Quantitative Aptitude - Arithmetic Ability Questions

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

 Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.

Given a secret is told to 3 persons in 3 minutes.
And if the process continues with each person saying 3 other persons,
In 30 minutes => the rocess for 10 times because each person time is 3 minutes.
Required number of persons = 3 + 32 + 33 + .... + 39 + 310

Now it forms a geometric progression,

Sum to n terms is S = arn - 1r - 1

Here a = 3, r = 3

 
=> 59048 x 3 /2

= 88572.

For this we have to find the LCM of 24, 36 and 30

LCM of 24, 36 and 30 = 360 sec

360/60 min = 6 minutes.

I=(p x r x t)/100

Total strength of the class is given by

15 + 44 + 4 + 3 - 1 = 65

When 16nis divided by 9, we have

 1619, remainder = 7  

1629, remainder = 4 

1639, remainder = 1 

1649, remainder = 7 

1659, remainder = 4 

1669, remainder = 1 

So, we have cyclicity of 3 factors i.e 7,4,1. 

Hence only 3 remainders are possible.