Official Quant thread for CAT 2013

MBA CAT 2024
28485
@DeAdLy said:
the average marks of 3 students x,y,z is 48. Another student k joins the group the new average becomes 44. If another student p joins who has 3 more marks than k , the average of y,z,k,p becomes 43. How many marks x got in exam ??1.43 2.353.394.47
x+y+z = 3.48 = 144
144+k = 4.44 = 176
So k = 32 and p = 35
Now y+32+z+35 = 172
So y+z = 105
Using this in the first equation, x = 39
28486

Q A Team of miners planned to mine 1800 tonnes of ore during certain number of days. Due to technical difficulties, In one third of the planned number of days, the team was able to achieve an output of 20 tonnes of ore less than the planned output. To make up for this, the team over achieved for the rest of the days by 20 tonnes. The end result was that the team completed the task, one day before the scheduled time. How many tons of ore did the team initially planned to ore per day ?


Ans ? Approach ?
28487
@DeAdLy said:
the average marks of 3 students x,y,z is 48. Another student k joins the group the new average becomes 44. If another student p joins who has 3 more marks than k , the average of y,z,k,p becomes 43. How many marks x got in exam ??1.43 2.353.394.47
39 ?

3*48 + k = 4*44
k = 32
so p = 35

now y + z + k + p/4 = 43
y + z = 105

x + 105/3 = 48
x = 39
28488
@ananyboss said:
Q A Team of miners planned to mine 1800 tonnes of ore during certain number of days. Due to technical difficulties, In one third of the planned number of days, the team was able to achieve an output of 20 tonnes of ore less than the planned output. To make up for this, the team over achieved for the rest of thedaysby 20 tonnes. The end result was that the team completed the task, one day before the scheduled time. How many tons of ore did the team initially planned to ore per day ?Ans ? Approach ?
n/3 (1800/n - 20) +(1800/n+20)(2n/3-1)=1800
1800 -20n/3 -1800/n+40n/3 -20 =1800
20n/3 -1800/n -20=0
n/3 -90/n -1=0
n^2 -3n -270=0
n=18
100 units per day

28489

The expression X^2+kX+9 = 0 becomes positive for what values of K(given that X is real)

28490
@Subhashdec2 said:
n/3 (1800/n - 20) +(1800/n+20)(2n/3-1)=18001800 -20n/3 -1800/n+40n/3 -20 =180020n/3 -1800/n -20=0n/3 -90/n -1=0n^2 -3n -270=0n=18100 units per day
Why not
n/3 (1800/n) -20 + (1800/n+20)(2n/3-1)=1800 !!! ??
28491
@vbhvgupta said:
The expression X^2+kX+9 = 0 becomes positive for what values of K(given that X is real)
-6
28492
@ananyboss said:
Why notn/3 (1800/n) -20 + (1800/n+20)(2n/3-1)=1800 !!! ??
for n/3 days they are under their target by 20 on each day
not the total target
28493
@vbhvgupta said:
The expression X^2+kX+9 = 0 becomes positive for what values of K(given that X is real)
what do u mean by expression becoming positive
28494
@ChirpiBird said:
-6
@Subhashdec2 said:
what do u mean by expression becoming positive
THis is what given in Q

ans |K|
28495
@Subhashdec2 said:
for n/3 days they are under their target by 20

That's what the confusion is !!
In one third of planned number of days, Team was able to achieve 20 tons less than what is planned !
--> Never mentioned each day failure !!

Anyway you got it right. That make me learn Thanks :)
28496

OA for Last problem set is C n D . Last one of the day....




28497
@vbhvgupta said:
THis is what given in Qans |K|
my bad :banghead:
bas positive karna h, equal to zero nhi karna.

28498
28499
@vbhvgupta said:
option a

a > 9/2
28500
@vbhvgupta said:
a
28501
@sbharadwaj said:
option aa 9/2
@ChirpiBird said:
a
E
28502
@vbhvgupta said:
E
@ChirpiBird

Sure.? Lemme check.. Tag me if u post the soln
28503
@vbhvgupta said:
option a ?

9 - a/2/a/2
18 - 2a/a

so for a 9/2..above eqn satisfies
28504
All those guys having Arun Sharma Quant book 5th edition, I have doubts in geometry. Kindly solve these questions and pls tag me with the answers:
1. Page no=384 , Q.no=14
2. Page no=385 , Q.no=22
3. Page no=385 , Q.no=24