SSC CGL 201911)A loan has to be returned in two equal yearly instalments each of ₹44,100. If the rate of interest is 5% p.a.. compounded annually, then the total interest paid is:

Correct Option: C

₹ 6,200

SSC CGL 201912)A sum amounts to ₹14,395.20 at 9.25 % p.a. simple interest in 5.4 years. What will be the simple interest on the same sum at 8.6 % p.a. in 4.5 years?

Correct Option: A

₹3,715.20

SSC CGL 202013)A certain sum amounts to Rs.280900 in 2 years at 6% per annum, interest compounded annually. The sum is:

Correct Option: B

Rs.250000

Sum = p; Time(t) = 2 years; r = 6%; Amount = 280900; \(Amount={P(1+\frac{r} {100})^t} \);

\(280900=P(1+\frac{6}{100})^2\) = Rs. 250000

SSC CGL 202014)The compound interest on a certain sum at the end of two years is Rs.408. The simple interest on the same sum for the same time is Rs.400. The rate of interest per annum is:

Correct Option: B

4%

S.I. for 2 years = Rs. 400; so S.I. for 1 year =\({400\over2 }= 200\); \({P\times r\times t\over100}=200\); ⇒ Pr = Rs. 20000; According to the question, \({(P+200)\times r\times 1\over100}= 208\); ⇒ Pr+200r = 20800; ⇒ 20000+200r=20800; ⇒ r = 4% p.a.

SSC CGL 202015)If in 13 years a fixed sum doubles at simple interest, what will be the interest rate per year? (Correct to two decimal places)

Correct Option: B

7.69%

Principal = Rs. x (let); Interest = Rs. x; ⇒ Rate = \({Interest \times 100\over Principal \times Time }= {{x\times100}\over{x\times 13}} =7.69\)%per annum

SSC CGL 202016)A man buys two watches ‘A’ and ‘B’ at a total cost of ₹800. He sells both watches at the same selling price, and earns a profit of 18% on watch ‘A’ and incurs a loss of 22% on watch ‘B’. Whatare the cost prices of the two watches? (correct to two places after decimal).

Correct Option: D

A = ₹318.37 and B = ₹481.63

Let C.P. of watch A be Rs. x. C.P. of watch B = Rs.(800 - x); According to the question, \(x\times{118\over100}=(800-x)\times({100-22\over100})\); ⇒ x = Rs. 318.37; C.P. of watch B = 800 - 318.37 = Rs. 481.63

SSC CGL 202017)A sum lent out at compound interest amounts to Rs.1,250 in one year and to Rs.1,458 in 3 years at a certain rate percentage p.a. What is the simple interest on the same sum for \(5{2\over5}\) years at the same rate of interest?

Correct Option: C

Rs. 500

\(P(1+{r\over100})=1250\)___(1); \(P(1+{r\over100})^3=1458\)_____(2); equation(2) ÷ equation(1); \((1+{r\over100})^2={1458\over1250}\); ⇒ r = 8%;

Calculate P by substituting value of r in equation (1) P = \(31250\over27\); Simple interest = \({31250\over27}\times8\times{27\over5}\over100\) = Rs. 500

SSC CGL 202018)The simple interest on a sum of Rs. 50,000 at the end of two years is Rs.4,000. What would be the compound interest on the same sum at the same rate for the same period ?

Correct Option: B

Rs. 4,080

Principal(p) = 50000 ; Interest(i) = 4000 ; Time(t) = 2 years ; Rate = \({S.I.\times100\over P\times T} = {4000\times100\over50000\times2} \) = 4% per annum ; Compound interest = \(P[(1+{R\over100})^T-1] =50000[(1+{4\over100})^2 -1]\) = Rs. 4080

SSC CGL 202019)The rate of simple interest on a sum of money is 5% p.a. for the first 4 years, 8% p.a. for the next 3 years and 10% p.a. for the period beyond 7 years. If the simple interest accrued by the sum over a period of 10 years is Rs.1,850, then the sum is :

Correct Option: C

Rs. 2,500

Simple interest = \(\frac{Principle\times Time\times Rate}{100}\);

Simple interest for 10 years = 1,850;

Simple interest for 4 years at 5% + Simple interest for 3 years at 8% + Simple interest for 3 years at 10% = 1850;

\(\frac{p \times 5 \times 4}{100} + \frac{p \times 8 \times 3}{100} + \frac{p \times 10 \times 3}{100}\) = 1850;

p(20 + 24 + 30) = 185000;

p = 185000/74 = 2500

SSC CGL 202020)Amit borrowed a sum of Rs. 25,000 on simple interest. Bhola borrowed the same amount on compound interest(interest compounded yearly). At the end of 2 years, Bhola had to pay Rs. 160 more interest than Amit. The rate of interest charged per annum is:

Correct Option: C

8%

For 2 years, Difference between C.I. and S.I. = \(P({R\over100})^2\); ⇒ \(160 = 25000\times({R\over100})^2\); ⇒ R = 8% per annum

showing 11 - 20 results of 30 results