Quantitative Aptitude - Arithmetic Ability Questions

Let the length of the 1st train = L mts

Speed of 1st train = 48 kmph

Now the length of the 2nd train = L/2 mts

Speed of 2nd train = 42 kmph

Let the length of the bridge = D mts

Distance = L + L/2 = 3L/2

Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)

Time = 12 sec

=> 3L/2x25 = 12

=> L = 200 mts

Now it covers the bridge in 45 sec

=> distance = D + 200

Time = 45 sec

Speed = 48 x5/18 = 40/3 m/s

=> D + 200/(40/3) = 45

=> D = 600 - 200 = 400 mts

Hence, the length of the bridge = 400 mts.

Taking Log to both sides 

 we get

 n = 8           

From given data,

R : M = 5 : 7

M : A = 5 : 7

R : M : A = 25 : 35 : 49

25 + 35 + 49 = 109

Let the place be at a distance of 'd' kms

From the given data,

$$\begin{gathered} d5 - 1 + d5 + 1 = 7560 kms \\ d4 + d6 = 54 \\ \end{gathered}$$

5d/12 = 5/4 => d = 3 kms.

Hence, the place is 3 kms far.

Let the initial speed of Akhil be '4p' kmph

Then speed after decrease in speed = '3p' kmph

We know that, 

Change in speed == Change in time

According to the given data,

$$\begin{gathered} 243p - 244p = 35 - 2560 \\ => p = 12 \end{gathered}$$

Hence, the initial speed of Akhil = 4p = 4 x 12 = 48 kmph.

Let the speed of boat in still water is ‘x’ km/hr & that of stream is ‘y’ km/hr.
Then, ATQ
(x+y)/y = 9/1 9y = x + y
x = 8y
y = 3 km/hr
So, x = 24 km/hr
Upstream speed = 24-3 = 21 km/hr
Hence, distance travelled upstream in 5 hours = 21*5 = 105 km.

Let he mixes the oils in the ratio = x : y

Then, the cost price of the oils = 60x + 65y

Given selling price = Rs. 68.20

=> Selling price = 68.20(x+y)

Given profit = 10% = SP - CP

=> 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y)

=> 6x + 6.5y = 8.20x + 3.20y

=>2.2x = 3.3y

=> x : y = 3 : 2

PW = 100×Amount100+R×T