Practice MCQ Questions and Answer on Interest

91.

A man takes a loan of some amount at some rate of simple interest. After three years, the loan amount is doubled and rate of interest is decreased by 2%. After 5 years, if the total interest paid on the whole is Rs. 13,600, which is equal to the same when the first amount was taken for $11 \frac{1}{3}$ ÃÂÃÂ years, then the loan taken initially is:

(A) Rs. 13,600
(B) Rs. 12,500
(C) Rs. 10,000
(D) Rs. 12,000

Solution: $\begin{aligned} \frac{x \times r \times 3}{100} + \frac{2 x \times (r - 2)}{100} \times 5 = \frac{x \times r}{100} \times \frac{34}{3} \\ \frac{2 x \times (r - 2) \times 5}{100} = \frac{x \times r}{100} \times \frac{34}{3} - \frac{x \times r \times 3}{100} \\ \frac{2 x \times (r - 2) \times 5}{100} = \frac{(34 r - 9 r) \times x}{300} \\ 6 (r - 2) \times 5 = 25 r \\ \frac{36 x}{100} + \frac{2 x \times 10}{100} \times 5 = 13600 \\ \frac{136 x}{100} = 13600 \\ x = Rs . 10000 \end{aligned}$

92.

Rs. 2,500, when invested for 8 years at a given rate of simple interest per year, amounted to Rs. 3,725 on maturity. What was the rate of simple interest that was paid per annum?

(A) 6%
(B) 6.125%
(C) 6.25%
(D) 5.875%

Solution: $\begin{aligned} Simple interest = 3725 - 2500 = 1225 \\ SI = \frac{P \times r \times t}{100} \\ 1225 = \frac{2500 \times r \times 8}{100} \\ r = \frac{49}{8} \\ r = 6.125 % \end{aligned}$

93.

Sum Rs. 20000 and Rs. 40000 are given on simple interest at the rate of 10% and 15% per annum respectively for three years. What will be the total simple interest?

(A) Rs. 36000
(B) Rs. 32000
(C) Rs. 24000
(D) Rs. 28000

Solution: Simple interest on 20000 for 3 years at 10% rate $SI = \frac{20000 \times 3 \times 10}{100} = 6000$ Simple interest on 40000 for 3 years at 15% rate $SI = \frac{40000 \times 3 \times 15}{100} = 18000$ Total interest = 6000 + 18000 = 24000

94.

In what time will Rs. 3300 becomes Rs. 3399 at 6% per annum interest compounded half-yearly?

(A) 6 months
(B) 1 year
(C) $1 \frac{1}{2}$ years
(D) 3 months

Solution: $\begin{aligned} P = R s . 3300 \\ A = R s . 3399 \\ R = 6 % per annum \\ Let the time be n years . \\ Compound interest is taken half - yearly . \\ A = P \times {[1 + (\frac{R}{2} \times 100)]}^{2 n} \\ 3399 = 3300 {(1 + \frac{3}{100})}^{2 n} \\ {(1.03)}^{2 n} = \frac{3399}{3300} \\ {(1.03)}^{2 n} = {(1.03)}^{1} \\ T h u s, 2 n = 1 y e a r \\ n = \frac{1}{2} year = 6 months \end{aligned}$

95.

A sum becomes its double in 10 years. Find the annual rate of simple interest.

(A) 8%
(B) 5%
(C) 10%
(D) 20%

Solution: Let Sum be Rs. x and Let S.I. = Rs. x $\begin{aligned} Time = 10 years \\ ∴ Rate = \frac{S .I . \times 100}{Principal \times Time} \\ = \frac{x \times 100}{x \times 10} \\ = 10 % per annum . \end{aligned}$

96.

Ramakant invested amounts in two different schemes A and B for five years in the ratio of 5 : 4 respectively. Scheme A offers 8% simple interest and bonus equal to 20% of the amount of interest earned in 5 years on maturity. Scheme B offers 9% simple interest. If the amount invested in scheme A was Rs. 20000, what was the total amount received on maturity from both the schemes?

(A) Rs. 50800
(B) Rs. 51200
(C) Rs. 52800
(D) Rs. 58200

Solution: Let the amounts invested in schemes A and B be Rs. 5x and 4x respectively. then, 5x = 20000 ⇒ x = 4000 ∴ Amount invested in scheme B = Rs. 16000 Total interest received on maturity $= Rs . [120 % of (\frac{20000 \times 8 \times 5}{100}) + (\frac{16000 \times 9 \times 5}{100})]$ $\begin{aligned} = Rs . (120 % of 8000 + 7200) \\ = Rs . (9600 + 7200) \\ = Rs . 16800 \\ ∴ Total amount \\ = Rs . (20000 + 16000 + 16800) \\ = Rs . 52800 \end{aligned}$

97.

A certain sum of money amount to Rs 2200 at 5% interest Rs 2320 at 8% interest in the same period of time. The period of time is = ?

(A) 2 years
(B) 4 years
(C) 5 years
(D) 3 years

Solution: $\begin{aligned} P + SI = \frac{P \times R \times T}{100} + P \\ \Rightarrow 2200 = \frac{P \times 5 \times T}{100} + P \end{aligned}$ ⇒ 2200 × 100 = 5PT + 100P ......... (i) $\Rightarrow 2320 = \frac{P \times 8 \times T}{100} + P$ ⇒ 2320 × 100 = 8PT + 100P ⇒ 2320 × 100 = 3PT + 5PT + 100P ............(ii) Value of equation (i) put equation (ii) ⇒ 2320 × 100 = 3PT + 2200 × 100 ⇒ 3PT = 120 × 100 ⇒ PT = 4000 Value of PT in equation (i) ⇒ 2200 × 100 = 5 × 4000 + 100P ⇒ 220000 - 20000 = 100P $\begin{aligned} \Rightarrow P = \frac{200000}{100} \\ \Rightarrow P = Rs 2000 \\ Using this formula \\ (∵ SI = \frac{P \times R \times T}{100}) \\ ∴ 200 = \frac{2000 \times 5 \times T}{100} \\ \Rightarrow T = \frac{200}{100} = 2 years \\ A l t e r n a t e : \\ (8 - 5) % = 2320 - 2200 \\ \Rightarrow 3 % = 120 \\ \Rightarrow 1 % = 40 \\ \Rightarrow 5 % = 200 \\ Principal = 2200 - 200 \\ = Rs . 2000 \\ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 200 = \frac{2000 \times 5 \times T}{100} \\ \Rightarrow T = \frac{200}{100} = 2 years \end{aligned}$

98.

The simple interest on a certain sum is one-eighth of the sum when the number of years is equal to half of the rate percentage per annum. Find the simple interest (in Rs.) on Rs. 15,000 at the same rate of simple interest for 8 years.

(A) 5,800
(B) 5,000
(C) 6,000
(D) 5,250

Solution: $\begin{aligned} Let sum = P \\ \frac{P \times r \times t}{100} = \frac{1}{8} P \\ r t = \frac{100}{8} = \frac{25}{2} . . . . . (i) \\ Since t = \frac{r}{2} \\ From equation (i) \\ r \times \frac{r}{2} = \frac{25}{2} \\ r = 5 % \\ If P = 1500 \\ t = 8 years \\ S .I . = \frac{P \times r \times t}{100} \\ = \frac{15000 \times 8 \times 5}{100} \\ = Rs . 6, 000 \end{aligned}$

99.

With a given rate of simple interest, the ratio of principal and amount for a certain period of time is 4 : 5. After 3 years with the same rate of interest, the ratio of the principal and amount becomes 5 : 7. The rate of interest is = ?

(A) 4%
(B) 6%
(C) 5%
(D) 7%

Solution: $\begin{aligned} \frac{Principal}{Amount} = \frac{4 \times 5}{5 \times 5} \\ = \frac{20}{25} \\ After three year \\ \frac{P}{A} = \frac{5 \times 4}{7 \times 4} \\ = \frac{20}{28} \\ In three year S .I . \\ = 28 x - 25 x \\ = 3 x \\ So, the required interest will be \\ 3 x = \frac{20 x \times R \times 3}{100} \\ R = 5 % \end{aligned}$

100.

If the simple interest for 6 years be equal to 30% of the principal, it will be equal to the principal after

(A) 10 years
(B) 20 years
(C) 22 years
(D) 30 years

Solution: $\begin{aligned} Let sum = Rs . x \\ Then, \\ S .I . = 30 % of Rs . x \\ = Rs . \frac{3 x}{10} \\ Time = 6 years . \\ ∴ Rate = (\frac{100 \times 3 x}{10 \times x \times 6}) % \\ = 5 % \\ Now, sum = Rs . x \\ S .I . = Rs . x \\ Rate = 5 % \\ ∴ Time = (\frac{100 \times x}{x \times 5}) years \\ = 20 years \end{aligned}$

«
1
2
...
7
8
9
10
11
12
13
...
21
22
Next ›
Last »

Indian Polity Mcq IAS GK Question India GK World GK GK questions GK Quiz Biology GK Science GK Physics GK Chemistry GK General Awareness Computer GK Political GK Indian Political GK Economics GK History GK Sports GK Geography GK Teaching Aptitude GK CTET/TET GK B Ed Entrance GK Banking GK SSC GK UPSC GK Indian Army GK Hindi Grammar GK Hindi Grammar MCQ Reasoning questions Bihar GK Rajasthan GK Jharkhand GK MP GK UP GK Chhattisgarh GK Haryana GK HP GK Maharashtra GK Electronics GK Agriculture GK States/Capitals GK Books Name & Author Arthimetic Verbal Reasoning Non Verbal Reasoning English

Practice MCQ Questions and Answer on Interest

General Knowledge