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Official Quant Thread for CAT 2011 [Part 8]

  • Discussion in CAT
  • Created by @pkaman
@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
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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.
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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
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> 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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> 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
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> 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 LCCJH......option 3
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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
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
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> 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
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> 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
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> 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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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
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
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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
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
  • 1 Like  
Commenting on this post has been disabled.
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
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> 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 from x^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)
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> 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= 1 4x^6 + 6x^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) 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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