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If x and y are nonnegative integers and 5x + 2y = 32, how many values

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If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 26 Dec 2016, 10:02
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A
B
C
D
E

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Re: If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 26 Dec 2016, 10:20
1
2
Bunuel wrote:
If x and y are nonnegative integers and 5x + 2y = 32, how many values of x are there?

A. 1
B. 2
C. 3
D. 4
E. 6


5x + 2y = 32
5x = 2(16 – y)
x = 2n, y = 16 – 5n
x,y >= 0 -------> n = 0, 1, 2, 3 --------> x = 0, 2, 4, 6
Answer D
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Re: If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 30 Sep 2018, 16:17
1
Bunuel wrote:
If x and y are nonnegative integers and 5x + 2y = 32, how many values of x are there?

A. 1
B. 2
C. 3
D. 4
E. 6



we are looking foe the number of x may take . Trail and Error is the best in my sense.

we know that both x and y are integers and they are non-negative in nature.

even - even = even, which is divisible by 2. Plug even values for x only.

5x + 2y = 32.

1. x = 0

5*0 + 2y = 32 = y = 16.

2. x=2

5*2 + 2y = 32

y = 11

3. x = 4

5*4 + 2y = 32

y = 6

4. x = 6

5*6 + 2y = 32

y = 1.


Therefore the equation is true when we put these values of x. one may plug the value of y and look for the integer value of x instead of y.

The best answer is D.
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If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 30 Sep 2018, 20:35
Bunuel wrote:
If x and y are nonnegative integers and 5x + 2y = 32, how many values of x are there?

A. 1
B. 2
C. 3
D. 4
E. 6


CONCEPT: Find one solution i.e. values of x and y and for each next solution remember that x always changes by co-efficient of y (i.e. 2 in this case) and y always changes by co-efficient of x (i.e. 5 in this case) and

First solution is x = 0 and y = 16
Second solution is x = 2 and y = 11
Third solution is x = 4 and y = 6
Third solution is x = 6 and y = 1

Answer: Option D
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Re: If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 02 Jun 2020, 17:11
:roll: :| I am not able to understand any of the solutions for the given problem. Can someone explain it?
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If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 02 Jun 2020, 19:45
MridulaSivanandan wrote:
:roll: :| I am not able to understand any of the solutions for the given problem. Can someone explain it?


This is a liner equation question and you have to find possible values of x which satisfy the linear equation: 5x + 2y = 32

What this means is - how many non-negative values can x take, which satisfy this equation.
Also keep in mind y can take non-negative values only. i.e. integer values of 0, 1, 2, 3, ...

If we replace x in the equation by each of 0, 2, 4 and 6, we get corresponding values of y as 16, 11, 6 & 1.
In this manner:

when x=0, 5x = 0 and thus 2y=32, y=16 ---> 16 is not an integer and thus x can take value of 0.
when x=1, 5x = 5 and thus 2y=27, y=13.5 ---> 13.5 is not an integer and thus, x cannot take value of 1.
(taking only even values of x)
when x=2, 5x = 10 and thus 2y=22, y=11
when x=4, 5x = 20 and thus 2y=12, y=6
when x=6, 5x = 30 and thus 2y=2, y=2

Thus, x can take 4 values.
Ans: D

Hope this helps!

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If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 06 Jun 2020, 04:19
1
Bunuel wrote:
If x and y are nonnegative integers and 5x + 2y = 32, how many values of x are there?

A. 1
B. 2
C. 3
D. 4
E. 6


Solution:

Simplifying, we have:

5x = 32 - 2y

5x = 2(16 - y)

x = 2(16 - y)/5

The expression (16 - y) must be a multiple of 5, in order for the entire expression on the right side of the equation to be a non-negative integer. Thus, we see that y can be 1, 6, 11, or 16. Since there are 4 possible values for y, there are also 4 possible values for x.

Alternate Solution:

Notice x and y are nonnegative integers, we see that in order for 5x + 2y = 32 to hold, x can’t be greater than 6 (since 5 * 7 = 35 already). Therefore, x can only be any integers from 0 to 6, inclusive. Furthermore, if x is odd, then 5x is odd and 2y = 32 - 5x will be odd. However, since 2y is always even regardless of what y is, we see that x can’t be odd. On the other hand, if x is even, then 5x is even and 2y = 32 - 5x will be even. Therefore, x can only be 0, 2, 4 or 6,

Answer: D
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Re: If x and y are nonnegative integers and 5x + 2y = 32, how many values  [#permalink]

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New post 05 Jul 2020, 11:50
The maximum value of x is 6 because 5x=30
Since 2y is always an even number 5x also has to be an even number
Therefore x can be 0,2,4,6
There are four values fore x

Answer D
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Re: If x and y are nonnegative integers and 5x + 2y = 32, how many values   [#permalink] 05 Jul 2020, 11:50

If x and y are nonnegative integers and 5x + 2y = 32, how many values

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