For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

Question

(A) 33
(B) 49
(C) 50
(D) 99
(E) 100

Archit3110 · Answer

pair series will go on from x,y = (2,49), (4,48),(6,47)...... (98,1)
total such pairs 49
IMO B

KSBGC · Answer

Bro, x and y can't be 0.

x and y are positive integers.

Berlin92 · Answer

Since x cannot be 0 y must be smaller than 50, therefore there are 49 different possible values of y. 
Answer B is correct since pairs are ordered?

jackharrywa · Answer

I'm supposing the appropriate solution is 16. Since it is x+2y=100 this suggests x is even. so x=2,4, 6, 8, 10, ..., 30, 32. There are 16 requested sets.

Berlin92 · Answer

Why do you stop at 32?

x could also have the value 98 and y=1, by your logic that's 49 possible values of x

jackharrywa · Answer

yes, you are saying absolutely right, it is 49 possible value.

joneyb · Answer

Here is how i am going to solve it...

x + 2y = 100 for positive pair of x and y..

Lets assume that y = 49 then 2y = 98. Then x will be 2. y can't be 50 because then 2y will be 100 and x will become zero. As per problem statement x cant be zero.

now lets reduce the value of y

case 1 - if y = 48 then x becomes 4
case 2 - if y = 47 then x becomes 6
case 3 - if y = 46 then x becomes 8 
and so on..

so y can have values from 49 to 1. It will change x values accordingly as well.

correct answer is B.

jfranciscocuencag · Answer

Hello!

It can be solved with arithmetic progression.

nth = a + d(n-1)

nth = 98 
 a = 2

Applying the formula  n = 49 .

49.

B

ScottTargetTestPrep · Answer

We see that the smallest integer value for y is 1 (and x will then be 98), and the largest value of y is 49 (and x will then be 2). Since y can be any integer from 1 to 49, there are 49 values for y and hence 49 ordered pairs of (x, y).

Answer: B

stne · Answer

x + 2y = 100
x=100-2y
x=2(50-y)
Since x and y have to be positive integers, smallest value of y = 1, and largest value of y=49, hence there are 49 such pairs.

Answer B.

Hope this helps .

BlueRocketAsh · Answer

99+1=100 => No values for x and y 
 98+2=100 => x=98 y=1 
 97+3=100 => No values for x and y 
 96+4=100 => x=96 y=2 
.
.
 50+50=100 => x=50 y=25 
.
.
.
 4+96=100 => x=4 y=48 
 3+97=100 => No values for x and y 
 2+98=100 => x=2 y=49 
 1+99=100 => No values for x and y

Here, y goes from 1 to 49 continuously. Therefore there are 49 pairs
Answer B

GMATGuruNY · Answer

Consider the following equation:
2x + 3y = 30.

If x and y are nonnegative integers, the following solutions are possible:
x=15, y=0
x=12, y=2
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following:
The value of x changes in increments of 3 (the coefficient for y).
The value of y changes in increments of 2 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

Greatest positive option for y = 49.
Since y can decrease only in increments of 1 -- the implied coefficient for x -- the following options are yielded for y:
49, 48, 47...3, 2, 1
Implication:
y can be any positive integer between 1 and 49, inclusive -- a total of 49 options.

B

ZIX · Answer

Let's use logic here.

Equation: x + 2y = 100

This means that 2y would always be even and therefore can add only even values to x. (2, 4, ... 98)

And if 2y is always even then x should always be even too for them to equal 100. (2 + 98; 4 + 96 ... 98 +2)

Therefore, total even values possible = 100/2 = 50. However, we subtract one value from it: when x =100 even though it is even. As when x = 100, 2y must = 0, which we can't use here since 0 is not a positive integer.

So total values possible = 50 - 1= 49

For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

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For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

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