Math, asked by santoshyadav73, 7 months ago

alo
ZUP
2
Ariftook a loan of *80,000 from a bank. If the rate of interest is 10% per annum, find the
difference in amounts he would be paying after 15 years, if the interest is compounded
annually and compounded half yearly.​

Answers

Answered by shivanksingh045
0

Here, Principal (P) = Rs. 80000, Time (n) = 1\ \frac{1}{2} years, Rate of interest (R) = 10%

Amount for 1 year (A) = P\left(1+\frac{R}{100}\right)^n

= 80000\left(1+\frac{10}{100}\right)^1

= 80000\left(1+\frac{1}{10}\right)^1

= 80000\left(\frac{11}{10}\right)^1

= Rs. 88,000

Interest for \frac{1}{2} year = \frac{88000\times10\times1}{100\times2}

= Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

(ii) Here, Principal (P) = Rs. 80,000

Time(n) = 1\ \frac{1}{2} year = 3 years (compounded half yearly)

Rate of interest (R) = 10% = 5% (compounded half yearly)

Amount (A) = P\left(1+\frac{R}{100}\right)^n

= 80000\left(1+\frac{5}{100}\right)^3

= 80000\left(1+\frac{1}{20}\right)^3

= 80000\left(\frac{21}{20}\right)^3

= 80000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

Answered by yakingulia5a1279
0

Answer:

(i) Here, Principal (P) = Rs. 80000, Time (n) = 1\ \frac{1}{2}1

2

1

years, Rate of interest (R) = 10%

Amount for 1 year (A) = P\left(1+\frac{R}{100}\right)^nP(1+

100

R

)

n

= 80000\left(1+\frac{10}{100}\right)^180000(1+

100

10

)

1

= 80000\left(1+\frac{1}{10}\right)^180000(1+

10

1

)

1

= 80000\left(\frac{11}{10}\right)^180000(

10

11

)

1

= Rs. 88,000

Interest for \frac{1}{2}

2

1

year = \frac{88000\times10\times1}{100\times2}

100×2

88000×10×1

= Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

(ii) Here, Principal (P) = Rs. 80,000

Time(n) = 1\ \frac{1}{2}1

2

1

year = 3 years (compounded half yearly)

Rate of interest (R) = 10% = 5% (compounded half yearly)

Amount (A) = P\left(1+\frac{R}{100}\right)^nP(1+

100

R

)

n

= 80000\left(1+\frac{5}{100}\right)^380000(1+

100

5

)

3

= 80000\left(1+\frac{1}{20}\right)^380000(1+

20

1

)

3

= 80000\left(\frac{21}{20}\right)^380000(

20

21

)

3

= 80000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}80000×

20

21

×

20

21

×

20

21

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

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