Quantitative Aptitude - Arithmetic Ability Questions

SP of first article = 1000

Profit = 20%

CP = 100100+P×SP= 5000/6 = 2500/3

SP of second article = 1000

Loss = 20%

CP = 100100-L×SP= 5000/4 = 1250

Total SP = 2000

Total CP = 25003+ 1250 = 6250/3

CP is more than SP, he makes a loss.

Loss = CP-SP = (6250/3)- 2000 = 250/3

Loss Percent =250362503×100= 4%

When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.

When B runs 900 m, distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 m.

In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.

In a race of 600 m, the number of meters by which A beats C

= (600 x 212.5)/1000 = 127.5 m.

Let the numbers be x, x+1, x+2 

Then x + x+1 + x+2 = 42

           =>  3x + 3 = 42

           => 3x = 39 

          =>  x = 13

The numbers would be 13, 14, 15 

Therefore, Sum of the digits of middle number = 1+4 = 5

P(black ball)=3/12

P(red ball)=5/12

P(black or red)=3/12+5/12=2/3

Let X = Time taken for each of Type B Problems(100 Problems)
And 2X = Time taken for each of Type A problems(50 problems)

Total time period = 3hrs = 3 x 60min = 180 minutes

100X + 50(2X) = 180
100X + 100X = 180
200X = 180 min

X = 180/200
X = 0.90 min

By convertiing into seconds,

X = 0.90 x 60 seconds
X = 54 sec

So, time taken for Part A problems is = 54 x 2 x 50 = 5400 seconds
= 5400/60sec = 90 minutes.

From above diagram
AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )
angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1
100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2