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[2014] CL Mock CAT 15 Scores and Discussion (DON'T OPEN UNLESS ATTEMPTED)Discussion related to CL Mock CAT 15.
Mock 15 can be attempted in unproctored mode right? It's not showing up in my online mocks, anyone else facing the same problem?
An Eternal Fire Burns Within.
http://www.pagalguy.com/discussions/2014-mock-score-repository-25116679/28946263
This is the Official CAT 2014 Quantitative preparation group. Solve questions & discuss answers/solutions shortcuts &...
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AIMCAT 1505-New: Scores And Discussion ~ (Non-Invigilated: Oct 21-24, 2014)TIME's mock CAT
totally disappointed ...dunno what went wrong in section 1 with 8 wrong attempts.....52+46 = 98....can anyone guess my %?
Hey puys, As we all are geared up to prepare ALL NEW CAT 2014, let us refresh our quant concepts bit by bit. All we n...
How to find number of solutions of equation
involving difference of perfect squares?
Before solving questions based on this concept, let us understand the absolute basics from where we build upon.
Equations involving difference and addition of two numbers, say x and y, will give solution over the set of integers only when either both x and y are odd or both are even. Let us understand this in 3 cases:
1. If x - y =Odd and x + y = Even
Then adding these two equations,
We will get 2 x = odd
(We all know: 2 X any number= even, odd+even=odd)
Which implies x cannot be an integer.
Similarly y cannot be integer.
2. If x - y =Even and x + y = Even
Then adding we will get 2 x = even
which implies x will be an integer.
Similarly y will be an integer.
3. If x - y =Odd and x + y = Odd
Then adding we will get 2 x = even
which implies x will be an integer.
Similarly y will be an integer.
NOTE: Thus while solving these questions given below we will keep in mind such equations will have solutions over the set of integers only when either both x and y are odd or both are even.
We will categorize such problems into 4 parts
CASE 1. Odd Number written as difference of two perfect squares
Example : Find the number of positive integer solutions of equation x2 - y2 = 85 ?
Sol : We can write given equation as
(x- y)(x +y )= 85
Now we want x and y both to be integers and we are given odd number on the
right hand side of equation .So clearly both factors will be odd
For finding factors of any number we decompose that number into prime factors
Since 85 = 5. 17
Factors of 5 are 50, 51 (2 in number)
Factors of 17 are 170, 171 (2 in number)
Thus total factors of 85 are obtained by multiplying the factors of 5 and 17
respectively
(i.e. 50170=1 , 50171=
17, 51170=5 , 51171= 85)
Thus total no of factors of 85 are 4 .
But our aim is that we have to write 85 as a product of two factors and Obviously difference of two positive integers is less than their sum i.e. x-y < x+y.So clearly there are two ways we can write 85 as a product of two factors i.e.
85 = 1 x 85 and 85 = 5 x 17
Thus the two possibilities are :
Either (x- y)= 1 and (x +y )= 85
Or (x- y)= 5 and (x +y )= 17.
So in above example answer is
(No of factors of 85)/2
i.e. 4/2 = 2
Answer : 2
NOTE:
Infact we can find the numbers also by solving the above equations x =43 ,y =42
and x=11 ,y = 6.But we are generally asked to find number of solutions only in
such questions
GENERALIZATION
So in such cases we can generalize no of ways in which we can write a
given number as a product of two factors is : (No of factors on the right
hand side of given)/2
Questions for Practice:
1. Find the number of positive integer solutions of equation x2
- y2 = 105 ?
Answer : 4
2. Find the number of positive integer solutions of equation x2
- y2 = 1815 ?
Answer : 6
3. Find the number of positive integer solutions of equation x2
- y2 = 1547 ?
Answer : 3
4. Find the number of positive integer solutions of equation x2
- y2 = 845 ?
Answer : 3
5. Find the number of positive integer solutions of equation x2
- y2 = 1215 ?
Answer : 10
CASE 2. Even Number which is not a multiple of 4 written as difference of two perfect squares
Example : Find the number of positive integer solutions of equation x2 - y2 = 66?
Sol : We can write given equation as
(x- y)(x +y )= 66
Now we want x and y both to be integers and we are given an even number on the
right hand side of equation which is not a multiple of 4.We know such equations
will have solutions over the set of integers only when either both x and y are odd
or both are even.
But none of the above case is possible
Case 1 : If both x and y are odd then (x- y) is also even
and (x +y ) is also even i.e. both are multiple of 2
Which implies that Left Hand Side is a multiple of 4 but right hand side
is not .Hence no solution is possible in case both are odd.
So clearly both factors on the Left Hand Side cannot be odd
Case 2 : If both x and y are even then (x- y) is also even
and (x +y ) is also even i.e. both are multiple of 2
Which implies that Left Hand Side is a multiple of 4 but right hand side
is not .Hence no solution is possible in case both are even.
So clearly both factors on the Left Hand Side cannot be even
Answer: Zero
Generalisation: Thus every number of the form 4k + 2 (where k is any integer ) can never be written as difference of perfect squares.
Questions for Practice:
1. Find the number of positive integer solutions of equation x2
- y2 = 1098 ?
Answer : Zero
2. Find the number of positive integer solutions of equation x2
- y2 = 3462 ?
Answer : Zero
3. Find the number of positive integer solutions of equation x2
- y2 = 3782 ?
Answer : Zero
4. Find the number of positive integer solutions of equation x2
- y2 = 84546 ?
Answer : Zero
5. Find the number of positive integer solutions of equation x2
- y2 = 1213514 ?
Answer : Zero
CASE 3. Even Number which is a multiple of 4 written as difference of two perfect squares
Example : Find the number of positive integer solutions
of equation x2 - y2 = 56?
Sol : We can write given equation as
(x- y)(x +y )= 56 (1)
Now we want x and y both to be integers and we are given an even number
which is a multiple of 4 on the right hand side of equation .We know such
equations will have solutions over the set of integers only when either both x
and y are odd or both are even.
So clearly both factors will be even
Let us take (x- y) =2p
(x + y)= 2q
So finally putting these in (1) we get
4 p q = 56
p q = 14
So our question reduces to writing 14 as a product of two factors
For finding factors of any number we decompose that number into prime
factors
14 = 21. 71
Factors of 2 are 20, 21 (2 in number)
Factors of 7 are 70, 71 (2 in number)
Thus total factors of 14 are obtained by multiplying the factors of 2 and 7 respectively
(i.e. 2070=1 , 2071=
7, 2170=2 , 2171= 14)
Thus total no of factors of 14 are 4 .
So clearly there are 4 ways we can write 36 as a product of two factors
14 = 1 x 14
14 = 2 x 7
Answer: 2
GENERALIZATION
So in such cases we can generalize no of ways in which we can write a
given number as a product of two factors is : (No of factors on the right hand
side of given after dividing by 4) /2.
Questions for Practise
1. Find the number of positive integer solutions of equation x2
- y2 = 67500 ?
Answer : 10
2. Find the number of positive integer solutions of equation x2
- y2 = 588 ?
Answer : 3
3. Find the number of positive integer solutions of equation x2
- y2 =10780 ?
Answer : 6
4. Find the number of positive integer solutions of equation x2
- y2 = 168 ?
Answer : 4
5. Find the number of positive integer solutions of equation x2
- y2 =720 ?
Answer : 9
CASE 4. Perfect square written as difference of perfect squares
Example : Find the number of positive integer solutions
of equation x2 - y2 = 144?
Sol : We can write given equation as
(x- y)(x +y )= 144 (1)
Now we want x and y both to be integers and we are given an even number which is a multiple of 4 on the right hand side of equation .We know such equations will have solutions over the set of integers only when either both x and y are odd or both are even.
So clearly both factors will be even
Let us take (x- y) =2p
(x + y)= 2q
So finally putting these in (1) we get
4 p q = 144
p q = 36
So our question reduces to writing 36 as a product of two factors
For finding factors of any number we decompose that number into prime
factors
36 = 22. 32
Factors of 22 are 20, 21 , 22
(3 in number)
Factors of 32 are 30, 31, 32 (3 in
number)
Thus total factors of 36 are obtained by multiplying the factors of 22 and 32 respectively
(i.e. 2030=1 , 2031=3, 2032=9 , 2130=2, 2131=6 , 2132=18, 2230=4 , 2231=12, 2232=36)
Thus total no of factors of 36 are 3 X 3 =9 .(Also note factors of perfect square are always odd )
Note : Both the factors cannot be 6 because if both (x- y)= 6 and (x +y )=6 Then x = 6 and y= 0 which is not possible as we are talking of positive integers. Since we cannot take both the factors 6 therefore remaining 8 factors can be written in 8/2 = 4 ways as a product of two factors.
So clearly there are 4 ways we can write 36 as a product of two factors
36 = 1 x 36
36 = 2 x 18
36 = 3 x 6
36 = 4 x 9
GENERALIZATION:
So in such cases we can generalize no of ways in which we can write a
given number as a product of two factors is:
{(No of factors on the right hand side of given after dividing by 4) - 1 }/2.
Questions for Practice
1. Find the number of positive integer solutions of equation x2
- y2 = 900 ?
Answer : 4
2. Find the number of positive integer solutions of equation x2
- y2 = 225 ?
Answer : 4
3. Find the number of positive integer solutions of equation x2
- y2 =324 ?
Answer : 2
4. Find the number of positive integer solutions of equation x2
- y2 = 122500?
Answer :7
5. Find the number of positive integer solutions of equation x2
- y2 =19600 ?
Answer : 13
-Wordpandit
bhai how to approach
x2 - y2 = 225
with the formula
(No of factors on the right hand side of given after dividing by 4) - 1 }/2
(as 225 is not divisible by 4) ?
@psycho.munna @harshcat91 how u are getting 10 for x^2-y^2= 1215? factors will be 6*2/2 =6( 3^5* 5). can u tell me where m making the mistake?
.Can someone help me out with the following problems with some clear and simple approcah to these qustions 4 balls ar...
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Can someone explain the approach to such questions please?
1) 902. Ans = 4k^2 +2.where k =15
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Bhai this is called TCS. Experience uncertainty.
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Hi Puys.... QADI 41
VALR 34
Total 75
I know it's pretty low score but what would be my percentile???
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Read more at www.englishgrammarlesson.blogspot.in
https://www.youtube.com/channel/UCwA7FhAHLqqmjlVjhiwG8GQhttps://www.youtube.com/channel/UCwA7FhAHLqqmjlVjhiwG8GQ
Took this today..Weirdest QA section ever..VA was Okaiish
OA - 128 (56/46/10) QA-54 (22/19/3) VA-74(34/27/7)..Expected %ile...?
Yup it can be. When I took it on Monday, it was showing
Okay then I'll email them to rectify the problem.
@ferlonso sir results of mock 15 are out? I mean solutions. I have taken non proctored mode and there are no solutions and analysis activated
@ani6 No, the window is not yet over