Quantitative Aptitude - Arithmetic Ability Questions

Let 's' be the number of small packets and 'b' the number of large packets sold on that day.

Therefore, s + b = 5000 ... eqn (1)

Each small packet was sold for Rs.150.
Therefore, 's' small packets would have fetched Rs.150s.

Each large packets was sold for Rs.250.
Therefore, 'b' large packets would have fetched Rs.250b.

Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)

Or 150s + 250b = 10,50,000 ... eqn (2)

Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)

Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000
Or b = 3000

We know that s + b = 5000
So, s = 5000 - b = 5000 - 3000 = 2000.

2000 small packets were sold.

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375

12% of 70,000

=> 12 x 70,000/100

=> 8400

Hence, 12% of 70,000 = 8,400

Since floor starts form the 1 therefore 2 will be the 1st floor.
Form 1 to 10 only 4,6,8,9,10 are not prime.
Form 11 to 20 only 14 ,16,18 are not prime also non of their digits are prime.
From 21 to 30 each number has a prime number so it must no be consider.
In between 40 to 49 we have to consider only 40,44,46,48,49.
50 to 59 follow either of the condition so we have too neglect it.
In between 60 to 68 60,64,66,68.
total option = 17.

Let the cost price of A as well as B is 100 Rs.
Then, ATQ:
Selling Price of A = 100+40 = 140
And Selling Price of B = 140 - (140*0.2) = 140 - 28 = 112
Total selling price = 140+112 = 252 Rs.
Total Cost price = Rs. 200
So by taking cost price = 100 Rs.
Total profit = 52 Rs.
Total profit will be Rs. 156 when cost price = (100/52)*156 = Rs. 300.

Except Horse, remaining all belong to female category.

Hind - Female Deer

Ewe - Female Sheep

Bitch - Female Dog. 

Literally any percentage means divide by 100.

So 125% = 125100. That can be reduced to 5/4.

To convert to a decimal, you can do long division, dividing 5 into 4.