@gaurav_sharma said in Quant Boosters - Gaurav Sharma - Set 5:
Q26) John and David work on alternate days with John starting on the first day. John does 12.5% of the total work on the last day and finishes the work. If John alone can do the work in 6 days then which of following can be the number of days in which David alone can do the whole work?(a) 24 days(b) 12 days(c) 8 days(d) 10 days
Need solutioninlinemathsinline mathsinlinemaths

If two numbers are a, b, then the new number you are putting in the place of these two numbers is = ab + a + b = (a + 1)(b + 1) - 1 = c (say)
Now when you take c and d next, then the replacement number becomes = cd + c + d
= (c + 1)(d + 1) - 1
= (a + 1)(b + 1)(d + 1) - 1.
So final number remaining will be (1 + 1)(2 + 1)(3 + 1)....(39 + 1)(40 + 1) - 1 = 41! - 1.

@Vikrant-Garg
Probability of catching the bus = Probability of getting faulty machines in the first 2 tests (this part is clear right?)
This is possible in 2 ways - if we get DD or if we get GG as the result of first 2 tests.
In your doubt, if we get DG or GD we will require to do one more test, which would make the mechanic miss the bus. So DG and GD are not favourable cases.
So total number of ways = 4!/(2! x 2!) = 6 ways.
Favourable cases = 2
Probability = 2/6 = 1/3
Is it clear ?

p = prob (overnight born = boy| Next day chosen = boy)
= prob (overnight born = boy AND Next day chosen = boy) / prob (Next day chosen = boy)
= .5 * 4/T / (.5*4/T + .5 * 3/T) = 2/3.5 = 4/7 ~ .57