1. case 1 ->
3W & 2M ->4C3*4C2=24 ways
Now three cases are possible ..
Now total possible cases are 5*4=20 (5 for President post and 4 for V.P)
But both cannot be male so we can have 2 cases where such thing happens (V-M1 & VP-M2 and vice versa)
So we have 18*24 here
2. case 2->
4M & 1 M->4 ways
And in each case we will have the required formation
So 4*20=80
So total 432+80=512
2. maximum 2 women is same as minimum 3 men
And also 1 women can hold the top two post . it means at least one M should be in top two . So it is the same as above and asking for Male so 512 again !
woman days required = 20*20
since man = 2woman =>a woman leaving and a man joining = net addition of 1 woman
hence 20+21+22+........(20+n)
=>20*(n+1)+n(n+1)/2=400
=>(n+1)(40+n)=800
trial and error shows n=14
=> work finishes on 15th Day.
ATDH.
QUANTEXPERT Quant Question of the Day (www.quantexpert.in)
QUESTION 1)
If a train runs at 20 km/hr, it reaches its destination late by 10 min. But if it runs at 30 km/hr, it is late by 2 min. only. At some speed, it reaches on time. What is the time required if it reaches on time?
(a) 12 min (b) 8 min (c) 14 min (d) 15 min
QUESTION 2)
Ram Singh deposits Rs.150 on the first of every month starting from 1st Jan1985, in the recurring deposit scheme of a bank which allows simple interest @ 6% p.a. on the sum standing to his credit at the end of each month. What is the amount, Mohan is entitled to on 31st Dec, 1985?
(a) Rs 1818 (b) Rs 1800 (c) Rs 1450 (d) Rs 1400
@ chillsir
sir formula for number of non- integral solution is C(n+r-1 ,r ) or is it C (n+r-1,r-1)
in book i have seen first one i.e C(n+r-1,r) . Which is correct???
Since 3N is divisible by 3, we can say that (a1 + a2 + a3 + .. + a7) is divisible by 3.
=> N is divisible by 3
So, sum of digits of N should be a multiple of 3
Only option is 9, i.e., option (1)
NOTE:- Question says which of the following is possible. Although sum should be of form 3n and 9 satisfies that
21 1/3=64/3
Now say we had an overall average of 22 then the numbers are 1 to 43
So now the sum should be 64/3*41 which is not integer so the number must be in the form 3n+2
Hit & trial around 21 ...
41,38,35,32 ... or 44,47,50
1.Sum of 44 is 990
Now 64*3n/3 is always even so we need the sum to be odd possibility 50 ,38 & 41
lets check ...
64*16=1024
25*51=1275
So now it's impossible ...
38*39/2=19*39=741
36*64/3=768 (not possible)
41*42/2=861
39*64/3=832
possible so difference is 29 and the numbers are 14 & 15 so answer=210
Nice sum
my approach
for a>b>c
from 0 to 9 pick any three numbers say 3 9 0 now u cn arrng thm in a>b>c way as 930
so if u pick any three nums from 0 to 9 we can have a num satisfyn the cndtn
=>10C3 = 120 ways
for a9C3 = 84 ways
total 204 ways
Question says,we need to have abc in increasing or decreasin order.In you cases you are taking all three digit no's.Isnt it ?
Suja
it says we have to arrange it in ascendin, descendin
so,10C3 is the number of ways to select 3 digits,
and as you know there is only one way in which these digits will be in ascendin or decending order rite ?
so,
2*10C3 ways for both ascendin,descendin order. and there are cases when we have first digit as 0 a