Quantitative Aptitude - Arithmetic Ability Questions

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.

Let rate = R% and time = R years.

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

=> R = 6.

Given in the question that, there are 7 boys and 6 girls. 

Team members = 5

Now, required number of ways in which a team of 5 having atleast 3 girls in the team = 

$$\begin{gathered} 6C3 x 7C2 + 6C4 x 7C1 + 6C5 \\ = 6x5x43x2x1 x 7x62x1 + 6x5x4x34x3x2x1 x 7 + 6x5x4x3x25x4x3x2x1 \\ = 420 + 105 + 6 \\ = 531. \end{gathered}$$

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

Flat rate = 12%
n = 4 × 4
= 16

Effective rate =2n/(n+1) × flat rate

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.

Total number of sweets

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

= 960 + 100 = 1060

T = (100 x I)/ (P x r)