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Mathivanan Palraj
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GRE 1: Q169 V154
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The Testmaker here is trying to confuse the reader to complicate the math. However, if we conceptually understand the nature of range, we can see the math is fairly simple.
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The problem begins with the notion of 100 multiples of 7 greater than 70. Naturally, the first multiple would be 77, the second multiple would be 84, etc. In other words, the Nth multiple of N greater than 70 could be calculated with the equation 70 + 7(N). To calculate the range, we would subtract the 100th multiple from the 1st multiple:

[(70 + 7(100)] - [70 + 7(1)]

7(100) - 7(1)

700 - 7
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There is an even easier solution, however. As we can see, the two values of 70 cancel. This underscores a fundamental notion of range: if we add the same value to every element in a set, this doesn’t change the range. If we recognized this concept from the very beginning, the math to this problem becomes even simpler. We simply need to subtract the 100th multiple of 7 from the 1st multiple of 7 to get our answer (700 – 7 = 693).
The answer is “A”.
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)The elements of the set are (70+1x7) , (70+2x7), ………………., (70+100 x 7).
This will be an A.P with positive common difference. So, first term is the lowest and last term is the highest.
First term = 77 & last term = 770
Range = 770 – 77 = 693
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alimad
What is the range of a set consisting of the first 100 multiples of 7 that are greater than 70?

A. 693
B. 700
C. 707
D. 777
E. 847

The first multiple of 7 that works is 11 x 7 = 77.

The 100th multiple of 7 (after 77) is 77 + 99 x 7 = 770.

So the range is 770 - 77 = 693.

Answer: A
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