Quantitative Aptitude Quiz for IBPS | IPPB – Set 168

1)A portion of $6600 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from the portion earning a 5% return is twice that of the other portion, what is the total income from the two investments after one year?
A)200
B)270
C)250
D)280

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Answer B)270
5x + 3y = z (total)
x + y = 6600
5x= 2(3y) [ condition given] 5x – 6y = 0
x + y = 6600
5x -6y = 0
Subtract both equations and you get x = 3600 so y = 3000
3600*.05 = 180
3000*.03 = 90
z (total) = 270

2)Nishu 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?
A)6400
B)6500
C)7200
D)7500

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Answer A)6400
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).
Then, ( x X 14 X 2 ) /100 + [(13900 – x) X 11 X 2] /100= 3508
28x – 22x = 350800 – (13900 X 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 – 7500) = Rs. 6400.

3)A sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to how much?
A)780
B)992
C)848
D)700

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Answer 2)992
S.I = Rs. (920 – 800) = Rs. 120; P = Rs. 800, T = 3 yrs
use SI=Px R x T/100 so, R = Si x 100 /Px t = ( 100 X 120 ) / 800 X 3 = 5%
New rate = (5 + 3) % = 8%
New S.I. = Rs. (800 X 8 X 3)/100 == Rs. 192.
New amount = Rs. (800 + 192) = Rs. 992

4)A certain sum is invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T)?
A)39 years
B)41 years
C)45 years
D)50 years

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Answer D)50years
We have, A1 = Rs. 400, A2 = Rs. 200, R1 = 10%, R2 = 4%
Time (T) = [A1 – A2] x 100 divide by A2R1 – A1R2
= [400 – 200]x 100 divide by [200 x 10 – 400 x 4]= 20000/400 = 50 Years.

5)David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
A)Rs 5000
B)Rs 6500
C)Rs 8000
D)Rs 10000

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Answer A)5000
Let x, y and z be the amounts invested in schemes A, B and C respectively. Then,
add individual interest to get total using Si= pxrxt/100
[x x 10 x 1]/100 + [y x 12 x 1]/100 + [z x 15 x 1]/100 = 3200
10x + 12y + 15z = 320000…. (i)Now, z = 240% of y =(12/5)y……… (ii)And, z = 150% of x =(3/2)x so,x=(2/3 )z = (2/3) x value of z from ii
x= (2/3) x (12/5) y = (8/5)y………..(iii)
From (i), (ii) and (iii), we have :
16y + 12y + 36y = 320000
64y = 320000
y = 5000
Sum invested in Scheme B = Rs. 5000

6)In a library 5 percent books are in English, 10 percent of the remaining are in hindi and 15 percent of the remaining are in Sanskrit. The remaining 11628 books are in French. Then find the total number of books in the library?
A)8000
B)12000
C)16000
D)20000

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Answer c) 16000
Let total books are A, then (95/100)*(90/100)*(85/100)*A = 11628

7)A student has to get 40 percent marks to pass an examination. He got 60 marks but fails by 20 marks. Find the maximum marks of the examination?
a)150
b)200
c)300
d)400

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Answer b) 200
(40/100)*M – 20 = 60 (M is the maximum marks)

8)One type of liquid contains 20 percent milk and in other liquid it contains 30 percent milk. If 4 parts of the first and 6 parts of the second are taken and formed a new liquid A. Find the percentage of milk in third liquid?
A)26
B)28
C)29
D)35

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Answer A) 26
milk = 20 and water = 80 (in 1st liquid)
milk = 30 and water = 70 (in 2nd liquid)
milk in final mixture = 20*4 + 30*6 = 260
so (260/1000)*100 = 26%

9)1000 sweets need to be distributed equally among the school students in such a way that each student gets sweet equal to 10% of total students. Then the number of sweets, each student gets?
A)10
B)12
C)14
D)16

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Answer – a) 10
No of students = T. Each student gets 10% of T. So, T students get T^2/10 sweets.
T^2/10 = 1000. We get T =10

10)If the price of an article is increased by 15%, then by how much the household should decrease their consumption so as to keep his expenditure same?
A)13(1/23) %
B)13(2/23)%
C)11(1/23)%
D)11(2/23)%

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Answer a)13(1/23) %
Decrease in expenditure = (15/115)*100 = 300/23 %