Quantitative Aptitude - Arithmetic Ability Questions

6P4 = 6! / (6-4)! = 360

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

Let C.P =Rs. x

Then as given, (753 - x ) = (x - 455)

=>2x = 1208 => x = 604

 There fore S.P= 150% of 604 => Rs. 150100×604= Rs.906

Area of square = 40 x 40

= 1600 sq.cm

Given that the areas of Square and Rectangle are equal

=> Area of rectangle = 1600 Sq.cm

We know that, Area of rectangle = L x B

Given L = 64 cm

Breadth of rectangle = 1600/64 = 25 cm

Perimeter of the rectangle = 2(L + B) = 2(64+25) = 178 cm.

Let their present ages be 13x and 17x.
Then, 13x-4/17x-4 = 11/15.
Solving this, we get x = 2.
Required ratio = 13x2 + 6/17x2 + 6 = 32/40 = 4/5.

Given total number of boys in the school = 1500

In that 35% are Hindu, 30% are Sikh, 15% are Muslims

=> 35 + 30 + 15 = 80%

Remaining Boys = 100% - 80% = 20% of 1500

=> 20 x 1500/100 = 300 boys are Christians. 

Area of the square = 12×diagonal2=  1/2 * 3.8 * 3.8 = 7.22 sq.m