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Intern  B
Joined: 06 Sep 2016
Posts: 3

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Hi Bunuel,

Shouldn't the answer be 90+28-4 instead of 90+28-2 since, we are counting 385 & 770 in both the cases two times i.e. 2 times in total 90 cases and 2 times in the other total 28 cases.
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Joined: 01 Mar 2017
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womsuko wrote:
wouldn't it be 1001-116? because it is inclusive

I wonder as well, can somehow explain that? (For instance if the problem included a trap answer such as 884 and 885)
Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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melin94 wrote:
womsuko wrote:
wouldn't it be 1001-116? because it is inclusive

I wonder as well, can somehow explain that? (For instance if the problem included a trap answer such as 884 and 885)

But there are 1000 numbers from 1 to 1000, inclusive, not 1001, in't it?
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Joined: 18 Feb 2017
Posts: 8

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1
smkashyap wrote:
Sash143 wrote:
Bunuel wrote:
Official Solution:

What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884
B. 890
C. 892
D. 910
E. 945

# of multiples of 11 in the given range $$\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90$$;

# of multiples of 35 in the given range $$\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28$$;

# of multiples of both 11 and 35 is 2 ($$11*35=385$$ and 770);

So, # of multiples of 11 or 35 in the given range is $$90+28-2=116$$. Thus numbers which are not divisible by either of them is $$1000-116=884$$.

Hi Bunuel,
I had come across a similar question "How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer.

But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770).

There is one critical aspect of your question ("How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?") that is different from the question discussed in this forum.

If you can't spot it yet, its "but not both".

Your question asks us not to consider multiples of both 12 and 13.

The question discussed in this forum has no such restrictions. We are merely removing repetitions of multiples of both 11 and 35.

Before I sign off, here's the cliched "Hope it helps!"

So there are 90 multiples of 11 and 28 multiples of 35. You're absolutely correct in mentioning that there's 2 sets of repetitions for (385 and 770). However, if you subtract 4, you'll take both sets out. You only need to subtract 2 to still count the one set and eliminate the redundancy, not wipe them out completely.
Intern  B
Joined: 22 Jan 2016
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Kindly seeking assistance

What is the difference between this question and M12-36 that asks about the number of positive integers less than 200 that are multiples of 13 or of 12 but not of both? In the solution of M12-36, it subtracts 156 (being the number of the common multiple of both 12 and 13), as follows:
# of multiples of 13....15-1
# of multiples of 12 ...16-1
Total 29.

So the common multiple was subtracted from both 15 and 16 whereas the two common multiples in question M07-14 were only subtracted once as in 90+28-2=116. How come it wasn't 90-2+28-2?

I hope you understand my question.
Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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Tamuka wrote:
Kindly seeking assistance

What is the difference between this question and M12-36 that asks about the number of positive integers less than 200 that are multiples of 13 or of 12 but not of both? In the solution of M12-36, it subtracts 156 (being the number of the common multiple of both 12 and 13), as follows:
# of multiples of 13....15-1
# of multiples of 12 ...16-1
Total 29.

So the common multiple was subtracted from both 15 and 16 whereas the two common multiples in question M07-14 were only subtracted once as in 90+28-2=116. How come it wasn't 90-2+28-2?

I hope you understand my question.

M12-36
How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?

So, here we don't want numbers which are both multiples of 13 and 12.
# of multiples of 13 is 15
# of multiples of 12 is 16
# of multiples of both 13 and 12 is 1. This one multiple (156) is counted both in 15 and 16, so to get the number of multiples of 13 or multiples of 12 but not both, we should subtract 1 from both sets: (15 - 1) + (16 - 1) = 29.

M07-14
Here we need numbers which are divisible by 11 or by 35. So, numbers which are divisible by both should be counted.

# of multiples of 11=90;

# of multiples of 35=28;

# of multiples of both 11 and 35 is 2. These two multiples are counted both in 90 and 28 but we want them to be counted only once, hence 90 + 28 - 2 = 116.

Hope it's clear.
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Joined: 22 Jan 2016
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Thanks Bunuel for the assistance, its clear now.
Manager  B
Joined: 23 Jun 2016
Posts: 91

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Hi Bunuel

In a previous question in this set, I saw the approach where to find the number of multiples, we divided the total number by the given number and rounded it off.

in this case, # of multiple of 11 = 1001/11 or 91 (not 90)
Off course this is wrong answer.

can you suggest if i am applying the approach incorrectly or should we avoid using this approach all together?
Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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sevenplusplus wrote:
Hi Bunuel

In a previous question in this set, I saw the approach where to find the number of multiples, we divided the total number by the given number and rounded it off.

in this case, # of multiple of 11 = 1001/11 or 91 (not 90)
Off course this is wrong answer.

can you suggest if i am applying the approach incorrectly or should we avoid using this approach all together?

Where did I use that approach?

The formula is below:
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.

How many multiples of 4 are there between 12 and 96, inclusive?

Last multiple of 4 IN the range is 96;
First multiple of 4 IN the range is 12;

$$\frac{96-12}{4}+1=22$$.

OR: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

$$\frac{30-(-5)}{5}+1=8$$.

OR: How many multiples of 7 are there between -28 and -1, not inclusive?

Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

$$\frac{-7-(-21)}{7}+1=3$$.
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Intern  B
Joined: 28 Oct 2017
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385 (11 x 35) and 780 (next multiple) is counted both in the multiples of 11 and multiples of 35. So should we not subtract these numbers twice?
So 90+28-4?
Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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Sriharsh wrote:
385 (11 x 35) and 780 (next multiple) is counted both in the multiples of 11 and multiples of 35. So should we not subtract these numbers twice?
So 90+28-4?

We need numbers which are divisible by 11 or by 35. So, numbers which are divisible by both should be counted.

# of multiples of 11=90;

# of multiples of 35=28;

# of multiples of both 11 and 35 is 2. These two multiples are counted both in 90 and 28 but we want them to be counted only once, hence 90 + 28 - 2 = 116.

Hope it's clear.
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Intern  B
Joined: 04 Aug 2018
Posts: 6

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This is one is based on Arithmetic Progression as well .
Nth term (last term)= first term + (number of terms-1) difference between terms

This will give number of terms in both cases easily.

Caution:- :- Always watch out for common terms if more than two numbers are mentioned. (tricky part). LCM will be handy in such case.
Intern  B
Joined: 30 Apr 2018
Posts: 18

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Bunuel wrote:
Official Solution:

What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884
B. 890
C. 892
D. 910
E. 945

# of multiples of 11 in the given range $$\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90$$;

# of multiples of 35 in the given range $$\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28$$;

# of multiples of both 11 and 35 is 2 ($$11*35=385$$ and 770);

So, # of multiples of 11 or 35 in the given range is $$90+28-2=116$$. Thus numbers which are not divisible by either of them is $$1000-116=884$$.

Hi Bunuel,

I follow the approach and my answer is right too, except that my answer is not exactly right - i had to approximate

The reason= the formula i used for '# of terms between a range' is same but with different lower an upper range.

I have an understanding that:
# of terms in the range a to b (inclusive) with an even interval n = [(b-a)/n]+1
Using the same, i calculated
multiples of 11 between 1 to 1000
OR
# of terms in the range 0 to 1000 (inclusive) with an even interval 11 (as 0+11 = 11, the first no. in the range)

= [(1000-0)/11]+1 = 92 ; however the right count is 90

where is my understanding not right? TIA Re: M07-14   [#permalink] 13 Sep 2018, 06:12

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