Quantitative Aptitude - Arithmetic Ability Questions

4b=6c⇒b=32c and 3a=4b⇒a=43b=43(32c)=2c 

a+b+c=2729 ⇒2c+32c+c=2729 ⇒c=629

 ∴a2+b2+c2=(a+b+b)2-2(ab+bc+ca) 

=(2729)2-22c×32c+32c×c+c×2c 

 =729×29-2×132 c2

= 87

142 x 142 x 11260.25

=> 142 x 142 x 135.5 = 568

Total Combination of getting a card from 52 cards = C152

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= C151

Total combination of getting 2 cards from 52 cards without replacement= (C152×C151) 

There are total 4 Ace in stack. Combination of getting 1 Ace is = C14

Because there is no replacement, So number of cards after getting first Ace = 3

Combination of getting an another Ace = C13

Total Combination of getting 2 Ace without replacement=C14×C13

Now,Probability of getting 2 cards which are Ace =C14×C13C152×C151 = 1/221.

Let the number of boys = b

Let the number of girls = g

From the given data,

81b + 83g = 81.8(b + g)

81.8b - 81b = 83g - 81.8g

0.8b = 1.2g

b/g = 1.2/0.8 = 12/8 = 3/2

=> g : b = 2 : 3

Hence, ratio between the number of girls to the number of boys = 2 : 3.

Total strength of the class is given by

15 + 44 + 4 + 3 - 1 = 65

It takes four of those little squares -- each one 1/2 inch on a side -- to cover one square inch. That's because each one is 1/2 x 1/2, or 1/4 of a square inch.
In a rectangle that is 48 inches by 48 inches, there are 2304 square inches.(1 ft = 12 inches)
Since it takes 4 little squares to cover 1 sq inch, then it would need 2304 x 4 squares (1/2 inch on a side each) to cover a rectangle 4 feet by 4 feet.

That is 9,216 of those squares.

8000 × 33.1% = 2648

729 = 9 x 9 x 9

=> Cube root of 729 = 9

Now, required square of 9 = 9 x 9 = 81.