official quant thread for cat08

[racer_jo]

Hi,
Plz solve this one:
Devarsh distributed Rs100,190 times,Rs200,220 times and Rs300,180 times to all the boys in an orphanage one by one.Some naughty boys went more than once to get the money.Out of them 30,40,60 and 20 boys got Rs800,Rs700,Rs600 and Rs300 respectively.What would have been the maximum number of boys in the orphanage?

1. 250
2. 235
3. 240
4. 285
5.350

Kindly suggest a general approach to solving such problems where we need to find 'at least' or 'at most' of the quantities..

Regards,
racer_jo

[The Last Romeo]

Originally Posted by priya_dream

In triangle ABC, produce a line from B to AC,meeting at D and from C to AB,meeting at E. Let BD and CE meet at X. Let triangle BXE have area a, triangle BXC have area b, and triangle CXD have area c. Find the area of the quadrilateral AXED in terms of a,b and c ?

a) bc(2a+b+c)/(b^2-ac)
b) ac(a+2b+c)/(b^2-ac)
c) abc/(b^2-ac)
d) ab(a+b+2c)/(b^2-ac)

A slightly different approach to the one suggested by mistique

Lets consider the area of the triagle to be A and also lets consider it to be a equilateral traingle.
now BD would divide the triangle into two equal parts
i.e. A/2 & A/2
now consider in triangle BEC the line CE which meets BD at X, since this point would be the centroid it would divide the triangle BEC in the ratio of 2:1
hence the area of triangle BXE would become 1/3 * A/2=A/6=a
and area of triangle BXC would become 2/3 *A/2 = A/3=b
similarly c=A/6
hence the rquired area of AEXD=A-(A/3 + A/6 + A/6)
=A-(2A/3)
=A/3
now put the corresponding value of a, b, c into the given equations and the only option that satifies is answer no 2. which also gives area=A/3

[Topgear]

Originally Posted by implex

Your method is wrong Here is why !
suppose the question is How many ways can six indistinguishable letters be put into three mail boxes ?

if we use your method it becomes x+y+z=6
then we get 5c2/3!=10/6 ?
but answer is 2 !

similarly here also
x+y+z=14
x=1 gives (y,z)=(1,12) ,(2,11), (3,10),(4,9) (5, (6,7)
x=2 gives (y,z)=(2,10) (3,9) (4,,(5,7)(6,6)
x=3 gives (y,z)=(3, (4,7) (5,6)
x=4 gives (y,z)=(4,6) (5,5)

we need not check further as all further cases have been taken care of !!

Thanks a lot..

[v_v_n]

Originally Posted by Topgear

Thanks a lot..

hi implex can u explain ans in following case of grouping example?

suppose 12 identical things need to be group into 3 distinct type

case 1 if no limit
case 2 if lower limit
case 3 if upper limit
case 4 if both lower and upper limit are there.

[Topgear]

Originally Posted by implex

Your method is wrong Here is why !
suppose the question is How many ways can six indistinguishable letters be put into three mail boxes ?

if we use your method it becomes x+y+z=6
then we get 5c2/3!=10/6 ?
but answer is 2 !

similarly here also
x+y+z=14
x=1 gives (y,z)=(1,12) ,(2,11), (3,10),(4,9) (5, (6,7)
x=2 gives (y,z)=(2,10) (3,9) (4,,(5,7)(6,6)
x=3 gives (y,z)=(3, (4,7) (5,6)
x=4 gives (y,z)=(4,6) (5,5)

we need not check further as all further cases have been taken care of !!

Thanks a lot..

But a bit of doubt,

No. of ways of 6 identical letters into three boxes should be = 5C2 = 3! ways (123) + 2! ways (141) + 1 way (222)

While the division into groups should be 123, 222 and 141.. so 3 ways..(when boxes will also be identical)..

How 2! ways ?? Not getting this.. Kindly Explain once again..

[holy_luv]

Originally Posted by racer_jo

Hi,
Plz solve this one:
Devarsh distributed Rs100,190 times,Rs200,220 times and Rs300,180 times to all the boys in an orphanage one by one.Some naughty boys went more than once to get the money.Out of them 30,40,60 and 20 boys got Rs800,Rs700,Rs600 and Rs300 respectively.What would have been the maximum number of boys in the orphanage?

1. 250
2. 235
3. 240
4. 285
5.350

Kindly suggest a general approach to solving such problems where we need to find 'at least' or 'at most' of the quantities..

Regards,
racer_jo

hi,
total amount donated: 117000
total amount given to naughty boys:94000
balance:23000

now try to finish all rs.300 (180 times) and then rs.200 (220 times) for the amount receivd by naughty boys. You can then use rs.100 and the left over of rs.300 and rs.200 for distributing the balance amount. In this way, you 'll get the max no. of boys.

is the answer 350?

[vivek417]

Originally Posted by The Last Romeo

A slightly different approach to the one suggested by mistique

Lets consider the area of the triagle to be A and also lets consider it to be a equilateral traingle.
now BD would divide the triangle into two equal parts
i.e. A/2 & A/2
now consider in triangle BEC the line CE which meets BD at X, since this point would be the centroid it would divide the triangle BEC in the ratio of 2:1
hence the area of triangle BXE would become 1/3 * A/2=A/6=a
and area of triangle BXC would become 2/3 *A/2 = A/3=b
similarly c=A/6
hence the rquired area of AEXD=A-(A/3 + A/6 + A/6)
=A-(2A/3)
=A/3
now put the corresponding value of a, b, c into the given equations and the only option that satifies is answer no 2. which also gives area=A/3

I have confusion in 2 of the lines:-

area of triangle BXE would become 1/3 * A/2=A/6=a
and area of triangle BXC would become 2/3 *A/2 = A/3=b

If the medians cut in the ratio 2:1, then is there any formula or some concept which says that the area of the 2 parts would be in the ratio 2:1?

Thanks
Vivek

[The Last Romeo]

Originally Posted by vivek417

I have confusion in 2 of the lines:-

area of triangle BXE would become 1/3 * A/2=A/6=a
and area of triangle BXC would become 2/3 *A/2 = A/3=b

If the medians cut in the ratio 2:1, then is there any formula or some concept which says that the area of the 2 parts would be in the ratio 2:1?

Thanks
Vivek

Hi Vivek,
It will would have been easier to understand if i could have presented the diagram before you but i'll still try.
Now consider the trg BEC, the line BX (median) cuts the one of the sides i.e. EC of the said trg BEC in the ratio of 2:1
Now consider the triagles EBX and XBC, they would be similar triagles due to the SAS (BX is common, Ex and XC are propotional and <EBX = <CBX =30)
hence you could say that if the sides are prpoptional hence the area would also be propotional i.e. in the ratio 2:1
Now since area of triangle EBC = A/2
therefore area of trg EBX = 1/3 * A/2
area of trg BXC = 2/3 * A/2
Please let me know if i have cleared the air else i'll try once more

[racer_jo]

Originally Posted by holy_luv

hi,
total amount donated: 117000
total amount given to naughty boys:94000
balance:23000

now try to finish all rs.300 (180 times) and then rs.200 (220 times) for the amount receivd by naughty boys. You can then use rs.100 and the left over of rs.300 and rs.200 for distributing the balance amount. In this way, you 'll get the max no. of boys.

is the answer 350?

Yes,the answer is 350. Can u explain it in another way..?i didnt quite understand this method
Regards,
racer_jo

[IIMAmitabh]

can any body answer this

48 48 Marks: 0.33/1

There are 80 balls which look identical. All the balls are of same weight except one which is heavier. In how many minimum weighing with a bean balance can you find the heavier ball?

Choose one answer.
a. 5 b. 4 You're a genius! c. 7 d. 6

Originally Posted by The Last Romeo

Hi Vivek,
It will would have been easier to understand if i could have presented the diagram before you but i'll still try.
Now consider the trg BEC, the line BX (median) cuts the one of the sides i.e. EC of the said trg BEC in the ratio of 2:1
Now consider the triagles EBX and XBC, they would be similar triagles due to the SAS (BX is common, Ex and XC are propotional and <EBX = <CBX =30)
hence you could say that if the sides are prpoptional hence the area would also be propotional i.e. in the ratio 2:1
Now since area of triangle EBC = A/2
therefore area of trg EBX = 1/3 * A/2
area of trg BXC = 2/3 * A/2
Please let me know if i have cleared the air else i'll try once more