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The number of baseball cards that John and Bill had was in the ratio o

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The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 06 Apr 2015, 07:19
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The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 13 Apr 2015, 07:49
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Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

The way most people would look at this problem is that it’s an algebra problem. The ratio of two numbers is 7:3, and after an exchange of 15 cards, the ratio is now 3:2. I can set up two equations and solve for the two unknowns in this equation, which will give me the number of cards Bill has and the number of cards John has. After that, it’s simply a question of subtracting the two in order to answer the question. Let’s run through the algebra because it’s somewhat time-consuming but otherwise fairly basic (the white-and-gold approach).

The initial ratio, before the gift, can be describes as J / B = 7 / 3.

The final ratio, after the gift, would then be J – 15 / B + 15 = 3 / 2.

Note that we are defining J and B to be the initial values of John and Bill, so we’ll have to keep that in mind for the final calculation.

Cross-multiplying the first equation gives us 3 J = 7 B. This should make sense as John has many more cards than Bill.

Cross-multiplying the second equation gives 2 (J – 15) = 3 (B + 15),

We can expand this to 2J – 30 = 3B + 45.

Finally we can move the constants to one side and get 2 J = 3 B + 75

You can use either the elimination method or the substitution method to solve for the two variables. I prefer the elimination method so I’d multiply the first equation by 2 and the second equation by 3 to isolate J.

6 J = 14 B

6 J = 9 B + 225

Since the left hand sides are the same, we can simplify to 14 B = 9 B + 225.

Subtracting 9 B from both sides gives 5 B = 225.

Dividing 225 by 5 gives 45.

If B is 45, and 3 J = 7 B, then 3 J must be 315, and so J is equal to 105.

We’re still not done, because these are the initial values: 105 and 45. If John gave Bill 15 cards, then the new totals would be 90 for John and 60 for Bill, which is where the 3/2 ratio comes in. The difference in cards is 30 after the gift, so the answer is B.

Other people see this ratio problem and don’t even think about the algebra, they solve it using the underlying concept (the blue-and-black approach). To illustrate this concept, suppose I had 199 cards and you had 101 cards. Since no simplification is possible, the ratio of our cards would be 199:101. But if you then gave me one card, our ratio would suddenly be 2:1. This reduced fraction does not change the fact that I still have 200 cards and you have 100. Simply because the fraction can be simplified, that does not mean that the totals have changed in any way.

Let’s apply that same logic here. The ratio was 7:3. After the gift, the new ratio is 3:2, but the total number of cards has stayed the same. This means that if I can get a new ratio that’s in the same proportions as the old ratio, the problem will seem much simpler. The ratio 7:3 has 10 total parts. The ratio of 3:2 has only 5 total parts, so they are not in the same proportions. However, if I recognize that I can simply multiply 3:2 by 2 to get a ratio of 6:4, I discover a shortcut that can help on ratio problems.

If the ratio used to be 7:3 then became 6:4 after a transfer of 15 cards, then each unit of the ratio must represent 15 cards. This would mean that 7 would drop to 6 and 3 would increase to 4 because of the same 15 card transfer. Thus the old ratio was (7×15): (3×15), or 105:45. The new ratio is similarly (6×15): (4×15), or 90:60. The difference in cards after the gift is still 30, answer choice B, but for some it’s much easier to see using a little logic than a lot of algebra.

On the GMAT, similar to the chameleon dress, your perspective is what’s going to dictate how you approach problems. Not every question will have a shortcut or an instant solution, but every problem can be approached in multiple ways. The only limit is your understanding of the concepts and your skill at analyzing the presented problem. Hopefully, on test day, these strategies will help you avoid feeling blue (and black).
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 06 Apr 2015, 07:35
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\(\frac{J}{B}\)=\(\frac{7}{3}\)
3J=7B

\(\frac{(J-15)}{(B+15)}\)=\(\frac{3}{2}\)
2J-30=3B+45

solve for either J or B-- substitute into equation 1
J= 105
B= 45
After switch:
J=105-15=90
B=45+15=60
Answer: 30
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 06 Apr 2015, 08:16
Yes, +1 for B. Straight forward question.

30 cards will be more.

Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.

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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 06 Apr 2015, 10:08
Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.



Old ratio 7:3
new ratio after xfer of 15 cards. 3:2 or 6:4 . so basically a movement of 15 caused 7units to 6units and 3units to 4units..
that means old ratio=7×15/3×15 and new ratio is 6×15/4×15 .
answer 90-60=30
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 06 Apr 2015, 16:01
shriramvelamuri wrote:
Yes, +1 for B. Straight forward question.

30 cards will be more.

Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.

ditto, I came up with pretty much same thing, the only difference is that I prefer X and Y as variables.
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 12 Jul 2017, 07:24
Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.


J/B = 7/3
-> 3J = 7B
-> 6J = 14 B

(J-15)/ (B+15) = 3/2
-> 2J-30 = 3B +45
-> 2J = 3B +75
-> 6J = 9 B + 225
-> 14 B = 9 B +225
-> 5B = 225
-> B = 45

J = (7/3) * 45 = 105

So, after gift, difference between John and bills card = (J-15) - (B+15) = (105 - 15) - (45 +15) = 90 - 60 = 30

Answer B
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 12 Jul 2017, 07:42
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There is a shortcut of solving such questions: lookout for the Percentages of either John or Bill

Let the asked which is John had how many more baseball cards than Bill= X

So if I take the percentage for John in 7:3 then 70% of cards are with John

after John can 15 cards then john:Bill are in ratio of 3:2 .So John now has 60% of the total cards.

it means than (70%-60%)(total cards)=15

which means total cards=150

so for calculating the asked value which is 20% of the total(calculating the ratio from 3:2 which could be written into 60:40)

and 20% of 150=30


See....easy...peasy

This way of solving with percentages would help in solving ratio and proportion questions in which solution or water is involved


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The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 12 Jul 2017, 19:46
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Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

My head is spinning from the OA explanation (Veritas is great, but this time, seriously?) and GMATOak 's impressive but NOT easy-peasy looking method. Think I'll stick to one measly variable, x, and a couple of ratios. They served just fine. Contrary to OA's prediction, I took just 1:40 with double checking.

\(\frac{J}{B}\) = \(\frac{7x}{3x}\)

\(\frac{7x - 15}{3x +
15}\) = \(\frac{3}{2}\)

2(7x - 15) = 3(3x + 15)

14x - 30 = 9x + 45

5x = 75

x = 15 --> that's the multiplier for the original ratio.

John had 7*15 = 105 cards, he gave 15 away, now he has 90.

Bill had 3*15 = 45 cards, he received 15 more, now he has 60.

90 - 60 = 30. John has 30 more cards than Bill.

Answer B
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The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 12 Jul 2017, 20:36
Algebra was simple but some how i got stuck and then tried this approach and it seems to have worked.

I plugged in a value from answer choice and arrived at the correct answer through elimination. The calculations are simple so it did not take much time.

Let us pick the middle from the given answer choices=45

Therefore John has 45 more caps than bill.

Let bill have X caps than john will have X+45 caps. Set up the equation-Take 2nd ratio.
J/B=X+45/X=3/2
X(Bill)=90.
No of caps John has is 90+45=135
The earlier ratio will be J+15/B-15=150/75=2 so this is wrong since earlier ratio is 7/2

Next plugin=30
Equation=X+30/X=3/2
X(Bill)=60
J=60+30=90

Earlier Ratio=J/B=105/45=7/3
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 08 Feb 2018, 08:19
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Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.


One option is to solve the question using 2 variables

Let J = # of cards that John INITIALLY had
Let B = # of cards that Bill INITIALLY had

The number of baseball cards that John and Bill had was in the ratio of 7:3
So, we can write: J/B = 7/3
Cross multiply to get: 3J = 7B
Rearrange to get: 3J - 7B = 0

After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2.
So, John now has J - 15 cards and Bill now has B + 15 cards
We can write: J/B = (J - 15)/(B + 15) = 3/2
Cross multiply to get: (2)(J - 15) = (3)(B + 15)
Expand: 2J - 30 = 3B + 45
Rearrange to get: 2J - 3B = 75

So, we now have the following system:
3J - 7B = 0
2J - 3B = 45

Solve to get: J = 105 and B = 45
NOTE: these are the number of cards they INITIALLY had

After the gift, John had how many more baseball cards than Bill?
AFTER THE GIFT, John had 90 cards and Bill had 60 cards
So, John had 30 more cards than John.

Answer: B

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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 01 Jul 2019, 17:34
Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.


We can let the original number of baseball cards John and Bill had be 7x and 3x, respectively. So we can create the equation:

(7x - 15)/(3x + 15) = 3/2

2(7x - 15) = 3(3x + 15)

14x - 30 = 9x + 45

5x = 75

x = 15

Therefore, John and Bill originally had 105 and 45 cards, respectively. After John gave Bill 15 cards, John now had 90 cards and Bill had 60 cards. So John had 30 more cards than Bill.

Answer: B
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Re: The number of baseball cards that John and Bill had was in the ratio o  [#permalink]

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New post 08 Aug 2019, 16:44
Bunuel wrote:
The number of baseball cards that John and Bill had was in the ratio of 7:3. After John gave Bill 15 of his baseball cards, the ratio of the number of baseball cards that John had to the number that Bill had was 3:2. After the gift, John had how many more baseball cards than Bill?

A. 15
B. 30
C. 45
D. 60
E. 90

Kudos for a correct solution.





Could someone tell me what is wrong with this calculation?

7J/3B=(3J/2B)+15

(7J/3B)-(3J/2B)=15

(14J/6B)-(9J/6B)=15

5J/6B=15

5J=75B

J=15B

Then I am unsure. I tried substituting B for 2, obtaining the solution by coincidence.

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Re: The number of baseball cards that John and Bill had was in the ratio o   [#permalink] 08 Aug 2019, 16:44
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