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30 Aug 2025, 04:37
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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)!
Difficulty:
95%
(hard)
Question Stats:
44% (02:08) correct
56%
(02:40)
wrong
based on 100
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Not Attempted Yet
Consider the equation \(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)=\(\frac{4}{3}\), where x, y and z are positive integers. Which of the following could be the value of x+y+z?
I.5
II.7
III.13
A) I only
B) II only
C) III only
D) I and II only
E) II and III only
I.5
II.7
III.13
A) I only
B) II only
C) III only
D) I and II only
E) II and III only
28 Sep 2025, 06:41
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We can only get a sum of 4/3 in the given condition as 1+1/3, but here, there are two numbers only, so we've to split one of them into half.
For this, we have 2 options
1 can be written as 1/2 + 1/2
OR
1/3 can be written as 1/6 + 1/6
Therefore, we have 2 cases here
Case 1: 1/2 + 1/2 + 1/3 = 4/3
x+y+z = 7
Case 2: 1/6 + 1/6 + 1/1 = 4/3
x+y+z = 13
Hence, option E
For this, we have 2 options
1 can be written as 1/2 + 1/2
OR
1/3 can be written as 1/6 + 1/6
Therefore, we have 2 cases here
Case 1: 1/2 + 1/2 + 1/3 = 4/3
x+y+z = 7
Case 2: 1/6 + 1/6 + 1/1 = 4/3
x+y+z = 13
Hence, option E
findingmyself
Updated on: 31 Aug 2025, 14:19
Originally posted by gmatophobia on 31 Aug 2025, 14:06.
Last edited by gmatophobia on 31 Aug 2025, 14:19, edited 1 time in total.
Last edited by gmatophobia on 31 Aug 2025, 14:19, edited 1 time in total.
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findingmyself
\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)=\(\frac{4}{3}\)
\(\frac{yz + xz + xy}{x*y*z} = \frac{4}{3}\)
Observation
1. x * y * z is a multiple of 3. Therefore, one of the three numbers must be a multiple of three.
2. yz + xz + xy is a multiple of 4, hence the value is even. This is possible when all three numbers are even, or when two numbers are even and one is odd.
Let's look at the available options
1. x + y + z = 5
Assume x = 3. Therefore y = 1 and z = 1. Note, it doesn't matter which of the variable is assigned the value 3 as the remaining sum (y + z), will be adjusted between the remaining two variables.
In this case, all the three numbers are odd. Hence, the sum yz + xz + xy cannot be even.
Hence, we can conclude that \(x + y + z \neq 5\)
We can eliminate A and D.
2. x + y + z = 7
As x * y * z must be a multiple of 3, let's assume x = 3
Possible values of x, y and z =
- x = 3, y = 1 , z = 3 → Ignore, as all three values are odd.
- x = 3, y = 2, z = 2 → Let's calculate the sum of the reciprocals
\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{3} + \frac{1}{2} + \frac{1}{2} = 1 + \frac{1}{3} = \frac{4}{3} \)
The value matches, x + y + z can be 7.
The possible answer is between B and E. Eliminate C.
2. x + y + z = 13
Possible values of x, y and z =
- x = 3, y = 1 , z = 9 → Ignore as all three values are odd
- x = 3, y = 2, z = 8 → 1/3 = 0.33 ; 1/2 = 0.5 ; 1/8 = 0.125 ⇒ Sum is less than 1.33. Hence, we can ignore this combination.
- x = 3, y = 3, z = 7 → Ignore as all three values are odd
- x = 3, y = 4, z = 6 → 1/3 = 0.33 ; 1/4 = 0.25 ; 1/6 = 0.167 ⇒ Sum is less than 1.33. Hence, we can ignore this combination.
- x = 3, y = 5, z = 5 → Ignore as all three values are odd
- x = 6, y = 1, z = 6 → 1/6 = 0.167 ; y = 1/1 = 1 ; z = 1/6 = 0.167 ⇒ Sum = 1 + 0.33 = 1.33. Matches & we don't have to try out more combinations.
Hence, x + y + z can be 13.
Option E
