CAT Questions

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p=d×1002R2 

Where d is difference, r is rate of interest. 

=> (48 x 100 x 100) / 8 x 8 = Rs.7500

Let the total distance be 'x' km.

Time taken to cover remaining 40% of x distance is   t1=40×x100×48

But given time taken to cover first 60% of x distance is   t2=t1+2060hrs

 $$\begin{gathered} t2=60×x100×48 \\ \end{gathered}$$

∴60×x100×48= 40×x100×48+2060

 x=80 km.

Non collision can happen only when all the cars move in same direction.

That is either all cars moving in clockwise direction or all cars moving in anti clockwise direction.
P(all cars not colliding)

= P(all cars moving in the same direction)

= P(all cars moving in clockwise direction)  +  P(all cars moving in anticlockwise direction)

= [p(A --> C) x p(C --> B) x p(B --> A)]  +  [p(A --> B) x p(B --> C) x p(C --> A)]

= [1/2 x 1/2 x 1/2] + [1/2 x 1/2 x 1/2]

= 1/8 + 1/8

= 1/4

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.

Marks in English  = 85×7 – 83×6

= 595 – 498

= 97

100
65
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65 ------- 35
100 ------ ? => 53.84 %

An object at rest may have Potential energy.