Quantitative Aptitude - Arithmetic Ability Questions

P1+r1004P1+r1005=11+r100 =100100+r=100120=56 

First letter can be posted in 4 letter boxes in 4 ways. Similarly second letter can be posted in 4 letter boxes in 4 ways and so on.

Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024

From the above formula i.e (m1 x t1/w1) = (m2 x t2/w2)

so, [(34 x 8 x 9)/(2/5)] = [(M x 6 x 9)/(3/5)]

so, M = 136 men

Number of men to be added to finish the work = 136-34 = 102 men.

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

Hence number of poles required = 280 / 5 = 56

Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5

Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively

They can do work for 12 days.

=> Total work = 12 x 16x = 192x

Now, the required time taken by Q to complete the job alone = 162x8x = 24days.

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

Let, Ages of Vijaya and Simran – 1 year ago was 4A and A years correspondingly.
Then, [(4A + 1) + 6] – [(A + 1) + 6] = 9
3A = 9  => A = 3

Needed Ratio = (4A + 1) : (A + 1) = 13 : 4

Total Items = 7
Discounted = 1
Percentage = 1/7 x 100
= 14.285%