Quantitative Aptitude - Arithmetic Ability Questions

Since floor starts form the 1 therefore 2 will be the 1st floor.
Form 1 to 10 only 4,6,8,9,10 are not prime.
Form 11 to 20 only 14 ,16,18 are not prime also non of their digits are prime.
From 21 to 30 each number has a prime number so it must no be consider.
In between 40 to 49 we have to consider only 40,44,46,48,49.
50 to 59 follow either of the condition so we have too neglect it.
In between 60 to 68 60,64,66,68.
total option = 17.

Ratio in which the profit will divide – 23:15. Let theprofit be P

Now, [(23/38) - (15/38)] x (60/100) x P = 500

P = 3958.34

let, speed of Ali=X mph and Faizer=Y mph

Now, relative speed when Ali is walking int the same direction with Faizer= (X-Y)mph
hence, according to question, 27/(X-Y)=9 ; solving X-Y=3; --- (1)

Now relative speed when walking in the opposite direction= (X+Y) mph
hence, according to question, 27/(X+y)=3; solving X+Y=9; --- (2)

solving 1 and 2 we get X=6 and Y=3, where X is the speed of Ali.

Let the required number be 'p'

36 of what number is 18 implies 36% of p = 18

=> 36 x p/100 = 18

=> p = 1800/36

=> p = 50.

Hence, 36% of 50 is 18.

Three digit number will have unit’s, ten’s and hundred’s place.

Out of 5 given digits any one can take the unit’s place.

This can be done in 5 ways. ...              (i)

After filling the unit’s place, any of the four remaining digits can take the ten’s place.

This can be done in 4 ways. ...              (ii)

After filling in ten’s place, hundred’s place can be filled from any of the three remaining digits.

This can be done in 3 ways. ... (iii) 

So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60

Let the number of males be 'M'
Let the number of females be 'F'
From the given conditions,
(F-15)=2M.....(i)
F=5(M-45).....(ii)
By solving the equations (1) & (2) we get M = 80
Then substitute M = 80 in (1)
F = 2M + 15
F = 2 x 80 + 15
=> F = 175
Therefore, number of males = 80 and females = 175 in the party.

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375

Sol : In this kind of questions we find the work force required to complete the work in 1 day (or given unit of time) then we equate the work force to find the relationship between the efficiencies (or work rate) between the different workers.

Therefore, 6B+8G = 6 days

 => 6(6B+8G)= 1 day  (inversely proportional)

 => 36B+48G =1         ( by unitary method)

Again  14B+10G = 4days

 => 56B+40G =1

so, here it is clear that either we employ 36B and 48G to finish the work in 1 day or 56B and 40G to finish the same job in 1 day. thus , we can say

 => 36B+48G = 56B+40G

 => G= 2.5B

Thus a Girl is 2.5 times as efficient as a boy.

Now, since  36B+48G = 1

 => 36B+48(2.5 B)=1

 =>156B=1

i.e., to finish the job in 1 day 156 boys are required or the amount of work is 156 boys-days

Again    1G+1B=2.5B+1B=3.5B

Now, since 156 boys can finish the job in 1 day

so 1 boy can finish the job in 156 days

Therefore, 3.5 boys can finish the job in 1×1563.5= 47 days.