Official Quant thread for CAT 2013

MBA CAT 2024
25588
@pakkapagal said:
sir mera bhi doubt clear kr dijiye na
see what i have done is i have divided the whole problem into 2 cases - 1. 2 cuts lie on same half of stick 2. when 2 cuts lie on different halves

after that when they lie on different halves - i again have 2 cases - 1. distance between them is greater than half of stick 2. distance between them is less than half of stick

hope now you got it :)

Team BV--Pratik Gauri
25589
@Highway66 said:
@ScareCrow28 nope10 straight lines, no two of which are parallel nd no three of which pass through any common point, are drawn on a plane. The total number of regions (including finite and infinite regions) into which the plane would be divided is1. 56 2. 255 3. 1024 4. not uniqueP.s. answer with approach
P=L+I+1
here p are the regions, L no of lines given ,I is intersection of these lines

l=10 , I= C(10,2)=45

So , P=45+10+1=56 :)

Team BV--Pratik Gauri
25590
@albiesriram said:
@albiesriram said: unsolved yet.
25591

A question from my side to all of you ..


IF 15X^2 + 8Y^2 DIVIDED BY 22 = XY AND X,Y ARE TWO NOS BELONGING TO THE SET OF THE FIRST 100 NATURAL NOS , HOW MANY DISTINCT ORDERED PAIRS (X,Y) ARE POSSIBLE ?

Team BV--Pratik Gauri
25592
@albiesriram said:
let the diameter be 10a+b

let oh be x
so
(10a+b/2 +x)(10a+b/2 -x) = (10b+a/2)^2
x = root (11*9(a-b)(a+b))/2
as x is a rational number of the form p/q
a+b=11
a-b=1
a=6 b=5
diameter 10*6+5=65
25593
@bodhi_vriksha said:
A question from my side to all of you ..IF 15X^2 + 8Y^2 DIVIDED BY 22 = XY AND X,Y ARE TWO NOS BELONGING TO THE SET OF THE FIRST 100 NATURAL NOS , HOW MANY DISTINCT ORDERED PAIRS (X,Y) ARE POSSIBLE ?Team BV--Pratik Gauri

53 ? not sure..
25594
@rnishant231 said:
53 ? not sure..
Correct 😃 Try everyone 😃 It is an easy question ..


Team BV--Pratik Gauri
25595
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?

a) 60
b) 62
c) 64
d) can't be determined.
25596
@bodhi_vriksha said:
see what i have done is i have divided the whole problem into 2 cases - 1. 2 cuts lie on same half of stick 2. when 2 cuts lie on different halvesafter that when they lie on different halves - i again have 2 cases - 1. distance between them is greater than half of stick 2. distance between them is less than half of stick hope now you got it Team BV--Pratik Gauri
got it 😃 thanks sir
25597
@albiesriram said:
OH^2 = OC^2 - CH^2
Let AB=10a+b , CD=10b+a
Oh^2= (10a+b/2)^2-(10b+a/2)^2
OH^2= 99/4(a^2 - b^2)
a+b=11 , a-b=1
a=6 and b=5

so AB=65 :)


Team BV--Pratik Gauri
25598
@bodhi_vriksha said:
A question from my side to all of you ..IF 15X^2 + 8Y^2 DIVIDED BY 22 = XY AND X,Y ARE TWO NOS BELONGING TO THE SET OF THE FIRST 100 NATURAL NOS , HOW MANY DISTINCT ORDERED PAIRS (X,Y) ARE POSSIBLE ?Team BV--Pratik Gauri
53 ?
25599
@nole said:
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
cbd ?
25600
@Narci no
25601
@Highway66 said:
@ScareCrow28 nope10 straight lines, no two of which are parallel nd no three of which pass through any common point, are drawn on a plane. The total number of regions (including finite and infinite regions) into which the plane would be divided is1. 56 2. 255 3. 1024 4. not uniqueP.s. answer with approach
56
25602
@nole said:
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
a1,a2...a5 in ascending order of weights
a5+a4=121
a5+a3=120
a4=a3+1
a1+a2=110
a1+a3=113
a3=a2+3
a4=a2+4
a5=117-a2
a2=110-a1
a3=113-a1
a4=114-a1
a5=a1+7
a1,110-a1,113-a1,114-a1,a1+7
a5 has to be atleast 61
a1 can be at max 54
possible
54,56,59,60,61


61 hona chea

not sure though
25603
@Subhashdec2 answer is not 61
25604
@nole said:
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
let the weights wen arranged in increasing order A B C D E(A lightest)....
wen pairs are taken then each weight appears 4 times.
total=1156 kg

4(A+B+C+D+E)=1156
(A+B+C+D+E)=289
A+B=110
D+E=121
so, C=58
D E can be 60 61 or
D E can be 59 62

now pla with eqn ...... 62 satisfies...
25605
@Subhashdec2 said:
a1,a2...a5 in ascending order of weightsa5+a4=121a5+a3=120a4=a3+1a1+a2=110a1+a3=113a3=a2+3a4=a2+4a5=117-a2a2=110-a1a3=113-a1a4=114-a1a5=a1+7a1,110-a1,113-a1,114-a1,a1+7a5 has to be atleast 61a1 can be at max 54possible54,56,59,60,6161 hona cheanot sure though
59 + 60 =119 nai h koi weight
25606
@Narci said:
53 ?
bhai iska solution post karo. na.......
25607
@jai.tuteja answer is 62.

@Dexian : can u explain what u did after getting value of c ?