Quantitative Aptitude - Arithmetic Ability Questions

log4928 = 12log77×4

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

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

 P(E)=n(E)n(S)=636=16

A                   B                 C

3x                 4x                5x 

(3x+x)          2x               (5x+x)         (B Gives 1/4 to A and 1/4 to C) 

=4x              2x                 6x 

(4x+x)          2x                 5x              ( C Gives 1/6 to A) 

=5x              2x                 5x

Therefore, 5 : 2 : 5

We have to rearrange the equation to make R the subject.

Start by cross multiplying by (r + R); V (r + R) = 12R

Multiply out the bracket Vr + VR = 12R

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

Let the numbers be p & q

q-p/4 = 4q/6

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

=> 2q/6 = p/4 

=> p/q = 4/3

1 hectare = 10,000 Sq.m

So, Area = (1.5 x 10000) Sq.m

Depth = 5/100 m = 1/20 m

Therefore, Volume = Area X Depth = 15000 x (1/20)= 750 cu.m

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