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31 Jul 2019, 01:16
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Difficulty:
35%
(medium)
Question Stats:
66% (01:28) correct
34%
(01:16)
wrong
based on 133
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If the result of squaring a number n is less than twice the number then the value of n must be
A. negative
B. positive
C. between -1 and +1
D. greater than 1
E. between 0 and 2
A. negative
B. positive
C. between -1 and +1
D. greater than 1
E. between 0 and 2
31 Jul 2019, 10:30
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This is a great question to learn the technique of breaking down the question statement and converting it to an inequality, before solving it.
The number that we have on hand is ‘n’. Squaring it will yield ‘\(n^2\)’; twice the number would be ‘2n’.
The question statement says that the result of squaring a number is less than twice the number. This means, \(n^2\) < 2n.
In any question on inequalities, always try to keep the RHS of the inequality as zero, while trying to express the LHS as a product or a quotient of variables/expressions.
So, \(n^2\)< 2n can be rewritten as \(n^2\) – 2n < 0. The LHS can be simplified and written as,
n(n-2) < 0. Here, we have a product on the LHS which is less than 0. This means that the product is negative.
This is possible only in two ways:
n is negative and (n-2) is positive. This means,
n < 0 and (n-2) > 0 i.e. n < 0 and n > 2. Clearly, that cannot happen at the same time. So, we can rule out this possibility.
n is positive and (n-2) is negative. This means,
n > 0 and (n-2) < 0 i.e. n > 0 and n < 2 i.e. 0<n<2. So, the value of n MUST be between 0 and 2.
The correct answer option is E.
An alternative approach would be to try the options. Options A and B are very generic and hence can be eliminated straight away, because clearly ANY negative value or ANY positive value will not satisfy the given inequality.
When we try option C, we can pick n = 0; when we do this \(n^2\) = 0 and 2n = 0. Clearly, \(n^2\) is not less than 2n. Option C can be ruled out.
Trying option D, if n = 2, \(n^2\) = 4 and 2n = 4. Again, \(n^2\) is not less than 2n. Option D can be ruled out.
Answer option E has to be the answer.
Although the back solving approach works just fine in this question, I’d prefer the analysis method because it’s from these type of questions that I can start learning the analysis of the question stem, and I can slowly graduate towards the difficult questions once I build confidence.
Hope this helps!
The number that we have on hand is ‘n’. Squaring it will yield ‘\(n^2\)’; twice the number would be ‘2n’.
The question statement says that the result of squaring a number is less than twice the number. This means, \(n^2\) < 2n.
In any question on inequalities, always try to keep the RHS of the inequality as zero, while trying to express the LHS as a product or a quotient of variables/expressions.
So, \(n^2\)< 2n can be rewritten as \(n^2\) – 2n < 0. The LHS can be simplified and written as,
n(n-2) < 0. Here, we have a product on the LHS which is less than 0. This means that the product is negative.
This is possible only in two ways:
n is negative and (n-2) is positive. This means,
n < 0 and (n-2) > 0 i.e. n < 0 and n > 2. Clearly, that cannot happen at the same time. So, we can rule out this possibility.
n is positive and (n-2) is negative. This means,
n > 0 and (n-2) < 0 i.e. n > 0 and n < 2 i.e. 0<n<2. So, the value of n MUST be between 0 and 2.
The correct answer option is E.
An alternative approach would be to try the options. Options A and B are very generic and hence can be eliminated straight away, because clearly ANY negative value or ANY positive value will not satisfy the given inequality.
When we try option C, we can pick n = 0; when we do this \(n^2\) = 0 and 2n = 0. Clearly, \(n^2\) is not less than 2n. Option C can be ruled out.
Trying option D, if n = 2, \(n^2\) = 4 and 2n = 4. Again, \(n^2\) is not less than 2n. Option D can be ruled out.
Answer option E has to be the answer.
Although the back solving approach works just fine in this question, I’d prefer the analysis method because it’s from these type of questions that I can start learning the analysis of the question stem, and I can slowly graduate towards the difficult questions once I build confidence.
Hope this helps!
