What annual Instalment will discharge a loan of Rs 6450 due in 4 years at a 5% rate of interest Find the SI?

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/-

₹ 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.