Official Quant thread for CAT 2013

@joyjitpal said:
there are 7 children standing in the line,not all of whom have the same number of cakes with them.if the first child distributes his cakes to the remaining 6 children such that he doubles their respective no. of cakes,then he will be left with four cakes.instead,if the 2 child takes away 2 cakes from each of the remaining 6 children,then he will be left with 2 cakes less than the no. of cakes that the first child initially had.what is the total no. of cakes that are there with the third child to the seventh child? 11141215
10 cakes?
Let the children from 1st to 7th have x1,x2,x3,x4,x5,x6,x7 no. of cakes initially.

First case:
x1 - (x2 + x3 + x4 + x5 + x6 + x7) = 4

Second case:
x2 + 2*6 = x1 - 2 => x1-x2 = 14

Subtracting both equations, we get
(x3 + x4 + x5 + x6 + x7) = 10

So 10 cakes.
@joyjitpal said:
a teacher asked a student to add nine 2-digit no. such that each no. is 10 more than the previous no. but the student has added all the no.,which are formed by interchanging the digits of each of the given no.surprisingly,the sum he got was the same as the sum the teacher has asked for.which of the following is the difference of the digits in the least of the no. the student was asked to add? a) 2 b) 3 c)4d) 5
Let first be "ab", which is 10*a+b. The others are 10*(a+1)+b, 10*(a+2)+b, ...10*(a+8)+b

The sum should be 10*a+b + 10*(a+1)+b...10*(a+8)+b =
90a+10*(0+1+...8)+9b => 90a+9b+360

The incorrectly done sum is 10b+a + 10b+(a+1) + 10b+(a+8) => 90b + 9a + 36

So 90a+9b+360=90b+9a+36
81a+324=81b
a+4=b

This leads to solutions (a,b) => (1,5),(2,6),(3,7),(4,8), and (5,9)
But since we need 9 2 digit numbers spaced at 10 each, smallest number will be 15
Series will be => 15,25,35,45,55,65,75,85,95
Difference of digits = 5-1 = 4
@joyjitpal said:
a teacher asked a student to add nine 2-digit no. such that each no. is 10 more than the previous no. but the student has added all the no.,which are formed by interchanging the digits of each of the given no.surprisingly,the sum he got was the same as the sum the teacher has asked for.which of the following is the difference of the digits in the least of the no. the student was asked to add? a) 2 b) 3 c)4d) 5
4. 10+20+...+90 gives 450. And 50*9=450. So if the second digit is 5, it satisfies the condition.
(Sort of a logical trial and error)

@joyjitpal said:
there are 7 children standing in the line,not all of whom have the same number of cakes with them.if the first child distributes his cakes to the remaining 6 children such that he doubles their respective no. of cakes,then he will be left with four cakes.instead,if the 2 child takes away 2 cakes from each of the remaining 6 children,then he will be left with 2 cakes less than the no. of cakes that the first child initially had.what is the total no. of cakes that are there with the third child to the seventh child? 11141215
I am getting 10 :splat:
@soumitrabengeri said:
11 cakes?Let the children from 1st to 7th have x1,x2,x3,x4,x5,x6,x7 no. of cakes initially.First case:x1 - (x2 + x3 + x4 + x5 + x6 + x7) = 4Second case:x2 + 2*6 = x1 - 3 => x1-x2 = 15Subtracting both equations, we get(x3 + x4 + x5 + x6 + x7) = 11So 11 cakes.
Second case mai why minus 3 and not minus 2 :O
@joyjitpal said:
a teacher asked a student to add nine 2-digit no. such that each no. is 10 more than the previous no. but the student has added all the no.,which are formed by interchanging the digits of each of the given no.surprisingly,the sum he got was the same as the sum the teacher has asked for.which of the following is the difference of the digits in the least of the no. the student was asked to add? a) 2 b) 3 c)4d) 5
(10x+y)*9+360=(10y+x)*9+36

==> 81y-81x=360-36
==> y-x=4

c)4

@ankita14 said:
Second case mai why minus 3 and not minus 2
Sorry..should be 2..was solving another frnds doubt at the same time and got distracted
@joyjitpal said:
there are 7 children standing in the line,not all of whom have the same number of cakes with them.if the first child distributes his cakes to the remaining 6 children such that he doubles their respective no. of cakes,then he will be left with four cakes.instead,if the 2 child takes away 2 cakes from each of the remaining 6 children,then he will be left with 2 cakes less than the no. of cakes that the first child initially had.what is the total no. of cakes that are there with the third child to the seventh child? 11141215
x1-(x2+x3+...+x7)=4

x2+12=x1-2
==> x1-x2=14

x3+x4+..+x7=14-4=10

@ankita14 me 2 :p
@joyjitpal said:
a teacher asked a student to add nine 2-digit no. such that each no. is 10 more than the previous no. but the student has added all the no.,which are formed by interchanging the digits of each of the given no.surprisingly,the sum he got was the same as the sum the teacher has asked for.which of the following is the difference of the digits in the least of the no. the student was asked to add? a) 2 b) 3 c)4d) 5


bhai what does this mean ? unable to understand language





@sujamait said:
bhai what does this mean ? unable to understand language
smallest number ke digits ka difference
say 21,31,41... hota to 2-1

ek Q ...


In an over 7 runs were scored..only 0,1,2,3,4,6 runs on a ball are possible.what is the probability of atleast one dot ball(0 rum) by the batsmen.

@sujamait said:
ek Q ...In an over 7 runs were scored..only 0,1,2,3,4,6 runs on a ball are possible.what is the probability of atleast one dot ball(0 rum) by the batsmen.
119/120 hai Kya?
@ankita14 said:
119/120 hai Kya?
method ankita ?
OA was something like 1/3^4.2^4
@sujamait said:
method ankita ?OA was something like 1/3^3.2^4 I guess...not sure about denominator though..I'd also got something else....
Oh sorry galat hai :mg:
@sujamait said:
method ankita ?OA was something like 1/3^4.2^4
But how can the probability be so low. 2,1,1,1,1,1 is the only case where not a single 0 exists :/
In how many ways can sachin attempt all the Questions with atleast 2 mistakes of a test having 10 Questions ?
@ankita14 said:
But how can the probability be so low. 2,1,1,1,1,1 is the only case where not a single 0 exists :/

exactly!!
actually I'm not satisfied with their OA...
they have given explanation like..
possibilities,
6,1 - prob of dot ball = 2/3
4,3 - . .... = 2/3
4,2,1 - ... =3/6=1/2
...and other cases...then multiplying these probabilities///



@sujamait said:
In how many ways can sachin attempt all the Questions with atleast 2 mistakes of a test having 10 Questions ?
1013?
2^10 ways to attempt all questions, 1 way to commit 0 mistakes, 10 ways to commit 1 mistake. 1024-11=1013
@ankita14 said:
But how can the probability be so low. 2,1,1,1,1,1 is the only case where not a single 0 exists :/
this makes six cases right?? as delivrys are diffrnt
m getng 6/551
@jaiswalicsi said:
this makes six cases right?? as delivrys are diffrnt m getng 6/551
Yea 6. I'm getting more than 551 though.
@sujamait said:
In how many ways can sachin attempt all the Questions with atleast 2 mistakes of a test having 10 Questions ?
1013??

2W8C, 3W7C, 4W6C, 5W5C, 6W4C,7W3C,8W2C,9W1C,10W...
and solve...
:)