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The SI on certain sum of money for 15 months at rate of 7.5% per annum exceed the SI on same sum at 12.5% per annum for 8 months by Rs. 3,250/-, find the sum.
- ARs.1,60,000/-
- BRs.20,000/-
- CRs.1,70,000/-
- DRs.18,000/-
Answer not set
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0
CCommunity
Solution
Let the principal = Rs. $P$
$S.I. = \frac{P \times R \times T}{100}$
Acc. to ques,
=> $(\frac{P \times 7.5 \times 15}{100 \times 12}) - (\frac{P \times 12.5 \times 8}{100 \times 12}) = 3250$
=> $\frac{112.5P}{1200} - \frac{100P}{1200} = 3250$
=> $12.5P = 3250 \times 1200 = 3900000$
=> $P = \frac{3900000}{12.5} = 3,12,000$
Let the principal = Rs. $P$
$S.I. = \frac{P \times R \times T}{100}$
Acc. to ques,
=> $(\frac{P \times 7.5 \times 15}{100 \times 12}) - (\frac{P \times 12.5 \times 8}{100 \times 12}) = 3250$
=> $\frac{112.5P}{1200} - \frac{100P}{1200} = 3250$
=> $12.5P = 3250 \times 1200 = 3900000$
=> $P = \frac{3900000}{12.5} = 3,12,000$
0
CCommunity
Solution
Let the principal = Rs. $P$
$S.I. = \frac{P \times R \times T}{100}$
Acc. to ques,
=> $(\frac{P \times 7.5 \times 15}{100 \times 12}) - (\frac{P \times 12.5 \times 8}{100 \times 12}) = 3250$
=> $\frac{112.5P}{1200} - \frac{100P}{1200} = 3250$
=> $12.5P = 3250 \times 1200 = 3900000$
=> $P = \frac{3900000}{12.5} = 3,12,000$
Let the principal = Rs. $P$
$S.I. = \frac{P \times R \times T}{100}$
Acc. to ques,
=> $(\frac{P \times 7.5 \times 15}{100 \times 12}) - (\frac{P \times 12.5 \times 8}{100 \times 12}) = 3250$
=> $\frac{112.5P}{1200} - \frac{100P}{1200} = 3250$
=> $12.5P = 3250 \times 1200 = 3900000$
=> $P = \frac{3900000}{12.5} = 3,12,000$