Aptitude

Showing 11 - 20 of 116 questions

11

Sum of present ages of A, B and C is 92 years. If 4 years ago, the ratio of their ages were 1:2:3 respectively, find A’s present age.

a

18.5 year

b

14.8 years

c

17.3 years

d

20.3 years

Sum of present ages of A, B and C is = 92 years

Therefore , Sum of their ages 4 years ago = 92 – (4 * 3)= 80 years.

4 years ago ratio of the ages of A , B and C was = 1:2:3

Therefore, A’s age four years ago = 1/6 * 80 = 13.3 years.

So , A’s present age =13.3 + 4 = 17.3 years

12

Mayur takes thrice as long to row a distance against the stream as to row the same distance in favor of the stream. The ratio of the speed of the boat(in still water) and the stream is:

a

3:1

b

1:3

c

2:1

d

1:1

Let speed in still water be x and speed of stream be y

(Speed in still water) : (Speed of stream)=x:y

Distance is same, then time taken downstream is t, and time taken to travel upstream is 3t.

d=d

[(x+y) / t] = [(x-y) / (3t)]

x+y= 3x-3y

2x = 4y

x/y = 4/2 = 2/1

x:y = 2:1

13

If 120 km is done by train and the rest by car, then it takes 8 hours for a 600 km journey. Of 200 km is done by train and the rest by car, then it takes 20 minutes more. Then what would be the ratio of the speed of the train to that of the car?

a

3:4

b

5:4

c

4:3

d

3:5

Let speed of the train be x km/hr and of the car be y km/hr.

Then, 120/x + 480/y = 8

=> 1/x + 4/y = 1/5    --- (i)

And, 200/x + 400/y = 25/3

=> 1/x + 2/y = 1/24    --- (ii)

Solving (i) and (ii), we get x = 60 and y = 80.

Ratio of speeds = 60: 80 = 3:4

14

2 years ago A’s age was 6 times of B’s age. 6 years after the ratio between the ages of A and B becomes 10 : 3. What is A’s present age?

a

44

b

38

c

42

d

34

Older ratio = 6:1

Let 2 years ago ages of B and A were x and 6x respectively.

[ (6x+8)/ (x+8)] = 10/3

18x +24 = 10x + 80

8x = 56

x=7 years

Therefore , 2 years ago , A’s age was 6 * 7 = 42 years

Therefore, A’s present age = 42 + 2 = 44

15

Find the product of

(8.77 xx 8.77xx 8.77+ 4.23xx 4.23 xx 4.23)/(8.77 xx 8.77 -8.77 xx 4.23+ 4.23 xx 4.23)   ?

a

11

b

4.54

c

8.77

d

13

= ((8.77)^3 + (4.23)^3) / ((8.77)^2 - (8.77 xx 4.23)+ (4.23)^2)

= (a^3 + b^3)/ (a^2 - ab+ b^2)

= (a+b)

= 8.77+4.23

= 13

16

A guy bought 10 pencils for Rs. 50 and sold them for Rs. 60.What is his gain in terms of percentage?

a

10%

b

5%

c

20%

d

12%

"Gain%"=("Gain"/"C.P")*100=20%

17

3 years back average age of A and B was 32. Today average age of A,B and C is 30. What is the age of C?

a

5

b

10

c

15

d

20

3 years back average age of A and B is

((A-3)+(B-3))/2 = 32

Hence A+B=70

Today average of A,B and C is 30 i.e.

(A+B+C)/3=30

A+B+C=90

Hence C=A+B+C-A-B=20

18

What is the unit digit in 2^30?

a

3

b

4

c

1

d

2

Cyclicity of 2 = 4
D = d * Q + R
30 = 4 * 7 + 2
30/4 = Remainder 2, 22 = 4

Notes:
Cyclicity of 2, 3, 7, 8 => 4
Cyclicity of 4, 9 => 2
Cyclicity of 1, 5, 6 => 1

19

If each edge of a cube is increased by 50%, find the percentage increase in its surface area.

a

125%

b

150%

c

175%

d

110%

Let the edge = a cm

So increase by 50% = a+a/2 = 3a/2

Total surface Area of original cube = 6a^2

TSA of new cube = 6(3a/2)^2

6 xx 9a^2/4 = 13.5a^2

Increase in area = 13.5a^2 - 6a^2 = 7.5a^2

7.5a^2 Increase % = (7.5a^2)/(6a^2) xx 100 = 125%

20

I walk a certain distance and ride the car back taking a total time of 33 Minutes. I could walk both sides in 45 min. How long would it take me to ride both ways?

a

19 mins

b

20 mins

c

21 mins

d

22 mins

Time taken in walking a certain distance from X and Y = 22 1/2 Min
Time taken in riding the same distance = 33 - 22 1/2 = 10 1/2 Min
It will take 21 Min (10 1/2 * 2) to ride both ways.