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What annual instalment will discharge a debt of ₹6450 due in 4 years at 5% simple interest ?
Let each instalment be y Then,
⇒ | y + | y × 5 × 1 | + | y + | y × 5 × 2 | + | y + | y × 5 × 3 | + y = 6450 | ||||||
100 | 100 | 100 |
⇒ | y + | y | + | y + | y | + | y + | 3y | + y = 6450 | ||||||
20 | 10 | 20 |
⇒ | 21y | + | 11y | + | 23y | + y = 6450 |
20 | 10 | 20 |
⇒ | 21y + 22y + 23y + 20y | = 6450 |
20 |
⇒ y = | 6450 × 20 | = ₹ 1500 |
86 |
Second method to solve this question : Here , A = ₹ 6450 , T = 4 years , R = 5%
Equal instalment = | A × 200 |
T[200 + (T - 1) × R] |
Equal instalment = | 6450 × 200 |
4[200 + (4 - 1) × 5] |
Equal instalment = | 6450 × 200 |
4 × 215 |
Equal instalment = | 6450 × 200 |
4 × 215 |
Equal instalment = | 6450 × 50 |
215 |
Equal instalment = ₹ 1500
21.
The amount to be paid, when principal = ₹ 2,000/-, Rate of simple interest (R) = 5%, T = 2 years, is
- ₹ 3,200/-
- ₹ 2,400/-
- ₹ 2,200/-
- ₹ 2,200/-
C.
₹ 2,200/-
100 S.I. = PRT
100 x S.I. =
₹ 2,000 x 5 x 2S.I. = ₹ 200/-Required amount = Principal + Interest
= 2000 + 200 = ₹ 2200/-
Correct Answer:
Description for Correct answer:
By using formula, Installment = \( \Large \frac{6450 \times 100}{4 \times 100+ \left(3+2+1\right) } \times 5 \) =\( \Large \frac{6450 \times 100}{4 \times 100+ \left(3+2+1\right) } \times 5 \) =\( \Large \frac{6450 \times 100}{430} \) Installment = Rs.1500 Hence value of installment= Rs.1500
Note: We have explained formula in previous questions.
Part of solved Simple and compound interest questions and answers : >> Aptitude >> Simple and compound interest
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