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

ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

Here as given Q is in the middle and it satisfies all other conditions given in the question.

Multiplying two negatives (or any even multiple) results in a positive. But multiplying three negatives (or any odd multiple) gives a negative. If the result of multiplying 6 negatives is odd, the largest number of negative integers will be the largest odd number (i.e.5)

The correct pattern is +1, -2, +3, -4, +5, . . . .
So,129 wrong and must be replaced by 123 + 5 = 128

Except 437 all others are prime numbers.

Truck and Ship are entirely different. But some Goods are carried by some Trucks and some Goods are carried by some Ships.