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rxs0005
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If the average of 5x and 8y is greater than 90, and x is twice y 02 Mar 2011, 05:32
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D
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45% (medium)

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68% (01:39) correct 32% (01:34) wrong based on 98 sessions

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If the average of 5x and 8y is greater than 90, and x is twice y, what is the least integer value of x ?

A. 4
B. 5
C. 20
D. 21
E. 23



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D
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If the average of 5x and 8y is greater than 90, and x is twice y 02 Mar 2011, 05:51
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\(\frac{5x+8y}{2}>90\)

\(5x+8y>180\)

\(5x+\frac{8x}{2}>180\)

\(10x+8x>360\)

\(18x>360\)

\(x>20\) OR \(x>=21\)

Least integral value of x = 21

Ans: "D"
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If the average of 5x and 8y is greater than 90, and x is twice y 02 Mar 2011, 06:01
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what is wrong with the below reasoning

lets say i solve by eliminating x first

so i get 10y + 8y > 180

or 18y> 180

or y > 10

since i know y > 10 i can rewrite original equation as

x >(180 - 8y) / ( 5 )

to get least value of x we can make y the max and y > 10 so i keep y = 20

x > ( 180 - 160 ) / 5 = 4 for integer x it will be x >=5
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If the average of 5x and 8y is greater than 90, and x is twice y 02 Mar 2011, 08:13
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rxs0005
what is wrong with the below reasoning

lets say i solve by eliminating x first

so i get 10y + 8y > 180

or 18y> 180

or y > 10

since i know y > 10 i can rewrite original equation as

x >(180 - 8y) / ( 5 )

to get least value of x we can make y the max and y > 10 so i keep y = 20

x > ( 180 - 160 ) / 5 = 4 for integer x it will be x >=5


Your way of solving for y instead of x is also fine. You have got that y > 10 and it is also given that x = 2y, so x > 20 and the least integer greater than 20 is 21, so answer is 21.

Your approach of putting an arbitrary value for y just gives us one more condition for x (that it should be greater than 5) but does not help us find lowest value of x such that x = 2y and average is greater than 90. To get minimum value of x, you need to work with minimum acceptable value of y, which is anything slightly greater than 10. if you put y = 10 in your equation you will get that x > (180 - 80)/(5) or x > 20 s x is at least 21.

Overall, you need not go further than solving for y, as we already are given that x = 2y
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If the average of 5x and 8y is greater than 90, and x is twice y 02 Mar 2011, 18:46
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rxs0005
what is wrong with the below reasoning

lets say i solve by eliminating x first

so i get 10y + 8y > 180

or 18y> 180

or y > 10

since i know y > 10 i can rewrite original equation as

x >(180 - 8y) / ( 5 )

to get least value of x we can make y the max and y > 10 so i keep y = 20

x > ( 180 - 160 ) / 5 = 4 for integer x it will be x >=5

Since you know y > 10 and that x = 2y, you can say that x/2 > 10 or that x > 20.
You should not substitute an inequality in another inequality. e.g.
x > 2 (we know smallest integral value of x is 3)
y > x (we know)
This doesn't mean that the least integral value of y is 3. It is possible that it may be 10 or 100 etc. All we know is that y > 2, how much greater we don't know.
On the other hand,
x > 2 (we know smallest integral value of x is 3)
y = x
Now we know that the smallest integral value of y is 3.
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If the average of 5x and 8y is greater than 90, and x is twice y 02 Mar 2011, 18:46
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rxs0005
If the average of 5x and 8y is greater than 90 , and x is twice y , what is the least integer value of x

4

5

20

21

23

Substitution can be used in the following way:
Always start with the equation: x = 2y. It is more straight forward to manage as compared to the inequality.
Substitute y = x/2, not the other way because you need to find the minimum value of x. So you can get rid of y.

Now go on to the inequality. So 8y = 8x/2 = 4x
Now average of 5x and 4x is greater than 90. Average of 5x and 4x is obviously 4.5x. So,
4.5x > 90
or x > 20
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If the average of 5x and 8y is greater than 90, and x is twice y 04 Mar 2011, 01:50
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VeritasPrepKarishma
You should not substitute an inequality in another inequality. e.g.
Hey Karishma
There is substitution within inequality. Pls take a look at this one. This is from GMATPrep.
inequalities-question-100853.html

If y is greater than 110 percent of x, is y greater than 75?
(1)\(x>75\)
(2) \(y - x = 10\)

Pls verify.
1) Sufficient.
y > 1.1 x and x > 75. Min(x) = 75
y > 1.1 * 75
y > 82.5

2) Insufficient. Substituting (x=y-10) in y > 1.1x
y > 1.1(y-10)
y < 110
Hence A.
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If the average of 5x and 8y is greater than 90, and x is twice y 04 Mar 2011, 05:37
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VeritasPrepKarishma
You should not substitute an inequality in another inequality. e.g.
Hey Karishma
There is substitution within inequality. Pls take a look at this one. This is from GMATPrep.
inequalities-question-100853.html

If y is greater than 110 percent of x, is y greater than 75?
(1)\(x>75\)
(2) \(y - x = 10\)

Pls verify.
1) Sufficient.
y > 1.1 x and x > 75. Min(x) = 75
y > 1.1 * 75
y > 82.5

2) Insufficient. Substituting (x=y-10) in y > 1.1x
y > 1.1(y-10)
y < 110
Hence A.

Yes, this is correct. Here you are looking for a range of values. You can say that y > 82.5 but can you find the minimum value of y? Since in the previous question, you are trying to find the minimum value of 20, substituting an inequality in another doesn't give it to you. As I said while explaining it 'All we know is that y > 2, how much greater we don't know.' Now if the question is, 'Is y>2?' then of course there are no issues. Because of this problem, if I have an equation, I would rather substitute that in the inequality. It will give me more defined results.
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If the average of 5x and 8y is greater than 90, and x is twice y 18 Aug 2015, 09:47
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rxs0005
what is wrong with the below reasoning

lets say i solve by eliminating x first

so i get 10y + 8y > 180

or 18y> 180

or y > 10

since i know y > 10 i can rewrite original equation as

x >(180 - 8y) / ( 5 )

to get least value of x we can make y the max and y > 10 so i keep y = 20

x > ( 180 - 160 ) / 5 = 4 for integer x it will be x >=5

however, there is a condition that x is twice y.
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If the average of 5x and 8y is greater than 90, and x is twice y 13 Jul 2016, 10:45
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Bunuel bb

How is the answer to this A?
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If the average of 5x and 8y is greater than 90, and x is twice y 28 Apr 2017, 05:17
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I guess the OA for this question is D.
Can bunnuel or someone else confirm?
THanks.
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If the average of 5x and 8y is greater than 90, and x is twice y 28 Apr 2017, 07:01
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VeritasPrepKarishma
guialain
I guess the OA for this question is D.
Can bunnuel or someone else confirm?
THanks.

Yes, it is (D).

Bunuel, VeritasPrepKarishma
Can you please correct the OA?
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If the average of 5x and 8y is greater than 90, and x is twice y 28 Apr 2017, 12:37
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yup, official answer is wrong, just like solve an equality that has limited values of X. therefore, X cannot be 4
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If the average of 5x and 8y is greater than 90, and x is twice y 28 Apr 2017, 20:59
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I think there is a typo error in the question. The question stem should be as follows:
If the average of 5x and 8y is greater than 9, and x is twice y, what is the least integer value of x ?
A. 4
B. 5
C. 20
D. 21
E. 23



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