If |x|

Question

If  |x| < 100  and  |y| < 100 , then what is the number of ordered pairs of integers (x, y) satisfying the equation  4x + 7y = 3  ?

A. 26
B. 27
C. 28
D. 29
E. 30

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GMATinsight · Accepted Answer

|x| < 100 i.e. -100 < x < 100 |y| < 100 i.e. -100 < y < 100 4x + 7y = 3 Let's find first solution using simple trial and error y = 3 and x = 6 RULE : In such linear equation in two variables, the value of x changes by coefficient of y (7 in this case) and value of y changes by coefficient of x (4 in this case) Now, x changes by 7 and y changes by 4 Since x has larger variation (7) in comparison to variation of y (4) so number of solutions will as many as possible values of x in given range because there will be less possible values of x in the given range -100 to 100 than the values of y among 200 numbers (-100 to +100) the values of x RULE : nth term of an Arithmetic progression, T_n = a+(n-1)*d where a is teh first term and d is the common differece i.e. 2nd term.- 1st term or 3rd term - 2nd term Positive values of x = {6, 13, 20....} nth term, 6+(n-1)*7 < 100 i.e. n-1 < 13.4 i.e. n = 14 Values of x less than 6 = {-1, -8, -15} nth term, -1+(n-1)*(-7) > -100 i.e. (n-1) < 99/7 i.e. (n) < 15.5 i.e. n = 15 So total possible values of x = 14 + 15 = 29 Answer: Option D Bunuel : The question must mention that x and y are integers. You might want to mention it.

sambitspm · Answer

See the attachment

nick1816 · Answer

7y = 3-4x

Assume x = k mod 7

0 mod 7 = (3-4k) mod 7

3-4k = 7; k = -1

x= 7a-1

-100<7a-1<100

-99< 7a < 101

-14.xyz < a < 14.abc

'a' can take 29 values.

NitishJain · Answer

D.
Given: -100 < x < 100 , -100 < y < 100

Mine was a lengthy approach and took a long time. I will go through other's solutions to get an alternate approach.

What I did: 4x+7y = 3 or x = (3 - 7y)/4
y = 1 , x = -1
y = 5, x = -4

Hence values of y increase with difference 4 so: .... .-7 -3, 1, 5, 9, 13 ...

We need to find extreme values and then will use formula:: a nth term = first term + (n-1)d
Here d= 4 and we will get n.

Moreover, the value for numerator (3 - 7y) should not exceed 400 as 400/4 =100

So extreme values for y = -55 & 57

so lets first term = -55, last term = 57
using formula stated above: 57 = -55 + (n-1)*4

112 = (n-1)*4
28 = n-1 or  n=29

ma29 · Answer

Since |x| < 100 and |y| < 100, and 4x+7y=3

-100<x<100 , so -100<(3-7y)/4<100

after solving this, -56.7<y<57.5

If y=(1,5) then X=(-1,-8) and If y=(-3,-7) then X=(6,13)

So, find the values of y (-56.7<y<57.5 ) for y=(1 & -3 and difference 4)

a+(n-1)d<57.5 = 1+(n-1)*4<57.5, after solving this n<15.1 so n=15

and -56.7<a+(n-1)d = -56.7<-3+(n-1)*(-4), after solving this n<14.4 so n=14

so total possible values of y= 15+14= 29

minirana · Answer

Hi Nitish,

I also worked out the problem in the same manner , what i am interested to know is how did you get the 'y' values? Was it trial and error ?

Thanks,
Mini.

karishmabatra · Answer

To find the value of x, you have written x=3-7y/3. Shouldn't we divide it by 4?

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If |x| < 100 and |y| < 100, then what is the number of ordered pairs o

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