Aptitude and Reasoning Questions

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in C16, ways.

Now select two distinct number out of remaining 5 numbers which can be done in C25 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

 C16×C25×4!2!=720

 =>  n(E) = 720

 ∴PE=72064=59

Given, y/(x - z) = (y + x)/z = x/y
yz = xy + x2 - yz - xz ....(1)
Also xy = yx-z ⇒x2 -xz = y2 ....(2)
Using (1) and (2), we get yz = xy - yz + y2
2yz = xy + y2
2z = x + y
Only option (B) satisfies the equation.

As Microphone makes sound louder similarly Microscope makes the object magnified.

Hence, Microphone : Loud :: Microscope : Magnify.

Given number of girls are twice that of boys.

--> G=2B and G+B = 60 

Therefore,  G=40 and B= 20

Given Laxmi ranked 27th from the top and 9 boys are ahead of Laxmi.

Therefore, There are 17 girls ahead of Laxmi. 

Therefore, Number of girls who ranked after Laxmi = 40 - (17+1) = 22.

Given total number of cows = k
Now, 1st son share = k/2
2nd son share = k/4
3rd son share = k/5
4th son share = 9

(k) + (k/4) + (k/5) + 9 = k
=> k - (19k/20) = 9
=> (20k-19k)/20 = 9
=> k = 180.

From the above fig we can observe that distance between Laxmi and her office is 15-11 = 4 kms.

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

 A = { T5, T6 }

 B = { T1, T2, T3, T4, T5, T6 }

Therefore, Required probability = PAB=PA∩BPB=2868=13