Quantitative Aptitude - Arithmetic Ability Questions

Let S.P = x then C.P = 

Selling price received by subbu = (108% of x)

                                            = 

Gain= 

Hence gain% =  = 20%

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

A : B = 100 : 75

B : C = 100 : 96.

A : C = (A/B*B/C) = (100/75*100/96) = 100/72  =100:72     

   A beats C by (100 - 72) m = 28 m.

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

therfore, on 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

Concentration of glucose are in the ratio = 12:35:45

Quantity of glucose taken from A = 1 liter out of 2

Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit

Quantity of glucose taken from C = 0.8 lit

So, total quantity of glucose taken from A,B and C = 3.6 lit

So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit

Ratio of glucose to alcohol = 3.6/2.4 = 3:2

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9C4 ways =126

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

a^3 + b^3 = (a+ b) * (a^2 + b^2 - ab)

[ (10.3)^3 + (1)^3 ] / [(10.3)^2  + (1)^2 - (1 * 10.3)]

 => a + b = 10.3 + 1 = 11.3 

At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.

FV=P(1+r/n)^nt