Crack It Out..!!

[bhavin_3]

Hey Guyz hw abt crackin dis out.. A Step To Cat..!!

. a group of new students who total age is 221
> years
> >joins a class, becoz if which the strength of the
> >class goes up by 50% but the avg age of the class
> >comes down by 1 year. whats the new avg of the
> class?
> >if it is known to be a natural number after the new
> >group of students have joined, given that the
> original
> >strength of the class was a digit number greater
> than
> >30
.


2.how many 4 digit numbers that are divisible by 4
> can
> >be formed by using the digits 0 to 7 if no digit is
> to
> >occur more than once
> >Optionz orite givin don worry choose frm below
> >520, 370, 345, 510

[bhavin_3]

Hey Guyz c'mon give it a try atleast. . Make da brain more active

[kartik]

Lol man.. first of all.. the question itself is not clear... Try to edit it with a proper formatting and options :P

[bhavin_3]

If i give da option itz damn simple.. wtz in frm u... so frst kept it a lil harder and second 1 u can check it out der are 4 optionz given..

[A.J.]

Answer for the first question is 16.

Let the original no. of students to be x and original average age to be A.
When new students join strength goes up by 50% which means no of students is 1.5x.......intial total age is Ax and new total age is Ax + 221 and new avg is (Ax+221)/1.5x which is A-1.

Solving for A we get A=442/x +3....now x is an integer as also A and x>30.. nearest value of x so that 442/x is an integer is x=34...and next value is 52 (442 is a multiple of 2,13 and 17)..putting x=34 we get A=13 +3 =16....

2.
The answer to this one is 370.

[kartik]

For Q 2 .) 360 kaise ???

[A.J.]

It's 370.

Check for yourself.........it meets all the conditions specified.

[bhavin_3]

Originally Posted by kartik

For Q 2 .) 360 kaise ???

Hey der kartik itz 370



Solution of second problem
Any number more than 2 digit is divisible by 4 if
last 2 digits are divisible by 4
so in ur case only 14 cases r there where we get
number divisible by 4
ie. 04,12,16,20,24,30,36,40,52,56,60,64,72,76.
out of these number 4 number are having 0 in them so
the total number of combinations are: 6*5*4=120 ---1
n rest 10 case combinations are: 5*5*10= 250 ---2
add 1 & 2
ans= 370


Kewl 1 Anupam... More to come..!!

[A.J.]

Thanks mate.......this one was quite simple actually.

Though I hope you won't be giving us Mensa stuff to solve next time 'round.

[miral.shah2003]

please can u make question simpler by editin it proper...