APTITUDE ANSWERS
1.SOLUTION:
The product of the first three prime numbers = 385
The product of the last three prime numbers = 1001
In the above products, the second and the third prime numbers occur in common.
∴ The product of the second and third prime numbers = HCF of the given products.
HCF of 385 and 1001 = 77
∴Largest of the given primes =1001/77=13
ANSWER: d)13
2.SOLUTION:
(A+B+C) earn per day Rs.120⇒ (1)
(A+C) earn per day Rs.80⇒ (2)
(B+C) earn per day Rs.66 ⇒ (3)
From (1) and (2),
we find that C earns Rs.40
From (3) we get,
That B earns 26 since C earns Rs.40
ANSWER: b)40
3.SOLUTION:
For one rotation, distance moved = 5 metres
For 12 rotations, distance moved = 12 × 5 = 60 metres
Distance covered in 1 minute = 60 metres
Total distance covered by the wheel = 930 minutes
Total time taken =930/60=15.5 minutes
ANSWER: b)15.5 minutes
4.SOLUTION:
Amount of liquid left after n operation, if x is the capacity of the container
from which y units are taken out each time is
FORMULA: [x(1-(y/x))n] units
Here,y=4 and n=2
ANSWER: c)7 lts
5.SOLUTION:
‘a’ = 25
‘d’ = 5
n/2[2a+(n-1)d] ≥1000
n2 + 9n – 400 ≥ 0
n2 + 25n – 16n – 400 ≥ 0
n(n + 25) – 16 (n + 25) ≥ 0
(n – 16)(n + 25) ≥ 0
n ≥ 16
After 16 years, the man will have savings of Rs.1000/- with him.
ANSWER: d)16 years and 1000
6.SOLUTION:
Total number of students in the school = 728
Number of students who come by bus = 162
Number of students who come on bicycle = 384
Adding these two, we get 546.
728 – 546 = 182 students come by walk.
% of students who come by walk to the school=(182/728)*100=25%
ANSWER: b)25%
7.SOLUTION:
C.P of 11 oranges = 7 rupees
C.P of 1 orange=7/11 rupees
S.P of 7 oranges=11 rupees
S.P pf 1 orange=11/7 rupees
Profit=11/7-7/11=72/77
Profit%=(72/77)/(7/11)*100=146.938%
ANSWER: d)146.938%
8.SOLUTION:
Let ‘a’ be the side of the square base.
∴Altitude of the cuboid = 2a.
Volume = Area of the base × Height
= a2 × 2a = 54
= 2a3 = 54
= a3 = 27
∴a = 3cm.
ANSWER: a)3
9.SOLUTION:
Let a son’s share be x rupees
The share of a daughter = 3x rupees.
Total amount father left = Rs.1750, to be divided among 2 daughters and 4
sons.
2 × 3x + 4x = 1750
10x = 1750
x=175
Each son’s share = Rs.175 and each daughter’s share = Rs.525
ANSWER: c)175 and 525
10.SOLUTION:
Let the numbers have x in ten’s place and y in the unit place
∴10x+y = 3xy – 8……….(1)
x-y = 2 …………...(2)
From (2) x= y+2
In (1), 10(y+2)+y = 3y(y+2)-8
10y + 20 + y = 3y2+6y-8
3y2-5y-28 = 0
3y2-12y+7y-28 = 0
3y(y-4) + 7(y-4) = 0
(y-4) (3y+7) = 0
y=4 or -7/3
( y = -7/3 is deleted)
∴ x = y+2 => x = 6
∴The number is ‘64’.
ANSWER: a)64
11.SOLUTION:
Let ‘a’ be the number of men in the front row.
a2 + 31 = 161610
No. of men in the front row = 127
a2 = 161610 – 31 = 16129
∴a = 127
ANSWER: b)127
12.SOLUTION:
Pipe A fills in an hour =1/6 of the cistern.
Pipe B empties in one hour 1/8 of the cistern.
In 1 hour, the portion of the cistern filled =1/6-1/8=1/24
∴Time taken to fill the cistern = 24 hours.
ANSWER: d)24 hrs
13.SOLUTION:
Speed of stream = 3 km/hr
Speed of boat in still water = speed of stream + speed of boat upstream
= 3 + 4 = 7 km/hr.
Distance to be covered during the return journey downstream = 6 km
Speed of the boat downstream = Speed of the boat in still water + speed
of the Stream= 7 + 3 = 10 km/hr.
Time taken for the return journey = Distance downstream/Speed of boat
Downstream
=6/10 hrs=6/10*60=36 mins.
ANSWER: c)36 mins.
14.SOLUTION:
Speed of A =10/60
Distance run by A is 6 min =(10/60)*6=1000 mtrs.
A has a start of 1000 + 100 = 1100 mtrs. In order to overtake A, B should gain
1100 mtrs. But B gains 2000 mtrs in 60 min.
The time taken by B to gain 1100 mtrs. =(60/2000)*1100=33 mins
ANSWER: a)33 mins
15.SOLUTION:
ANSWER: c)both a and b
16.SOLUTION:
Let x be the total distance
x/35-x/40=1/4
x=70
ANSWER: d)70
17.SOLUTION:
ANSWER: a)20
18.SOLUTION:
Assume the initial wages to be Rs.100/Then wages after reduction of 40% = Rs.60
After 40% increase on the reduced wages, new wages=(140/100)*60=Rs.84
Percentage of decrease on his initial wages = 100 – 84 = 16%
ANSWER: d)16%
19.SOLUTION:
If A scores 60 points, then B scores 50
If A scores 60 points, then C scores 45
When B scores 50 points, C scores 45. When B scores 90 points
C scores=(40/50)*90=81 points
Thus B can give C 9 points in a game of 90 points.
ANSWER: a)9
20.SOLUTION:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
ANSWER: c)85
21.SOLUTION:
A's 1 hour's work =
1
;
4
1
;
3
1
(A + C)'s 1 hour's work = .
2
(B + C)'s 1 hour's work =
1 1
7
+
= .
4 3
12
7 1
1
= .
12 2
12
(A + B + C)'s 1 hour's work =
B's 1 hour's work =
B alone will take 12 hours to do the work.
ANSWER: c)12 hrs
22.SOLUTION:
A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.
B's share = Rs.
25000 x
3
= Rs. 7,500.
10
ANSWER: a)Rs.7500
23.SOLUTION:
ANSWER: c)7:3
24.SOLUTION:
Largest 4-digit number = 9999
88) 9999 (113
88
---1199
88
---319
264
--55
--Required number = (9999 - 55) = 9944.
ANSWER: a)9944
25.SOLUTION:
Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.
= 36 x
5
18 m/sec
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Time taken =
360
= 36 sec.
10 sec
ANSWER: c)36 sec
26.SOLUTION:
C.P. = Rs.
100
x 392 = Rs.
122.5
1000
x 392 = Rs. 320
1225
Profit = Rs. (392 - 320) = Rs. 72.
ANSWER: c)Rs.72
27.SOLUTION:
Dividend on Rs. 20 = Rs.
9
9
x 20 = Rs. .
100
5
Rs. 12 is an income on Rs. 100.
Rs.
9
is an income on Rs.
5
100 9
x
= Rs. 15.
12
5
ANSWER: b)15
28.SOLUTION:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) =
n(E)
9
= .
n(S) 20
ANSWER: d)9/20
29.SOLUTION:
Angle traced by hour hand in 12 hrs = 360º.
Angle traced by hour hand in 5 hrs 10 min. i.e.,
ANSWER: c)155 deg
30.SOLUTION:
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
ANSWER: b)Saturday
31
hrs =
6
360 31 º
x
= 155º.
12
6