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18 May 2025, 01:24
Kudos
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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:
43% (02:24) correct
57%
(02:40)
wrong
based on 97
sessions
History
Date
Time
Result
Not Attempted Yet
If x + y = 20 and 2x + 3y < 54, which of the following must be correct
I. 3x > 17
II. 2y < 30
III. y - x < 8
A. i only
B. II only
C. I and II
D. II and III
E. I, II and III
I. 3x > 17
II. 2y < 30
III. y - x < 8
A. i only
B. II only
C. I and II
D. II and III
E. I, II and III
18 May 2025, 01:33
Kudos
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Bunuel
Substituting x = 20 - y into 2x + 3y < 54, we get:
40 - 2y + 3y < 54
y < 14
Substituting y = 20 - x into y < 14, we get:
20 - x < 14
x > 6
Let's check the options:
I. 3x > 17 gives x > 5.6... Since we know that x > 6, then this option is correct (any x more than 6 will for sure be more than 5.6).
II. 2y < 30 gives y < 15. Since we know that y < 14, then this option is correct (any y less than 14 will for sure be less than 15).
III. y - x < 8. Subtract x > 6 from y < 14 to get y - x < 8. So, this option is correct.
Answer: E.
This question belongs to a type of question that tends to be very confusing for many. Therefore, I recommend practicing similar questions from the following collection: Trickiest Inequality Questions Type: Confusing Ranges.
Hope it's clear.
18 May 2025, 01:42
Kudos
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Bunuel
x + y = 20
2x + 3y < 54
Subtract thrice of (1) from (2):
-x < -6 => 3x > 18
Hence, I is true
Subtract twice of (1) from (2):
y < 14 => 2y < 28
Hence, II is true
We already have:
-x < -6
y < 14
Adding: y-x<8
Hence, III is true
Ans E
