MKU QUANTITATIVE APTITUDE : Unit -IV SMTJN61

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Contents

4 Simple Interest

4.1 Important Facts and Formulae

4.2 Solved Problems

Chapter 4 Simple Interest

4.1 Important Facts and Formulae

1. Principal : The money borrowed or lent out for a certain period is called the principal or the sum.

2. Interest : Extra money paid for using other’s money is called interest.

3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Let Principal \(= P\), Rate \(= R\%\) per annum (p.a.) and Time \(= T\) years. Then,

(i) \(SI = \left (\dfrac {P \times R \times T }{100}\right )\)

(ii) \(P=\left ( \dfrac {100 \times SI}{R \times T} \right ) \); \(R=\left ( \dfrac {100 \times SI}{P \times T} \right ) \); \(T=\left ( \dfrac {100 \times SI}{P \times R} \right ) \);

4.2 Solved Problems

1. Find the simple interest on Rs. 68,000 at \(16 \dfrac {2}{3}\%\) per annum for 9 months.

Solution. Given \(P=\rs 68000, R = 16 \dfrac {2}{3}\% = \dfrac {50}{3}\%\) and \(T = 9 \) months \(=\dfrac {9}{12}\) years \(= \dfrac {3}{4} \) years.
\(SI= \dfrac {P \times R \times T }{100} = \rs \left ( 68000 \times \dfrac {50}{3} \times \dfrac {3}{4} \times \dfrac {1}{100} \right ) = \rs 8500 \)

2. Find the simple interest on Rs. 3000 at \(6 \dfrac {1}{4}\%\) per annum for the period from 4th Feb., 2005 to 18th April, 2005.

Solution. Given \(P=\rs 3000\), \(R= 6\dfrac {1}{4}\% p.a = \dfrac {25}{4}\% p.a\)
Time \(= (24+31+18) \) days = 73 days \(= \dfrac {73}{365} \) year \(=\dfrac {1}{5}\) year.
\(SI= \rs \left ( 3000 \times \dfrac {25}{4} \times \dfrac {1}{5} \times \dfrac {1}{100} \right ) = \rs 37.50 \)

Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted.

3. A sum at simple interests at \(13\dfrac {1}{2}\%\) per annum amounts to Rs.2502.50 after 4 years find the sum.

Solution. Let sum be \(\rs x\). Then,
\(SI=\rs \left ( x \times \dfrac {27}{2} \times 4 \times \dfrac {1}{100} \right ) = \rs \dfrac {27x}{50} \)
Amount \(= \rs \left ( x+ \dfrac {27x }{50} \right ) = \rs \dfrac {77x}{50}\)
\(\dfrac { 77x}{50} = 2502.50 \Leftrightarrow x = \dfrac { 2502.50 \times 50 }{77} = 1625\) Hence, sum = Rs.1625.

4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple interest rate is increased by 8%, it would amount to bow much ?

Solution. \(SI = \rs (920 - 800) = \rs 120\); \(P = Rs. 800,\) \(T = 3\) years.
\(R = \left ( \dfrac {100 \times 120}{800 \times 3} \right )\% = 5\%\).
New rate \(= (5 + 3)\% = 8\%\).
New \(SI = \rs \left ( \dfrac { 800 \times 8 \times 3}{100}\right ) =\rs 192.\)
New amount \(= \rs (800+192) = \rs 992.\)

5. Adam borrowed some money at the rate of 6% p.a. for the first two years, at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. If he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Solution. Let the sum borrowed be \(x\). Then,
\(\left ( \dfrac { x\times 2\times 6}{100 }\right ) + \left ( \dfrac { x\times 9\times 3}{100} \right ) + \left ( \dfrac { x\times 14\times 4}{100}\right ) = 11400\)
\(\Leftrightarrow \left ( \dfrac { 3x}{25} + \dfrac { 27x}{100} + \dfrac { 14x }{ 25}\right ) = 11400 \)
\(\Leftrightarrow \dfrac {95x}{100 }= 11400 \Leftrightarrow x = \left ( \dfrac { 11400\times 100}{95}\right ) = 12000.\)
Hence, sum borrowed = Rs. 12,000.

6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in \(3 \dfrac {1}{2}\) years. Find the sum and rate of interests.

Solution. SI for \(1 \dfrac {1}{2}\) years = Rs.(1164-1008) = Rs.156.
SI for 2 years = Rs.\(\left ( 156 \times \dfrac { 2}{3} \times 2\right )=Rs.208\)
Principal = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and SI = 208.
Rate \(= \left ( \dfrac { 100 \times 208}{800 \times 2}\right )\% = 13\%\)

7. At what rate percent per annum will a sum of money double in 16 years.

Solution. Let principal = P. Then, SI = P and T = 16 years.
Rate \(= \left ( \dfrac { 100 \times P}{P \times 16}\right )\% = 6 \dfrac {1}{4} \%\) p.a.

8. The simple interest on a sum of money is \(\dfrac {4}{9}\) of the principal. Find the rate percent and time, if both are numerically equal.

Solution. Let sum = Rs. \(x\). Then, SI = Rs. \(\dfrac { 4x}{9}\)
Let rate = R% and time = R years.
Then, \(\left ( \dfrac { x\times R \times R}{100}\right )= \dfrac { 4x}{9}\) or \(R^2 = \dfrac { 400}{9}\) or \(R = \dfrac {20}{3} = 6 \dfrac { 2}{3}.\)
Rate \(= 6 \dfrac {2}{3}\%\) and Time \(= 6 \dfrac {2}{3}\) years = 6 years 8 months.

9. The simple interest on a certain sum of money for \(2 \dfrac {1}{2}\) years at 12% per annum is Rs. 40 less than the simple interest on the same sum for \(3 \dfrac {1}{2}\) years at 10% per annum. Find the sum.

Solution. Let the sum be Rs. x Then, \(\left (\dfrac { x\times 10\times 7}{100\times 2} \right )) âĂŞ\left ( \dfrac {x\times 12\times 5}{100\times 2}\right ) = 40\)
\(\Leftrightarrow \dfrac {7x}{20}-\dfrac {3x}{10}=40 \Leftrightarrow x = (40 \times 20) = 800.\)
Hence, the sum is Rs. 800.

10. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.

Solution. Let sum = P and original rate = R. Then, \(\left [\dfrac { P\times (R+2)\times 3}{100}\right ] - \left [\dfrac { P\times R\times 3}{100}\right ] = 360.\)
\(\Leftrightarrow 3PR + 6P - 3PR = 36000 \Leftrightarrow 6P=36000 \Leftrightarrow P=6000\).
Hence, sum = Rs. 6000.

11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?

Solution. Let each Instalment be Rs. \(x\). Then,
\(\left ( x+ \dfrac { x\times 12\times 1}{100} \right ) + \left (x+ \dfrac {x\times 12\times 2}{100}\right ) + x = 1092\)
\(\Leftrightarrow \dfrac {28x}{25} + \dfrac {31x}{25} + x = 1092 \Leftrightarrow (28x+31x+25x)=(1092\times 25)\)
\(\Leftrightarrow x= \left ( \dfrac { 1092\times 25}{84}\right ) = \rs .325. \) Each instalment = Rs. 325.

12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.

Solution. Let the sum lent at 8% be Rs. x and that at 6% be Rs. \((1550 - x)\).
\(\left [ \dfrac {x\times 8\times 1}{100}\right ] + \left [\dfrac { (1550-x)\times 6\times 1}{100}\right ]=106\)
\(\Leftrightarrow 8x + 9300 âĂŞ6x=10600 \Leftrightarrow 2x = 1300 \Leftrightarrow x = 650.\)
Money lent at 8% = Rs. 650.
Money lent at 6% = Rs. (1550 - 650) = Rs. 900.