x, a, z, and b are single digit positive integers. x = ¼ a.

Question

x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal

A. 33
B. 36
C. 44
D. 63
E. 66

VeritasPrepRon · Accepted Answer

Without doing any algebra, you can reason out that the first part of the subtraction will be exactly four times bigger than the second part of the subtraction. Since all of the numbers are integers, it means that a and be will be divisible by 4 (while x and z can be anything). Once you subtract 1/4 from a and b, you'll be left with 3/4 of what they were orignally, which must be divisible by 3. The answer will necessarily be divisible by 3, so it elmininates answer choices a, b, d and e. Answer choice C is the only impossible outcome.

This type of question can lead people down rabbit holes, trying to calculate every possible number, especially with a premise as tempting as "single digit positive integers". There can't be that many choices, right? Doing this the long way may work out well, but you can't beat conceptual understanding for giving you the right answer in the most efficient way.

Hope this helps!
-Ron

mau5 · Answer

(10a+b)-(10x+z) = 10(a-x)+(b-z) = 10(4x-x)+(4z-z) = 3*10(x+z). Thus the sum has to be a multiple of 3.Only option C is not a multiple of 3.

C.

pranbo · Answer

From the given equations, a = 4x, b = 4z

(10a+b) - (10x+z) = 40x+4z - 10x-z = 30x +3z = 3(10x+z)

Therefore the resultant should be divisible by 3 which isn't possible only in the case of C.

Hence C is the answer

eddyki · Answer

another way, not as short as the methods mentioned above but also helpfull:

"x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal"

since they are all single digit positive integers, x and z could be 1 or 2, a and b would become 4 or 8.

if you look at the answer choies you can see a pattern in the single digit's. if you evaluate the possibility for the single digit of (10a + b) – (10x + z), you get:

single digit: b-z
if z=1, b=4
if z=2, b=8.

so 4-1=3 or 8-2=6

I hope its understandable

PareshGmat · Answer

Did in the same way. Just like to add.... instead of 44, if the option had been 48 or some multiple of 3, then again all yielded values would had to be checked

Temurkhon · Answer

did by algebra use

a=4x, b=4z

10a+b-10x-z
10(a-x)+b-z, by substitution get 30x+3z => 3(10x+z). So, should be multiple of 3

Only C does not fit

chetan2u · Answer

Hi,
almost everyone has done by considerimg non multiple of 3..
But if it doe snot strike in the stress of the exam, another method would be..

1) firstly we are not given that all are different integers..
2) since each is a positive single digit integer and x=a/4... a can take only 4 and 8 as value, and x can take 1 and 2..
3) same as 2) for b and z.

so in (10a + b) – (10x + z) ..

10a+b can be
44,48,84,88..

10x+z can be
11,12,21,22..

lets see teh choices 
A. 33.. 44-11 possible
B. 36..48-12
C. 44.. no combination possible
D. 63..84-21
E. 66..88-22

ans C

Nevernevergiveup · Answer

Original Explanation:

Solution: C

Spot the pattern here.

All four variables must represent integers, so if x = ¼ a and z = ¼ b, then both a and b must be either 4 or 8.

Now try all four possibilities.

If a = 4 and b = 4, then (10a+b) – (10x+z) = 44 – 11 = 33. 
If a = 4 and b = 8, then (10a+b) – (10x+z) = 48 – 12 = 36. 
If a = 8 and b = 4, then (10a+b) – (10x+z) = 84 – 21 = 63. 
If a = 8 and b = 8, then (10a+b) – (10x+z) = 88 – 22 = 66.

(C) is impossible.

Dheeraj_079 · Answer

10a+ b -10x-1-z ====> substituting x = a/4 and z = b/4, we get

30a/4 + 3b/4 =====> The value must be a multiple of 3.
Of all the options only 44 is not a multiple of 3

Hence answer is C

bumpbot · Answer

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x, a, z, and b are single digit positive integers. x = ¼ a.

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x, a, z, and b are single digit positive integers. x = ¼ a.

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