Quantitative Aptitude - Arithmetic Ability Questions

If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.

The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.

We have 4 eights and 4 sevens.

We want 3 eights and 2 sevens.

C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens) 

C(4,3) x C(4,2) = 24

Therefore there are 24 different ways in which to deal the desired hand.

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9C4 ways =126

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

Given the average age of couple and their son = 40

=> Sum of age (H +W +S) = 40×3

Let the age of daughter in law at the time of marriage = D years

Now after 10 years,

(H + S +W) + 3×12 + D +12 + 10 = 38×5

178 + D = 190

D = 190 -178 = 12 years

Because each vertice of the square is the center of a circle
(1/4) part of the total area of each circle inside of the square.
(3/4) part of the total area of each circle outside of the square.
Thus total area outside the square is 434×227×3.5×3.5 = 115.3955 sq cm.

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

therfore, on 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

Given Word is PROMISE.

Number of letters in the word PROMISE = 7

Number of ways 7 letters can be arranged = 7! ways

Number of Vowels in word PROMISE = 3 (O, I, E)

Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways

Now, the number of ways of arrangements so that three vowels should not come together

= 7! - (5! x 3!) ways = 5040 - 720 = 4320.

Let the three numbers be x, y, z.

From the gien data,

x = 2y ....(1)

x = z/2 => z = 2(2y) = 4y .....(From 1)  ...........(2)

Given average of three numbers = 56

Then,

$$\begin{gathered} x + y + z3 = 56 \\ 2y + y + 4y3 = 56 (From 1 & 2) \\ 7y = 56 x 3 \\ y = 24 \end{gathered}$$

Now,

x = 2y => x = 2 x 24 = 48

z = 4y = 4 x 24 = 96

Now, the highest number is z = 96 & smallest number is y = 24

Hence, required sum of highest number and smallest number

= z + y

= 96 + 24

= 120.