Question

subject - maths
chapter 11.compound interest
please tell me answer... all questions... ​

Answer 1

Answer with step-by-step explanation:

1. To find compound interest on Rs.3000 at 5% p.a. for 3 years:

P = 3000 Rs.

r = 5%

n =  2 years

Formula for amount:

$A = P(1 + \frac{r}{100})^n$

   $= 3000(1 + \frac{5}{100})^2$

   $= 3000(\frac{105}{100})^2$

   $=3,307.5$

Compound interest formula:

$CI = A - P$

      = 3307.5 - 3000

      = 307.5 Rs.

2. To find the compound interest on Rs.4000 at 5% p.a. for 2 years:

P = 4000 Rs.

r = 5%

n =  2 years

Formula for amount:

$A = P(1 + \frac{r}{100})^n$

   $= 4000(1 + \frac{5}{100})^2$

   $= 4000(\frac{105}{100})^2$

   $=4410$

Compound interest formula:

$CI = A - P$

      = 4410 - 4000

      = 410 Rs.

3. To find the compound interest on Rs. 1250 at 5% p.a. for 3 years:

P = 1250 Rs.

r = 5%

n =  3 years

Formula for amount:

$A = P(1 + \frac{r}{100})^n$

   $= 1250(1 + \frac{5}{100})^3$  

   $= 1250(\frac{105}{100})^3$

   $=1447.03$

Compound interest formula:

$CI = A - P$

      = 1447.03 - 1250

      = 197.03 Rs.

4. Difference between CI and SI in Rs.12000 at 9% p.a. for 2 years:

P = Rs.12000

R = 9%

T/n = 2 years

Simple interest formula:

$SI = \frac{P*T*R}{100}$

    $= \frac{12000 * 9 * 2}{100}$

    = 2160

Formula for amount:

$A = P(1 + \frac{r}{100})^n$

   $= 12000(1 + \frac{9}{100})^2$  

   $= 12000(\frac{109}{100})^2$

   $=14257.2$

Compound interest formula:

$CI = A - P$

     $= 14257.2 - 14000$

    $=257.2$

Difference in simple interest and compound interest:

2160 - 257.2 = 1902.8 Rs.

5. To find compound interest on Rs.40000 at 7% p.a. for 2 years:

P = 40000 Rs.

r = 7%

n =  2 years

Formula for amount:

$A = P(1 + \frac{r}{100})^n$

   $= 40000(1 + \frac{7}{100})^2$

   $= 40000(\frac{107}{100})^2$

   $=45796$

Compound interest formula:

$CI = A - P$

      = 45796 - 40000

      = 5796 Rs.

6. To find the principle and then compound interest at 10% p.a. for 2 years:

Simple Interest = Rs. 200

R = 10%

T = 2 years

$SI = \frac{P*T*R}{100}$

$P = \frac{SI *100}{T*R}$

$P = \frac{200*100}{2*10}$

$P = \frac{20000}{20}$

$P = 1000}$

Formula for amount:

$A = P(1 + \frac{r}{100})^n$

   $= 1000(1 + \frac{10}{100})^2$

  $= 1000( \frac{110}{100})^2$

  $=1210$

Compound interest formula:

$CI = A - P$

     $= 1210 - 1000$

      = 210 Rs.

7. Find the profit gained by Mr.Sharma form Mr.Gupta:

P = Rs.20000

R = 18%

T = 2 years

• For Mr.Sharma:

$SI = \frac{P*R*T*}{100}$

    $= \frac{20000*18*2}{100}$

    $= 7200$

Amount = 20000 + 7200 = 27200 Rs.

• For Mr.Gupta:

Amount compounded annually = $P(1+\frac{r}{100})^n$

                                                    = $20000(1+\frac{18}{100})^2$

                                                    = $20000(\frac{118}{100})^2$

                                                    = 27848 Rs.

Profit = 27848 - 27200 = 648 Rs.

8. Find compound interest on Rs.15000 at 5%, 8% and 10% for 3 years:

P = Rs.15000

$R_1 = 5%$

$R_2 = 8%$

$R_3 = 10%$

T = 3

Amount = $P(1 + \frac{R_1}{100})(1 + \frac{R_2}{100})(1 + \frac{R_3}{100})$

             = $15000(1 + \frac{5}{100})(1 + \frac{8}{100})(1 + \frac{10}{100})$

             = $15000( \frac{105}{100})(\frac{108}{100})( \frac{110}{100})$

             = 18711

Compound interest formula:

$CI = A - P$

     $= 18711 - 15000$

     $= 3711$

9. To find compound interest on Rs.10400 at 4.5% p.a. for 3 years:

P = 10,400 Rs.

R = 4.5%

T = 3 years

Amount = $P(1+\frac{r}{100})^n$

             = $10400(1+\frac{4.5}{100})^3$

             = $10400 (\frac{104.5}{100})^3$

             = 10400 × 1.141166

             = 11868.13

$CI = A - P$

     $= 11868.13 - 10400$

     $=1468.13$

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Points to remember:

• Amount formula: $A = P(1 + \frac{r}{100})^n$
• Compound interest formula: $CI = A - P$  

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Hope it helps ☺