Quantitative Aptitude - Arithmetic Ability Questions

Let S.P = x then C.P = 

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

                                            = 

Gain= 

Hence gain% =  = 20%

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

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.

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 

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

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x C36 ways = 6x5x4 = 120 ways.