Quantitative Aptitude - Arithmetic Ability Questions

Let the rquired number be 'P'

According to the question,

$$\begin{gathered} 100 x P100 = 20 \\ => P = 20 x 100100 = 20 \end{gathered}$$

Hence, 100 percent of the number 20 is 20.

Now, you can also try for

20 percent of what number is 100

$$\begin{gathered} 20 x P100 = 100 \\ => P = 100 x 10020 = 500 \\ => P = 500 \end{gathered}$$

Hence, 20 percent of the number 500 is 100.

Let the total applicants be 100

Then, 20% got 85 marks

i.e 20 x 85 = 1700

and 25% got 95 marks

 i.e 25 x 95 = 2375

Now, the remaining applicants are 55 and let the average marks scored by them be x.

Therefore, 2375 + 1700 + 55x  =  60×100 

6000 - 4075 = 55x

55x=1925

x=35.

1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability = 

$$\begin{gathered} 5C214C2 + 3C214C2 \\ = 1091 + 391 \\ = 1391 \\ = 17 \end{gathered}$$

0.46 = 46/100 = 23/50.

Sum of the numerator and denominator is 23 + 50 = 73

Assume Hebah has Rs. M

Since 25% more money at Poonam

=> Money at Poonam = M + (25% of M)

=> Money at Poonam = M + 0.25M = Rs. 1.25M

Money with Navaneet is thrice the money with Poonam,

=> Money at Navaneet = 3(1.25M) = Rs. 3.75M

Now, sum of all Money = (M + 1.25M + 3.75M) = Rs. 6M

But given the average of the money is Rs. 350

=> 6M/3 = 350

=> 2M = 350

=> M = 350/2 = Rs. 175

=> Amount of money Navaneet has = Rs. 3.75M = 3.75 x 175 = Rs. 656.25.

NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions.

And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions.

Therefore, Can't be determined.