# Simple interest & Compound interest Part-3

- A sum of Rs.12000 is borrowed at a rate of 12% per annum for 3 year. Find the SI and on this sum find also the amount to be paid at the end of 3 years.

A.Rs 14000

B.Rs 18000

C.Rs 15000

D.Rs 17000

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Ans: (D)

Sol. P=Rs.12500 T=3years R=12%

I=PTR/100

12500 x 3 x 12/100=Rs.4500

Amount to be paid after 3 years= Rs.12500 +4500=Rs.17000

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2. What will be the amount and CI if Rs.5000 is invested at 4% per annum for 2 years?

A.Rs.6500

B.Rs.5408

C.Rs.750

D.Rs.7000

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Ans:(B)

- P=Rs.5000 R=4% n=2years

A=P(1+R/100)^{n}

^{ } 5000(1+4/100)^{2}

5000(104/100 x 104/100)=Rs.5408

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3. Calculate CI interest on Rs.2000 over a period of 1 year at 10% per annum if interest is compounded half yearly .

A.Rs.360

B.Rs.420

C.Rs.205

D.Rs.180

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Ans:(C)

As interest is compounded half yearly ,so there will be 2 conversion periods in a year.

Rate of interest =1/2 x 10%=5%

A=P(1+R/100)^{n}

2000(1+5/100)^{2}

2000(105/100 x 105/100)=Rs.2205

CI=A-P=Rs.2205 –Rs.2000=Rs.205

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4. The population of a village is 5270. It is found that the rate of increase in population is 6% per annum . Find the population after 2years.

A.Rs.5851

B.5650

C.Rs.5525

D.Rs.5921

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Ans: (D)

Sol. P=5270 R=6% T=2years

Population after 2 years A=P(1+R/100)^{n}

^{ }5270(1+6/100)^{2}

5270(106/100 x 106/100)=5921

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5. A rubber ball is dropped from a certain height . It is found to rebounce only 8% of its previous height if it is dropped from the top of 20m tall building to what height would it raise after bouncing on the ground two times?

Hint: The ball rises to a height of 80% at the first bounce. So at each bounce the loss is 20%. So take R=20%

A. 14.8m

B. 10.8m

C.12.8

D.2.8m

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Ans:(C)

Sol. P=20 R=20% n=2

A=P(1+R/100)^{n}

20(1-20/100)^{2}

20(100-20/100)^{2}

20(80/100)^{2}

^{ }20 x 80/100 x 80/100=12.8m

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6. Sai charan borrows Rs.18000 from a bank to renovate his house. He borrows the money at 10% per annum simple interest over 8 years. What are his monthly repayments ?

A.Rs.347.50

B.Rs.360

C.Rs.367.50

D.Rs.420

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Ans: (C)

P=Rs.18000 R=12% T=8years

SI=PTR/100

18000 x 12 x 8/100=Rs.17280

A=P+I=18000+17280=Rs.35280

Number of months =12 x 8=96=Rs.367.50

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7. A T.V was brought at aprice of Rs.25000. After 1 year the value of was depreciated by 5%. Find the value of the T.V after 1year?

A.Rs.23,750

B.Rs.22,460

C.Rs.30,000

D.Rs.20,000

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Ans: (A)

Sol. Cost of TV=Rs.25000=P

Depreciation = 5%=R, I=1

25000 x 5 x 1/100=Rs.1250

Value after 1 year =Rs.25000-250=Rs.23750

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8. Find the interest on Rs.6000 at 6% per annum, for 2years compounded annually.

A.Rs.500.80

B.Rs.850.50

C.Rs.741.60

D.Rs.960

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Ans: (C)

Sol: P=Rs.6000 R=6% n=2years

Amount in CI=P(1+R/100)^{n}

6000(1+6/100)^{2}

6000(106/100 x 106/100)=Rs.6741.60

Interest =Rs.6741.60-Rs.6000=Rs.741.60

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9. Find the SI on the principal of Rs.8500 at 12% per annum for 4years.

A.Rs.3780

B.Rs.4080

C.Rs.3580

D.Rs.2540

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Ans: (B)

P=Rs.8500 R=12% T=4yrs

SI=PTR/100=8500 x 12 x 4/100= Rs.4080

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10. In a laboratory the count of bacteria in a certain experiment was increasing at the rate of 1.5% per hour. Find the count of 2 hours if the count was initially 56,000?

A.65652

B.57692

C.58482

D.4964

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Ans: (B)

Sol: P=56000 R=1.5% n=2

A=P(1+R/100)^{n}

56000(1+1.5/100)^{2}

56000(100+1.5/100)^{2}

56000(101.5/100 x 101.5/100)=57692

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