Arithmatical - Simple Interest
DIRECTIONS : Problems based on Simple Interest.
1. |
A man took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was |
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A. Rs.2000 |
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B. Rs.10,000 |
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C. Rs.15,000 |
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D. Rs.20,000 |
Solution |
Principal | = Rs. (100×5400 / 12×3) |
‹=› Rs. 15,000. |
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2. |
The price of a T.V set worth Rs.20,000 is to be paid in 20 instalments of Rs. 1000 each. If the rate of interest be 6% per annum, and the first instalments be paid at the time of purchase, then the value of the last instalments covering the interest as well will be |
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A. Rs.1050 |
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B. Rs.2050 |
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C. Rs.3000 |
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D. Rs.19000 |
Solution |
Monet paid in cash | = Rs. 1000. |
Balance amount | = Rs. (20000 - 1000) |
‹=› Rs. 19000. |
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3. |
A lent Rs.5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs.2200 in all from both of them as interest. The rate of interest per annum is |
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A. 5% |
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B. 7% |
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C. 8% |
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D. 10% |
Solution |
Let the rate be R% p.a. |
Then, | (5000xRx2/100) + (3000xRx4/100) |
‹=›100R+120R= 2200 |
‹=›R=(2200/220) |
Rate ‹=›10%. |
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4. |
A money lender finds that dues to a fall in the annual rate of interest from 8% to 7x3/4%, his yearly income diminishes by Rs.61.50. His capital is |
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A. Rs.22,400 |
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B. Rs.23,800 |
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C. Rs.24,600 |
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D. Rs.26,000 |
Solution |
Let the capital be Rs. x. |
Then, | = (x × 8×1/100) - (x × 31/4×1/100) |
= 61.50. |
‹=›32x - 31x =6150×4 |
‹=›x= 24600. |
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