Official Quant thread for CAT 2013

MBA CAT 2024
9121
@scrabbler said:
Think set theory. For 2 sets, total is (1 at a time ) 0 (2 at a time). For 3 sets, (1 at a time) - (2 at a time) + (3 at a time). For 4 sets, (1 at a time) - (2 at a time) + (3 at a time) - (4 at a time) and so on...alternate +/- is the key to remember.Also in this case the coefficients will be 4C4, 4C3, 4C2 etc as that gives the number of ways in which we can select 4, 3, 2...so that also makes life a little bit easier...regardsscrabbler
sorry couldnt get it :(
9122

4²² - C(4, 1)*3²² + C(4, 2)*2²² - 4 is correct for the last question

For detailed explanations read about Principle of Inclusion - Exclusion (derangements is one of the example of this principle)
http://onlinemathcircle.com/wp-content/uploads/2011/05/21-pie.pdf
http://www.math.cornell.edu/~putnam/inclExcl.pdf

In a school, there are 4 more students in Class A than in Class B, and 4 more students in Class B than in Class C. As a reward for good behaviour a teacher decides to give out some sweets. Every student in class C got 5 more sweets than every student in Class B, and every student in Class B got 3 more sweets than every student in Class A. As it turns out, Class A got 10 more sweets than Class B, and Class B got 14 more sweets than Class C. How many sweets were given out in all?
9123
@sujamait said:
random putting...din get any gud method x=0,y=0x=0,y=-1x=-1,y=0x=-1,y=-1
bhai x, y mein quadratic equation bana lo and say write y in terms of x, play with the determinant and you are done :)
9124
@chillfactor :can u explain inclusion and exclusion rule that you used in this sir ?? dat 4 room q
9125
@chillfactor said:
Yup, 4 is correct.x(x + 1) = 4y(y + 1)4x² + 4x = 16y² + 16y(4y + 2)² - (2x + 1)²= 3(4y - 2x + 1)(4y + 2x + 3) = 3 = 1*3 = 3*1 = (-1)(-3) = (-3)(-1)So, we will get 4 solutionsYup, 343 is correct.Twelve female workers and ten male workers are to be assigned to work in one of four different departments of a company. In how many ways can this be done if each department gets at least one worker?

Females same males same
0F 4 M->10C4
1 F 3M->10C3*12C1*4!/3!
2F 2 M->10C2*10C2*4!/2!*2!
3F 1 M->12C3*10C1*4!/3!
4F 0M->12C4

Females different males different
0F 4M->10C4*4!
1F 3M->10C3*12C1*4!
2F 2M->10C2*12C2*4!
3F 1M->10C2*12C3*4!
4F 0M->12C4*4!



9126
@chillfactor said:
4²² - C(4, 1)*3²² + C(4, 2)*2²² - 4 is correct for the last questionFor detailed explanations read about Principle of Inclusion - Exclusion (derangements is one of the example of this principle)onlinemathcircle.com/wp-conten...www.math.cornell.edu/~putnam/i...In a school, there are 4 more students in Class A than in Class B, and 4 more students in Class B than in Class C. As a reward for good behaviour a teacher decides to give out some sweets. Every student in class C got 5 more sweets than every student in Class B, and every student in Class B got 3 more sweets than every student in Class A. As it turns out, Class A got 10 more sweets than Class B, and Class B got 14 more sweets than Class C. How many sweets were given out in all?
Getting 760.

Edit: 668 rather. I took total sweets of C > B > A for earlier solution. Haste!!!

regards
scrabbler
9127
@Angadbir said:
Each of the 12 cases would also have different distribution in themselves .So think about accounting that. Approach is in the right direction.@Brooklyn just a little bit of +/- mix-up with that soln there .
Can you give just one or two pointers as to how to account for distribution of 12 cases ?
9128
@rachit_28 said:
bhai x, y mein quadratic equation bana lo and say write y in terms of x, play with the determinant and you are done
yar thoda lengthy hai woh sab...kuch tricky sa soch rha tha me..w/o using quadr...



@chillfactor said:
4²² - C(4, 1)*3²² + C(4, 2)*2²² - 4 is correct for the last question

@chillfactor said:
In a school, there are 4 more students in Class A than in Class B, and 4 more students in Class B than in Class C. As a reward for good behaviour a teacher decides to give out some sweets. Every student in class C got 5 more sweets than every student in Class B, and every student in Class B got 3 more sweets than every student in Class A. As it turns out, Class A got 10 more sweets than Class B, and Class B got 14 more sweets than Class C. How many sweets were given out in all?

C=x-4 ,got y+8 sweets each
B=x , got y+3 sweets each
A=x+4 , got y sweets each

xy+4y = 10 + xy + 3x => 4y - 3x=10
xy + 3x = 14 + xy + 8x - 4y - 32 => 4y - 5x = -18

x = 14
y = 13

sweets given = 16*14 + 13*18 + 21*10 = 668


9129
@Brooklyn said:
didnt get u can u explain dis lil more
Yes I will, let me get my thoughts together :banghead:
9130
@chillfactor said:
4²² - C(4, 1)*3²² + C(4, 2)*2²² - 4 is correct for the last questionFor detailed explanations read about Principle of Inclusion - Exclusion (derangements is one of the example of this principle)onlinemathcircle.com/wp-conten...www.math.cornell.edu/~putnam/i...In a school, there are 4 more students in Class A than in Class B, and 4 more students in Class B than in Class C. As a reward for good behaviour a teacher decides to give out some sweets. Every student in class C got 5 more sweets than every student in Class B, and every student in Class B got 3 more sweets than every student in Class A. As it turns out, Class A got 10 more sweets than Class B, and Class B got 14 more sweets than Class C. How many sweets were given out in all?
Let students in B be b and sweets given to exery students be x
From question
A total sweets(AT)=(x-3)(b+4)
BT=xb
CT=(x+5)(b-4)
---
(x-3)(b+4)-xb=10
xb-(x+5)(b-4)=14
x=16, b=14
TS=(16-3)*(14+4)+(16*14)+(16+5)*(14-4)=668?
9131
@gnehagarg said:
Females same males same 0F 4 M->10C41 F 3M->10C3*12C1*4!/3!2F 2 M->10C2*10C2*4!/2!*2!3F 1 M->12C3*10C1*4!/3!4F 0M->12C4Females different males different0F 4M->10C4*4!1F 3M->10C3*12C1*4!2F 2M->10C2*12C2*4!3F 1M->10C2*12C3*4!4F 0M->12C4*4!
I think you are doing something else (read the question again). Question is asking that in how many ways those 22 persons can be divided in 4 different groups such that none of the group is empty
@scrabbler said:
Getting 760. regardsscrabbler
I guess some calculation mistake
9132

668 is correct for sweets question :thumbsup:


Isaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from 0 to 10. He never scores more in a later question than in any earlier question. How many dierent possible sequences of six marks can he achieve?
9133
@chillfactor said:
Solve for integers (x, y) such that:-x(x + 1) = 4y(y + 1)

(0, -1), (-1, 0), (0,0) , (-1, -1) => 4 solutions ??
9134
@chillfactor said:
In a school, there are 4 more students in Class A than in Class B, and 4 more students in Class B than in Class C. As a reward for good behaviour a teacher decides to give out some sweets. Every student in class C got 5 more sweets than every student in Class B, and every student in Class B got 3 more sweets than every student in Class A. As it turns out, Class A got 10 more sweets than Class B, and Class B got 14 more sweets than Class C. How many sweets were given out in all?

Students:
C - x
B - x+4
A - x+8

Sweets(each):
A - a
B - a+3
C - a+8

Total sweets:
A - a(x+8)
B - (a+3)(x+4)
C - (a+8)x

A - B = 10, which gives 4a - 3x =22;
B - C =14, which gives 4a - 5x =2;
Solving, we get: x=10; a=13

Total sweets: 234 + 224 + 210 = 668

9135

The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as m! —n Pr where m, n, r are distinct positive integers. What is the sum m + n + r ?
OPTIONS

1) 66
2) 67
3) 68
4) None

9136
@sujamait said:
The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as m! —n Pr where m, n, r are distinct positive integers. What is the sum m + n + r ?OPTIONS1) 66 2) 67 3) 68 4) None
67

30!*31P6
9137
@rachit_28 said:
Can you give just one or two pointers as to how to account for distribution of 12 cases ?
@Brooklyn

I'll play with a simpler example, and then you can take it further to larger problems like the last one.

Consider 5 different balls that are to be put inside 3 different boxes, such that no box is left empty. How many ways?

Total ways (includes any no. of boxes being empty) -> 3^5

From this figure, we ought to remove cases where 1 box is empty.
So select 1 box to be empty (3 ways) and distribute 5 balls in the rest.
-> 3.2^5 ways

Now here, we've had some repetitions. In 3.2^5, cases where 2 boxes go empty have been counted 2 times.
Example: Suppose you chose A to be empty, and distributed 5 balls in B and C. One case is such that B is left empty and C has 5 balls.
Now suppose you chose B to be empty, and distributed 5 balls in A and C. One case is such that A is left empty and C has 5 balls.

Both the above cases are same. and since they are being subtracted twice than actually required, we add these cases (where 2 boxes are empty)

So total ways = 3^5 - C(3,1)2^5 + C(3,2)1^5

This can be extended for the previous problem


9138
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?
175
100
90
none of the above
9139
@chillfactor said:
668 is correct for sweets questionIsaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from 0 to 10. He never scores more in a later question than in any earlier question. How many dierent possible sequences of six marks can he achieve?
Will take a shot on 16C6. Options would help :D

regards
scrabbler
9140
@sujamait said:
The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as m! —n Pr where m, n, r are distinct positive integers. What is the sum m + n + r ?OPTIONS1) 66 2) 67 3) 68 4) None
_ 6 _ 12 _ 18 _ 24 _ 30 _ 36 _

g1 + g2 + g3 + g4 + g5 + g6 + g7 = 25
=> 31C6 * 30!

required answer = 30 + 31 + 6 = 67