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BrentGMATPrepNow
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If x, y, and z are positive integers, x : y = 1 : 3 27 Mar 2017, 07:01
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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15% (low)

Question Stats:

84% (01:39) correct 16% (02:08) wrong based on 433 sessions

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If x, y, and z are positive integers, x : y = 1 : 3, and z : y = 2 : 5, then x + z COULD equal

A) 46
B) 50
C) 54
D) 62
E) 66

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E
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If x, y, and z are positive integers, x : y = 1 : 3 27 Mar 2017, 08:20
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GMATPrepNow
If x, y, and z are positive integers, x : y = 1 : 3, and z : y = 2 : 5, then x + z COULD equal

A) 46
B) 50
C) 54
D) 62
E) 66

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Hi

x/y = 1/3 ----> 3x=y

z/y = 2/5 ----> 5z = 2y

5z = 6x -----> z=6n, x=5n

z + x = 6n + 5n = n(6 + 5) = 11n

z + x should be a multiple of 11.

Answer E (66)
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If x, y, and z are positive integers, x : y = 1 : 3 Updated on: 13 Nov 2019, 17:05
Originally posted by BrentGMATPrepNow on 28 Mar 2017, 06:17.
Last edited by BrentGMATPrepNow on 13 Nov 2019, 17:05, edited 1 time in total.
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GMATPrepNow
If x, y, and z are positive integers, x : y = 1 : 3, and z : y = 2 : 5, then x + z COULD equal

A) 46
B) 50
C) 54
D) 62
E) 66
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Given: x : y = 1 : 3
y : z = 5 : 2

To COMBINE these ratios, their shared term (the y-value) must be the same in each ratio.

So, take x : y = 1 : 3 and multiply both parts by 5 to get an EQUIVALENT RATIO: x : y = 5 : 15
Likewise, take y : z = 5 : 2 and multiply both parts by 3 to get: y : z = 15 : 6
We can now COMBINE the ratios to get: x : y : z = 5 : 15 : 6
This means the ratio x : z = 5 : 6

Since x and z are positive integers, x + z must be a MULTIPLE of 5 + 6
In other words, x + z must be a MULTIPLE of 11

Answer: E
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If x, y, and z are positive integers, x : y = 1 : 3 27 Mar 2017, 08:50
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GMATPrepNow
If x, y, and z are positive integers, x : y = 1 : 3, and z : y = 2 : 5, then x + z COULD equal

A) 46
B) 50
C) 54
D) 62
E) 66

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x:y = 1:3 and y:z = 5:2
so we have x:y:z = 1:3:(2*3/5)
x:y:z = 5:15:6

now x+z = 5n+6n =11n (where n is the common term between x and z)
and x+z = 11n must be a multiple of 11

Hence option E is correct
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If x, y, and z are positive integers, x : y = 1 : 3 27 Mar 2017, 10:35
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Given: x,y, z>0
x/y=1/3
z/y=2/5

As y is common in both the ratios, I equated them.
y=3x and y=5z/2
3x=5z/2
x=5z/6

That means: z is divisible by both 2 and 3 (x is an integer). Therefore z contains at least 2 prime factors 2 and 3 as 5 is not divisible by 6.

x+z=5z/6 + z= 11z/6
z is divisible by 6 (2 and 3) and the answer is a multiple of 11.
Therefore the answer is 66 (E)
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If x, y, and z are positive integers, x : y = 1 : 3 28 Mar 2017, 03:23
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x : y: z
1 : 3
5: 2
To correct the relationship must multiply (x:y) in 5 (y:z) in 3.

x : y: z
5 : 15: 6

x+z=11.. Therefore the it is multpile of 11

Only 66 fits

Answer: E
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If x, y, and z are positive integers, x : y = 1 : 3 04 Oct 2019, 02:54
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GMATPrepNow
If x, y, and z are positive integers, x : y = 1 : 3, and z : y = 2 : 5, then x + z COULD equal

A) 46
B) 50
C) 54
D) 62
E) 66

*Kudos for all correct solutions
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This is how i solved

x : y
1 : 3'
y: z
5: 2
To correct the relationship must multiply (x:y) by 5 and (y:z) by 3. to get the common LCM of 5 and 3

New ratios

x : y: z
5 : 15: 6

x+z=11.. Therefore the it is multpile of 11

Only 66 fits

Answer: E
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If x, y, and z are positive integers, x : y = 1 : 3 04 Oct 2019, 08:51
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GMATPrepNow
If x, y, and z are positive integers, x : y = 1 : 3, and z : y = 2 : 5, then x + z COULD equal

A) 46
B) 50
C) 54
D) 62
E) 66

*Kudos for all correct solutions
Show more
\(x : y = 1 : 3\)
\(z : y = 2 : 5\)

Make y equall in both the equations....

\(x : y = 5 : 15\)
\(z : y = 6 : 15\)

So, Finally we have , \(x : y : z = 5 : 15 : 6\)

Now, \(x + z = 5 + 6 => 11\), Thus the correct answer must be a multiple of 11, among the given options only (E) follows...
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If x, y, and z are positive integers, x : y = 1 : 3 05 Oct 2019, 05:50
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Solution:

x : y = 1:3

y: z =5:2

To co-relate the ratios we should get the common ratio in the same propotion
Hence muptiply x:y by 5 and y:Z by 3
New ratio is:
x:y:z
5:15:6

x+z=11

Hence the answer must be a multiple of 11 and only 66 is the possible

IMO E
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If x, y, and z are positive integers, x : y = 1 : 3 22 Sep 2025, 18:08
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