Given

      125% ---- 3400

=> 100% ---- ?

=> ? = 3400x100/125 = 2720

=> Cost price of the article = Rs. 2720

Profit when article sold at Rs. 3265 = 3265 - 2720 = 545

Hence, Profit% = Gain x 100/cost price

=> P% = 545 x 100/2720

=> P% = 20%

Let the Efficiency of pavan be E(P)

Let the Efficiency of sravan be E(S)

Here Work W = LCM(25,20) = 100

Now, E(P+2S) = 100/25 = 4 ....(1)

E(2P+S) = 100/20 = 5 ....(2)

Hence, from (1) & (2) we get

E(S) = 1

=> Number of days Savan alone work to complete the work  = 100/1 = 100 days.

Let the interest rate be r%

We know that,

S.I = PTR/100

=> (1540 x 5 x r)/100  +  (1800 x 4 x r)/100 = 1788

=> r = 178800/14900 = 12%

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

Let the length be 'l' and breadth be 'b'.

b = l × 3/4__________(a)

2(l+b) = 1050

l+b = 525___________(b)

From equations (a) and (b),

l = 300m, b = 225 m

Area = l × b

= 300 × 225

= 67500 sq.m.

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
⇒ The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

∴ required probability = 26/52 = 1/2.

Here the series follows the pattern that, product of succesive prime numbers i.e,

 2 × 3 = 6

 3 × 5 = 15

 5 × 7 = 35

 7 × 11 = 77

 11 × 13 = 143

 13 × 17 = 221

 17 × 19 = 323

So here in place of 77 the number 63 is written.

So the wrong number is 63.

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

On adding (I) and (II),

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.