a/5 + 5/a 2 ?

Question

a/5 + 5/a > 2 ?

(1) a < 4
(2) a > 0.1

Shreeya24 · Answer

We can simplify the question stem to (a^2+25)/5a>2?
1) a<4, 
If a is negative, the stem will have negative value and it will be less than 2.
If a is 3 the stem will have positive value greater than 2.
Hence insuffecient.
2) a>0.1
No question of negative value. For a=0.1, itself the stem is greater than 2. hence suffecient. Answer is "B"

Bunuel · Answer

Let me ask you, for (2) what will happen if a = 5?

ENGRTOMBA2018 · Answer

\frac{(a^2+25)}{5a}-2 > 0  --->  \frac{(a^2+25-10a)}{5a} > 0  ---->  \frac{(a-5)^2}{5a} > 0

This expression can be plotted on the number lines as shown in the attached image. Thus, the expression becomes = 0 for a=5. You need be to wary of a=5. This is the reason why B is NOT sufficient. You need to combine both statements to come up with a unique expression.

Takeaway from this question: whenever you have a square of the form \((a-b)^2\), make sure to check the case when a=b.

Hope this helps.

Takeaway from this question: whenever you have a square of the form  (a-b)^2 , make sure to check the case when a=b.

Hope this helps.

D3N0 · Answer

on the number line, plotted shouldn't there be alternate change of signs?
I am getting singns as for a>5 +ve, for 0<a<5 -ve and for -5<a<0 +ve, a<-5 -ve (alternate change)

ENGRTOMBA2018 · Answer

No. Alternate signs are true for odd powers of variables but for even powers, (a-5)^2, this expression will ALWAYS be + for whatever value of a . Thus, there is no change of sign when a>5 or when 0<a<5.

When 0<a<5, lets say a=3, the expression (a-5)^2/5a is still > 0

You should always check the signs by plugging in some values.

Hope this helps.

D3N0 · Answer

Got it.. i made mistake while solving.. i solved it for a^2 - 5^2 = (a-5)(a+5)
thanks

bumpbot · Answer

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a/5 + 5/a > 2 ?

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a/5 + 5/a > 2 ?

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