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Vishal R
@vishal.das
11.3
k

@All CAT 2011 is over continue posting in the CAT 2012 thread
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html#post3048600

@All CAT 2011 is over continue posting in the CAT 2012 thread

http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html#post3048600

http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html#post3048600

- 1 Like

vishal rana
@vishal99999
204

Bhaiyo some questions from permutation and combination inme doubt hai jyada...
1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.
2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mang...

Bhaiyo some questions from permutation and combination inme doubt hai jyada...

1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.

2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated

any number of times.

3)There are 5 different blue color, 4 different blue color and 3 different golden color. In how many ways the same color that are different can be selected.

4)There are 10 white, 8 green , 6 red balls in a box. In how many ways can one or more ball can be put in the box.

1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.

2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated

any number of times.

3)There are 5 different blue color, 4 different blue color and 3 different golden color. In how many ways the same color that are different can be selected.

4)There are 10 white, 8 green , 6 red balls in a box. In how many ways can one or more ball can be put in the box.

Commenting on this post has been disabled.

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vishal rana
@vishal99999
204

Bhaiyo two question :
1)After ajay gave Rs500 to Raman, he had less money than Vijay. If vijay gives Rs501 to ajay. Ajay will have Rs500 more than vijay. Who had largest money and how much initially?
2)A,B,C had some sweets. The total nu,ber of sweets that A and C have are 1,000 . If B has 50...

Bhaiyo two question :

1)After ajay gave Rs500 to Raman, he had less money than Vijay. If vijay gives Rs501 to ajay. Ajay will have Rs500 more than vijay. Who had largest money and how much initially?

2)A,B,C had some sweets. The total nu,ber of sweets that A and C have are 1,000 . If B has 500 sweets, the total sweets that A and B have cannot be more than C by?

a)377

b)179

c)244

d)none

Plz tell the approach also

thankx

1)After ajay gave Rs500 to Raman, he had less money than Vijay. If vijay gives Rs501 to ajay. Ajay will have Rs500 more than vijay. Who had largest money and how much initially?

2)A,B,C had some sweets. The total nu,ber of sweets that A and C have are 1,000 . If B has 500 sweets, the total sweets that A and B have cannot be more than C by?

a)377

b)179

c)244

d)none

Plz tell the approach also

thankx

Commenting on this post has been disabled.

@jackoneill
4.2
k

> If a,b ,c are in AP.then A) 1/loga,1/logb etc are in HP?B)log a,lob,log c are in HP
3)log a,logb,logc are in GP
=> 2b=a+c
=> log(2b) = log(a+c)
I think something is wrong in the question. It shd be none of the above??

If a,b ,c are in AP.then A) 1/loga,1/logb etc are in HP?B)log a,lob,log c are in HP

3)log a,logb,logc are in GP

=> 2b=a+c

=> log(2b) = log(a+c)

I think something is wrong in the question. It shd be none of the above??

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Sumit Jamwal
@sujamait
9.1
k

> If all letters of the word "CHCJL" be arranged in an English dictionary, what will be the 50th word ?
ans option:
1.HCCLJ
2.LCCHJ
3.LCCJH
4.JHCLC
5.none of the above
getting 3
till J it will be 48 words
after this 2nd word with L is LCCJH
Suja ::

If all letters of the word "CHCJL" be arranged in an English dictionary, what will be the 50th word ?

ans option:

1.HCCLJ

2.LCCHJ

3.LCCJH

4.JHCLC

5.none of the above

getting 3

till J it will be 48 words

after this 2nd word with L is LCCJH

Suja

- 1 Like

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Varun Tyagi
@varun.tyagi
14.2
k

> If all letters of the word "CHCJL" be arranged in an English dictionary, what will be the 50th word ?
ans option:
1.HCCLJ
2.LCCHJ
3.LCCJH
4.JHCLC
5.none of the above
Words starting with C_ _ _ _ in 4! = 24 ways
Words starting with H_ _ _ _ in 4!/2! = 12 ways
Words starti...

If all letters of the word "CHCJL" be arranged in an English dictionary, what will be the 50th word ?

ans option:

1.HCCLJ

2.LCCHJ

3.LCCJH

4.JHCLC

5.none of the above

Words starting with C_ _ _ _ in 4! = 24 ways

Words starting with H_ _ _ _ in 4!/2! = 12 ways

Words starting with J_ _ _ _ in 4!/2! = 12 ways

Total 48 words

Now 49th word is LCCHJ

50th word is

- 1 Like

Commenting on this post has been disabled.

Supraja Banukumar
@SuprajaB
6

If all letters of the word "CHCJL" be arranged in an English dictionary, what will be the 50th word ?
ans option:
1.HCCLJ
2.LCCHJ
3.LCCJH
4.JHCLC
5.none of the above

ans option:

1.HCCLJ

2.LCCHJ

3.LCCJH

4.JHCLC

5.none of the above

Commenting on this post has been disabled.

/* *\
@naga25french
81.7
k

> In a triangle ABC, AB = 3, BC = 4 and CA = 5. Point D is the midpoint of AB, point E is on segment AC and point F is on segment BC. If AE = 1.5 and BF = 0.5 then Angle DEF =??
Plz help
abc is a right angled triangle with angle a =53 and angle b = 37
now ade is an isoceles triangle...

In a triangle ABC, AB = 3, BC = 4 and CA = 5. Point D is the midpoint of AB, point E is on segment AC and point F is on segment BC. If AE = 1.5 and BF = 0.5 then Angle DEF =??

Plz help

abc is a right angled triangle with angle a =53 and angle b = 37

now ade is an isoceles triangle with ad=ae=1.5 and the so equal angle = 63.5

similarly cfe is also an isoceles triangle with cf=ce=3.5 and the equal angle = 71.5

angle def = 180-aed-fec = 180- 63.5- 71.5 = 45

- 1 Like

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King
@king.casanov
253

> In a triangle ABC, AB = 3, BC = 4 and CA = 5. Point D is the midpoint of AB, point E is on segment AC and point F is on segment BC. If AE = 1.5 and BF = 0.5 then Angle DEF =??
Plz help
it is useful to remember sin 3/5= 37degree and sin 4/5=53degree
so in given question angle cfe=a...

In a triangle ABC, AB = 3, BC = 4 and CA = 5. Point D is the midpoint of AB, point E is on segment AC and point F is on segment BC. If AE = 1.5 and BF = 0.5 then Angle DEF =??

Plz help

it is useful to remember sin 3/5= 37degree and sin 4/5=53degree

so in given question angle cfe=angle cef (and cef+cfe+37=180degree)

hence cef is obtained

similarly in triangle ade, angle a=53degrees

also angle aed=angle ade so angle afed can be obtaind..

now angle aed+ angle def + angle cef= 180degrees

so we obtain angle def

- 1 Like

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King
@king.casanov
253

> Bhaiyo some questions from permutation and combination
1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.
2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated
any n...

Bhaiyo some questions from permutation and combination

1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.

2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated

any number of times.

3)There are 5 different blue color, 4 different blue color and 3 different golden color. In how many ways the same color that are different can be selected.

4)There are 10 white, 8 green , 6 red balls in a box. In how many ways can one or more ball can be put in the box.

1. By Partition Method: we have to insert 4 partitions in 10 boxes(since 4 partitions of a set results into 5 groups) Thus no. of ways= 14C3=1092.

2.Question is not very clear, does it means u are selecting 6 mangoes out of 10 with replacement? 10^6 is it the ans?

3. clarify the question please...5diff. blue and then again 4 diff. blue??

4. there are in all 24 balls (i.e., 10 white,8 green, 6red similar balls) thus selecting 1 or more out of it= (2^24 -24)/ 10!8!6!

do tell the answers and as I am not too used to posting on this thread please let me know what all clarifications u need on my part thanks..

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Deepak
@Deepakmr95
489

In a triangle ABC, AB = 3, BC = 4 and CA = 5. Point D is the midpoint of AB, point E is on segment AC and point F is on segment BC. If AE = 1.5 and BF = 0.5 then Angle DEF =??
Plz help

Plz help

- 1 Like

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@modavasu
75

If a,b ,c are in AP.then A) 1/loga,1/logb etc are in HP?B)log a,lob,log c are in HP
3)log a,logb,logc are in GP

3)log a,logb,logc are in GP

- 1 Like

Commenting on this post has been disabled.

vishal rana
@vishal99999
204

Bhaiyo some questions from permutation and combination
1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.
2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated
any num...

Bhaiyo some questions from permutation and combination

1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.

2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated

any number of times.

3)There are 5 different blue color, 4 different blue color and 3 different golden color. In how many ways the same color that are different can be selected.

4)There are 10 white, 8 green , 6 red balls in a box. In how many ways can one or more ball can be put in the box.

Need help in these questions (Plz post approach also as I have a lot of confusion in such (P&c;)questions)

Thankx

1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.

2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated

any number of times.

3)There are 5 different blue color, 4 different blue color and 3 different golden color. In how many ways the same color that are different can be selected.

4)There are 10 white, 8 green , 6 red balls in a box. In how many ways can one or more ball can be put in the box.

Need help in these questions (Plz post approach also as I have a lot of confusion in such (P&c;)questions)

Thankx

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Hemant Yadav
@chillfactor
46.7
k

> If F(x, n) be the number of ways of distributing "x" toys to "n" children so that each child received at the most 2 toys then F(4, 3) =
P.S I'm getting the answer as 4C3=4, but the answer is given as 6..Plz help
Suppose they got 2 - a, 2 - b, 2 - c, then
(2 - a) + (2 - b) + ...

If F(x, n) be the number of ways of distributing "x" toys to "n" children so that each child received at the most 2 toys then F(4, 3) =

P.S I'm getting the answer as 4C3=4, but the answer is given as 6..Plz help

Suppose they got 2 - a, 2 - b, 2 - c, then

(2 - a) + (2 - b) + (2 - c) = 4

a + b + c = 2

=> C(4, 2) = 6 ways

2)x^12 fromx^6 from first and x^6 from second

=> -4 * (-1)^6 * C6

=> -4 * 9C3

i am getting this?

Bold part is typo I think it shouldn't be there!!!

Its:-

x^6 from first and x^6 from second

First one:- (1 - x^6)^4 = 1 - 4(x^6) + 6(x^12) - 4(x^18 ) + x^24

Second :- (1 - x)^(-4)

Coefficient of x^n in (1 - x)^(-r) = C(n + r - 1, r - 1)

So, Coefficient of x^6 in (1 - x)^(-4) = C(6 + 4 - 1, 4 - 1) = C(9, 3)

Multiply them to get (-4)*C(9, 3)

- 1 Like

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Neha Garg
@gnehagarg
350

> 1.(1- x^6)^4= 1 *4*x^6 + *6*x^12 4x^18 + x^24 - ...
There is a formula in Binomial Theorem which is written as
*_The coefficient of x^r in (1-x)^n is_*
_*= (-1)^r * nCr *_
In our case when 'n' is negative the formula becomes
*_(-1)^r * Cr_*
The soln in our case will be like this.. ...

1.(1- x^6)^4= 14x^6 +6x^12 4x^18 + x^24 - ...

There is a formula in Binomial Theorem which is written asThe coefficient of x^r in (1-x)^n is= (-1)^r * nCr

In our case when 'n' is negative the formula becomes(-1)^r * Cr

The soln in our case will be like this..

1) x^12 from (1-x)^-4

=> (-1)^12 * 15C12

=> 15C3

2) x^12 from x^6 from first and x^6 from second

=> -4 * (-1)^6 * C6

=> -4 * 9C3

3) x^12 from the first series

=> 6

Total = 15C3 -4*9C3 + 6 = 125

3. The first digit can either be 6 or 7..so 2 ways for the first digit.

Rest of the digits can be arranged in 6!/2 ways

Total ways = 2*6!/2 = 720 ways

2) x^12 from x^6 from first and x^6 from second

=> -4 * (-1)^6 * C6

=> -4 * 9C3

i am getting this?

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