SI = \({PNR \over {100}}\)

= \({17600*({3 \over {4}})*({9 \over {2}}) \over {100}}\) = Rs.594 . Hence (a).

= \({17600*({3 \over {4}})*({9 \over {2}}) \over {100}}\) = Rs.594 . Hence (a).

SI for 1 ^{st} year = (918 – 876) = Rs.42

SI for 3 years = 42 * 3 = Rs.126

Principal = (876 – 126 ) = Rs.750 Hence (b).

SI for 3 years = 42 * 3 = Rs.126

Principal = (876 – 126 ) = Rs.750 Hence (b).

times itself?

If a sum becomes 5 times itself in 8 years , the interest will be 4 times that sum.

For the sum to become 17 times itself, the interest has to become 16 times the sum.

SI for n consecutive years follows AP. It will take (8 * 4) = 32 years. Hence (d).

For the sum to become 17 times itself, the interest has to become 16 times the sum.

SI for n consecutive years follows AP. It will take (8 * 4) = 32 years. Hence (d).

SI = \({PNR \over {100}}\)

= \({ 40000*({7 \over {2}})* 8 \over {100}}\) = Rs.11200

Amount = P + SI = 40000 + 11200 = Rs.51200 . Hence (c).

= \({ 40000*({7 \over {2}})* 8 \over {100}}\) = Rs.11200

Amount = P + SI = 40000 + 11200 = Rs.51200 . Hence (c).

Difference between SI and CI after 2 years = Rs.98 = \({PR{}{}{}{}{}{}{}{}{}{}{}{} \over {100}}\)

P = Rs.20000

SI after 4 years at 7% per annum SI = \({ 20000*4* 7 \over {100}}\) = Rs.5600 . Hence (b).

P = Rs.20000

SI after 4 years at 7% per annum SI = \({ 20000*4* 7 \over {100}}\) = Rs.5600 . Hence (b).

The simple interest for 2 years is Rs.4200 and the compound interest for the second year is Rs.2415 for the same principal at the same rate. Find the principal amount.

SI for 2 years = Rs.4200 SI for 1 year = Rs.2100 = PR …..(i) CI for 2^{nd} year = Rs.2415 = PR + PR^{2} => PR^{2} = 2415 – 2100 à Rs.315 …..(ii) Divide equation ii and i R = \({315 \over {2100}}\) à R = 15% Substitute R value in equation i P * \({15 \over {100}}\) = 2100 = Rs.14000. Hence (c).

The ratio of compound interest to simple interest on a certain sum at 16% per annum for 2 years is:

SI = \({PNR \over {100}}\) = \({P*2*16 \over {100}}\) = 0.32P CI = P [(1+ \({R \over {100}}\) ) ^{n} - 1] = P [(1+ \({16 \over {100}}\) ) ^{2} - 1] = 0.3456P Ratio of CI to SI = \({0.3456P \over {0.32P}}\) = 27: 25 Hence (b).

A invested an amount of Rs.15200 divided in two different banks A and B at the simple interest rate of 16% p.a. and 10% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs.3424, what was the amount invested in bank B?

Let the invested amount in bank A and B be x and 15200-x respectively. SI for 2 years at 16% p.a. = \({x*2*16 \over {100}}\) = \({32x \over {100}}\) SI for 2 years at 10% p.a. = \({(15200-x)*2*10 \over {100}}\) = \({(15200-x)*20 \over {100}}\) Total SI earned in 2 years = Rs.3424 \({32x \over {100}}\) + \({(15200-x)*20 \over {100}}\) = 3424 à x = Rs.3200 Amount invested in bank B = 15200 – x = 15200 – 3200 = Rs.12000 Hence (c).

CI – SI = Rs.2736 (after 3 years)

Difference between SI and CI after 3 years = PR^{3} + 3PR ^{2} = Rs.2736

R = 4%

P = \({2736*100*100*100 \over {R+300R}}\)

= \({2736*100*100*100 \over {16(300+4)}}\) = Rs.562500 . Hence (d).

Difference between SI and CI after 3 years = PR

R = 4%

P = \({2736*100*100*100 \over {R+300R}}\)

= \({2736*100*100*100 \over {16(300+4)}}\) = Rs.562500 . Hence (d).

Divide Rs.5046 between P and Q such that P’s share at the end of 11 years is equal to Q’s share at the end of 13 years at 5% rate of compound interest. Find the share of P.

Let the shares of P and Q be Rs.x and Rs.y respectively.

Both P and Q get the same amount

x *(1 + \({5 \over {100}})\) ^{11} = y *(1 + \({5 \over {100}})\) ^{13}

\({x \over {y}}\) = (1 + \({5 \over {100}})\) ^{2}

= \({441 \over {400}}\)

Share of P = \({441 \over {841}}\) * 5046 = Rs.2646 Hence (a).

Let the sum of money lent out be Rs.x

SI for 1^{st} year = \({PNR \over {100}}\) = \({x*4 \over {100}}\) = \({x \over {25}}\)

Amount = x + \({x \over {25}}\) = \({26x \over {25}}\)

Returned amount after first year = Rs.8400

Balance amount = ( \({26x \over {25}}\) – 8400)

Second year’s interest = ( ( \({26x \over {25}}\) – 8400) * 8)/100 = \({46 \over {125}}\) of the first year’s interest

(( \({26x \over {25}}\) – 8400) * 8)/100 = \({46 \over {125}}\) * \({x \over {25}}\) = Rs.12500 . Hence (d).

SI for 1

Amount = x + \({x \over {25}}\) = \({26x \over {25}}\)

Returned amount after first year = Rs.8400

Balance amount = ( \({26x \over {25}}\) – 8400)

Second year’s interest = ( ( \({26x \over {25}}\) – 8400) * 8)/100 = \({46 \over {125}}\) of the first year’s interest

(( \({26x \over {25}}\) – 8400) * 8)/100 = \({46 \over {125}}\) * \({x \over {25}}\) = Rs.12500 . Hence (d).

Rs.5720 is divided into three parts such that their interest after 1, 2 and 3 years respectively are equal, the rate of interest being 12% per annum in all cases. Find the difference between largest two parts and smallest two parts.

Let a, b and c be the 3 parts invested for 1, 2 and 3 years respectively. Let the interest be d \({a*1*12 \over {100}}\) = \({b*2*12 \over {100}}\) = \({c*3*12 \over {100}}\) = d a = \({25d \over {3}}\), b = \({25d \over {6}}\), c = \({25d \over {9}}\) a: b: c = \({25d \over {3}}\): \({25d \over {6}}\): \({25d \over {9}}\) a: b: c = 6: 3: 2 => 5720 6x + 3x + 2x = 5720 à x = 520 Largest two parts are 6x and 3x. and smallest two parts are 3x and 2x. Difference between largest and smallest two parts = 9x – 5x = 4x = 4*520 = Rs.2080. Hence (c).

CI is compounded half yearly = p((1+ \({r \over {200}}\) ) ^{2n} – 1)

Amount after the first year = 50000(1+ \({10 \over {200}})\)^{2} = Rs.55125

Amount after second year = 55125 (1+ \({10 \over {100}}\) ) = Rs.60637.5

^{ } Total interest = 60637.5 – 50000 = Rs.10637. 5 . Hence (a).

Amount after the first year = 50000(1+ \({10 \over {200}})\)

Amount after second year = 55125 (1+ \({10 \over {100}}\) ) = Rs.60637.5

Let the instalment be Rs.x

Amount at the end of 1^{st} year = 7800 * \({106 \over {100}}\)

x = \({7800 \over {{1 \over {1+({6 \over {100}})}}+{1 \over {1+({6 \over {100}})}}}}\) = Rs.4255.32 . Hence (b).

Amount at the end of 1

x = \({7800 \over {{1 \over {1+({6 \over {100}})}}+{1 \over {1+({6 \over {100}})}}}}\) = Rs.4255.32 . Hence (b).

SI for 2 years = \({45000*2*9 \over {100}}\) = Rs.8100

Amount he has to return = 45000 + 8100 = Rs.53100

He invested Rs.45000 at 8% CI

Amount = 45000(1+\({8 \over {100}}\) )^{2} = Rs.52488

Loss = 53100 – 52488 = Rs.612 Hence (d).

Amount he has to return = 45000 + 8100 = Rs.53100

He invested Rs.45000 at 8% CI

Amount = 45000(1+\({8 \over {100}}\) )

Loss = 53100 – 52488 = Rs.612 Hence (d).