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

Question

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

A. 9
B. 10
C. 11
D. 12
E. 20

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firas92 · Accepted Answer

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)

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yashikaaggarwal · Answer

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

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sambitspm · Answer

See the attachment
My answer = B

N22J · Answer

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.

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NitishJain · Answer

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

naveenban · Answer

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

GMATWhizTeam · Answer

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.

yashikaaggarwal · Answer

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

satya2029 · 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

gradschool2021 · Answer

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.

firas92 · Answer

Edited. Thank you

Kinshook · Answer

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

GMATinsight · Answer

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

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

SInce x and y both have to be Integers so y must be ODD for 5y to be odd so that 103-5y is even for x to be integer

at y = 1, x = 49

RULE: 
- The value of x changes (increases or decreases) by coefficient of y (i.e. 5 in this case and value of y changes (increases or decreases) by coefficient of x (i.e. 2 in this case and
- Value of x will decrease for increase in the value of y (for x and y to be positive integers) so the number of solution can be counted by possible values of x

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

Archit3110 · Answer

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

jhaamod10 · Answer

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.

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EgmatQuantExpert · Answer

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

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

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

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