Dear Reader,

Below you will find important formulas of simple and compound problems. Following that you will find solved questions. These questions will help you in preparation for all bank exams including IBPS, SBI, RRB and other banks.

**Question 1**

Veer invested an amount of Rs.9000 for 2 years at compound interest rate 15% per annum. How much amount will Veer obtain as interest?

a)Rs.2902.50 b)Rs.2900.50 c)Rs.2899.50 d)Rs.2899

a)Rs.2902.50 b)Rs.2900.50 c)Rs.2899.50 d)Rs.2899

**Answer :**a)Rs.2902.50

**Solution:**

When interest is compounded Annually, we have to use the following formula:

Amount = P x [1+ (R/100)]

^{n}where P = principal, R = rate of interest and n = time(years)
Here P = Rs.9000, R = 15%, n = 2 years.

Then, Amount= Rs.9000 x [1 + (15/100)]

^{2}= 9000 x (23/20)^{2}= 23805/2 = Rs.11902.5
The amount obtained by the way of interest in compound interest = Amount - principal = Rs.(11902.5 - 9000) = Rs.2902.50

Hence the required answer is Rs.2902.50

**Question 2**

Shagi deposits Rs.1500 each on 1st January and 1st July of a year at the rate of 8% compound interest calculated on half-yearly basis. How much amount he would have at the end of the year?

a)Rs.2150.50 b)Rs.3140.40 c)Rs.3182.40 d)Rs.2152.50

a)Rs.2150.50 b)Rs.3140.40 c)Rs.3182.40 d)Rs.2152.50

**Answer :**c) Rs.3182.40

**Solution:**

When interest is compounded Half-yearly: Amount = P x [1+ (R/2)/100 ]

The total amount for the investment on 1st january is:

Amount1 = Rs. 1500 x [1+ (8/2)/100]

= Rs. 1500 x [1 + (4/100)]

= Rs. 1500 x [26/25]

^{2n}The total amount for the investment on 1st january is:

Amount1 = Rs. 1500 x [1+ (8/2)/100]

^{2x1}= Rs. 1500 x [1 + (4/100)]

^{2}= Rs. 1500 x [26/25]

^{2}
The total amount for investment on 1st july is:

(Here n= 1/2 year since it starts from 1st july to end of the same year)

Amount2 = Rs. 1500 x [1+ (8/2)/100]

= Rs. 1500 x [1+ 4/100 ]

= Rs. 1500 x [26/25]

(Here n= 1/2 year since it starts from 1st july to end of the same year)

Amount2 = Rs. 1500 x [1+ (8/2)/100]

^{[2 x(1/2)]}= Rs. 1500 x [1+ 4/100 ]

= Rs. 1500 x [26/25]

The total amount at the end of the year = amount1 + amount2

= 1500 x [26/25]

= 1500 x [26/25] x [(26/25) + 1]

= 1500 x 26/25 x 51/25

= 3182.40

= 1500 x [26/25]

^{2}+ 1500 x [26/25]= 1500 x [26/25] x [(26/25) + 1]

= 1500 x 26/25 x 51/25

= 3182.40

Hence Rs.3182.40 is the required answer.

**Question 3**

What is the difference between the compound interests on Rs.10,000 for 2 years at 5% per annum compounded yearly and half-yearly?

a)Rs.6.00 b)Rs.6.25 c)Rs.6.50 d)Rs.6.75

a)Rs.6.00 b)Rs.6.25 c)Rs.6.50 d)Rs.6.75

**Answer :**b)Rs.6.25

**Solution:**

Here, P = Rs.10,000, n = 2 years, and R = 5%

Amount invested for compound interest (yearly) = Rs. 10000 x [1 + 5/100] = Rs.10000 x 21/20 = Rs.10,500

Amount invested for compound interest (Half-yearly) = Rs. 10000 x [1 +(5/2)/100]

^{2}= 10000 x (41/40)^{2}= Rs.10,506.25
Difference = Rs.(10506.25-10500) = Rs.6.25

**Question 4**

If a sum of Rs.8000 lended for 20% per annum at compound interest then the sum of the amount will be Rs.13824 in:

a) 2 years b) 1year c) 3years d) 4years

**Answer :**c)3years

Solution :

Let Principal = P, Rate = R% per annum, Time = n years.

When interest is compounded Annually total amount can be calculated by using the formula,

Total Amount = P(1 + R/100)

Total Amount = P(1 + R/100)

^{n}
Given that, P = Rs.8000, R = 20% per annum

We have to find the time period during which the amount will be Rs.13824

We have to find the time period during which the amount will be Rs.13824

i.e., Rs.13824 = 8000 x (1 + 20/100)^n

13824/8000 = (120/100)

(24/20)^3 = (12/10)

(12/10)^3 = (12/10)

Therefore, n = 3.

Hence the required time period is 3 years.

13824/8000 = (120/100)

^{n}(24/20)^3 = (12/10)

^{n}(12/10)^3 = (12/10)

^{n}Therefore, n = 3.

Hence the required time period is 3 years.

**Question 5**

Find the compound interest on a principal amount of Rs.5000 after 2 years, if the rate of interest for the 1st year is 2% and for the 2nd year is 4%.

a) Rs.304 b) Rs.314 c) Rs.324 d) Rs.334

**Answer :**a)Rs.304

Solution :

When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively.

Then, Amount (= Principal + Compound interest) = P(1 + R1/100)(1 + R2/100)(1 + R3/100).

Here R1 = 2% R2 = 4% and p = Rs.5000, we have to find CI (compound interest).

CI = 5000(1 + 2/100)(1 + 4/100) - 5000

= 5000 x (102/100)(104/100) - 5000

= 5000 x (51/50) x (52/50) - 5000

= 5000 x (51 x 52/2500) - 5000

= 5000 x (2652 / 2500) - 5000

= 5304 - 5000 = 304

Hence the required compound interest is Rs.304.

Then, Amount (= Principal + Compound interest) = P(1 + R1/100)(1 + R2/100)(1 + R3/100).

Here R1 = 2% R2 = 4% and p = Rs.5000, we have to find CI (compound interest).

CI = 5000(1 + 2/100)(1 + 4/100) - 5000

= 5000 x (102/100)(104/100) - 5000

= 5000 x (51/50) x (52/50) - 5000

= 5000 x (51 x 52/2500) - 5000

= 5000 x (2652 / 2500) - 5000

= 5304 - 5000 = 304

Hence the required compound interest is Rs.304.

**Question 6**

What sum(principal) will be amount to Rs.34536.39 at compound interest in 3 years, the rate of interest for 1st, 2nd and 3rd year being 5%, 6% and 7% respectively?

a) Rs.25576 b) Rs.29000 c) Rs.28012 d) Rs.24000

**Answer :**b)Rs.29000

Solution :

Let Rs.P be the required sum.

34536.39 = p(1 + 5/100)(1 + 6/100)(1 + 7/100)

= p (105/100) x (106/100) x (107/100)

p = 34536.39 x 100 x 100 x 100 / 105 x 106 x 107

p = Rs.29000

Hence the required amount is Rs.29000

34536.39 = p(1 + 5/100)(1 + 6/100)(1 + 7/100)

= p (105/100) x (106/100) x (107/100)

p = 34536.39 x 100 x 100 x 100 / 105 x 106 x 107

p = Rs.29000

Hence the required amount is Rs.29000

**Question 7**

What will be the amount if sum of Rs.10,00,000 is invested at compound interest for 3 years with rate of interest 11%, 12% and 13% respectively?

a) Rs.14,04,816 b) Rs.12,14,816 c) Rs.11,35,816 d) Rs.16,00,816

**Answer :**a)Rs.14,04,816

Solution:

Here, P = Rs.10,00,000 R1 = 11 R2 = 12 R3 = 13.

Therefore, Amount after 3 years

= p(1 + R1/100)(1 + R2/100)(1 + R3/100)

= 10,00,000 x(1 + 11/100)x(1 + 12/100)x(1 + 13/100)

= 10,00,000 x (111/100) x (112/100) x (113/100)

= 111 x 112 x 113

= 1404816

Hence the total amount after 3 years is Rs.14,04,816.

Therefore, Amount after 3 years

= p(1 + R1/100)(1 + R2/100)(1 + R3/100)

= 10,00,000 x(1 + 11/100)x(1 + 12/100)x(1 + 13/100)

= 10,00,000 x (111/100) x (112/100) x (113/100)

= 111 x 112 x 113

= 1404816

Hence the total amount after 3 years is Rs.14,04,816.

**Question 8**

A man lent out Rs.9600 at 9/2 % per annum for a year and 9 months. At the end of the duration, the amount he earned as S.I was:

a) Rs. 567 b) Rs.756 c) Rs.874 d) Rs.784

**Answer :**b) Rs.756

Solution :

Given that, principal = P = Rs.9600, R = 9/2 % and T = 1 year and 9 months = 1 + 9/12 year = 7/4 years.

Now, we have to find the S.I for 7/4 years.

S.I = PRT/100 = Rs. 9600 x 9/2 x 7/4 x 1/100 = 12 x 9 x 7 = 756

Hence, the required S.I amount is Rs.756

Given that, principal = P = Rs.9600, R = 9/2 % and T = 1 year and 9 months = 1 + 9/12 year = 7/4 years.

Now, we have to find the S.I for 7/4 years.

S.I = PRT/100 = Rs. 9600 x 9/2 x 7/4 x 1/100 = 12 x 9 x 7 = 756

Hence, the required S.I amount is Rs.756

**Question 9**

A man borrowed Rs.33600 at 25/4 % per annum on September 2012 and he paid back in May 2013. Find the amount he paid as S.I.

a) Rs.2075 b) Rs.2575 c) Rs.1575 d) Rs.1975

**Answer :**c) Rs.1575.

Solution :

Given that, principal = P = Rs. 33600 and R = 25/4 %.

Time duration = From September 2012 to May 2013 = 9 months = 9/12 year = 3/4 year.

S.I = PRT/100 = Rs. 33600 x 25/4 x 3/4 x 1/100 = 21 x 25 x 3 = Rs.1575.

Hence, the answer is Rs.1575.

Given that, principal = P = Rs. 33600 and R = 25/4 %.

Time duration = From September 2012 to May 2013 = 9 months = 9/12 year = 3/4 year.

S.I = PRT/100 = Rs. 33600 x 25/4 x 3/4 x 1/100 = 21 x 25 x 3 = Rs.1575.

Hence, the answer is Rs.1575.

**Question 10**

How much time will it take for a sum of Rs. 9000 to yield Rs. 1620 as S.I at 4 1/2 % per annum?

a) 1 year b) 2 years c) 3 years d) 4 years

**Answer :**d) 4 years.

Solution :

Given that, Principal = P = Rs. 9000, S.I = Rs. 1620 and rate R = 4 1/2 % = 9/2 %

We have to find T.

T = S.I x 100/PR = 1620 x 100/9000 x (9/2)

= 162x2 / 9x9 = 4

Therefore, required time is 4 years.

We have to find T.

T = S.I x 100/PR = 1620 x 100/9000 x (9/2)

= 162x2 / 9x9 = 4

Therefore, required time is 4 years.

**Question 11**

At what rate of simple interest, a sum of Rs.8540 amounts to Rs.9710 in 3 years?

a) 3 1/2 % b) 2 1/4 % c) 3 3/4 % d) 4 1/2 %

**Answer :**d) 4 1/2 %

Solution :

Given that, principal = P = Rs. 8540

Final amount = Rs. 9710

Then, interest = S.I = Rs.9710 – Rs.8540 = Rs. 1170.

Time = T = 3 years.

Required rate = R = S.I x 100 / PxT = 1170 x 100 / 8540 x 3 = 1950/427

= 4.566 = 4.5 (approximately) = 9/2 = 4 1/2 %

Hence, the answer is 4 1/2 %.

Final amount = Rs. 9710

Then, interest = S.I = Rs.9710 – Rs.8540 = Rs. 1170.

Time = T = 3 years.

Required rate = R = S.I x 100 / PxT = 1170 x 100 / 8540 x 3 = 1950/427

= 4.566 = 4.5 (approximately) = 9/2 = 4 1/2 %

Hence, the answer is 4 1/2 %.