Official Quant thread for CAT 2013

MBA CAT 2024
8621
@veertamizhan said:
If the difference in the time taken to cover a certain distance at 6 kmph and at 10kmph is 30 minutes, find the distance.
Speed becomes 5/3 => time becomes 3/5 => 2/5 of time is 30 min => time is 75 min at 6kmph = 7.5 km.

regards
scrabbler
8622
@veertamizhan said:
I am horrible at math. Any suggestions?
What Pratyush said :P

Also, lose the negative attitude - saying "I am bad at maths" (even if it is only to yourself) is pretty much a self-fulfilling prophecy (I did that to myself with geometry for a long time, assumed I was bad and hence never improved)

regards
scrabbler
8623
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
8624
@jain4444 said:
you got the right place then just be regular on this thread Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:A. 10 B. 12C. 14 D. 16
1/a+1/b+1/c=1/6
2*1/6+7(1/a+1/b)=1
1/a+1/b=2/21; 1/c=1/6-2/21=1/14; c=14
8625
@rkshtsurana said:
how v ll proceed if instead of alternate..every 3rd contestant is asked to step up...there was a method that we convert in base 3 and then ve do right shift..can u explain ?
I think you are talking about this problem:-

The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010?

The logic that you mentioned goes like this:-
Convert 150 in base 3
150 in base 3 is 12120
Now remove first two digits and append them to the end of the remainder number.
So, new number becomes 12012
It is 140 is base 10
So, contestant number 140 will be the winner.
(I don't know the reason why it works)

But somehow I'm not sure if its the correct way to go about it.
For example has the number been 159, then in base 3 its 12220
After removing first digits and appending them at the end it will become 22012 which is 221 in base 10 (which doesn't make any sense)
So, I'm not sure about this method.

Alternatively:-
Backtracking will also help.
If any contestants position in a round is n, then in the previous round its position will be 3n/2 or [3n/2] + 1
So, positions of winner are 1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, ...

Means for any n (no of contestants) between 62 and 92 winner will be contestant no 62 and so on

Since here we have 150 contestants winner will be 140

But here the contestants are not standing around the circle (for that you can read about Josephus problem)
8626
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
38
8627
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
38
8628
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
38?
ab+ac+bc=472
a^2+b^2+c^2+3(ab+bc+ca)=1916
(a+b+c)^2=500+2(472)
a+b+c=38
8629
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
38?
8630
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
38?

ab + bc + ca = 472
(a+b)(b+c) + (b+c)(c+a) + (c+a)(a+b) = 1916
ab + ac + b^2 + bc + bc + ab + c^2 + ac + ac + bc + a^2 + ab = 1916
3(ab + bc + ca) + (a^2 + b^2 + c^2) = 1916
(a^2 + b^2 + c^2) = 500

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
= 500 + 944
= 1444

a+b+c = 38
8631
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
ab + bc + ca = 472 ...........(1)

(a + b)(b + c) + (b + c)(c + a) + (c + a)(a + b) = 1916
=> (a² + b² + c²) + 3(ab + bc + ca) = 1916 .............(2)

(2) - (1) will give
(a² + b² + c²) + 2(ab + bc + ca) = 1444
(a + b + c)² = 38²
(a + b + c) = 38
8632
@gautam22 said:
An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.On average, how many sides will have been hit once the ball has travelled 1000m ?
yaar, this is quant thread for cat 2013, not for mensa society , i bet chill sir won't break a sweat, but for us mortals :splat:
8633
@gautam22 said:
An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.On average, how many sides will have been hit once the ball has travelled 1000m ?
Yar.. options milenge??
Options nai milenge to kya ans 132 ho sakta hai??

8634
@gautam22 said:
An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.On average, how many sides will have been hit once the ball has travelled 1000m ?
This quest will go un answered 😃 Only God (read @chillfactor ) may solve it
8635
@gautam22 said:
My digital watch shows hours and minutes only. For what fraction of a complete day is at least one "1" showing on the display?
Iska ans 1/8 ho sakta hai kya??
8636
@sowmyanarayanan said:
The sum of the products of the sides of a triangle with sides a, b and c taken two at a time is 472. The sum of the products of the sides of the triangle with sides a + b, b + c and a + c taken 2 at a time is 1916. Find the perimeter of the triangle having its sides as a, b and c.
ab+bc+ca=472
(a+b)(b+c)+(b+c)(a+c)+(a+c)(a+b)=1916
3(ab+bc+ca)+a^2+b^2+c^2=1916
a^2+b^2+c^2=500
(a+b+c)^2=500+944=1444
a+b+c=38?
8637
@gautam22 said:
A sphere 5.5 metres in diameter is filled with 1m diameter hemi-spheres.(1) What is the theoretical maximum amount of hemi-spheres that can be crammed into the big sphere given that the following condition is met:Each hemi-sphere's flat side (which I'll now refer to as its 'disc') has a central point (indicated by the white point shown in the hemisphere diagram to the right). The point must not 'see' another hemisphere's disc. (2) By cramming them as efficiently as possible, a relatively small volume will be left. How large is this volume?
Is this the way to proceed? will Theta value help us in any way - PFA attchment
8638
@gautam22 said:
sir options nahi hain..net pe dekha tha...approach batao
Approach btaunga to tum log forum se nikal doge 😛 I won't tell
8639
@gautam22 said:
My digital watch shows hours and minutes only. For what fraction of a complete day is at least one "1" showing on the display?
19/32?

Edit: 5/8 I think....calculation galti in above answer...

regards
lakesidey
8640
@gautam22 said:
kisi ka kuch nahi pata sir saare net pe dekhe hain.....kisi ne kuch ans post kara hai kisi ne kuch.....to m not sure
My approach:
00:00 to 23:59

Here, Total conditions= 24(00-23) * 60(00-59)
Favourable ways:
Minutes section:
01_______1 way
10-19____10 ways
21,31,41 and 51___1 way each
So total= 15 ways

Hour section:
01___1 way
10-19____10 ways
21__1 way
Total= 12 ways

So, fraction = 12*15/(24*60) = 1/8??