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How many multiples of 7 between 100 and 150? [#permalink]
10 May 2012, 04:45

1

This post was BOOKMARKED

Hey everyone. As you can see, I need help with something so simple. I'm sure the answer is right in front of me and I can't see it. They use the above question in Manhattan Number Properties. They start listing multiples: 105, 112,119,126, 133, 140, 147. The thing that is not clear to me is how I know to start with 105 instead of say, 103 or 101.

Re: How many multiples of 7 between 100 and 150? [#permalink]
10 May 2012, 05:22

2

This post received KUDOS

Expert's post

The first multiple of 7 that is greater than 100, is 105. (70+35 = 105). The one before that is 98, and that's less than 100.

This is the most reliable way to solve the question (listing). Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Re: How many multiples of 7 between 100 and 150? [#permalink]
10 May 2012, 05:37

bb wrote:

The first multiple of 7 that is greater than 100, is 105. (70+35 = 105). The one before then is 98, and that's less than 100.

This is the most reliable way to solve the question (listing). Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Re: How many multiples of 7 between 100 and 150? [#permalink]
20 May 2012, 19:58

Expert's post

mdavidm531 wrote:

gmatz101 wrote:

Hey everyone. As you can see, I need help with something so simple. I'm sure the answer is right in front of me and I can't see it. They use the above question in Manhattan Number Properties. They start listing multiples: 105, 112,119,126, 133, 140, 147. The thing that is not clear to me is how I know to start with 105 instead of say, 103 or 101.

Re: How many multiples of 7 between 100 and 150? [#permalink]
19 Mar 2014, 12:47

bb wrote:

The first multiple of 7 that is greater than 100, is 105. (70+35 = 105). The one before that is 98, and that's less than 100.

This is the most reliable way to solve the question (listing). Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Hope this helps.

Why does the (last - first) / increment rule not always work here? I know it doesn't (for example, count the number of multiples of 7 between 11 and 24. There are two (14, 21), but (last - first) / increment gives you 1.

Also, the Manhattan GMAT book lists (last - first) / increment + 1 as the rule to use here, but that doesn't always work either. For example, count the number of multiples between 9 and 24. There are two: 14, and 21. But (last - first) / increment + 1 is gives you 3 multiples.

It appears that in these sorts of problems involving the number of multiples between two numbers: we cannot strictly rely on formulas and instead need to think them through on a case by case basis, right?

Re: How many multiples of 7 between 100 and 150? [#permalink]
19 Mar 2014, 20:23

onewayonly wrote:

bb wrote:

The first multiple of 7 that is greater than 100, is 105. (70+35 = 105). The one before that is 98, and that's less than 100.

This is the most reliable way to solve the question (listing). Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Hope this helps.

Why does the (last - first) / increment rule not always work here? I know it doesn't (for example, count the number of multiples of 7 between 11 and 24. There are two (14, 21), but (last - first) / increment gives you 1.

Also, the Manhattan GMAT book lists (last - first) / increment + 1 as the rule to use here, but that doesn't always work either. For example, count the number of multiples between 9 and 24. There are two: 14, and 21. But (last - first) / increment + 1 is gives you 3 multiples.

It appears that in these sorts of problems involving the number of multiples between two numbers: we cannot strictly rely on formulas and instead need to think them through on a case by case basis, right?

Thanks.

The formula is NOT just "last - first," but the actual formula is "last MULTIPLE in range - first MULTIPLE in range." Here is the complete formula:

# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1.

Using one of your examples, you stated: "count the number of multiples of 7 between 11 and 24. There are two (14, 21), but (last - first) / increment gives you 1."

This would be solved like this: \frac{21 - 14}{7}+1 = 2

Using your other example, you stated: "count the number of multiples between 9 and 24. There are two: 14, and 21. But (last - first) / increment + 1 is gives you 3 multiples." I'm assuming you meant "count the number of multiples of 7" between the range of 9 and 24 (you didn't specify the multiple in this example).

This would be solved using the exact same numbers as shown in the above example. Once again, the answer would be 2.

Re: How many multiples of 7 between 100 and 150? [#permalink]
21 Mar 2014, 02:40

@onewayonly The last and first term refer to the terms of an arithmetic sequence, and you have to make sure that all the terms are part of an arithmetic sequence, where the difference between successive terms is fixed. In the case of counting multiples, you have to start with the first multiple in the range and the last multiple in the range, we can't just use the first and last term of the given range.

Re: How many multiples of 7 between 100 and 150? [#permalink]
24 Mar 2014, 04:16

For such questions : Last term is the last multiple within the range and the first term is the first multiple within the range. This series of multiples forms an AP sequence.

Lets understand the formula ((Last term - First term)/ spacing) +1 We get this formula from AP(Arithmetic Progression) formula for last term i.e Tn = a + (n - 1) d

Lets take the first term as 'a' and the nth term as 'b'

Using the formula for nth term of an AP, we get : b = a + (n - 1)d. We need the value of 'n'. Solving we get : n = ((b - a)/d)+1 where, b = last term , a = first term, and d = common difference.

I hope this helps.

gmatclubot

Re: How many multiples of 7 between 100 and 150?
[#permalink]
24 Mar 2014, 04:16