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Max possible integer value

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by GMATGuruNY » Fri Feb 12, 2016 6:22 am
uniyal01 wrote:If -13 < 7a + 1 < 29 and 19 < 2 - b < 23, what is the maximum possible integer value of a + b?

Thanks.
-13 < 7a + 1 < 29

Subtract 1:
-14 < 7a < 28

Divide by 7:
-2 < a < 4

Thus:
a < 4.

19 < 2 - b < 23

Subtract 2:
17 < -b < 21

Multiply by -1 and flip the inequalities:
-17 > b > -21

Thus:
b < -17.

Add together a < 4 and b < -17:
a + b < 4 + (-17)
a + b < -13.

Thus, the greatest possible integer value for a+b = -14.
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by MartyMurray » Fri Feb 12, 2016 6:29 am
uniyal01 wrote:If -13 < 7a + 1 < 29 and 19 < 2 - b < 23, what is the maximum possible integer value of a + b
-13 < 7a + 1 < 29 --> -14 < 7a < 28 --> -2 < a < 4

So a can be any number less than 4, such as 3.99999999.

19 < 2 - b < 23 --> 17 < -b < 21

When you multiply an inequality by a negative number, reverse the direction of the inequality.

-17 > b > -21

So the maximum value of b is just below -17, something along the lines of -17.000000001.

The question asks for the maximum possible INTEGER value of a + b.

The maximum value of a + b = (3 + some decimal) + (-17 - some decimal) = -14 + some decimal - some decimal.

There is no way the decimals can get a + b to -13. So the answer is -14.
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