Quantitative Aptitude - Arithmetic Ability Questions

Days remaining 124 – 64 = 60 days

Remaining work = 1 - 2/3 = 1/3

Let men required men for working remaining days be 'm'

So men required = (120 x 64)/2 = (m x 60)/1 => m = 64

Men discharge = 120 – 64 = 56 men.

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

Let wholesaler dealers marked price = 100%, Retailer's C.P = 80%
And the retailer sells at 5% less than the marked price => S.P = 95%
If S.P of 95% of the retailer costs Rs.19 to customer,so its C.P of 80% will cost 80 x 19/95 = 16
Profit made by the retailer = 19-16 =  Rs.3

Given total number of cows = k
Now, 1st son share = k/2
2nd son share = k/4
3rd son share = k/5
4th son share = 9

(k) + (k/4) + (k/5) + 9 = k
=> k - (19k/20) = 9
=> (20k-19k)/20 = 9
=> k = 180.

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

Let the sum be Rs. 100. Then,

S.I. for first 6 months = Rs.[ (100 x 10 x 1)/(100 x 2) ]= Rs.5

S.I. for last 6 months =Rs.[(102 x 10 x 1)/(100 x 2) ] = Rs.5.25

So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25

Effective rate = (110.25 - 100) = 10.25%

Given, y/(x - z) = (y + x)/z = x/y
yz = xy + x2 - yz - xz ....(1)
Also xy = yx-z ⇒x2 -xz = y2 ....(2)
Using (1) and (2), we get yz = xy - yz + y2
2yz = xy + y2
2z = x + y
Only option (B) satisfies the equation.

Except 437 all others are prime numbers.