4! ( 4C4 + 5C4 + 6C4 +...+14C4) Sum : 4! (15C5) = 72072 Average : 72072/11 = 6552

98. 4! ( 4C4 + 5C4 + 6C4 +...+14C4) Sum : 4! (15C5) = 72072Average : 72072/11 = 6552

Let the free luggage allowance be x kg.M + S = 1050... (1)x + 2M = 2400 ... (2)x + 2S = 900 ... (3) from (1) , (2) and (3) x = 600M = 900

5! = 120 last two digits of 6! => 6 x 20 = ___20 last two digits of 7! => 7 x 20 = ___ 40 last two digits of 8! => 8 x 40 = ___ 20 last two d...

Let the original length of the rectangle be : L and the breadth be B Original Area: LB Length is increased by 20% so new length : 120L/100 = 6L/5 Bre...

Check the notes section of TG's facebook page.

In order to divisible by 6. sum of digits should be a multiple of 3 and last digit should be even. so, when 0 is not included2346 => 4! numbers s...

No! It is 1. 100% probable.

Alternative Approach:

45a x 45b = 21600 ab = 21600/45 Which is not possible as a, and b are positive integers .

ABCD – DBCA 1000A + 100B + 10C + D – ( 1000D + 100B + 10C + A) 999A – 999D 999( A-D) 999 is not divisible by 7 . so ( A – D) should be a multiple of 7...

Volume of pyramid : 1/3 of area of base x height So 1/3 x 2562 x 40 - 1/3 x 144 x 30

PFA the solution: