Quantitative Aptitude - Arithmetic Ability Questions

Let A’s investment be= 7a
Let B’s investment be= 6a
Let C’s investment be= 5a
Their investment in 2ndyear:= (125% of 7a): (120% of 6a) : (115% of 5a) = 175 : 144 : 115

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.

Mass of zinc ore = 8000 kg

Mass of metal = 60% of 8000 = 4800 kg

Mass of silver in metal = 3/4 x 4800/100 = 36 kg

Mass of zinc = 4800 - 36 = 4764 kg.

Let
x = peacocks
y = deers

2x + 4y = 200
x + y = 72
y = 72 - x
2x + 288 - 4x = 200
x = 44 peacocks
y = 28 deers

Let present ages of Sushanth = w, Krish = x, Rishi = y and Rohit = z , then
w + x + y = 43 x 3 = 129 ---(i) and
w + y + z = 49 x 3 = 147 ---(ii)
Subtracting (i) from (ii), (w+y+z) - (w+x+y) = 147 - 129 , z - x = 18 ---(iii)
Given Rohit's age=z=54, so from (iii), x = 54 - 18 = 36.
Therefore, Krish age is 36 years.

With slight modifications, the basic formula can be made to deal with compounding at intervals other than annually.

Since the compounding is done at six-monthly intervals, 4 per cent (half of 8 per cent) will be added to the value on each occasion.

Hence we use r = 0.04. Further, there will be ten additions of interest during the five years, and so n = 10. The formula now gives:

V = P(1 + r)10 = 5,000 x (1.04)10 = 7,401.22

Thus the value in this instance will be £7,401.22.

In a case such as this, the 8 per cent is called a nominal annual

rate, and we are actually referring to 4 per cent per six months.

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 = 

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