Quantitative Aptitude - Arithmetic Ability Questions

Let rate = R% and time = R years.

Then, (1200 x R x R) / 100 = 432

=> R = 6.

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.

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

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

363650.

sin 46. cos 44+ cos 46. sin 44

sin 46. sin (90 -44)+ cos 46. cos (90 -44)

= sin^2 46 + cos^2 46

= 1

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

Flat rate = 12%
n = 4 × 4
= 16

Effective rate =2n/(n+1) × flat rate

Let the amount at 6% be x
Amount at 10% be (10000 –x)
At 9.2 % interest, SI = 10000 X 9.2 X 1/100 = 920
Amount = 10920
⇒ 6x/100 + (10000 – x)10/100 = 920
⇒ 6x – 10x + 100000 = 92000
⇒ 4x = 8000
⇒ x = 2000
Money lent at 10 % = 10000 – 2000 = 8000