Dear Readers, SBI is conducting Online preliminary Examination for the recruitment of Clerical Cadre. preliminary Examination of SBI Clerk was scheduled from June/July. To enrich your preparation here we have providing new series of 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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1) The compound interest on a certain sum at 20 % per annum for 3 years is Rs. 1456. The simple interest on the same sum for double the time at half the rate percent per annum is:
2) What total will be amount to Rs. 51516 at compound interest in 2 years, the rate of interest for 1st& 2nd year being 6% and 8% respectively?
3) Kumar wants to take Rs. 45000 at rate of interest 4% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his profit in the above transaction?
4) A sum of money for the first two years at a rate of interest is 7% p.a., for the next two years it is 8% p.a. and 9% p.a. for the period exceeding four year; all at simple interest. If a person earns an interest of Rs. 15180 by the end of the 8 years, what is the amount at the end of the period of investment?
5) The difference between simple and compound interests together annually on a certain sum for 2 year at 5% per annum is Rs 4. Find the sum?
6) If Rs.18000 amounts to 22579.20 in 2 yrs compounded annually, then find the rate of interest per annum?
7) A certain national bank offers 20 % interest rate compounded annually. Selvi deposits Rs. 35000 every year in his account. If she does not withdraw any amount, then how much balances will his account show after four year?
8) Find the CI on 6 months on Rs.12350 at 12% per annum compounded quarterly.
9) The sum that amounts to Rs. 2704 in 2 yrs at 4% per annum CI compounded annually is,
Answers:
1) Answer: b
Let the sum be Rs. P.
[P(1+R/100)3-P]=1456 [P(1+20/100)3-P]=1456 [P(12/10)3-P]=1456P[(12/10)3-1]=1456
P[(1728/1000)-1]=1456
P[728/1000]=1456
P=1456000/728 =2000
SI=PNR/100
=2000*6*10/100
= 1200
2) Answer:d
Let Rs. P be the required sum.
51516 = P (1 + 6/100) (1 + 8/100)
51516 =P (106/100) (108/100)
51516*(100/106)*(100/108) = P
P= 45000
Hence the required amount is Rs. 45000
3) Answer:a
Amount of money Kumar taken at S.I at 4% p.a. for two years = Rs.45000
He lend the same amount for C.I at 4% p.a. for two years
Kumar’s income = C.I – S.I
= p [1 + r/ 100]n – p – [pnr/100]
= p {[1 + r/ 100]2 – 1} – [pnr/100
= 45000{[1+ (4/100)]2-1}-[45000*2*4/100]
= 45000{[104/100]2-1}- 3600
= 45000{[26/25)2 – 1} – 3600
= 45000[676/625 – 1} – 3600
= 3672 – 3600
= Rs.72
4) Answer:c
Let the principal be x
According to the question,
X*7*2/100 +X*8*2/100 +X*9*4/100 = 15180
14X/100 + 16X/100+36X/100 = 15180
66X/100 = 15180
66X = 1518000
X =23000
Total amount end of the period = 15180 + 23000
= Rs. 38180
5) Answer:b
The difference between S.I and C.I together on a certain sum for 2 years is,
Diff = Sum*(r/100)2
4 = Sum*(5/100)2
4*25/(100*100) = Sum
Sum = Rs. 1600
6) Answer:d
Given P = Rs.18000
Amount = 22579.20
n = 2years
Let the rate be R% per annum. Then
[18000(1+R/100)2] = 22579.20(1+R/100)2 = 18000 / 22579.20
(1+R/100)2 = (28/25)2
1+R/100 = 28/25
(100+R)/100 = 28/25
100+R = 2800/25
100+R = 112
R = 12 %
7) Answer:a
For the 1st year interest will be,
= > 35000*(20/100) = Rs. 7000
35000+7000+35000 = 77000
= > 77000*(20/100) = 15400
77000+ 15400+ 35000 = 127400
= > 127400*(20/100) = 25480
127400 + 25480 + 35000 = 187880
= > 187880*(20/100) = 37576
187880 + 37576 = Rs. 225456
8) Answer:b
Given P = Rs.12350; n=6 months = 2 quarters
R=3% per quarter
Amount = Rs [p(1+R/100)n]
=Rs. [12350(1+3/100)2]
=Rs. [12350(103/100)2]
=Rs. [12350*103/100*103/100]
=Rs. (12350*1.03*1.03)
=Rs.13102.115
Compound interest = amount – principal
=Rs. (13102.115 – 12350)
=Rs. 752.115
(or)
12350*(3/100) = Rs. 370.5
12720.5*(3/100) = Rs. 381.615
C.I = 370.5 + 381.615 = Rs. 752.115
9) Answer:c
Amount = P(1+r/100)n
2704 = P(1+4/100)2
P = Rs [2704/ (1+4/100)2]
P =Rs. (2704 / (104/100)2]
P =Rs. (2704 *(100/104) * (100/104)]
P =Rs. (27040000/10816)
P =Rs. 2500
10) Answer:c
Amount = P (1+R/100)n
=4000(1+10/100)2
=4000(110/100)2
=4000*110/100*110/100
=40*11*11
=Rs.4840
Hence the required answer is Rs. 4840
Daily Practice Test Schedule | Good Luck
| Topic | Daily Publishing Time |
| Daily News Papers & Editorials | 8.00 AM |
| Current Affairs Quiz | 9.00 AM |
| Logical Reasoning | 10.00 AM |
| Quantitative Aptitude “20-20” | 11.00 AM |
| Vocabulary (Based on The Hindu) | 12.00 PM |
| Static GK Quiz | 1.00 PM |
| English Language “20-20” | 2.00 PM |
| Banking Awareness Quiz | 3.00 PM |
| Reasoning Puzzles & Seating | 4.00 PM |
| Daily Current Affairs Updates | 5.00 PM |
| Data Interpretation / Application Sums (Topic Wise) | 6.00 PM |
| Reasoning Ability “20-20” | 7.00 PM |
| English Language (New Pattern Questions) | 8.00 PM |
| General / Financial Awareness Quiz | 9.00 PM |
This post was last modified on June 12, 2021 6:43 pm