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> 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->...

1. case 1 ->

3W & 2M ->4C3*4C2=24 waysNow three cases are possible ..(5 for President post and 4 for V.P)

Now total possible cases are 5*4=20

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 !

bhai can you please expand the bolded part..

> 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.
20+21+22+23+...nt...

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.

20+21+22+23+...nth term =400

=>n/2=400

=>n(n+39)=800 can you please tell me where i am wrong

> 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...

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 minQUESTION 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

1)d/20=t+1/6

d/30=t+1/30

on solving we get t=14 minutes

> *@ 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???
*
it is C (n+r-1,r-1)

@ 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???

it is C (n+r-1,r-1)

- 1 Like

> 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 b...

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

chill sir, can you explain how you can infer that a1+a2+a3+...+a7 is divisible by 3 from 3N is divisible by 3... ? each of a1,a2, a3.... a7 is multiplied by a different number?

> 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 w...

21 1/3=64/3

Now say we had an overall average of 22 then the numbers are 1 to 43So 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

subha bhai, can you please expand the bolded part..

> 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 a*9C3 = 84 ways
total 204 ways*
why can't we select 0 in ...

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

why can't we select 0 in the second case when 0 can be selected in the case a>b>c even if 0 is selected when a

> 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 ::
okay there's a mistake there... if i take it this way.. the last digit can be filled in 0-7 ie 8 ways the first one can be filled in 2-9 i.e.,8 way...

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

okay there's a mistake there... if i take it this way.. the last digit can be filled in 0-7 ie 8 ways the first one can be filled in 2-9 i.e.,8 ways and the middle one can be filled in 1-8 i.e., 8 ways... so 8.8.8... what's wrong here? hope you are getting what i am trying to get clarified??

> 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 ...

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

bhai, i have a little doubt in this question... since a 3 digit number is needed.. _ _ _ i can fill the last space in 10 ways the first space can be filled in 7 ways and the second in 9 ways i.e., 10*9*7 *2 for ascending and descending.. now where is my understanding wrong? please help