Simple Interest/Compound Interest Questions for IBPS PO, SBI PO, NIACL, NICL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams

- Find the compound interest on Rs36,000 at a rate in which Rs216 becomes Rs343 in 3years and the time is 2years?

A) Rs12000

B) Rs12500

C) Rs13000

D) Rs13500

E) Rs14200

View Answer

**Option C**

Solution:

first we find the rate

3âˆš216 : 3âˆš343

6 : 7

(+1)

1/6*100= 16(2/3) %

Now R = 16(2/3) %=1/6

6â€¦â€¦â€¦â€¦â€¦.7

6â€¦â€¦â€¦â€¦â€¦.7

36 49

Â (13) =13000

- If a principal becomes triple in 3years on C.I. then find in how many years it will be 27 fold?

A) 39years

B) 9years

C) 18years

D) 27years

E) 10years

View Answer

**Option B**

Solution:

in C.I principal increase like

1â€¦.3….9â€¦..27

…3…..3….3

= 9years

- If a principal becomes amount of rs14500 at 14(2/7)% rate of interest in 3years at simple interest. Find the S.I on principal?

A) Rs4250

B) Rs4300

C) Rs4400

D) Rs4350

E) Rs4270

View Answer

**Option D**

Solution:

R = 14(2/7)% = 1/7

S.I remains same in all years soâ€¦

(P)7 + 1+1+1= 10(A)

10-7 = 3S.I

10 =14500

1=1450

3=4350

- If the difference between C.I and S.I is rs256 at 20% rate of interest in 3years. Find the amount on C.I?

A) Rs4320

B) Rs2500

C) Rs3456

D) Rs3200

E) Rs3478

View Answer

**Option C**

Solution:

S.I in 3years = 20*3= 60%

C.I in 3years = 5â€¦â€¦â€¦â€¦â€¦6

. Â Â Â Â Â Â Â Â Â Â Â 5â€¦â€¦â€¦â€¦…6

. Â Â Â Â Â Â Â Â Â Â Â 5â€¦â€¦â€¦..â€¦.6

. Â Â Â Â Â Â Â Â Â Â 125 Â Â Â Â Â Â 216

. Â Â Â Â Â Â Â Â Â Â Â Â Â Â (91)

91/125*100= 72.8

Difference = 72.8 â€“ 60= 12.8

12.8%= 256

100% = 2000

Now P = 2000

Means in C.I â€¦. 125 =2000

1 = 16

216 = 3456

- A sum becomes 8000 in 3years and 10000 in 6years at C.I. Find the sum ?

A) Rs6400

B) Rs6500

C) Rs6000

D) Rs7000

E) Rs7200

View Answer

**Option A**

Solution:

x : y = y : z

x : 8000 = 8000 : 10000

x = 6400

- Find the C.I on rs9000 at 15% rate of interest for 3years?

A) Rs4645.87

B) Rs4680.87

C) Rs4685.87

D) Rs4687.87

E) Rs4356.77

View Answer

**Option D**

Solution:

15% of 9000 = 1350

1350â€¦â€¦â€¦â€¦â€¦1350â€¦â€¦â€¦â€¦â€¦â€¦1350

. Â Â Â Â Â Â Â Â Â Â Â 202.5â€¦â€¦â€¦â€¦â€¦.202.5

. Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 202.5

. Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 30.37

= 4687.87

- Find the compound interest on 18000 at 20% rate of interest in 1(1/2) years, if compounded half yearly ?

A) Rs5958

B) Rs4916

C) Rs5780

D) Rs3500

E) Rs6724

View Answer

**Option A**

Solution:

in half yearly we make rate half and time double.

So R =20/2 = 10%

T = 3/2 * 2= 3years

So 10% of 18000 = 1800

1800â€¦â€¦â€¦â€¦â€¦â€¦1800â€¦â€¦â€¦â€¦â€¦â€¦â€¦.1800

. Â Â Â Â Â Â Â Â Â Â Â Â Â 180â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦180

. Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 180

. Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 18

= 5400 + 540 + 18 = 5958

- Find the difference between S.I and C.I on Rs 5000 if rate of interest for first year is 10% and 2nd year is 15% and 3rd year is 20%?

A) Rs300

B) Rs320

C) Rs330

D) Rs340

E) Rs360

View Answer

**Option D**

Solution:

S.I = 10+15+20 = 45%

C.I .. Â Â 10…….11

. Â Â Â Â Â Â 20……23

. Â Â Â Â Â Â Â 5…….6

. Â Â Â Â 1000…..1518

. Â Â Â Â Â Â Â (518)

= 518/1000*100 = 51.8 %

= difference = 51.8 â€“ 45 = 6.8%

= 6.8% of 5000 = 340

- If the principal become 6 fold on S.I in 10 years then find in how many years it will be 12 fold?

A) 24years

B) 22years

C) 12years

D) 20years

E) 25years

View Answer

**Option B**

Solution:

P â€¦â€¦â€¦â€¦â€¦â€¦â€¦..6P

6P â€“ P = 5P interest

5P = 10years

P = 2years

11 P = 22years

- If the compound interest on a sum at 25% rate of interest is Rs900 then find theÂ S.I of 3years at same rate?

A) Rs1000

B) Rs1100

C) Rs1300

D) Rs1200

E) Rs1500

View Answer

**Option D**

Solution:

S.I = 25*3 = 75%

C.I = 25% =1/4

4â€¦â€¦â€¦â€¦..5

4â€¦â€¦â€¦â€¦â€¦5

16…………25

25 -16 = 9

9 = 900

16 = 1600 = principal

So 75% of 1600 =1200

question no 4 ,

plz explain the method

C.I in 3years = 5â€¦â€¦â€¦â€¦â€¦6

. 5â€¦â€¦â€¦â€¦â€¦6

. 5â€¦â€¦â€¦..â€¦.6

. 125 216

. (91)

skip this method

use basic method

si 20*3 = 60

ci 20% for 3 year when u calculated value comes = 72.8

differ 72.8-60=12.8

12.8 is value of 256

256/12.8*100=2000

this is P =2000

now u calculate Ci on 2000 for 20% in 3 year u find answer

no

ryt hai qs

solution bhi ryt hai

concept hota hai one less hota hai

solution diya hai achche se pdo aa jayga

try this

R=(n-1)*100/T where n=no of times

find out R frm here then

T=(n-1)*100/R put value of R frm abv

ðŸ™‚

bro…here principal means amount …;)

are dese quiz enough for maths?

yes

bt u have to practice more

1 to 12

mtlb difference 12-1 =11 so here 11 comes

1 to 6

difference 6-1=5

10/5=2

11*2= 22 ans everything is correct here question and solution i hope u understand

always wlc bro

CI For 3 year there are 2-3 approaches:

1st =3r.3r^2.r^3

2nd=effective method

3rd=percentile

you can use any approach ,

By Effective method = 20+20(20*20/100)=44

again consider 44 as first value and 20 is another =44+22(44*22/100)=72.8

p =100

=>si =60 (3yrs)

=>ci =72.8 (3yrs)

=>12.8 = 256 =>100 =2000

now 72.8 =1456

=>totl =2000+1456 =3456

thanx

SI = 2250

CI = 2590

Diff = 340

no prob its easy

See –

For Si =10+15+20=45%

For CI=

use effective method For first 2 year = 10+15+(10*15/100)=25+1.50=26.50 ,

again use the same, For remaining =

26.5+20+(26.5*20/100)=46.5+5.30=51.80

Diff of CI and SI==51.80-45=6.80%

6.80% of 5000=340

all done but i have a doubt on Q 10 i think this question is incompleteany mod plz help to understand this question

Re-check Qo no 10.

Info is not properly mentioned /

all done bt qus 10-time is not given for ci

same problem

Q8> however i gt the answer correct frm diffrent method this method seems new to me..pls explain @Shubhra ma’am (hw 10..11

20…23

5…6)

C.I .. 10â€¦â€¦.11

. 20â€¦â€¦23

. 5â€¦â€¦.6

. 1000â€¦..1518

. (518)

= 518/1000*100 = 51.8 %

= difference = 51.8 â€“ 45 = 6.8%

= 6.8% of 5000 = 340

pls explain 5 th q.. how to cal

tq mam

ty ðŸ™‚

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