Quantitative Aptitude - Arithmetic Ability Questions

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!

Total number of cases=p+q

Favourable cases=p

Probability of that event is=p/(p+q)

B.G. =(T.D.)^2/P.W.

      = Rs. (160*160)/1600 = Rs. 16.

Except 437 all others are prime numbers.

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)

Given total number of passengers in the bus = 45

First average weight of 45 passengers = 52 kgs

Average weight of 5 passengers who leave bus = 48

Average weight of 5 passengers who joined the bus = 54

Therefore, the net average weight of the bus is given by

$$\begin{gathered} = 45 x 52 - 5 x 48 + 5 x 5445 \\ = 237045 \\ = 1583 \\ = 52.66 kgs. \end{gathered}$$

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

% Change in Curved Surface Area is given by 

%Change in radius + %Change in height + %Change in radius × %Change in height 100  

⇒+20 + 15 + 20×15100 = 38%