GMAT Interests Questions With Solutions PDF, Download Now

Compound Interest after 2 years = 1414.4

$$P(1+\frac{8}{100})^2 - P$$ = 1414.4

=> $$P(1.08)^2 - P$$ = 1414.4

=> 0.1664P = 1414.4

=> P = 8500

Therefore amount = 8500 + 1414.14 = 9914.14

Let the rate of interest be r
Simple Interest = $$\frac{P t r}{100}$$

10800 = $$\frac{22500}{100}$$ x 4 x r

r = $$\frac{10800}{22500}$$ x $$\frac{100}{4}$$ = 12%

Compound Interest = 22500(1 + $$\frac{12}{100}$$)2 - 22500
= 22500 x $$\frac{112}{100}$$ x $$\frac{112}{100}$$ - 22500
= 28224 - 22500
= 5724

Let x be the amount lent for 6%
16800 - x is the amount lent for 8%
$$\frac{2x6}{100}$$ + $$\frac{2(16800-x)8}{100}$$ + 16800 = 19000
x = 12200

Amount of $100 for 1 year when compounded half-yearly = $$[100\times(1+3/100)^2 ]$$=$106.09
Effective rate =(106.09-100) %=6.09.

Interest = (Principal * interest * No. of years )

Interest = (30,000*3*5)/100

Interest = 4,500

x% of 22,500 = 4,500

x = (4500/22,500)*100

x = 20

Let the rate of interest on the principal be R.

So, 19800*R*3/100 = 7128

So, R = 12%

Hence, the compound interest on the same amount for the same time period equals

19800*1.12^3 - 19800 = 27817.57 - 19800 = $8017.57

Let the principal be P.

Hence, compound Interest = $$P(1+8\%)^2$$ - P = 1.1664P - P = 0.1664P = 1414.4

P = 1414.4/ 0.1664 = 8500

Hence, Amount = 8500 + 1414.4 = 9914.4

Let the population of village Q before 1 year be x.
There has been 15% increase in poopulation and population now stands at 16537
Hence, 1.15x = 16,537
x = 14380
Population of village P = 2.5*14380 = 35,950

Let the amount invested in Scheme A be '$x'.
Then, the amount invested in Scheme B = $(16000-x)
Amount earned from Scheme A = $$x\times1.1\times1.1 = 1.21x$$.
Interest earned from Scheme A = $1.21x - $x = $0.21x

Interest earned from Scheme B = $$\dfrac{(16000-x)\times12\times2}{100} = 3840 - 0.24x$$
Total interest = 0.21x+3840-0.24x = 3504
3840-0.03x = 3504
0.03x = 336
x = 11200.
Amount invested in Scheme B = 16000-11200 = $4800.

Let x be the principal amount invested by him.
Amount earned by him as SI = 900
900 = x*0.06*3
x = 5000
Compound interest he could have earned = 5000*(1.06)^3 - 5000 = 955.08
Amount earned more = 955.08 - 900 = 55.08

Let the principal be P.

Hence, CI = $$P(1+16\%)^2 - P$$ = 1.3456P - P = 0.3456P = 3041.28

P = 3041.28/0.3456 = 8800

Let the amount invested in the simple interest scheme be $x, then the amount invested in the compound interest scheme will be $(60k-x).

Total amount withdrawn at the end of two years from the simple interest scheme $$=x\times\left(1+\dfrac{58\times2}{100}\right)=2.16x$$

Total amount withdrawn at the end of two years from the compound interest scheme$$=\left(60k-x\right)\times1.2^2=\left(60k-x\right)\times1.44$$

At the end of two years, the total amount withdrawn from the simple interest investment was 75% of the amount withdrawn from the compound interest scheme.

$$2.16x=0.75\times\left(60k-x\right)\times1.44$$

$$2.16x=1.08\times\left(60k-x\right)$$

$$2x=60k-x$$

$$x=20k$$

Thus, the total amount withdrawn at the end of two years will be $$=2.16\left(20000\right)+\left(60000-20000\right)\times1.44$$

Total amount withdrawn =$100,800

Nilesh invested 40,000 for 4 years, 50,000 for 3 years, 60,000 for 2 years, and 70,000 for one year.
Hence, his total investment including time (in thousands) is 40*4+10*3+10*2+10*1 = 160+30+20+10 = 220

Suresh invested 85,000 for 2 years.
Hence, his total investment, including time (in thousands) is 85*2 = 170

Total earnings from the workshop are 1,95,000.
Nilesh's share is $$195,000 * \frac{220}{390} = 110,000$$

Given that, Simple interest = compound interest

so, $$\frac{PTR}{100}=Q[(1+\frac{r}{100})^{t}-1]$$

Let the amount invested in A be x. So, the amount invested in B will be (6100 - x)

$$\frac{(6100-x)4\times10}{100}=x[(1+\frac{10}{100})^{2}-1]$$

On solving, we get x = 4000.

Amount invested in scheme A = $x
Amount invested in scheme B = $3x
Given that,
$$\frac{3x \times 2 \times 9}{100} - x[(1 + \frac{20}{100})^2 - 1] = 1200$$
$$\frac{54x}{100} - \frac{11x}{25} = 1200$$
$$x = 12000$$

Hence, amount invested in scheme A = $12,000

Simple Interest for 1 year = $$\frac{1920}{2}$$ = Rs. 960
Compound Interest - Simple Interest = 2112 - 1920 = Rs. 912
Interest on Rs. 960 for 1 year = Rs. 912
$$\therefore$$ Rate $$= \frac{192 \times 10}{960 \times 1}$$
= 20% per annum
Principal = $$\frac{960 \times 100}{20 \times 1}$$
= Rs. 4800
New rate = 24% per annum
S.I. = $$\frac{4800 \times 2 \times 24}{100}$$
= Rs. 2304
Compound Interest = $$4800[(1 + \frac{24}{100})^2 - 1]$$
= $$4800[(1.24)^2 - 1]$$
= Rs. 2580.48
Total interest = Rs. 4884.48

it is given that the initial amount invested in scheme A is Rs P at 10%per Annum

S.I. = $$\frac{P x R x T}{100}$$

=$$\frac{P\times10\times2}{100}$$

Now the total amount after 2 years is = 1.2P

New Rate of interest = 12% per annum

Time = 5 years

S.I for next 5 years when new principal amount is 1.2 P = $$\frac{1.2P\times12\times5}{100}$$

= 0.72P

total amount after 5 years at 12% per annun = 1.72P

Given that 1.72P - 1.2P = 1300

P = 1300/0.52

=$2500

let the initial amount invested be P .

So after two years with 12%per annum ,P will amount to = P + 0.24P = 1.24P

Now 1.24P is invested again at SI 15% for two years and hence it will give an interest amount of = ( 0.3 x1.24P) = 0.372 P

It is given that 0.372 P - 0.24 P =264

So P = 2000

Here, the principal amount is Rs 13033

R = 13%

Time = 3 years

For 3 years difference between CI and SI = $$\frac{P(R)^2}{(100)^2}\times(3 + \frac{R}{100})$$

= $$\frac{13033(13)^2}{(100)^2}\times(3 + \frac{13}{100})$$

= Rs 689.41

Let the rate of interest be R % per annum

Principal Amount = Rs 58750

Total Amount in 4 years = Rs 79900

Simple interest paid = 79900 - 58750 = Rs 21150

SI = (P x R x T )/ 100

21150 = $$\frac{58750 \times R \times 4}{100}$$

R = 9 % per annum

Let the principal amount be P and given that R = 7%/ annum and T = 4 years, S.I for 4 years = $3584

S.I. = $$\frac{P \times R \times T}{100}$$ = $$\frac{P \times 7 \times 4}{100}$$ = 3584

P = $12,800

Now , to calculate the C.I rate of interest has been changed to 4%/ annum and time period to 2 years

C.I. = P$$(1 + \frac{4}{100})^2$$ - P

= 12800(1.0816-1)

= 12800 x 0.0816

= $1044.48

Let $$p$$ be the principal amount. $$r$$= 15% $$n = 3$$ years.

SI = pnr/100 = 3*0.15*p = 0.450
Total amount accrued due to SI = p + 0.45p =1.45p

Amount accrued due to CI in 3 years = $$p\left(1+\frac{r}{100}\right)^{n}$$ = $$p(1.15)^{3}$$ = 1.520875p

Difference in amounts = 1.520875p-1.45p = 0.07087p
0.07087p = 453.6 => p = $6400

Option B is the right answer.

We know that, $$CI = P\left(1+\frac{r}{100}\right)^{n}- P$$

$$5.616 = 16.250 \left(1+\frac{r}{100}\right)^{2} - 16.250$$

Solving, we get, $$\frac{r}{100}=16$$

$$r = 16$$

Option B is the right answer.

Let amount invested by Raman (P) =  $$100x$$ at 20% interest fro 2 years

Compound interest earned = $$P[(1+\frac{r}{100})^T-1]$$

= $$100x[(1+\frac{20}{100})^2-1]$$

= $$100x[(\frac{6}{5})^2-1]$$

= $$100x\times(\frac{36-25}{25})$$

= $$100x\times\frac{11}{25}= $$44x$$

Now, Raman lent  $$44x$$ to Shubh at 7.5% for 2 years

Total amount under simple interest = $$P+\frac{P\times r\times t}{100}$$

=> $$44x+\frac{44x\times7.5\times2}{100}=3036$$

=> $$44x(1+\frac{15}{100})=3036$$

=> $$44x\times\frac{115}{100}=3036$$

=> $$x=3036\times\frac{100}{115\times44}=60$$

$$\therefore$$ Value of P = $$100\times60= 6,000$$

=> Ans - (A)

Scheme A : Simple interest =[(P x 3 x 12)/100)] + P = (36P/100)+P
Scheme B : Compund Interest = (P+400)(1 + (10/100))$$^2$$ = (P+400)(11/10)$$^2$$ = (P+400)(121/100)
(36P/100) + P + 304 = (P+400)[(121/100)]
36P + 100P+ 30400 = 121P + 48400
15P = 18000
P =1200
Option E is the correct answer.

Let the amounts invested in each scheme be $$100x$$

Scheme A : rate = 12% , time = 6 years

Interest earned = $$\frac{P \times R \times T}{100}$$

= $$\frac{100x \times 12 \times 6}{100} = 72x$$

Scheme 2 : rate = 8% , time = 8 years

Interest earned = $$\frac{100x \times 8 \times 8}{100} = 64x$$

=> Difference in interest = $$72x - 64x = 1280$$

=> $$x = 160$$

$$\therefore$$ Amount invested in each scheme = 100 * 160 = $16,000

Let principal amount = $$P$$

=> $$C.I. = P [(1 + \frac{R}{100})^T - 1]$$

=> $$2,862 = P [(1 + \frac{12}{100})^2 - 1]$$

=> $$2,862 = P [(\frac{28}{25})^2 - 1]$$

=> $$2,862 = P (\frac{784 - 625}{625})$$

=> $$2,862 = P \times \frac{159}{625}$$

=> $$P = \frac{2,862 \times 625}{159}$$

=> $$P = 18 \times 625 = 11,250$$

Amount in scheme A at 12% S.I.

= $$\frac{P \times 12 \times 2}{100} + P$$

= $$\frac{24P}{100} + P = \frac{124P}{100}$$

Amount in scheme B at 9% S.I.

= $$\frac{P \times 9 \times 2}{100} + P$$

= $$\frac{18P}{100} + P = \frac{118P}{100}$$

Acc to ques,

=> $$\frac{124P}{100} + \frac{118P}{100} = 21,780$$

=> $$242P = 21,780 \times 100$$

=> $$P = \frac{21,78,000}{242}$$

=> $$P = 9,000$$

Interest earned when a sum of $$X$$­ was invested for 3 years in scheme A at 6% S.I.

=> $$\frac{X \times 6 \times 3}{100} = 482.40$$

=> $$X = \frac{482.40 \times 100}{18} = 2,680$$

=> Amount invested in scheme B = $$2,680 - 680 = 2,000$$

Interest received when $2,000 was invested for 2 years in scheme B at 10% C.I.

= $$2,000 [(1 + \frac{10}{100})^2 - 1]$$

= $$2,000 [(\frac{11}{10})^2 - 1] = 2,000 [(\frac{121}{100}) - 1]$$

= $$2,000 \times \frac{21}{100}$$

= $$20 \times 21$$ = $$420$$

Interest earned when a sum of $1,200/­ was invested for 4 years in scheme A at 20% S.I.

= $$\frac{1,200 \times 20 \times 4}{100}$$

= $$12 \times 80 = 960$$

Amount received when $x was invested for 2 years in scheme B at 10% C.I.

= $$x (1 + \frac{10}{100})^2$$

= $$x (\frac{11}{10})^2 = \frac{121x}{100}$$

Acc to ques,

=> $$\frac{121x}{100} - 960 = 1460$$

=> $$\frac{121x}{100} = 1460 + 960 = 2420$$

=> $$x = \frac{2420 \times 100}{121}$$

=> $$x = 20 \times 100$$ = $$2,000$$

Amount under simple interest = $$\frac{P \times R \times T}{100} + P$$

Given : $$\frac{P \times 10 \times 4}{100} + P = 22,960$$

=> $$\frac{2P}{5} + P = 22,960$$

=> $$\frac{7P}{5} = 22,960$$

=> $$P = \frac{22,960 \times 5}{7} = 16,400$$

Now, new amount invested = $$16,400 + 600 = 17,000$$

$$\therefore$$ S.I. = $$\frac{17,000 \times 10 \times 4}{100}$$

= $$170 \times 40$$ = $$6,800$$

C.I. compounded half yearly = $$P [(1 + \frac{R}{200})^2T - 1]$$

= $$5000 [(1 + \frac{4}{200})^2 - 1]$$

= $$5000 [(\frac{51}{50})^2 - 1]$$

= $$5000 (\frac{2601}{2500} - 1) = 5000 (\frac{101}{2500})$$

= $$101 \times 2 = 202$$

C.I. compounded yearly = $$P [(1 + \frac{R}{100})^T - 1]$$

= $$5000 [(1 + \frac{4}{100})^1 - 1]$$

= $$5000 [(1 + \frac{1}{25}) - 1]$$

= $$5000 (\frac{26}{25} - 1) = 5000 (\frac{1}{25})$$

= $$200$$

$$\therefore$$ Required difference = 202 - 200 = $2

Let the amount invested in each scheme = $$P$$

Scheme A : $$C.I. = P [(1 + \frac{R}{100})^T - 1]$$

= $$P [(1 + \frac{30}{100})^2 - 1]$$

= $$P [(\frac{13}{10})^2 - 1] = P [(\frac{169}{100}) - 1]$$

= $$\frac{69P}{100}$$

Scheme B : $$C.I. = P [(1 + \frac{20}{100})^2 - 1]$$

= $$P [(\frac{12}{10})^2 - 1] = P [(\frac{144}{100}) - 1]$$

= $$\frac{44P}{100}$$

Acc to ques,

=> $$(\frac{69P}{100}) - (\frac{44P}{100}) = 450$$

=> $$\frac{25P}{100} = 450$$

=> $$P = 450 \times 4$$ = $$1,800$$

Let the principal = $$P$$

$$S.I. = \frac{P \times R \times T}{100}$$

Acc. to ques,

=> $$(\frac{P \times 18 \times 4}{100}) - (\frac{P \times 22 \times 2}{100}) = 427$$

=> $$\frac{72P}{100} - \frac{44P}{100} = 427$$

=> $$28P = 427 \times 100 = 42700$$

=> $$P = \frac{42700}{28} = 1525$$

Now, S.I. after 4 years at 18%

= $$\frac{1525 \times 18 \times 4}{100}$$

= $$61 \times 18 = 1098$$

$$\therefore$$ Required amount = 1098 + 1525

= $2,623

Simple interest remains same in every year.

=> Simple interest for 1 year = Rs. 80

Difference of C.I. and S.I. for two years = Rs.(170 - 160) = Rs. 10

=> Interest for 1 year on Rs. 80 = Rs. 10

=> $$\frac{80 \times r \times 1}{100} = 10$$

=> $$r = \frac{100}{8}$$ = 12.5%

We know that, in compound interest, amount A :

=> $$A = P (1 + \frac{R}{100})^T$$

=> $$A = 16800 (1 + \frac{8}{100})^2 (1 + \frac{10}{100})^2$$

=> $$A = 16800 \times (\frac{27}{25})^2 \times (\frac{11}{10})^2$$

=> $$A = 23710$$

$$\therefore$$ C.I. = 23710 - 16800

= $6,910

Amount when interest is compounded quarterly

=> $$A = P (1 + \frac{\frac{R}{4}}{100})^{4T}$$

= $$1,25,000 (1 + \frac{\frac{8}{4}}{100})^{4 \times \frac{3}{4}}$$

= $$1,25,000 (1 + \frac{1}{50})^3$$

= $$1,25,000 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}$$

= $$51 \times 51 \times 51 = 1,32,651$$

$$\therefore$$ C.I. = $$1,32,651 - 1,25,000$$

= $$7,651$$

Let amount invested in company X = $$100x$$

=> S.I. = $$\frac{Principal \times Rate \times Time}{100}$$

= $$\frac{100x \times 15 \times 2}{100} = 30x$$

$$\therefore$$ Amount invested in company Y = $$100x + 30x$$

= $$130x$$

$$\therefore A = P (1 + \frac{R}{100})^T$$

=> $$81536 = 130x (1 + \frac{12}{100})^2$$

=> $$81536 = 130x \times \frac{28}{25} \times \frac{28}{25}$$

=> $$x = \frac{81536 \times 25 \times 25}{130 \times 28 \times 28}$$

=> $$x = 500$$

$$\therefore$$ Amount invested in company X in the beginning = $$100 \times 500$$

= $$50,000$$

Required Amount =

$$ A = P (1 + \frac{R}{100})^T$$

= $$18600 (1 + \frac{8}{100}) (1 + \frac{15}{100})$$

= $$18600 \times 1.08 \times 1.15$$

= $$23,101.20$$

$$\therefore$$ C.I. = $$(23,101.20 - 18,600)$$

= $$4,501.20$$

C.I. = $$P[{1 + \frac{R}{100}}^T - 1]$$

= $$9650 [(1 + \frac{6}{100})^3 - 1]$$

= $$9650 [(1.06)^3 - 1] = 9650 [1.1910 - 1]$$

= $$9650 \times 0.191$$

= $$ 1,843.15 \approx 1,843$$

Let the amount invested in schemes A, B and C respectively be $$2x , 3x$$ and $$4x$$

Amount after 1 year in scheme :

A = $$\frac{120}{100} \times 2x = 2.4x$$

B = $$\frac{116}{100} \times 3x = 3.48x$$

C = $$\frac{115}{100} \times 4x = 4.6x$$

=> Required ratio = 2.4 : 3.48 : 4.6

= 60 : 87 : 115

Principal = $24,000

S.I. = $$\frac{24000 \times 14 \times 8}{100}$$

= $26,880

$$\therefore$$ Total amount = 24000 + 26880

= $50,880

Total amount under compound interest

= $$15000 (1 + \frac{11}{100})^3$$

= $$15000 \times 1.11 \times 1.11 \times 1.11$$

= $$20514.465$$

$$\therefore$$ Compound Interest = 20514.465 - 15000

= $5,514.4

Let rate of interest be $$r$$ %

=> Simple interest = $$\frac{19800 \times r \times 3}{100} = 7128$$

=> $$r = 12$$%

Now, C.I. = $$19800 [(1 + \frac{12}{100})^3 - 1]$$

= $$19800 (1.12^3 - 1)$$

= $$19800 * 0.405 = 8019$$

= $8017.57

Interest is compounded half yearly

=> rate of interest = 10/2 = 5%

and time period = 1 * 2 = 2

=> Amount = $$15000 (1 + \frac{5}{100})^2$$

= $$15000 * \frac{21}{20} * \frac{21}{20}$$ = $16,537.50

Principal amount = $8,000

Time period = 3 years and rate of interest = 15% under compound interest.

=> $$C.I. = P [(1 + \frac{R}{100})^T - 1]$$

= $$8,000 [(1 + \frac{15}{100})^3 - 1]$$

= $$8,000 [(\frac{23}{20})^3 - 1] = 8,000 (\frac{12167 - 8000}{8000})$$

= $$12167 - 8000 = 4,167$$

Principal amount = $16,500

Simple interest earned = $5,940 and time = 3 years

Let rate of interest = $$R \%$$

=> $$S.I. = \frac{P \times R \times T}{100}$$

=> $$5940 = \frac{16500 \times R \times 3}{100}$$

=> $$R = \frac{5940}{165 \times 3}$$

=> $$R = 12 \%$$

Now, compound interest accrued on the same amount at the same rate in the same period = $$P [(1 + \frac{R}{100})^T - 1]$$

= $$16,500 [(1 + \frac{12}{100})^3 - 1]$$

= $$16,500 [(\frac{28}{25})^3 - 1] = 16,500 (\frac{21952 - 15625}{15625})$$

= $$16,500 \times \frac{6327}{15625} = 6681.31$$

Let the sum invested in scheme A = $$100x$$

Time = 14 years and rate = 8% under simple interest

=> $$S.I. = \frac{P \times R \times T}{100}$$

= $$\frac{100x \times 8 \times 14}{100} = 112x$$

=> Amount invested in Scheme B = $$100x + 112x = 212x$$

Time = 2 years and rate = 10% under compound interest.

$$C.I. = P [(1 + \frac{R}{100})^T - 1]$$

=> $$6678 = 212x [(1 + \frac{10}{100})^2 - 1]$$

=> $$6678 = 212x [(\frac{11}{10})^2 - 1] = 212x (\frac{21}{100})$$

=> $$0.21x = \frac{6678}{212} = 31.5$$

=> $$x = \frac{31.5}{0.21} = 150$$

$$\therefore$$ Sum invested in scheme A = $$100 \times 150 = 15,000$$

Amount = $3000

Time = 2 years and rate = 8% under compound interest

=> $$C.I. = P [(1 + \frac{R}{100})^T - 1]$$

= $$3000 [(1 + \frac{8}{100})^2 - 1]$$

= $$3000 [(\frac{27}{25})^2 - 1] = 3000 (\frac{729}{625} - 1)$$

= $$3000 \times \frac{104}{625} = 104 \times 4.8$$

= $$499.20$$

Amount deposited = $24,000

Rate = 14 % and time = 8 years under simple interest

=> $$S.I. = \frac{P \times R \times T}{100}$$

= $$\frac{24000 \times 14 \times 8}{100}$$

= $$240 \times 112 = 26,880$$

$$\therefore$$Total amount will ms Suchi gets at the end of 8 years

= $$24000 + 26880 = 50,880$$