 SI 2 = P ×11/100×9/2 = 99P/200 SI2 SI1 = 421.50 = 99P/200 77P/200 421.50 ...

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Smriti invest...

Question

Smriti invested the same sum of money in two different banks at 11% p.a for $3 \frac{1}{2}$ years and $4 \frac{1}{2}$ years respectively. If the difference in interests for two periods is $412.50, then the sum of money invested in each bank is

3500

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3750

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4000

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4250

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None of these

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Solution

The correct option is B 3750
Principal amount = P
$S I 1 = P \times \frac{11}{100} \times \frac{7}{2} = \frac{77 P}{200} S I 2 = P \times \frac{11}{100} \times \frac{9}{2} = \frac{99 P}{200} S I 2 - S I 1 = 421.50 = \frac{99 P}{200} - \frac{77 P}{200} 421.50 \times 200 = 22 P P = 3750$

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Smriti invested the same sum of money in two different banks at 11% p.a for 312 years and 412 years respectively. If the difference in interests for two periods is $412.50, then the sum of money invested in each bank is

The correct option is B 3750Principal amount = P SI1=P×11100×72=77P200SI2=P×11100×92=99P200SI2−SI1=421.50=99P200−77P200421.50×200=22PP=3750

Smriti invested the same sum of money in two different banks at 11% p.a for $3 \frac{1}{2}$ years and $4 \frac{1}{2}$ years respectively. If the difference in interests for two periods is $412.50, then the sum of money invested in each bank is

The correct option is B 3750
Principal amount = P
$S I 1 = P \times \frac{11}{100} \times \frac{7}{2} = \frac{77 P}{200} S I 2 = P \times \frac{11}{100} \times \frac{9}{2} = \frac{99 P}{200} S I 2 - S I 1 = 421.50 = \frac{99 P}{200} - \frac{77 P}{200} 421.50 \times 200 = 22 P P = 3750$