Source: Official Guide Revised GRE 1st Ed. Arithmetic Exercises; #15

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Habtamu Anteneh

First we divide n'/3=2, cross multiplying them gives, n'=6. when we divide n''/5=1 cross multiplying gives n''=5. finally we add n'+n''= 6+5=11.

hope this helps!

Nov 16, 2017 • Comment

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Habtamu Anteneh

we have a ratio of 3:2. we have to add them so we have 5, now we multiply 2 (ratio of cats) by 20 (the total number), which gives us 40, finally we divide 40 by 5 (3+2)= 8. similarly, we can multiply 3 by 20/5= 12.

Nov 16, 2017 • Comment

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Aditya Om

Is there a general method other that this trial and error method?

Mar 30, 2017 • Comment

Cydney Seigerman, Magoosh Tutor

Hi Aditya,

Excellent question! Yes, we can derive a general formula to approach this question :) First, use the statements to write two equations in the form

dividend = (integer quotient)*(divisor) + remainder

n = 3p + 2

n = 5q + 1

where p and q are the quotients.

Now, we can plug in the values of p and q into the equations, starting with 0:

n = 3p + 2

p = 0 --> n = 3*0 + 2 = 2

p = 1 --> n = 3*1 + 2 = 5

p = 3 --> n = 3*2 + 2 = 8

p = 4 --> n = 3*3 + 2 = 11

etc.

n = 5q + 1

q = 0 --> n = 5*0 + 1 = 1

q = 1 --> n = 5*1 + 1 = 6

q = 2 --> n = 5*2 + 1 = 11

We can stop here, since we have found a common value for n, n = 11. Therefore, the least possible value for n for which both statements is true is n = 11 :)

Hope this helps!

Apr 1, 2017 • Reply

Tiffany Arnold

How was that formula derived? Does it come from (dividend/divisor) + remainder = quotient?

Mar 23, 2018 • Reply

David Recine, Magoosh Tutor

Actually, "(dividend/divisor) + remainder = quotient" wouldn't be correct. What's actually the case is that (dividend/divisor) = quotient. Any remainder will already be part of the quotient, so adding an integer to dividend/divisor will incorrectly double whatever integer is already naturally part of the quotient.

But if by "quotient" you're actually talking only about the integer portion of the quotient (as seen in Cydney's formula), then (dividend/divisor) = (integer quotient) + remainder.

And that:

(dividend/divisor) = (integer quotient) + remainder

Can be algebraically rearranged to get Cydney's formula:

dividend = (integer quotient)*(divisor) + remainder

Mar 23, 2018 • Reply

Michelle Nguyen

I guess I'm confused why the divisor is not also multiplied by the remainder.

In my mind, if (dividend/divisor) = (integer quotient) + remainder, to remove the divisor (equal to 1/divisor) from the denominator you'd have to multiply by the reciprocal, or:

(divisor/1)*[(integer quotient) + remainder]=dividend.

What am I missing that the divisor is only multiplied by the integer quotient and not the remainder?

Jun 19, 2018 • Reply

Sam Kinsman, Magoosh Tutor

Hi Emily,

Let me use an example to explain. Let's say we do this:

14/3 = 4, r2

In other words, 14 divided by 3 is equal to 4, with a remainder of 2.

What this is really saying is that if we take three 4s, and add 2, we get 14. In other words, 3 * 4 + 2 = 14

Now, if we start with our equation, and multiply both sides by 3, we will get: 3 * 4 + 2 = 14. So we are left with 14 = 14, which makes sense.

If we were to multiply the remainder (2) by 3 as well, we wouldn't get the right answer. We would get: (4*3) + (2*3) = 18. So we'd have 14 = 18, which is not right.

Keep in mind that this remainder formula doesn't really follow the standard math conventions. Usually, when we have a sum in an equation (q + r), and we multiply that side by c, we would have to multiply both q and r by c. But here, since r represents the remainder, it does not get multiplied.

You might also want to take a look at this blog post, which clarifies further: https://magoosh.com/gmat/2012/gmat-quant-thoughts-on-remainders/

I hope this helps!

Jun 19, 2018 • Reply

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Samia Aljuwayed

I've question about the substitution part(y = 0, 1, or 2) is it randomly?

Jan 7, 2017 • Comment

Sam Kinsman, Magoosh Tutor

Hi Samia,

What Priyanka was saying is that since x = (5y - 1)/3, where x and y are whole numbers, we know that 5y - 1 must be divisible by 3.

We want to find the smallest possible value of 5y-1. This value will be small when y is small. So we can start with the smallest possible value of y (which is 0), and move up from there until we find a value of y which will make 5y-1 divisible by 3.

If we plug in y=0, we get 5(0)-1 = -1, which is not divisible by 3. So this doesn't work.

If we plug in y=1, we get 5(1)-1 = 4, which is not divisible by 3. So this doesn't work.

If we plug in y=2, we get 5(2)-1 = 9, which is not divisible by 3. So this works.

If y = 2, then n = 11 (because, as Priyanka said, 5y + 1 = n).

Jan 11, 2017 • Reply

Cydney Seigerman, Magoosh Tutor

Hi Shubhangi,

Happy to clarify :) This idea comes from the solution posted by Priyanka:

"We have,

3x + 2 = n

5y + 1 = n (x & y being their respective quotient)

Therefore, 3x + 2 = 5y + 1

3x = 5y - 1

x = (5y - 1)/3"

When we solve for x in terms of y, we find that

x = (5y - 1)/3

Since x is an integer, we know that the right side, which is equal to x, must be an integer. So, (5y - 1) must be divisible by 3. On the other hand, if the expression (5y - 1) were not divisible by 3, then the expression (5y - 1)/3 would not be an integer. In that case, it could not be equal to x, since we know x is an integer.

Does that make sense? I hope this clears things up for you! :)

Jul 30, 2017 • Reply

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We have,

3x + 2 = n

5y + 1 = n (x & y being their respective quotient)

Therefore, 3x + 2 = 5y + 1

3x = 5y - 1

x = (5y - 1)/3

Now we know that, 5y - 1 is divisible by 3, so lets substitute y = 0, 1, or 2

Trying these values we find y = 2

3x = 5(2) - 1

3x = 9

x = 3

n = 3x + 2 = 9 + 2 = 11

n = 5y + 1 = 10 + 1 = 11

Thus 11 is the least possible value of n

Oct 8, 2013 • Comment

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