Practice Quantitative Aptitude Questions For SBI PO Prelims& NIACL 2017 (Application Problems& Simplification):
Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.
1). The ratio of the radii of two right circular cylinders (A and B) is 2 : 5. The ratio of the highest of cylinders A to B is 3 : 1. What is the ratio of the volumes of cylinders A to B?
a) 12:25
b) 9:25
c) 9:20
d) 3:5
e) 12:35
2). Raja gives 30% of his salary to his mother, 40% of the remaining salary he invests in an insurance scheme and PPF in the ratio of 4 : 3 and the remaining he keeps in his bank account. If the difference between the amount he gives to his mother and that he invests in insurance scheme is Rs.8400, how much is Raja’s salary?
a) Rs.60,000
b) Rs.62,000
c) Rs.64,000
d) Rs.65,000
e) Rs.54,000
3). C is 40% less efficient than A. A and B together can finish a piece of work in 10 days. B and C together can do it in 15 days. In how many days can A alone finish the same piece of work?
a) 18
b) 12
c) 14
d) 20
e) 15
4). In a bag there are 7 red balls and 5 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green in colour?
a) 29/44
b) 21/44
c) 27/44
d) 23/44
e) 19/44
5). Shyama invested Rs. P for 2 years in scheme A, which offered 11% pa simple interest. She also invested Rs. 600+P in scheme B, which offered 20% compound interest (compound annually)for 2 years. If the amount received from scheme A was less than that received from scheme B by Rs. 1216 then what is the value of P?
a) Rs. 1500
b) Rs. 1400
c) Rs. 2000
d) Rs. 1600
e) Rs. 1800
Directions (Q. 6-10): What approximate value will come in place of question mark (?) in the given questions? (You are not expected to calculate the exact value.)
6). 619.002 – 134.99 ÷ 14.998—(9.01)2=?
a) 720
b) 530
c) 650
d) 690
e) 490
7). 439.97 ÷ 15 .02 + 208.08 ÷ 8.01 — 16.01 =?
a) 120
b) 60
c) 100
d) 80
e) 40
8). 4? × √226 =245.998 ÷ 8.001 + 929.99
a) 4
b) 5
c) 2
d) 3
e) 1
9). ?%of(140.06 x 7.99 — 679.92) = 330.01
a) 70
b) 90
c) 80
d) 50
e) None of these
10). 40% of 859 + 86.01 ÷ 7.99 = ?
a) 398
b) 286
c) 412
d) 215
e) 355
Explanation With Answer Key
1). A)Let the radius of cylinder A be 2x and that of cylinder B be 5x.
Now, height of cylinder A = 3y and height of cylinder B = y
Now, volume of cylinder A = ￗr2h
= (22/7)x (2x)2 x 3y = (22/7) x 4x2 x 3y
= (22/7) x l2 x2y
Volume of cylinder B = ￗr2h
=( 22/7)×(5x)2 × y = (22/7) x 25 x2y
Reqd ratio = [ (22/7) x l2 x2y] / [(22/7) x 25 x2y] = 12/25
= 12:25
2). A)Let Raja’s salary be Rs. x.
Raja gives 30% of his salary to his mother.
Raja gives[(x × 30)/100 = Rs. 3x/10] to his mother
Remaining salary of Raja = x – (3x /10) = Rs. 7x /10
Investments of Raja in insurance and PPF is 40% of the remaining salary.
Insurance + PPF = (7x x 40)/ (10×100) = 7x/25
Remaining salary of Raja = 7x/10 – 7x/25
= (35x —14x)/50 = 21x /50
Raja’s investment in insurance scheme
= (7x/25) × (4/7) = 4x/25
Now, according to the question,
3x/10 – 4x/7 = 8400
or, (15x-8x) / 50 =8400
or, 7x = 8400 x 50
x = (8400x 50)/7 =1200 x 50 = Rs. 60000
3). B)Suppose total work = 60 units (LCM of 10 and 15)
(A + B)’s one day’s work = 60/10
= 6 units
And (B + C)’s one day’s work = 60/15 = 4 unit
According to the question, C : A = 60 : 100
or, C :A = 3 : 5
C/A = 3/5
Or, A = 5C/3
Again, A +B = 6units ..(i)
B+C = 4 units …(ii)
Putting the value of A in equation (i), we get
[ 5C/3 + B = 6 unit ] – [B + C = 4 units]
= 5C/3 – C=2units
Or, 2C/3 = 2 units
C = 3 units
Then A = 5C/3 = 5×3 / 5 = 5 units
Now, total work is 60 units Then A alone can do the work in
(60 / 5 =) 12 days
4). B)Total number of balls = 7 + 5 = 12
Now, three halls are picked randomly
Then, the number of sample space n(S)
=12C3 =(10 ×11 ×12) / (1×2×3) = 220
The number of events
n(E) =7C2 x5C1 = [(6×7)/2] x 5
= 21 x 5 = 105
105 21 P(E) = n(E)/n(S)=105/220 = 21/24
5). D)Amount received from scheme A
= P+ [(Px2x11)/100] =( 100P+22P)/100 = 122P/100
Amount received from scheme B
= (P + 600) [1+ (20/100)]2
= (P + 600) (6/5)2
= (P + 600) (36/25)
= 36P/25 + (600×36)/25 = 36P/25 + 864
Now, according to the question,
36P/25 + 864 – 122P/100 = 1216
Or, 36P/25 – 122P/100 = 1216-864
Or, (144P-122P)/100 = 352
Or, 22P = 352×100
P = (352×100) / 22 = Rs. 1600
6). B)? = 619.002 – 134.99 ÷ 14.998 –(9.01) = 620 – 135 ÷ 15 – (9)2
= 620 – 90 = 530
7). E)? = 439.97 ÷ 15.02 + 208.08 ÷ 8.01 –16.01 = 450 ÷ 15 + 208 ÷ 8 – 16
= 30 + 26 – 16 =30 + 10 = 40
8). D)4?x √226 = 245.998 ÷ 8.001 + 929.99
or, 4? x √225 = 248 ÷ 8 + 930
or, 4? x 15 = 31 + 930 = 961
or 4? = 960/15 = 64 = 43
? = 3
9). E)?% of (140.06 x 7.99 — 679.92) = 330.01
= [?x(140×8-680)] / 100 = 330
or, ? x (1120 — 680) = 330 x 100
or, ? x 440 = 33000
? = 33000 / 440 = 75
10). E)? 40% of 859 + 86.01 ÷ 7.99
= (40 x 860) / 100 + 86 ÷ 8
= 344 + 11 = 355
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