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Bunuel
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This will form an AP 312,314,316,.............,590
a=first term of the ap= 312
tn =last term of the ap =590
d= difference between consecutive terms=2
using formula tn= a +(n-1)d
=> 590 = 312+(n-1)2
=> 278/2 = n-1
=> n=140
option B
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Correct Answer: B (140)

Given:
- Range: 312 to 590, inclusive
- Condition: All numbers are even integers

Formula:
Number of terms in an AP = [(Last Term - First Term) / Difference] + 1

Calculation:
- First even integer = 312
- Last even integer = 590
- Difference between consecutive evens = 2

Count = [(590 - 312) / 2] + 1
= [278 / 2] + 1
= 139 + 1
= 140 (Option B)
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Bunuel
If building numbers in a range from 312 to 590, inclusive, are all even numbers, how many buildings are there?

A. 139
B. 140
C. 141
D. 142
E. 143
An intuitive / reasoning approach works well here, but you must watch out for a common trap on questions about inclusive sets.

Intuitively, about half the integers from 312 to 590 are even. How many total integers are there? 590 - 312 = 278, and half of 278 is 139.

However, since this is an inclusive set (the "edge" numbers 312 and 590 are included), we must add 1, so there are actually 279 integers. And since the set begins and ends with an even, that extra integer goes to the evens. So there are 140 evens (and 139 odds). The answer is B.

If you forget this rule, think of an easy example. How many numbers are there from 1 to 10? Obviously, 10. But 10 - 1 = 9. So, to find the number of consecutive integers in a range, you must find their difference and add 1.

Remember to add 1 when counting inclusive sets. If you can't remember this rule (or another math rule), use a simple example to remind yourself.
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