Quantitative Aptitude - Arithmetic Ability Questions

Total number of sweets

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

= 960 + 100 = 1060

$$\begin{gathered} Given 3.2% of 500 × 2. 4% of ? = 312 \\ Then, \\ 3.2 x 500100 x 2.4 x ?100 = 312 \\ 16 x 2.4x?100 = 312 \\ 9.6 x ?25 = 312 \\ 9.6 x ? = 7800 \\ ? = 78009.6 \\ ? = 812.5 \end{gathered}$$

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

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

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

363650.

L's 10 days work = 115×10=23

Remaining work = 1-23=13

Now,  work is done by K in one day = 1/18

1/3 work is done by K in  18×13 = 6 days

Cp1 = p2 - p1 ,
Cp2 = p3 - p2
..
..
..
Cp23 = p24 - p23
Sum of series = (p2-p1) + (p3-p2) + .....(p23-p22) + (p24-p23)
All terms get cancelled, except p1 = 2 and p24 = 89
So Sum = -p1 + p24
Sum of series = -2 + 89 = 87

Let the number be x. Then, x - 3x/5 = 50  <=>  x = (50 * 5) / 2 = 125..

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.