Quantitative Aptitude - Arithmetic Ability Questions

Let the distance be 'd' kms.

According to the given data,

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

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

We know that,

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

Number of permutations = 4P3 = 24

100 cm is read as 102 cm.

 A1 = (100 x 100)  and A2 =(102 x 102)   

(A2 - A1) = [(102 x 102)-(100 x 100) ]= 404    

Percentage error =[404/(100 x 100 )] x 100%= 4.04%

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

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

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.