If 2x + 5y = 103, then how many pairs of (x, y), where x and y are pos

2x+5y=103
Let's Put Y value as (1,2,3,....)
We know,
Odd - Odd gives even no.
If Y = even
Then 103-5(2) = 103-10 = 93 (which is neither divisible by 2 nor will leave perfect integer)
So Y values will be in odd no. Such as (1,3,5,.....)
=> 2X+5Y=103

=> Putting Y=1
=> 2x+5*1=103
=> 2x=98 => x=49

=> Putting Y=3
=> 2x+5*3=103
=> 2x=88 => x=44

=> Putting Y=5
=> 2x+5*5=103
=> 2x=78 => x=39

=> Putting Y=7
=> 2x+5*7=103
=> 2x=68 => x=34

=> Putting Y=9
=> 2x+5*9=103
=> 2x=58 => x=29

=> Putting Y=11
=> 2x+5*11=103
=> 2x=48 => x=24

=> Putting Y=13
=> 2x+5*13=103
=> 2x=38 => x=19

=> Putting Y=15
=> 2x+5*15=103
=> 2x=28 => x=14

=> Putting Y=17
=> 2x+5*17=103
=> 2x=18 => x=9

=> Putting Y=19
=> 2x+5*19=103
=> 2x=8 => x=4

So the pairs of (X,Y) are (4,19)(9,17)(14,15)(19,13)(24,11)(29,9)(34,7)(39,5)(44,3)(49,1) = 10 pairs.

IMO B

Posted from my mobile device
_________________

See the attachment
My answer = B

2x+5y= 103
No need to try values rather understand pattern.
Find least possible value for X,
It's 4 and y will be 19.
Now, X will increase by 5 and y will decrease by 2.

(4,19),(9,17)....(49,1)
10 pairs.

Can be applied for any equation of type ax+by=c, to find integral solutions.

Posted from my mobile device

\(2x + 5y = 103\)

\(x=\frac{103-5y}{2}\), so \(103-5y\) must be even for \(x\) to be an integer which means \(5y\) is odd and so \(y\) is odd

And since \(x\) and \(y\) are positive, we have \(5y<103\) or \(y<20.6\)

This means that \(y\) can take any odd value between \(1\) and \(19\) inclusive for each of which \(x\) has a corresponding value

So \(y=1,3,5,7,9,11,13,15,17,19\) that is \(10\) values

Answer is (B)

Posted from my mobile device
_________________

B.

2x + 5y = 103
Given: x,y >0
solve for x: x = (103-5y)/2

Lets put y = 1: x = 49 ok
Lets put y = 2: x = 93/2 NO
Lets put y = 3: x = 44 ok

this means only odd values of y will satisfy the requirement that x is an integer.

Lets put y = 19: x = 4 ok
Lets put y = 20: x = 3/2 NO

for y = 21, x <0

so values of y = {1,3,5,7,9,11,13,15,17,19} = 10 values
_________________

Ans. B
2x + 5y = 103.

We need 2x to end with 8 and 5y to end with 5, for the result to have unit digit 3.
X = 49, Y = 1. ( First value with the result 103 )
X = 44, Y = 3.
X = 39, Y = 5.
We can observe that an Arithematic Progression has been formed.
Last value pair would be (4,19) for (x,y)
If we solve A.P for either X or Y, we would get 10.
Option B - 10

Solution

• x and y are positive integers.
• \(2x + 5y = 103\)

o Or, \(x = \frac{(103 – 5y)}{2}⟹ x = \frac{ (102 – 4y)}{2} +\frac{(1-y)}{2} ⟹ x = 51 - 2y + \frac{(1 – y)}{2}\)
o Now y can be 1, in that case \( x = 51 - 2y + \frac{(1 – y)}{2} = 51-2*1 + \frac{1-1}{2} = 49\)
o Next value of y can be 3, in the case \(x = 51 -2*3 + \frac{(1-3)}{2} = 44\)
o Next value of y can be 5, in that case \(x = 51 -2*5 +\frac{(1- 5)}{2} = 39\)….. And so on.
• We can notice that value of x is decreasing by 5 and maximum possible value of x = 49.
• Since, x is a positive integer, minimum value of x can be 4

o In that case \(y = \frac{103 – 4*2}{5} = 19\)
• Thus, the values of x are in A.P. such that first term is 4, last term is 49 and the common difference is 5.

o Now, let us assume that the total possible positive integral values of x is n.

 Then, \(49 = 4 + (n – 1)*5 ⟹ n = \frac{45}{5} + 1 = 10\)
• Therefore, total possible values of x will be 10 and correspondingly y will also have 10 values.
• So, the required number of (x,y ) pairs = 10

Thus, the correct answer is Option B.
_________________

You might have to change your option From D to B. Since your solution represent B answer
_________________

Y=(103-2X)/5
Since 13-8=5 and the number will be divisible by 5
so when 2X has a unit digit 8 we will get Y positive integer
X=4,9,14,19,24,29,34,39,44,49
10 values
B:)

In order for 2X + 5Y = 103 to be true, 5Y must be odd therefore Y must be an odd number.

The largest number that Y could be for this would be 19 as 5*19 = 95. The smallest number that Y could be would be 1.

Therefore the number of pairs would be the count of odd numbers, inclusive, between 1 and 19.

Answer is B.

Edited. Thank you
_________________

Asked: If 2x + 5y = 103, then how many pairs of (x, y), where x and y are positive integers, are there ?

y = (103-2x)/5 ; x & y are positive integers
(x,y) = (4,19) is one solution
x is increased by 5 and y is decreased by 2 for each successive solution
(x,y) = {(4,19),(9,17),(14,15),(19,13),(24,11),(29,9),(34,7),(39,5),(44,3),(49,1)}; 10 solutions

IMO B

The equation that we have is 2x + 5y = 103

WHich we can re-write as \(x = \frac{103-5y}{2}\)



at y = 1, x = 49

5 can be subtracted 9 times from 49 to keep it positive and least

so total solutions = 1 (for x=49) + 9 (9 more solutions) = 10 SOlutions

Answer: Option B

2x+5y=103
given x& y are integers
so
x=103-5y/2
we know odd-odd = even so value of y has to be odd to get integer value of x
and maximum value of y can be 20 i.e total 10 odd and 10 even
so we get 10 odd values of y from 1 to 19 and 10 integer values of x whose unit digit would always be either (4,9) i.e ∆ of 5
OPTION B ; 10

2x+5y=103 ,x>0, y>0
Now x is integer if y is odd
So possible values are
(49, 1) ,(44,3) ,(39, 5),(34, 7)
(29, 9),(24,11), (19, 13), (14,15)
(9, 17) and (4, 19) total 10 pairs
Hence ans is B.

Posted from my mobile device

Solution

Given

• \(2x + 5y = 103\), x and y are positive integers.

To Find

• Pairs of (x,y) that satisfy the equation.

Approach and Working Out

• The expression can be rearranged as \(2x = 103 –5y.\)
• Thus \(x = (103-5y)/2\). Now since x and y are positive integers, the term \((103-5y)\) should be a multiple of 2. In other words, \(103-5y\) should be even.

o Since odd – 5y should be even, therefore 5y should be odd as well and thus y should be odd (since odd *odd = odd).
• Therefore, the problem has been reduced to finding the possible values of positive integer ‘x’ when ‘y’ is an odd positive integer and \(x =(103-5y)/2\), where x is a positive integer.

Now smallest positive odd integer is 1, when y = 1, we get \(x = (103-5)/2 = 49.\)

• Similarly, when \(y= 3, x = (103-15)/2 = 88/2 = 44.\)
• Since \(103 –5y >0\), therefore \(y < 103/5\). Therefore \(y <=20 \) (since ‘y’ is a positive integer)

o But since y is odd, therefore maximum possible value of \(y = 19\).

Number of pairs (x, y) that satisfy the equation are therefore all the pairs possible for each possible value of ‘y’.

• Least value of \(y = 1\), greatest \(y = 19\). ‘y’ has to be odd, therefore number of possible values of \( y = 1, 3, 5, 7, 9,… 19. = 10\) total possible values.

o We can find ‘x’ for each such possible value of ‘y’.

Correct Answer: Option B
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________