Official Quant thread for CAT 2013

MBA CAT 2024
30927
@vbhvgupta said:
a piece of work can be completed by 10 men and 6 women in 18 days. Men works 9 hours per day while women work 7.5 hours per day. Per hour efficiency of a woman is 2/3 of man efficiency. In how many days 10 men and 9 women complete the workk.
16 days.
30928
@vbhvgupta said:
7 Indian 4 chinese finish a job in 5 days. 7 japanese and 3 chinese finish the same job in 7 days. Given that the efficiency of each person of a particular nationality is same but different from others. 1 Indian 1 chinese and 1 japanese will complete the work in
eff. of 7I + 4C = 1/5
eff. of 7J + 3C = 1/7

adding both
eff of 7(I+j+C) = 12/35
I+j+C = 12/245

Hence no. of days = 245/12 ?
30929
@junefever said:
64? by simply applying Euler's method
question was what is the remainder when 2^96 is divided by 96. but 2 and 96 have common factor 2 .so can we apply euler ? m i missing something ?
30930
A group of people were analysed on three qualities - health, wealth and wisdom. It was found that 139 people were either healthy or wise but not wealthy. 23 people were neither healthy, wealthy and wise. 208 people were wealthy.If 199 residents have at least two of the three qualities then how many residents have exactly one quality?

Please post the approach as well ..
30931
@nramachandran 148
30932

Q2/Q5:http://www.quantexpert.co.in/questionoftheday.html

30933
@nramachandran said:
A group of people were analysed on three qualities - health, wealth and wisdom. It was found that 139 people were either healthy or wise but not wealthy. 23 people were neither healthy, wealthy and wise. 208 people were wealthy.If 199 residents have at least two of the three qualities then how many residents have exactly one quality?Please post the approach as well ..
Is it 148
30934
@td_bouncer said:
@nramachandran 148
Can u post the approach
30935
@hippophobia said:
Is it 146
Ans is 148 bro
30936
@nramachandran yah got it got.........do by venn diag

a+f+g=139

b+c+d+e=208

b+f+d+e=199

so, f=c-9

a+c+g=148(req ans)


30937
@catahead said:

2) f(x,y) = y^2 + x^3

So , f(3,10) = 10^2 + 3^3 = 127

3) Should be 96.

Use Mass Point Geometry.
30938
@catahead said:
2)10^2+3^3=127

30939
@Aizen said:
2) f(x,y) = y^2 + x^3So , f(3,10) = 10^2 + 3^3 = 1273) Should be 96.Use Mass Point Geometry.
5th wala kaise hua??
30940
@catahead said:
@Dexian said:
5th wala kaise hua??

For details on Mass Point Geo:
www.sanjosemathcircle.org/handouts/2007-2008/20071114.pdf

http://totalgadha.com/mod/forum/discuss.php?d=7992
30941
there are 28 identical looking coins all of which except one weigh same. using a common balance, what is the minimum weighings required to rnsure coin of different weight is ensured ?ďťż
3
4
5
6

30942
@raopradeep said:
there are 28 identical looking coins all of which except one weigh same. using a common balance, what is the minimum weighings required to rnsure coin of different weight is ensured ?3456
4 ??
30943

@raopradeep

there are 28 identical looking coins all of which except one weigh same. using a common balance, what is the minimum weighings required to rnsure coin of different weight is ensured ?

ďťż

3



4



5



6



Minimum 4 weighings...

30944
@ananyboss @Dexian oa is 4
@Dexian explain the process iam also getting 3 as minimum
30945

@heylady

Minimum 4 weighings...



approach batao yar

30946

@raopradeep

there are 28 identical looking coins all of which except one weigh same. using a common balance, what is the minimum weighings required to rnsure coin of different weight is ensured ?

ďťż

3



4



5



6



First weighing: Divide the coins in 2 groups, 14-14

Second weighing: Divide the heavier group in 7-7

Third weighing: Divide the heavier group in three sub groups 3-3-1

if both are equal, then single coin have higher weight.

If not,

fourth weighing: take the heavier sub group of 3 coins and divide it into three small groups 1-1-1

if both are equal, then third coin has higher weight.

If not, then higher weight one will be the answer


Hence, 4 weighing 😃