Ratio Problem

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gmatusa2010: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Jul 25, 2010 5:22 amFollowed by: 2 members; Thanked: 1 times

by gmatusa2010 » Mon Nov 29, 2010 1:45 am

The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21

I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.

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rishab1988: Master | Next Rank: 500 Posts; Posts: 332; Joined: Tue Feb 09, 2010 3:50 pmFollowed by: 7 members; Thanked: 41 times; GMAT Score:720

by rishab1988 » Mon Nov 29, 2010 1:59 am

My answer 13.

Here is my way

a/b=3/4
4a-3b=0 -(1)

a+k/b+k =4/5

5a+5k=4b+4k
5a-4b=-k -(2)

5a-4b-4a+3b=-k (subtracting 1 from 2)

a-b=-k

4a-4b=-4k -(3)

Using 1 and 3

-b=-4k
b=4k
a=3k

a+b+2k=117

9k=117
k=13

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Nov 29, 2010 4:23 am

gmatusa2010 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21

I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.

A great way to handle ratio problems is to use a ratio box. Let's call the two unknown values X and Y and make a box for the second ratio of 4 to 5. Here's what the box looks like:

__________X____Y____Total

Ratio:_____4____5_____9

Multiplier:

Actual:_______________117

In the top row, we put the ratio of X:Y = 4:5.
The last column is the total: 4+5 = 9.
In the bottom row, we put any known values. The sum of X and Y is 117, so this number is put under "Total" in the bottom row.

To determine the middle row -- the multiplier -- we divide the known actual value (total = 117) by its corresponding value in the top row (9): 117/9 = 13. The multiplier is placed in every column along the middle row:

__________X____Y____Total

Ratio:_____4____5______9

Multiplier:__13__ 13_____13

Actual:________________117

Now we can complete the box by multiplying every value in the top row by the multiplier:

__________X_____Y______Total

Ratio:_____4_____5_______9

Multiplier:__13____13_____13

Actual:____52____65_____117

So after k has been added to both values, X=52 and Y=65. Now we can plug in the answer choices, which represent the value of k. When the correct answer choice is subtracted from 52 and 65, the ratio will be 3:4.

Only answer choice B works: k=13.
52-13 = 39.
65-13 = 52.
39/52 = 3/4. Success!

The correct answer is B.

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Nov 29, 2010 12:57 pm

gmatusa2010 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21

I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.

What is the source of this problem? that's definitely not a revised problem... and it has more than one answer

My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Nov 29, 2010 1:22 pm

Night reader wrote:
gmatusa2010 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21

I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.
What is the source of this problem? that's definitely not a revised problem... and it has more than one answer

leave it to my two solutions- both appear to be legitimate, and we have two different answers

the 1st solution
Set of Equations:
I. (3x+k)/ (4x+k) = 4/5 and
II. x+2k=117 since x= x*3/7+x*4/7

I. (3x+k)*5=(4x+k)*4, x=k
II. x=117-2k <> x=117-2x <> 3x=117 <> x=39 and k=78 or k=39?

the 2nd solution
One Equation ` (3x+k)/(4x+k) = 4/5
4/5 of the final number means ratio of 4/9 and 5/9 for this number, hence
9x=117 (the sum of nominator and denominator), where x=13
x=k, k=13

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jaymw: Master | Next Rank: 500 Posts; Posts: 144; Joined: Sun Aug 29, 2010 9:17 amFollowed by: 4 members; Thanked: 40 times; GMAT Score:760

by jaymw » Mon Nov 29, 2010 7:20 pm

What is the source of this problem? that's definitely not a revised problem... and it has more than one answer

That problem can logically not have more than one answer!

You add k to both the numerator and the denominator. The higher the value of k the closer the fraction will be to one.

3/4 = 0.75
(3+1)/(4+1) = 0.8
(3+100000)/(4+100000) = almost 1

As GMATGuruNY explained very logically, your base is the 4:5 ratio of 52:65 because the sum of those numbers is 117. If you subtract THE SAME value from both the denominator and the numerator, the result will go against zero with increasing values of k.

Try 1.

(52-1)/(65-1)=51/65 = a little bit less than 4/5, but not enough to equal 3/4

Try 50.

(52-50)/(65-50)=2/15 = a lot less than 4/5, way less than the wanted 3/4

So the value for k must be in between 1 and 50 (which I only chose to explain the concept). Now, just plug in the answer choices.

Logically, there will only be one value for k for which the ratio becomes 3/4.

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gmatusa2010: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Jul 25, 2010 5:22 amFollowed by: 2 members; Thanked: 1 times

by gmatusa2010 » Mon Nov 29, 2010 8:11 pm

Actually thats not true. As I understand the problem, there can be multiple answers, just that theres only one right answer in the choices. Not to mention, this problem doesn't restrict integers. It says positive numbers thats it.

jaymw wrote:
What is the source of this problem? that's definitely not a revised problem... and it has more than one answer
That problem can logically not have more than one answer!

You add k to both the numerator and the denominator. The higher the value of k the closer the fraction will be to one.

3/4 = 0.75
(3+1)/(4+1) = 0.8
(3+100000)/(4+100000) = almost 1

As GMATGuruNY explained very logically, your base is the 4:5 ratio of 52:65 because the sum of those numbers is 117. If you subtract THE SAME value from both the denominator and the numerator, the result will go against zero with increasing values of k.

Try 1.

(52-1)/(65-1)=51/65 = a little bit less than 4/5, but not enough to equal 3/4

Try 50.

(52-50)/(65-50)=2/15 = a lot less than 4/5, way less than the wanted 3/4

So the value for k must be in between 1 and 50 (which I only chose to explain the concept). Now, just plug in the answer choices.

Logically, there will only be one value for k for which the ratio becomes 3/4.

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gmatusa2010: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Jul 25, 2010 5:22 amFollowed by: 2 members; Thanked: 1 times

by gmatusa2010 » Mon Nov 29, 2010 8:31 pm

GMATGuru,

Is my way correct? It's the fastest (if my logic is correct). But I concede that it is very easy for GMAC to design the answers to eliminate my method. This is not an official question, and the writer didn't go the extra mile to eliminate corner-cutters such as myself.

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gmatusa2010: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Jul 25, 2010 5:22 amFollowed by: 2 members; Thanked: 1 times

by gmatusa2010 » Mon Nov 29, 2010 8:41 pm

the 1st solution
Set of Equations:
I. (3x+k)/ (4x+k) = 4/5 and
II. x+2k=117 since x= x*3/7+x*4/7
I. (3x+k)*5=(4x+k)*4, x=k
II. x=117-2k <> x=117-2x <> 3x=117 <> x=39 and k=78 or k=39?

1) How did you get X+2k= 117? You wrote 3x+k and 4x+k right there. The problem says new number plus old number = 117. SO according to your equation its supposed to be 7x + 2k= 117 where X is an integer.

the 2nd solution
One Equation ` (3x+k)/(4x+k) = 4/5
4/5 of the final number means ratio of 4/9 and 5/9 for this number, hence
9x=117 (the sum of nominator and denominator), where x=13
x=k, k=13[/quote]

2) I like this method. Its actually similar to mine except you used the 4/5 ratio. I will have to study the intuition behind this solution. Good discussion!

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Nov 29, 2010 9:19 pm

gmatusa2010 wrote:the 1st solution
Set of Equations:
I. (3x+k)/ (4x+k) = 4/5 and
II. x+2k=117 since x= x*3/7+x*4/7
I. (3x+k)*5=(4x+k)*4, x=k
II. x=117-2k <> x=117-2x <> 3x=117 <> x=39 and k=78 or k=39?

1) How did you get X+2k= 117? You wrote 3x+k and 4x+k right there. The problem says new number plus old number = 117. SO according to your equation its supposed to be 7x + 2k= 117 where X is an integer.

the 2nd solution
One Equation ` (3x+k)/(4x+k) = 4/5
4/5 of the final number means ratio of 4/9 and 5/9 for this number, hence
9x=117 (the sum of nominator and denominator), where x=13
x=k, k=13 2) I like this method. Its actually similar to mine except you used the 4/5 ratio. I will have to study the intuition behind this solution. Good discussion!

jaymv: I've put in math what logic says; there are at least two answers.

gmatusa2010: the problem says 117=old numbers+k+k (or nominator of old number and denominator of old number - both added by k) - so we have 3/4 ratio = number1/number2 or number1/total and number2/total ~ 3/7 and 4/7; suppose number total = x, then x*3/7 and x*4/7 gives you x as a whole. Write down x+k+k=117 (!)

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Nov 29, 2010 9:25 pm

gmatusa2010 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21

I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.

The equation 7Q = 117 - 2K doesn't take into account that, according to the problem, (X+K)/(Y+K) = 4/5.

If the answer choices had included 48 as a possible value of K, would you have seen that K=48 is incorrect?

If K=48:
7Q = 117 - 2*48
7Q = 21
Since 7Q represents the sum of X+Y, and X/Y = 3/4, we would get that X=9 and Y=12.

These values satisfy the following conditions:
X/Y = 9/12 = 3/4.
X+Y+2K = 9+12+2*48= 117.

But (X+K)/(Y+K) = (9+48)/(12+48) = 57/60 = 19/20, not 4/5 as stated in the problem. So K=48 is incorrect.

As stated above, only K=13 satisfies all the conditions in the problem.

Hope this helps!

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gmatusa2010: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Jul 25, 2010 5:22 amFollowed by: 2 members; Thanked: 1 times

by gmatusa2010 » Mon Nov 29, 2010 9:38 pm

In step I, you defined X differently. YOu said 3x= old number 1 and 4x= old number 2. But in X+2k=117, you are saying X = old number 1 + old number 2.

Night reader wrote:
gmatusa2010 wrote:the 1st solution
Set of Equations:
I. (3x+k)/ (4x+k) = 4/5 and
II. x+2k=117 since x= x*3/7+x*4/7
I. (3x+k)*5=(4x+k)*4, x=k
II. x=117-2k <> x=117-2x <> 3x=117 <> x=39 and k=78 or k=39?

1) How did you get X+2k= 117? You wrote 3x+k and 4x+k right there. The problem says new number plus old number = 117. SO according to your equation its supposed to be 7x + 2k= 117 where X is an integer.

the 2nd solution
One Equation ` (3x+k)/(4x+k) = 4/5
4/5 of the final number means ratio of 4/9 and 5/9 for this number, hence
9x=117 (the sum of nominator and denominator), where x=13
x=k, k=13 2) I like this method. Its actually similar to mine except you used the 4/5 ratio. I will have to study the intuition behind this solution. Good discussion!
jaymv: I've put in math what logic says; there are at least two answers.

gmatusa2010: the problem says 117=old numbers+k+k (or nominator of old number and denominator of old number - both added by k) - so we have 3/4 ratio = number1/number2 or number1/total and number2/total ~ 3/7 and 4/7; suppose number total = x, then x*3/7 and x*4/7 gives you x as a whole. Write down x+k+k=117 (!)

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Nov 29, 2010 10:05 pm

and x can be 1,000,000*x/2,000,000*x, so what?
one side is fraction and the other side contains a whole number

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gmatusa2010: Master | Next Rank: 500 Posts; Posts: 134; Joined: Sun Jul 25, 2010 5:22 amFollowed by: 2 members; Thanked: 1 times

by gmatusa2010 » Mon Nov 29, 2010 10:16 pm

i dunno. theres definitely something funky about the algebra in method 1. if u back test your solutions it still doesnt work, so that should say something.

Ex: X= 117-2k, X=117- 2(78)= negative number? X= 117- 2(39) = 39, thats not divisible by 9. Pluggin X and K into the original expression makes even less sense.

Either that or your method is beyond me. Not a quantitative person or major in anyway.

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Tue Nov 30, 2010 3:43 am

gmatusa2010 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21

I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.

Everything is fine with your approach except a single flaw in your logic!
As you have found out three equations in three unknowns, which are

(1) 4x = 3y
(2) 5*(x + k) = 4*(y + k)
(3) (x + y + 2k) = 117

Now the values of x or y or k can only be uniquely obtained if and only if we consider all the three equations together. Otherwise if we take any two of them, we cannot solve them for sure.

For example, in your case you have considered 1st and 3rd equation only.
Let's see what we can achieve with these two equations. From the first equation we have y = (4x/3). Replace y with this value in the third equation. Then we have, (x + 4x/3 + 2k) = 117. Thus,

k = 117 - (7x/3)

As we have expected we cannot uniquely determine the value of k as it depends upon x.

Where have gone wrong then?
You've done a very common assumption which lead you in the wrong way! What is that? You assumed x and y to be integer and thus said that (x + y) is divisible by 7! In the question they were declared as numbers only! So basically you are assuming a new information and discarding a given information!

Conclusion: "The ratio of two positive numbers is 3 to 4 doesn't mean sum of the numbers is divisible by 7! For example 3/5 and 4/5 has a ratio 3 to 4, but their sum 7/5 is not divisible by 7!

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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