Quantitative Aptitude - Arithmetic Ability Questions

Let the distance be 'd' kms.

According to the given data,

$$\begin{gathered} d30 - d50 = 4060 hrs \\ => 6d = 300 \\ => d = 50 kms. \\ \end{gathered}$$

We know that,

Area of rectangle A = L x W = 5 x 3 = 15 sq.inches.

Number of permutations = 4P3 = 24

From the given data,

                       Day      Capacity

A              ->    12          2

B              ->     8           3             2

A + B + C ->     2            12

=> Capacity of C = 12 - 5 = 7

Ratio of capacity of A : B : C = 2 : 3 : 7

Difference of wages of C & B = 4/5 x 1540

= 4 x 308 = Rs. 1232

I = (600x11x5)/100 = 330 Paise = Rs. 3.30

least cost price      = 200*8 = 1600
greatest sold price = 425 * 8 = 3400
profit required = 3400- 1600 = 1800

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.

Let 'd' be the distance between A and B
K -time = d/10 + d/9 = 19d/90 hours
L -time = 2d/12 = d/6 hours
We know that, 10 min = 1/6 hours
Thus, time difference between K and L is given as 10 minutes.
=> 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
Thus,
d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.