28.
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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What annual instalment will discharge a debt of 6450 Rs. due in 4 years at 5% simple interest? [A]1500 Rs. [B]1835 Rs. [C]1935 Rs. [D]1950 Rs.
1500 Rs. Let each instalment be x. Then, $latex \because \left ( x+\frac{x\times 5\times 1}{100} \right ) + \left ( x+\frac{x\times 5\times 2}{100} \right ) + \left ( x+\frac{x\times 5\times 3}{100} \right ) + x = 6450&s=1$ $latex => \left ( x+\frac{x}{20} \right )+\left ( x+\frac{x}{10} \right )+\left ( x+\frac{3x}{20} \right )+x = 6450&s=1$ $latex => \frac{21x}{20}+\frac{11x}{10}+\frac{23x}{20}+x = 6450&s=1$ $latex => \frac{21x+22x+23x+20x}{20} = 6450&s=1$ $latex => \frac{86x}{20} = 6450&s=1$ $latex => x = \frac{6450\times 20}{86} = 1500 Rs.&s=1$
Hence option [A] is correct answer.
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Calculation:
Installment for first year = x
Installment for second year = 1.10x
Installment for third year = 1.20x
Installment for third year = 1.30x
Installment for final year = 1.40x
Total amount to be paid = (1 + 1.10 + 1.20 + 1.30 + 1.40) × x = 6450
∴ x = 6450 / 6 = 1075
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