Quantitative Aptitude - Arithmetic Ability Questions

Let x be the number.

Given: x = 296n + 75 [where n is quotient] 

 --> (378n) +( 372 + 1) 

 -->37(8n + 2) + 1

Hence, when x divided by 37, 1 is the remainder.

The number of letters in the given word CREATIVITY = 10

Here T & I letters are repeated

=> Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200

This question is a combination since order is not important.

Answer = 7C3 = 126

Given that to find 15 of 80

=> What % of 80 is 15

Let it be 'p'

=> p% of 80 = 15

=> $$\begin{gathered} p x 80100 = 15 \\ ⇒p = 15 x 10080 \\ ⇒p = 150080 = 18.75 \end{gathered}$$

Therefore, 15 is 18.75% of 80 => 18.75% of 80 = 15

Example::

Now, in the similar way we can find 15% of 80 =?

=> 15x80/100 = p

=> p = 4 x 3 = 12

Therefore, 15% of 80 = 12

T.D. =Ö(P.W.*B.G)

B.G. =(T.D.)2/ P.W.      

       = Rs.[(110x110)/ 1100]

       = Rs. 11.

B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121.

Total number of possible ways = 

9C5 = 9 x 8 x 7 x 6 x 55 x 4 x 3 x 2 x 1 = 126

Number of favorable cases = 

$$\begin{gathered} 4C2 x 5C3 + 4C3 x 5C2 + 4C4 x 5C1 \\ 4 x 32 x 1 x 5 x 4 x 33 x 2 x 1 + 4 x 3 x 23 x 2 x 1 x 5 x 4 2 x 1 + 1 x 5 \\ = 6 x 10 + 4 x 10 + 5 \\ = 60 + 40 + 5 \\ = 105 \end{gathered}$$

Therefore, required probability = 105/126 = 5/6

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. 

Area of each tile = (41 x 41) sq.cm 

Required number of tiles = 1517 x (902/41) x 41 = 814