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.

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 = 

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

A can complete the work in 12 days working 8 hours a day

=> Number of hours A can complete the work = 12×8 = 96 hours 

=> Work done by A in 1 hour = 1/96

B can complete the work in 8 days working 10 hours a day

=> Number of hours B can complete the work = 8×10 = 80 hours 

=> Work done by B in 1 hour = 1/80

Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480

 => A and B can complete the work in 480/11 hours

A and B works 8 hours a day.

Hence total days to complete the work with A and B working together 

= (480/11)/ (8) = 60/11 days

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

Let rate = R% and time = R years.

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

=> R = 6.

Let the present age of Ravali = x

=> Swarna's present age = x - 9

From the given data,

x + 7 = 2(x - 9 + 7)

=> x = 11 yrs

Required Ravali's age after 4 years = 11 + 4 = 15 yrs.

Annual salary of Arun = 7,68,000 Rs.
Monthly salary = 7,68,000/12 = Rs. 64,000
Spending on children = Rs. 12,000
Rest = 52,000
1/13th of the rest = 52,000/13 = Rs. 4,000 is spent on food.
Rs. 8,000 is spent in mutual funds.
Monthly savings = 64,000 - (12,000+4,000+8,000) = Rs. 40,000.

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.