Quantitative Aptitude - Arithmetic Ability Questions

$$\begin{gathered} 3 \\ 32 + 3 = 12 \\ 5 \\ 52 + 5 = 30 \end{gathered}$$

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.

When x=2. [(22 + 2) = 24 is divisible by 3]

So, the answer is 2

Let the Father be x years when he died

Average Age 10 years ago be A

Total Age 10 years ago = 6*A

Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60

Father Died and Baby was born => the Total number of people in the family is Same (6)
Baby born today so age of baby = 0

(6A +60 - x)/6 = 6A/6
=> A + 10 -(x/6) = A
=> x/6 = 10
=> x = 60

Therefore we can conclude that the father was 60 years old when he died.

required percentage = πr22πr×100 =250%

Let the number of candidates appeared from each state be x.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000