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by Mo2men » Thu Nov 03, 2016 1:43 am
If m^2<225 and n−m=−10, what is the difference between the smallest possible integer value of 3m+2n and the greatest possible integer value of 3m+2n?

A. -190
B. -188
C. -150
D. -148
E. -40
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Source: — Problem Solving |

by GMATGuruNY » Thu Nov 03, 2016 3:11 am
Mo2men wrote:If m^2<225 and n−m=−10, what is the difference between the smallest possible integer value of 3m+2n and the greatest possible integer value of 3m+2n?

A. -190
B. -188
C. -150
D. -148
E. -40
Substituting n=m-10 into 3m+2n, we get:
3m + 2(n-10) = 3m + 2n - 20 = 5m-20.
Implication:
To determine the least possible integer value and the greatest possible integer value for 3m+2n, we must calculate the least possible integer value and the greatest possible integer value for 5m-20.

Test the THRESHOLDS.
m² < 225 implies that -15 < m < 15.
Thus, the threshold values are m=-15 and m=15.

Least possible integer value:
If m=-15, then 5m-20 = 5(-15)-20 = -95.
Since m>-15, 5m-20 > -95.
Thus, the least possible integer value for 5m-20 is -94.

Greatest possible integer value:
If m=15, then 5m-20 = 5(15)-20 = 55.
Since m<15, 5m-20 < 55.
Thus, the greatest possible integer value for 5m-20 is 54.

(least possible integer value) - (greatest possible integer value) = -94 - 54 = -148.

The correct answer is D.
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by crackverbal » Thu Nov 03, 2016 3:25 am
Hi Mo2men,

The question here tests you purely on the concepts of inequalities. If your inequality concepts are crystal clear then the question here boils down to simple subtraction. Let me first elaborate on the two inequality concepts that are being tested here and then solve the question.

Concept 1 : If x^2 < a^2 where a is a constant then the range for x will be -a < x < a and
If x^2 > a^2 where a is a constant then the range for x will be x > a and x < -a

Concept 2 : Say the range of x is 4 < x < 12 and the range for y is -6 < y < 9, then we can find the range of x + y, x - y and x * y by placing the ranges for x and y one below the other and then add or subtract or multiply the extreme values. If we need to find the range for x + y then the 4 values we get by adding the extremes are -2, 21, 13 and 6. The min value here is -2 and the max value here is 21, so -2 < x + y < 21. the same process can be used to find the range of x - y and x * y. You just need to make sure that the inequality signs of both the given ranges are the same to use the max min concept.

Now coming to the question, we are given that m^2 < 225 and n - m = -10.

If m^2 < 225 then the range for m is -15 < m < 15.
Since m = n + 10, the range for n is -15 < n + 10 < 15 ------> -25 < n < 5 (Subtracting 10 throughout)

We are asked to find the difference between the smallest possible integer value of 3m+2n and the greatest possible integer value of 3m + 2n, so let us first find the ranges of 3m and 2n and then use the max min concept to add and find the max and min values of 3m + 2n.

If -15 < m < 15 then -45 < 3m < 45 (Multiplying throughout by 3)
If -25 < n < 5 then -50 < 2n < 10 (Multiplying throughout by 2)

Now adding the 4 extreme values we get -95,-35,-5,55.
The range of 3m + 2n is -95 < 3m + 2n < 55.

The smallest integer value of 3m + 2n is -94 and the greatest integer value of 3m + 2n is 54. The difference here is -94 - 54 = -148.

OA : D

To learn some more important properties of inequalities use the following link https://gmat.crackverbal.com/free-resour ... k-library/ to download our free Inequalities ebook.

Hope this helps!
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by Jay@ManhattanReview » Thu Dec 08, 2016 9:42 pm
Both the explanations are great.

The trap here is that the question asks for the two extreme integer values of (3m + 2n). This does not mean that m and n cannot be non-integers. Had one of the option been -140, the question would have another trap.

For -15 < m < 15, many may have assumes min. value of m = -14, thus, min value of (5m - 20) = 5*-14 - 20 = -90;

Similarly, for -15 < m < 15, many may have assumes max. value of m = 14, thus, max value of (5m - 20) = 5*14 - 20 = 50;

Thus the difference = -90 - 50 = -140, which is incorrect!

Hope this helps!

-Jay

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