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.
a) Rs. 1500
b) Rs. 1400
c) Rs. 2000
d) Rs. 1600
e) Rs. 1800
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