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1) How long will it take a sum of money invested at 5%p.a to increase its value by 60%?A) 12 B) 8 C) 10 D) 11
Answer & Explanation
Answer: 12 (Let the sum be x.
Then, amount=160/100x=8x/5
Therefore, S.I = (8x/5 - x) = 3x/5
Thus, sum=x, S.I=3x/5 and rate=5%
Therefore, Time=[(100*3x*1/(x*5)]= 12 yrs
)
2) If A lends $5000 to B at 10%p.a and B lends the same sum to C at 11.5%p.a, then the gain of B (in $) in a period of 3 yrs is:A) $235 B) $230 C) $220 D) $225
3) Avin borrowed $10000 from Sam at S.I. After 4 years, Sam got $1000 more than what he had given to Avin. What was the rate of interest per annum?A) 2% B) 3% C) 2.5% D) 2.25%
4) The difference between the interests received from two different bank on $1000 for 2 yrs is $10. The difference between their rates is:A) .45% B) .50% C) .75% D) 1%
Answer & Explanation
Answer: .50% (The difference between the interests received from two different bank on $1000 for 2 yrs is $10.
1000*R1*2/100 - 1000*R2*2/100= 10
Or 2000(R1-R2)=1000
Therefore, R1-R2=1000/2000=1/2 = 0.50%
)
5) $ 2189 are divided into three parts such that their amounts after 1,2 and 3 yrs respectively may be equal, the rate of S.I being 4%p.a in all cases. The smallest part is:A) $597 B) $587 C) $567 D) $577
Answer & Explanation
Answer: $597 (Let the parts be x,y,[2189-(x+y)]. Then,
x*1*4/100=y*2*4/100=[2189-(x+y)]*3*4/100
x/y=2 ; x=2y.
2y*1*4/100=[2189-3y]*3*4/100
44y=2189*12 ; y=(2189*12/44)=597
Therefore, the smallest part = $ 597
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