Exams

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

The path traced by planet is called Orbit. Similarly, the path traced by projectile is called Trajectory.

Hence, Planet : Orbit : : Projectile : Trajectory.

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

So they meet at => 4 + 2.5 = 6:30 pm

The given letter equation implies that, 

7 C in the R = 7 Colours in a Rainbow.

We know that,

The equivalent K =  L1 + L2L1K1+L2K2

where L1 & L2 are plate1 & Plate2 thicknesses, K1 & K2 are coefficient thermal conductivity of plate1 & Plate2 respectively.

But here L1 = L2 = L (say)

Then K = L+L/(L/K1 + L/K2)  =  2K1K2K1+K2

Manager's monthly salary

= Rs. (1900 x 25 - 1500 x 24) = Rs. 11,500

Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.