Hello Aspirants. Welcome to Online Maths in AffairsCloud.com. Here we are creating question sample in **Compound Interest**, which is common for all the IBPS,SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in exams !!

**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)^{2 }= 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 = 2^{2}

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%

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