Quantitative Aptitude - Arithmetic Ability Questions

75% of 450 + ? % of 560 = 600 

$$\begin{gathered} 75 x 450100 + ? x 560100 = 600 \\ 75 x 92 + 28p5 = 600 \\ 75x9x5 + 56p10 = 600 \\ 3375 + 56p = 6000 \\ 56p = 2625 \\ p = 46.875 \end{gathered}$$

Hence, in the given question ? = 46.875

Let the three parts M, N, O.
Then, M:N = 2:3 and N:O = 5:8
= (5 x 3/5) : (8 x 3/5) = 3:24/5

M:N:O = 2:3:24/5

= 10:15:24 => N = 98 x 15/49 = 30.

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.

This can be solved as

$$\begin{gathered} 20 1 - 4202 = 20 1 - 152 \\ = 20452 \\ = 20 1625 \\ = 645 \\ = 12.8 \end{gathered}$$

I
= ptr

Therefore, the total amount the owner received on January 17, 2001 was

$1295.57 + $10.47 =
$1306.04

Water required = 4000 x 140 = 560000 lit = 560 cu.m (1 cu.m = 1000 lit)

Volume of tank = 18 x 12 x 8 = 1728 cu.m

Water of this tank will suffice for = 1728/560 = 3 days.

n (A) = (60/100 * 96) = 288/5, n (B) = (30/100 * 96) = 144/5, n (A ∩ B) = (15/100 * 96) = 72/5.

n(A ∪ B) = n (A) + n (B) - n (A ∩ B) = 72.

So, people who had either or both types of lunch = 72.

Hence, people who had neither type of lunch = (96 - 72)= 24.

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

Now, required probability = 

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