# Simple Interest & Compound Interest Part-4

1. Find the CI on the Rs.13,500 for 2 years at 18% per annum compounded annually.

A.Rs.2200

B.Rs.2246.40

C.Rs.2400.40

D.Rs.2500

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Ans: (B)

Sol. Here P=Rs.13,500 R=8% n=2

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

13500(1+8/100)^{2}

13500(108/100 x 108/100)=Rs.15746.40

CI=A-P=Rs.15746.40-13500=Rs.2246.40

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2. What is the amount to be paid on a loan of Rs.10000 for 1^{1}/_{2} year at 12% per annum

compounded annually.

A.Rs.12126.50

B.Rs.13550.20

C.11910.16

D.Rs.12220.50

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Ans: (C)

Sol. Since loan is taken is taken 1^{1}/_{2} years, there are 3 half yearly in 1^{1}/_{2} years

Compounding has to be 3 times.(n=3)

Then rate of interest = half of 12%(1years)=6%

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

10000(1+6/100)^{3}

10000(106/100 x 106/100 x 106/100)=Rs.11910.16

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3. A TV was bought at a price of Rs.20000. After one year the year the value of the TV was depreciated by 2%. Find the value of the TV after 1 year.

A.Rs.19600

B.Rs.15400

C.Rs.16800

D.Rs.13200

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Ans: (A)

Sol. Here p=Rs.20000 depreciation(R)=2% on 20000 per year T=1year

PTR/100=20000 x 2 x 1/100=Rs.400

Value at the end of 1 year=Rs.20000-Rs.400=Rs.9600

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4. A machinery worth Rs.12500 depreciated by 5%. Find its value after 1 year

A.Rs.10875

B.Rs.12500

C.Rs.11875

D.Rs.11500

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Ans: (C)

Sol. P=Rs.12,500 R=5% T=1year

PTR/100=12500 x 5 x 1/100=Rs.625

Value at the end of the 1 year =Rs.12500-Rs.625=Rs.11875

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5. Find the population of a city after 2 years which is at present 10 lakh, if the rate of increase is 4%?

A.11.650 lakhs

B.12.750 lakhs

C.10.816 lakhs

D.13.312lakhs

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Ans: (C)

Here P=10lakhs R=4% n=2years

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

10(1+4/100)^{2}

10(104/100 x 104/100 )=10.816

Estimated population=10.816 lakhs

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6. The population of a city was 25,000 in the year 2000.It increased at the rate of 5% per year. Find the population at the end of 2002.

A.26,482.5

B.28,572.0

C.27562.5

D.27500

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Ans: (C)

Here P=25000 R=5% n=2002-2000=2years

Apply CI formula=25000(1+5/100)^{2}

25000(105/100 x 105/100)=27,562.5

So estimated population is 27,562.5

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7. A scooter was bought at Rs.48000.Its value depreciated at the rate of 5% per annum find its value after 1 year?

A.Rs.46,000

B.Rs.45,600

C.Rs.44,600

D.Rs.46,400

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Ans: (B)

Sol. P=Rs.48,000 depreciation(R)=5% n=1yr

Then applying PTR/100

48,000 x 5 x 1/100=Rs.2,400

Value at the end of 1yr =Rs.48,000-Rs.2,400=Rs.45,600

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8. In a laboratory the count of bacteria in an experiment was increasing at the rate of 1.5%per hour find the bacteria at the end of 2 hours if the count was initially 5,60,000.

A.504026

B.516136

C.523456

D.576926

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Ans: (D)

P=5,60,000 R=1.5% n=2hours

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

5,60,000(1+1.5/100)^{2}

5,60,000(101.5/100 x 101.5/100)=576926

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9. What amount is to be repaid on a loan of Rs.60,000 for 1^{1}/_{2}yrs at 8% per annum compounded half yearly.

A.Rs.69,490

B.Rs.68,491.48

C.Rs.67,491.84

D.Rs.68,000

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Ans: (C)

There are 3 half years in 1^{1}/_{2} that is n=3

R for one year=8% then half year =8/2=4%

A=P(1+R/100)^{n } P=Rs.60,000

60,000(1+4/100)^{3}

60,000(104/100 x 104/100 x 104/100)=Rs.67,491.84

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10. Calculate the amount to be paid on Rs.15,000 at 4% per annum for 3yrs at CI .

A.Rs.16,876.96

B.Rs.16,576

C.Rs.16,376.6

D.Rs.17,000

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Ans: (A)

P=Rs.15,000 R=4% n=3yrs

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

15,000(1+4/100)^{3}

^{ }15,000(104/100 x 104/100 x 104/100)=Rs.16.876.5

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