Quantitative Aptitude - Arithmetic Ability Questions

 $$\begin{gathered} = log 33 + log 23+ log 103log10×3×22 \\ \end{gathered}$$ 

=log33 12+log 23+log 10312log(10×3×22)  

=12log 33+3 log 2+12 log103log10+log3+log22  

=32log 3 + 2 log 2 + log 10log 3 + 2 log 2 + log 10 = 32  

Let 'B' be the nuber of bricks.

=> 10 x 4/100 x 5 x 90/100 = 25/100 x 15/100 x 8/100 x B

=> 10 x 20 x 90 = 15 x 2 x B

=> B = 6000

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.

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.

This can be solved as

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

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.

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

Now, required probability = 

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