Quantitative Aptitude - Arithmetic Ability Questions

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can  = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37 

let sum =x . s.i.=x

rate =(100*x)/(x*6)=50/3%

now ,sum =x, s.i= 3x, rate =50/3%

time =(100*3x)/(x*50/3) = 18 years

Let the numbers be p & q

q-p/4 = 4q/6

=> q-4q/6 = p/4

=> 2q/6 = p/4 

=> p/q = 4/3

We know that Time is inversely proportional to Efficiency

Here given time ratio of P & P+Q as

P/P+Q = 150/100 = 3:2

Efficiency ratio of P & Q as

P:Q = 2:1 ....(1)

Given Efficiency ratio of Q & R as

Q:R = 3:1 ....(2)

From (1) & (2), we get

P:Q:R = 6:3:1

P alone can finish the work in

22.5(3+1)/6 = 15 days.

At the moment the paint is 3 liters of paint and 3 liters of water. We need to add to this to make the new ratio 2 liters paint to 1.5 liters water. As this is equivalent to 4 : 3 we can see that we have the right amount of water, and just need 1 liter of paint to make it correct.

There are 15 dots in total,and to make a triangle we need to select any three of those dots.

So,15C3 = 455

log4928 = 12log77×4

= 12+122log72
= 12+log72
= 12 + m
= 1+2m2.

A has to pay = Present worth of Rs220 due 1yr hence  

= Rs.[(220 x 100)/100+(10 x 1)] = Rs.200 

A actually pays = Rs.110+PW  of Rs.110 due 2 yrs hence 

= [110+(110 x 100)] / [100+(10 x 2)] = Rs.192.66  

So, A gains Rs. (200-192.66) = Rs.7.34