Interest - Study Mode

[#171] A sum of Rs. 50,250 is divided into two parts such that he simple interest on the first part for $$7frac{1}{2}$$xa0years at $$8frac{1}{3}\% $$ xa0p.a. $$frac{5}{2}$$ is times the simple interest on the second part for $$5frac{1}{4}$$xa0years at 8% p.a. What is the difference (in Rs.) between the two parts?
Correct Answer

(A) 12,750

Explanation

Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{{ ext{.}}_1} = frac{5}{2} imes { ext{S}}{ ext{.I}}{{ ext{.}}_2} cr
& frac{{x imes 25 imes 15}}{{100 imes 3 imes 2}} = frac{5}{2} imes frac{{y imes 8 imes 21}}{{100 imes 4}} cr
& 25 imes 5 imes x = 5 imes 2 imes 21 imes y cr
& x:y = 210:125 = 42:25 cr
& = 42 - 25 = 17{ ext{ units}} cr
& 67{ ext{ units}} = 50250 cr
& 1{ ext{ unit}} = 750 cr
& 17{ ext{ units}} = 750 imes 17 = 12750 cr
& { ext{Difference between both parts}} = 12750 cr} $$

[#172] A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of the interest be 10% compounded annually, then the value of each installment is:
Correct Answer

(A) Rs. 121

Explanation

Solution: Let X = equal installment at the end of one year( rate% annually) .
Now 1st year, P =210, Interest = $$frac{{{ ext{PTR}}}}{{100}}$$xa0 = 210 * 0.1 = 21. Let X is to be paid as an equal installment. Now, at the beginning of 2nd year, P = 210 + 21 - X, Interest at the end of 2nd year, = (231 - X) * 0.1 = 23.1 - 0.1X. Hence,total installment, 2X = 210 + 21 + 23.1 - 0.1X, X = $$frac{{{ ext{254}}{ ext{.1}}}}{{2.1}}$$xa0 = 121.

[#173] There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?
Correct Answer

(C) Rs. 8,000

Explanation

Solution: $$A = P {left( {1 - frac{R}{{100}}}
ight)^n}$$ Where A = Value of goods after n years P = Initial Price R = Rate of depriciation $$eqalign{
& herefore P = frac{{5832}}{{{{left( {1 - frac{{10}}{{100}}}
ight)}^3}}} cr
& Rightarrow P = frac{{5832}}{{{{left( {1 - frac{1}{{10}}}
ight)}^3}}} cr
& Rightarrow P = frac{{5832}}{{{{left( {frac{9}{{10}}}
ight)}^3}}} cr
& Rightarrow P = 5832 imes frac{{10}}{9} imes frac{{10}}{9} imes frac{{10}}{9} cr
& Rightarrow P = 8000 cr} $$

[#174] A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
Correct Answer

(C) Rs. 698

Explanation

Solution: S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Therefore Principal = Rs. (815 - 117) = Rs. 698.

[#175] Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
Correct Answer

(A) Rs. 6400

Explanation

Solution: Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x ) Then, $$ {frac{{x imes 14 imes 2}}{{100}}} + $$ xa0 $$ {frac{{left( {13900 - x}
ight) imes 11 imes 2}}{{100}}} $$ xa0 xa0 $$ = 3508$$ ⇒ 28x - 22x = 350800 - (13900 x 22) ⇒ 6x = 45000 ⇒ x = 7500 So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400