Dear Readers, SBI is conducting Online preliminary Examination for the recruitment of Clerical Cadre. preliminary Examination of SBI Clerk was scheduled from March 2018. To enrich your preparation here we have providing new series of Simple Interest and Compound Interest – Quantitative Aptitude Questions. Candidates those who are appearing in SBI Clerk Prelims Exam can practice these Quantitative Aptitude average questions daily and make your preparation effective.
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Q.1) Ragu invested Rs.72000 in two different parts one at 20% SI and another one at 20% CI (Compounded annually). What is the difference between the amount invested in SI and CI, if he received total interest at the end of two years is Rs.30480?
Q.2) A person invested sum of the amount at the rate of 15% SI per annum for two years and received total amount of Rs.19500. He invested same sum at the rate x% per annum compounded annually for two years and he received interest Rs.2100 more as compared to the Simple Interest, then find the value of ‘x’
Q.3) Shivani invested Rs.18000 in two different parts and the interest received at the rate of 8% SI for two years is equal to the interest received at the rate of 10% interest for two years. What is the difference between the Sum invested in 10% and 8%?
Q.4) Ramu invested Rs.x amount at the rate of 20% per annum compounded annually. After one year, he withdraws Rs.2500 and the rate of interest is reduced by 15%. Find the value of ‘x’ if he received Rs.735 less at the end of two years.
Q.5) Akhilan invested some amount in SI at the rate of x% per annum for 2 years and Mathi also invested some amount at the rate of 20% CI per annum for 3 years and he received total interest Rs.2275. Find the value of x if Mathi invested Rs.625 less than Akilan and interest received is Rs.25 more than Akhilan
Q.6) Mukesh invested Rs.30000 in two different parts one at 15% SI and another one at 20% CI. At the end of two year he received total interest Rs.10750 then find the amount invested in SI?
Q.7) Priya invested Rs. x in SI at the rate of 20% per annum for two years and the same amount is invested in CI at the same rate of interest if she received Rs.860 more interest as compared to S.I, then find the value of x.
Q.8) Raji invested Some amount in SI at the rate of 20% per annum for 2 years and that total amount invested in CI at the same rate of interest for same period, he got total amount of Rs.4032. Find the sum invested by Raji
Q.9) Kamal invested Rs.2400 and Rs.3000, one at the rate of 10% SI per annum and another at x% SI per annum respectively. If he interchanges the rate of interest he got Rs.120 more at the end of two years, and then find the value of x
Q.10) Suresh invested Rs.80000 in two different parts one at the rate of 20% CI per annum and another one at the rate of 15% SI per annum. If he interchanges the rate of interest he got Rs.2875 less at the end of two years, and then find the difference of the two sums
Answers:
Q.1) Answer: c
Let us take one part x and another part 72000-x
Total interest = x*40/100 + (72000-x)*44/100=30480
10x+72000*11-11x=30480*25
X=30000
Required difference = 42000-30000=12000
Q.2) Answer: d
Let us take sum be y
Given,
130/100*y=19500
=>y=15000
S.I = 15000*(15/100)*2 =4500
C.I = 4500+2100 =6600
Total amount= 15000+6600 = 21600
15000(1+x/100)2=21600
(1+x/100)2=(6/5)2
1+x/100=6/5x/100=1/5=>x=20%
Q.3) Answer: b
Let us take one part be x and another one be 18000-x
Given,
(18000-x)*20/100=x*16/100
=>18000*5=9x
=>x=10000
Required difference = 10000-8000=2000
Q.4) Answer: c
Given,
44/100*x-735=x*20/100+(x-2500)*15/100
X=4000
Q.5) Answer: e
Let us take Mathi’s sum be y
2275=Y ((1+20/100)3-1)
2275*125/91=Y
=>Y=3125
Akhilan’s investment and interest are 3750 and 2250
3750*x*2/100=2250
=>x=30%
Q.6) Answer: d
Given,
(30000-x)*30/100+x*44/100=10750
900000-30x+44x=1075000
14x=175000=>x=12500
Required sum = 30000-12500=17500
Q.7) Answer: a
Given,
x*20/100*2+860=x((1+20/100)2-1)
44/100*x-40/100*x=860
4x/100=860=>x=21500
Q.8) Answer: d
Let us take amount invested in SI be x
Total amount of money in SI = 140/100*x
Given,
4032=140/100*x*(1+20/100)2
252x/125=4032
=>x=2000
Q.9) Answer: c
(3000*2x/100+2400*20/100)- (3000*20/100+2400*2x/100)=120
(60x+480)-(600+48x)=120
12x=240=>x=20%
Q.10) Answer: a
Let us take one part as x and another part as 80000-x
Given,
(44/100*x+(80000-x)*30/100)-(x*32.25/100+(80000-x)*40/100)=2875
14x+7.75x-940000=-287500
21.75x=652500
X=30000
Required difference = (80000-30000)-30000=20000
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This post was last modified on March 19, 2019 3:59 pm