Quantitative Aptitude - Arithmetic Ability Questions

LCM of 9,10,12,15 is 180
Smallest natural number to get remainder of 3 is (180 + 3) = 183

Total number of sweets

= 80 x 15 x 80/100 + 5 x 80 x 25/100

= 960 + 100 = 1060

S=R[(1+i)^n-1)/i]

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18
Multiples of 5 below 20 are 5, 10, 15, 20
Required number of possibilities = 10
Total number of possibilities = 20
Required probability = 10/20 = 1/2.

As per given data, P = 415600, R1= 25% incresed, R2 = 30% decreased

Population of the city at the end of the second year= 

363650.

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.

Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 – 20 = 60 litre water to be added.