If 1/55

Question

If  \frac{1}{55}<x<\frac{1}{22}  and  \frac{1}{33}<x<\frac{1}{11} , then which of the following could be the value of x?

(I) \frac{1}{54}

(II) \frac{1}{23}

(III) \frac{1}{12}

A) Only I
B) Only II
C) I and II
D) II and III
E) I, II, III

Source => NOVA.

KarishmaB · Accepted Answer

Two ranges of x are given. \frac{1}{55} and ................................................ <---------------------------- x ----------------------> 1/33 < x < 1/22 So all values such as 1/23, 1/24, 1/25.... 1/32 will lie within this range. Answer (B)

emockus · Answer

The question is asking about the value of x that fits both ranges.

The ranges are: -55<x<-22 and -33<x<-11

The only value that fits both ranges in -23; choice "B"

rohit8865 · Answer

II satisfies both condition

Ans B

Nunuboy1994 · Answer

The most efficient way to solve this problem is to just cross multiply the denominators all across the expression- for example

<  
  <

and then

1/23 < 1/22
22/(23)(22) < 23/(23)(22)

The most efficient way, then, is to just compare the denominators

* once you cross multiply two fractions such as -   <   - do not cross multiply again by the next fraction for example 
  <   cross multiplied by 1/22

grichagupta · Answer

I solved it the traditional way (taking LCM and comparing each fraction option with the ranges given and it took me almost 3 mins to conclude the right answer. Request you to advise the best approach to solve such problems.

Nunuboy1994 : I could not understand the cross-multiplication approach you are suggesting

Nunuboy1994  : I could not understand the cross-multiplication approach you are suggesting

guptarahul · Answer

If 155

C)I and II D)II and III E)I,II,III

cbh · Answer

For me the best way to resolve this was to write the numbers with negative exponents
1/55 = (55)^-1
1/33 = (33)^-1
1/22 = (22)^-1
1/11 = (11)^-1

So if 1/55<x<1/22 and 1/33<x<1/11
then

------55---------------22------------->
------------33---------------11------->
range that covers both conditions is (33)^-1<x<(22)^-1
then only B (23)^1 satisfies it

Inquisition · Answer

<    < x <    <

x must be between      and

Only II fits this criteria

Aman Bhargava · Answer

I did it like this:- First says 1/55 6 < 330x < 15 Second says 1/33 < x < 1/11 - again multiplying by 330 --> 10 < 330x < 33 Combining, we get 10 < 330x < 15 Divide above by 330, we get 1/33 < x < 1/22 Analysing answer choices, only ii satisfies above.

jabhatta2 · Answer

Hi  VeritasKarishma  - Isn't there overlap between the two ranges ?

I thought I could combine both the ranges and assume x can be ANY VALUE on this combined giant range

Here is what I put the range down as (After combining the two individual ranges into one large range given the overlap)

.......... (1/55)......................(1/33) ........................ (1/22).......................(1/11).........

.................. <---------------------------- x ----------------------------------------->

Thus i chose E

Where do you think is the break in my logic ?

Thank you !

KarishmaB · Answer

This is not correct. Notice the 'and' between the inequalities. x has to satisfy both the conditions.

For example, if you have 2 < x < 6 and 4 < x < 8, what values can x take? Can it be 3? No because it doesn't satisfy the second inequality. 
x can be 5 because it satisfies both inequalities. But it again cannot be 7 because it doesn't satisfy the first inequality.

Basshead · Answer

VeritasKarishma , is it fine to simply remove the fractions to solve?

55 < x < 22 AND 33 < x < 11

33 < x < 11

Only '23' is within 33 < x < 11, giving us answer B.

KarishmaB · Answer

basshead  - Yes you can. But note the thing about reciprocals. 
If both sides have the same sign, then when you take reciprocal, the inequality sign flips. 
But if the two sides have opposite signs, then the inequality sign does not flip. 
So before you take reciprocal, ensure you understand this very well. Else, it could lead to an error.

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If 1/55 < x < 1/22 and 1/33 < x < 1/11, then which of the following

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