Quant Quiz On Compound Interest Day 17 Bag

Quant Quiz On Compound Interest Day 17 Bag

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• The difference between compound interest compounded every 6 months and simple interest after 2 years is 248.10. The rate of interest is 10 percent. Find the sum
a) 12000
b) 14000
c) 16000
d) 18000
e) None of these

Answer – c) 16000
Explanation :
P*(1+5/100)^4 – P – P*(10/100)*2 = 248.10
P = 16000
• A person earns an interest of 240 on investing certain amount at Simple interest for 2 years at 5 percent amount. If the rate of interest is compounded annually then how much more interest will be gain by the person at same rate of interest and on the same sum.
a) 6
b) 8
c) 12
d) 10
e) None of these

Explanation :
240 = P*(5/100)*2, P = 2400
CI = 2400(1+5/100)^2 – 2400 = 246
So, 246 – 240 = 6
• Find the least number of years in which the sum put at 25% rate of interest will be more than doubled.
a) 2 years
b) 3 years
c) 4 years
d) 5 years
e) None of these

Answer – c) 4 years
Explanation :
Amount >= P*(1+25/100)^n
Amount = p*(5/4) ^n
For n = 4, (625/256) which is greater than 2.
• A sum of rupees 4420 is to be divided between rakesh and prakash in such a way that after 5 years and 7 years respectively the amount they get is equal. The rate of interest is 10 percent. Find the share of rakesh and prakash
a) 2000, 2420
b) 2420, 2000
c) 2480, 2420
d) 2210, 2210
e) None of these

Answer – b) 2420, 2000
Explanation :
Let the share of rakesh and prakash be R and P
R*(1+10/100)^ 5 = (4420 – R)*(1+10/100)^ 7
We get R = 2420, so P = 2000
• The simple interest on a certain sum of money for 4 years at 15 percent per annum is 600. Find the compound interest in the same sum at 10 percent interest for 2 years
a) 220
b) 200
c) 210
d) 120
e) None of these

Answer – c) 210
Explanation :
600 = p*4*(15/100), P = 1000
CI = 1000(1+10/100)^ 2 – 1000 = 210
• Find the effective annual rate of 10 percent per annum compounded half- yearly-
a) 10.5
b) 10.25
c) 11.25
d) 11.50
e) None of these

Answer – b) 10.25
Explanation :
Take principal as 100 and then calculate,
A = 100*(1+5/100)^ 2
A = 110.25
So effective rate is 10.25
• A sum of rupees 3200 is compounded annually at the rate of 25 paise per rupee per annum. Find the compound interest payable after 2 years.
a) 1200
b) 1600
c) 1800
d) 2000
e) None of these

Answer – c) 1800
Explanation :
Rate of interest is 25 paise per rupee per annum.
So for 100 rupees it is 2500 paise i.e. 25 percent
Now, CI = 3200(1+25/100)^ 2  – 3200 = 1800
• A sum of 3000 becomes 3600 in 3 years at 15 percent per annum. What will be the sum at the same rate after 9 years.
a) 5124
b) 5184
c) 5186
d) 5192
e) None of these

Answer – b) 5184
Explanation :
3600 = 3000*(1+15/100)^ 3
(1+15/100)^ 3 = 6/5
Amount = 3000*[(1+15/100)^ 3]^ 3
Amount = 3000*(6/5)^ 3 = 5184
• Priya saves an amount of 500 every year and then lent that amount at an interest of 10 percent compounded annually. Find the amount after 3 years.
a) 1820.5
b) 1840.5
c) 1920.5
d) 1940.5
e) None of these

Answer – a) 1820.5
Explanation :
Total amount  = 500*(1+10/100)^ 3 + 500*(1+10/100)^ 2 + 500*(1+10/100)
= 1820.5
• A man borrows 10000 rupees at 20 % compound interest for 3 years. If every year he pays 2000 rupees as repayment. How much amount is still left to be paid by the man?
a) 5000
b) 7000
c) 9000
d) 10000
e) None of these

Answer – d) 10000
Explanation :
Amount to be paid at the end of three years = 10000*(1+20/100)^ 3 = 17280
Amount paid as instalment by the man = 2000*(1+20/100)^ 2 + 2000*(1+20/100) + 2000 = 7280
So remaining amount = 10000

• On a certain sum of money, after 2 years the simple interest and compound interest obtained are Rs 800 and Rs 960 respectively. What is the sum of money invested?
A) Rs 1420
B) Rs 1325
C) Rs 1000
D) Rs 1405
E) Rs 1375

C) Rs 1000
Explanation:

Diff = 960-800 = 160
r = 2*Diff*100/SI
So r = 2*160*100/800 = 40%
Now 160 = Pr2/1002
• Rs 6000 becomes Rs 7200 in 3 years at a certain rate of compound interest. What will be the amount received after 9 years?
A) Rs 11,498
B) Rs 10,352
C) Rs 9,368
D) Rs 10,368
E) None of these

D) Rs 10,368
Explanation:
6000[1 + r/100]3 = 7200
So [1 + r/100]3 = 6/5
So 6000[1 + r/100]9 = 6000*(6/5)*(6/5)*(6/5)
• A man borrows Rs 4000 at 8% compound interest for 3 years. At the end of each year he paid Rs 500. How much amount should he pay at the end of 3rd year to clear the debt?
A) Rs 4254.5
B) Rs 3465.2
C) Rs 3485.2
D) Rs 4345.4
E) Rs 3915.6

E) Rs 3915.6
Explanation:

Amount after 1 yr = 4000[1 + 8/100] = 4320
Paid 500, so P = 4320 – 500 = 3820
Amount after 2nd yr = 3820[1 + 8/100] = 4125.6
So P= 4125.6-500 = 3625.6
Amount after 3rd yr = 3625.6[1 + 8/100] = 3915.6
• A sum of money is lent for 2 years at 20% p.a. compound interest. It yields Rs 482 more when compounded semi-annually than compounded annually. What is the sum lent?
A) Rs 25,600
B) Rs 20,000
C) Rs 26,040
D) Rs 40,500
E) None of these

B) Rs 20,000
Explanation:

P[1 + (r/2)/100]4 – P[1 + r/100]2 = 482
P[1 + 10/100]4 – P[1 + 20/100]2 = 482
Solve, P = 20,000
• The compound interest obtained after 1st and 2nd year is Rs 160 and Rs 172.8 respectively on a certain sum of money invested for 2 years. What is the rate of interest?
A) 10%
B) 8%
C) 8.5%
D) 9%
E) 9.2%

B) 8%
Explanation:

Difference in interest for both yrs = 172.8 – 160 = 12.8
So (r/100)*160 = 12.8
• A sum of money becomes Rs 35,280 after 2 years and Rs 37,044 after 3 years when lent on compound interest. Find the principal amount.
A) Rs 32,000
B) Rs 28,000
C) Rs 31,500
D) Rs 32,500
E) None of these

A) Rs 32,000
Explanation:

P[1 + r/100]3 = 37,044, and P[1 + r/100]2 = 35,280
Divide both equations, [1 + r/100] = 37044/35280 = 21/20
So P[21/20]2 = 35280
• The difference between compound interest earned after 3 years at 5% p.a. and simple interest earned after 4 years at 4% p.a. is Rs 76. Find the principal amount.
A) Rs 32,000
B) Rs 28,000
C) Rs 31,500
D) Rs 32,500
E) None of these

A) Rs 32,000
Explanation:

[P[1 + 5/100]3 – P] – P*4*4/100 = 76
P [9261/8000 – 1 – 16/100] = 76
• A sum of money is lent at simple interest and compound interest. The ratio between the difference of compound interest and simple interest of 3 years and 2 years is 35 : 11. What is the rate of interest per annum?
A) 20 3/4%
B) 17 2/5%
C) 18 2/11%
D) 22 1/5%
E) 24 5/6%

C) 18 2/11%
Explanation:

Difference in 3 yrs = Pr2(300+r)/1003
Difference in 2 yrs = Pr2/1002
So Pr2(300+r)/1003 / Pr2/1002 = 35/11 (300+r)/100 = 35/11
• A sum of money borrowed at 5% compound interest is to paid in two annual installments of Rs 882 each. What is the sum borrowed?
A) Rs 1650
B) Rs 2340
C) Rs 2630
D) Rs 1640
E) Rs 2640

D) Rs 1640
Explanation:

P = 882/[1 + 5/100] + 882/[1 + 5/100]2
• Rs 3903 is to be divided in a way that A’s share at the end of 7 years is equal to the B’s share at the end of 9 years. If the rate of interest is 4% compounded annually, find A’s share.
A) Rs 2475
B) Rs 1875
C) Rs 2175
D) Rs 1935
E) Rs 2028

B) Rs 1875
Explanation:

A’s share = (1 + 4/100)7
B’s share = (1 + 4/100)9
Divide both, B/A = (1 + 4/100)2 = 676/625
So A’s share = 625/(676+625) * 3903

• The difference between the total simple interest and the total compound interest compounded annually at the same rate of interest on a sum of money at the end of two years is Rs. 350. What is definitely the rate of interest per cent per annum?
A.9,300
B.7600
C.12000
E.None of these

Explanation :
Difference = Pr2/(100)2
= (350×100×100)/(P×r2)
P is not given
• Aswin invested an amount of Rs.9000 in a fixed deposit scheme for 2 years at CI rate 6% pa. How much amount will Aswin get on maturity of the fixed amount ?
A.Rs.11,230
B.Rs.10,250
C.Rs.10,112
E.None of these

Explanation :
Amount = 9000*106/100*106/100
= 9000*53/50*53/50
= 10,112
• A sum of money invested for 7years in Scheme 1 which offers SI at a rate of 8% pa. The amount received from Scheme 1 after 7 years invested for 2 years in Scheme 2 which offers CI rate of 10% pa. If the interest received from Scheme B was Rs.1638. What was the sum invested in Scheme 1 ?
A.Rs.7500
B.Rs.5000
C.Rs.8200
D.Rs.9000
E.None of these

Explanation :
SI =>Amount = x*8*7/100 + x = 56x+100x/100 = 156x/100 = 39x/25
CI=> 39x/25[(1+10/100)2 – 1] 1638 = 39x/25[121/100 – 1] = 39x/100[21/100] X = 1638*100*25/21*39 = 5000
• Rs.5200 was partly invested in Scheme A at 10% pa CI for 2 years and Partly invested in Scheme B at 10% pa SI for 4 years. Both the schemes earn equal interests. How much was invested in Scheme A ?
A.Rs.1790
B.Rs.2200
C.Rs.3410
D.Rs.2670
E.None of these

Explanation :
Amount invested in Scheme B = X
Amount invested in Scheme A = 5200 – x
X*10*4/100 = (5200-x)*21/100……………………[(1-10/100)2-1] = 21/100
40x/100 = (5200-x)*21/100
2x/5 = (5200-x)*21/100
200x = 5200*21*5 – x*5*21
200x = 546000 – 105x
305x = 546000
X = 1790
Scheme A = 5200 – 1790 = 3410
• The CI on Rs.7000 for 3 years at 5% for first year, 7% for second year, 10% for the third year will be
A.Rs.1800
B.Rs.1530
C.Rs.1651
D.Rs.2050
E.None of these

Explanation :
A = 7000*105/100*107/100*110/100
= 7000*1.05*1.07*1.1
= 8650.95 = 8651
CI = 8651-7000 = 1651
• Poorni and Priyanka have to clear their respective loans by paying 2 equal annual instalments of Rs.20000 each. Poorni pays at 10% pa of SI and Priyanka pays at 10% CI pa. What is the difference in their payments ?
A.200
B.300
C.400
D.500
E.None of these

Explanation :
D =[(20,000 *110/100*110/100) – 20,000]  – 20,000 *10*2/100
=[ 24200-20000]-4000
=4200 – 4000
D = 200
• A sum of Rs.3,50,500 is to be paid back in 2 equal annual instalments. How much is each instalment, if the rate of interest charged 4% per annum compound annually ?
A.Rs.1,90,450
B.Rs.1,65,876
C.Rs.1,76,545
D.Rs.1,85,834
E.None of these

Explanation :
Total value of all the 3 instalments
[X*100/104] + [X*100*100/104]  = 3,50,500
X*25/26 + x*625/676 = 3,50,500
X*25/26[1+25/26] = 3,50,500
X*25/26[51/26] = 3,50,500
X = 3,50,500*676/25*51 = 1,85,833.72 = 1,85,834
• At CI, if a certain sum of money is doubled in 4 years, then the amount will be four fold in how many years ?
A.7years
B.4years
C.10years
D.8years
E.None of these

Explanation : 2n = 2*4 = 8 years
• X has lent some money to A at 5% pa and B at 8% pa. At the end of the year he has gain the overall interest at 7% pa.In what ratio has he lent the money to A and B ?
A.2:1
B.1:2
C.3:1
D.1:4
E.None of these

Explanation :
5………………………….8
……………7…………….
1………………………….2
Ratio A : B = 1:2
• The difference between interest received by X and Y is Rs.315 on Rs.3500 for 3 years. What is the difference in rate of interest ?
A.1.5%
B.2%
C.3%
D.2.7%
E.None of these

Explanation :
3500*3/100(R1-R2) = 315
R1-R2 = 315*100/10500 = 3%

• What is the difference between the CI and SI on Rs.6500 at the rate of 4 pcpa in 2 yrs ?
A)9.4
B)10.4
C)15.4
D)14.5

B)10.4
Explanation :
SI = 6500*4*2/100 = 520
CI = 6500*(104/100)*(104/100) = 7030.4 = 530.4
530.4 – 520 = 10.4
• The CI on a certain sum at 10% pa for 2yrs is Rs.6548. What is SI on the sum of money at 7%pa for 4 yrs(approx) ?
A)8700
B)8731
C)8370
D)8470

B)8731
Explanation :
CI = p [ {1 + (10/100)}2 – 1] 6548 = p [(110/100) 2 – 1] 6548 =p (121 – 100)/100
P = 654800/21 =31,181
SI = 31181*7*4/100 = 8731
• A sum of money amount to Rs.900 in 3 yrs and Rs.1080 in 4yrs at CI. What is the rate of CI pa ?
A)10%
B)15%
C)20%
D)18%

C)20%
Explanation :
(1080-900/900)*100
180*100/900 = 20%
• Find the CI on Rs.4500 at 4% pa for 2 yrs compounded half yearly(approx) ?
A)963
B)369
C)639
D)360

C)639
Explanation :
CI = P [(1+(r/100)n – 1] = 4500[(1+2/100)4 – 1
= 4500[ (102*102*102*102/100*100*100*100) – 1] = 4500[1.082 – 1] = 4500*0.082 = 369
• In how many years will a sum of Rs.15625 at 8% pa compounded semi annually become Rs.17,576?
A)2(1/3)yrs
B)2yrs
C)1(1/2)yrs
D)1yr

C)1(1/2)yrs
Explanation :
15625*(1+(4/100))2n = 17576
15625*(104/100) 2n = 17576
(26/25) 2n = 17576/15625 = (26/27)3
2n = 3
N =3/2 = 1(1/2) yrs
• If 40% increase in an amount in 4 years at SI. What will be the CI of Rs.10,000 after 2 yrs at the same rate ?
A)1000
B)1100
C)2000
D)2100

D)2100
Explanation :
P = 100, SI = 40,  T = 4
R = 100*40 / 100*4 = 10%pa
CI = 10,000(110/100)= 10000*11*11/100 = 12100
12100 – 10000 = 2100
• A sum of money is borrowed and paid in 2 annual instalments of Rs.642 each allowing 4% CI. The sum borrowed was
A)1120
B)1211
C)1112
D)1121

B)1211
Explanation :
P = [642/(104/100)] + [642/(104/100)2)
= 642*(25/26) + 642 (25/26)2
= 617.3 + 593.6 = 1210.9 = 1211
• The least number of complete years in which a sum of money put out at 10% CI will be more than  doubled is
A)8yrs
B)6yrs
C)4yrs
D)7yrs

A)8yrs
Explanation :
P(110/100)n  > 2P
(11/10) n > 2P
1.1*1.1*1.1*1.1*1.1*1.1*1.1*1.1 = 2.14 > 2
N = 8
• A sum of money doubles itself at CI in 10yrs. In how many yrs will it become 4 times ?
A)10yrs
B)20yrs
C)15yrs
D)22yrs

B)20yrs
Explanation :
P(1+(R/100))10 = 2P
1+(R/100)10 = 2
(1+(R/100))n = 4 = 22
2*10 = 20yrs
• At what rate of CI pa will a sum of Rs.1000 becomes Rs.1040.4 in 2yrs ?
A)2%
B)1%
C)1.5%
D)3%

A)2%
Explanation :
1000*(1+(R/100)) 2 = 1040.4
(1+(R/100)) 2 = 1040.4/1000
(1+(R/100)) 2 = 10404/10000 = (102/100) 2
R = 2%