Official Quant thread for CAT 2013

MBA CAT 2024
26488
@bodhi_vriksha said:
One from my side : A square of unit diagonal is cut along one of the diagonals, and the two parts are moved towards each other along the line of the other diagonal. What is the maximum possible area of the overlapping region?Team BV--Pratik Gauri

i first tried to divide the diagonal into 2 parts. got overlapping area = 1/8.
then divided into 3... got overlapping area= 1/6

so 1/6.
26489
@vbhvgupta said:
Q 33 and 34
33 ? drop of 3 %
26490
@Narci said:
33 ? drop of 3 %
33 OA B
26491
@vbhvgupta said:
33 OA B
oopsy... ans option B nikaala aur yahan A likh diya
26492
@vbhvgupta said:
Q 33 and 34
1)b?
2)a?
26493
@ChirpiBird said:
i first tried to divide the diagonal into 2 parts. got overlapping area = 1/8. then divided into 3... got overlapping area= 1/6so 1/6.
Good approach 😃 This was what i was looking for .. this problem can be solved using differentiation but that is lengthy .. Dividing the figure is the best approach to use in such questions :)

Team BV--Pratik Gauri
26494
@iLoveTorres said:
1)b?2)a?
explain 34 OA C
26495

'x' kilograms of berries were found to contain 99% water.After 15 days,the sample of berries from the same lot was found to contain 98% water. What must be the weight of berries after these 15 days?

26496
@maroof10 said:
'x' kilograms of berries were found to contain 99% water.After 15 days,the sample of berries from the same lot was found to contain 98% water. What must be the weight of berries after these 15 days?
x kg = 99% water + 1% matter..
Matter = 0.01x

After 15 days:
2% X = 0.01x
=> X = 0.5x

Hence, Weight = 0.5 x ?

26497



Diagonal of the square is having unit length hence side equals 1/_/2

1/_/2 = dg+gh+hf =2x+y

now if we look at the pic,
req area is half of the square area -( 2* x*x/2) - (y*y)/2

square area is (2x+y)^/2

we have to remove this x,

x = (1/_/2 -y ) / 2

hence required area is

Area = ( (1/_/2 - y+y) )^2 /2 - ( 1/_/2 -y)^2/4 - y^2/2

simplify this u will get a nice expression where u can substitute the value y = _/2/6 to get the maximum area. bit lengthy simplification hein.. i am lazy hum..

when u substitute y = _/2/6 u will get 1/6 as the maximum area.

The symmetry of x,x and y,y are gauged by the fact that, the rate of movement is same every where. hence we have to assign the values according to their angles. perpendiculars move with same rate hence all the perpendicular distances are taken as X. other slope wala distance as y. correct me if im wrong anywhere..


26498
@maroof10 said:
'x' kilograms of berries were found to contain 99% water.After 15 days,the sample of berries from the same lot was found to contain 98% water. What must be the weight of berries after these 15 days?
x/2??
0.99x and 0.01x before
let y amt of water is evaporated
0.99x-y=0.98(x-y)
x=2y
y=.5x

0.49x and 0.01x
0.5x=x/2
26499
@vbhvgupta said:
explain 34 OA C
assume total population to be 200
NO of Male=110 male will earn = 110*50*5/100=275
female=90 female will earn = 90*50*8/100 = 360
% profit = 635*100/(50*200)=6.35%

Now the population grows by 10% and the ratio is the same 11:9
New population is 220 divide this in the ratio 11:9 you will get 121 and 99

Now male will earn = 121*50*5/100=302.5
female will earn = 99*50*8/100=396
% profit = 698.5*100/(50*220)=6.35%

Hence no change :)

PS @scrabbler bro aapke orally wale method bohat kaam aa rahe hai
26500
@ScareCrow28 answer is right....can u explain what is capital X?
26501
@vbhvgupta said:
Q 33 and 34
33.b
34. c

26502
@maroof10 said:
'x' kilograms of berries were found to contain 99% water.After 15 days,the sample of berries from the same lot was found to contain 98% water. What must be the weight of berries after these 15 days?
0.5x
26503
@maroof10 said:
@ScareCrow28 answer is right....can u explain what is capital X?
Sorry, "X" is the Total Weight of berries ( Water + Matter ) after 15 days
26504
@bodhi_vriksha said:
Good approach This was what i was looking for .. this problem can be solved using differentiation but that is lengthy .. Dividing the figure is the best approach to use in such questions Team BV--Pratik Gauri
Me too used this method but by midway i was doubtful about the hexagonal area, hence went for hectic procedure..
26505
@ScareCrow28 ok thanks..got it
26506
@bodhi_vriksha said:
Good approach This was what i was looking for .. this problem can be solved using differentiation but that is lengthy .. Dividing the figure is the best approach to use in such questions Team BV--Pratik Gauri
sir, i dont remember much of differentiation. Please post that method too.
#farzi Engineer! :mg:
26507
@maroof10 said:
'x' kilograms of berries were found to contain 99% water.After 15 days,the sample of berries from the same lot was found to contain 98% water. What must be the weight of berries after these 15 days?
The "berry solid" has not changed. It was 1% of earlier stuff and has now become 2% of the new stuff so the new volume must be 1/2 the earlier.

regards
scrabbler