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Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40

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Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 01:08
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Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?


(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.


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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post Updated on: 27 Mar 2020, 06:56
1

Solution


Step 1: Analyse Question Stem

    • Let x, y, and z be the number of stamps of 30cent, 35 cents, and 40 cents, respectively
    • Alice spent a total of $4.20 = 420 cents
      o \(30*x + 35*y + 40*z = 420\)
      o \(6x + 7y + 8z = 84\)…(i)
We need to find, whether any of x, y, and z is greater than 5 or not.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: The number of 35 cent stamps and 40 cent stamps are equal.
    • \(y = z\)
On substitute y = z in equation (i), we get
    • \(6x + 7y + 8y = 84\)
      o \(6x + 15y = 84\)
      o \(2x + 5y = 28\)
    • We know that x and y are integers. [Stamps cannot be in fraction.]
      o There are 2 possibilities
         Case 1: x = 4 and y =4
        • \(2*4 + 5*4 = 8+20 =28\)
        • In this case none of x, y, and z are greater than 5
         Case 2: x = 9 and y =2
          • \(2*9 + 5*2 = 18 + 10 = 28\)
          • In this case x is greater than 5.
We are getting contradicting results,
Hence, statement 1 is not sufficient, we can eliminate answer options A and D.
Statement 2: The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps is not more than the number of 40 cent stamps she bought.
    • \(x = y\), and z is not greater than y.
On substituting x = y in equation (i), we get
    • \(6y + 7y + 8z = 84\)
      o \(13y + 8z = 84\)
      o \(z = \frac{(84-13y)}{8}\) [this must be a non-negative integer; z can never be in fraction]
    • There is only one possibility.
      o \(y = 4\), then \(z =\frac{(84 - 52)}{8} = 4\)
      o \(x = y = z = 4\)
      o In this case none of x, y, and z are greater than 5. and this is the only possible case
Hence, the correct answer is Option B.
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Originally posted by GMATWhizTeam on 27 Mar 2020, 02:38.
Last edited by GMATWhizTeam on 27 Mar 2020, 06:56, edited 2 times in total.
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 03:07
Statement 1 alone is insufficient but statement 2 alone describes that she cannot buy any of the stamps more that 5.

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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 03:27
1
Let the number of stamps of 30 cent, 35 cent and 40 cent be a, b & c respectively
Given, 30a + 35b + 40c = 420
--> 6a + 7b + 8c = 84 ....... (1)

(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
--> b = c
From (1),
--> 6a + 15b = 84
--> 2a + 5b = 28
--> Possible values of (a, b, c) = {(4, 4, 4), (9, 2, 2)} --> Insufficient

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
--> a = c and b ≤ c
From (1),
--> 14a + 7b = 84
--> 2a + b = 12
--> Possible values of (a, b, c) = {(4, 4, 4), (5, 2, 5)}
--> None of the stamps are more than 5 --> Sufficient

Option B
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 06:47
1
The number of 30 cent stamp = a
The number of 35 cent stamp = b
The number of 40 cent stamp = c
--> 30a +35b +40c = 420

(1) b=c
If b=c=1, then a>5 (ok)
If b=c=4, then a=4 <=5 (no)
NOT SUFFICIENT

(2) a=c. b<= a or c
If a=c=4, then b=4 (ok)
If a=c=5, then b=2 (ok)
If a=c=6, then b=0, BUT b must be at least 1.(ok)
SUFFICIENT

FINAL ANSWER IS (B)

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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 08:20
2
Quote:
Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?

(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.


im assuming she bought at least 1 of each stamp;
30a+35b+40c=420 notice that b must be even
because 35 is not divisible by 420

(1) insufic

30a+35b+40c=420; b=c
b=c=2: 420-70-80=270
a=270/30=9
b=c=4: 420-140-160=120
a=120/30=4

(2) sufic

30a+35b+40c=420; b=c
a=c=1: 420-30-40=350
b=350/35=10>a
a=c=3: 420-90-120=210
b=210/35=6>a
a=c=5: 420-150-200=70
b=70/35=2<a

Ans (B)
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 09:34
1
Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?


(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought

1) 30 x + 35 y + 40 y =420 or, 30x + 75y = 420,or, 10x + 25y =140 , where x and y has to be positive integer . y can be 2, in that case x is 9. Again when y = 4, x is 5. Not sufficient.

2) 30x + 35y + 40x = 420 or, 70x +35y = 420, or 2x + y = 12, where y is equal or less than x. when x= 4, y =4. Again when x =5, y =2. x cannot be less than 3 as in that case y will be more than x.So, in both cases the number of stamps doesn't exceed 5. Sufficient.

B is the answer.
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 11:13
Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?

Let number of 30 cent stamps = x
number of 35 cent stamps = y
number of 40 cent stamps = z
where a, y and z all are positive integers.
So,
30x + 35y + 40z = 420
6x + 7y + 8z = 84

(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
y = z so 6x + 7y + 8z = 84 becomes
6x + 15z = 84
x = 4, z = 4 NO
x = 9, z = 2 YES

INSUFFICIENT.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
x = z and y ≤ z
Now
6x + 7y + 8z = 14x + 7y = 84
2x + y = 12
2*1 + 10 = 12 NO
2*2 + 8 = 12 NO
2*3 + 6 = 12 NO
2*4 + 4 = 12 NO
2*5 + 2 = 12 NO
2*6 + 0 = 12 YES

INSUFFICIENT.

Together 1 and 2.
x = y = z
6x + 7y + 8z = 21x = 84
x = 4
OR x = y = z = 4 NO
Only one case.

SUFFICIENT.

Answer C.
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 27 Mar 2020, 15:37
Quote:
Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?

(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.


Fun one!

Let's say that x is the number of 30 cent stamps, y is the number of 35 cent stamps, and z is the number of 40 cent stamps. So, in the question stem, we have an equation:

30x + 35y + 40z = 420 (it's 420 and not 4.20 because we're using cents rather than dollars!)

Divide everything by 10 to simplify the math:

3x + 3.5y + 4z = 42

The question is whether she bought more than 5 stamps of any of the three values. That is, is x, y, or z greater than 5?

Also note that we know x, y, and z are non-negative integers, since you can't buy a fractional or negative number of stamps.

Statement 1: The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

In other words, y = z. So, the equation simplifies as follows:

3x + 3.5y + 4z = 42
3x + 7.5y = 42

Okay, one of the values could be bigger than 5, for instance, if x = 14 and y = 0. So, it's possible to get a "yes" answer. Can we also get a "no," where both of the values are smaller than 5?

Let's try the biggest values possible that are still smaller than 5. So, if y = 4, then we have this:

3x + 7.5(4) = 42
3x + 30 = 42
3x = 12
x = 4

Therefore, x = 4, y = 4, z = 4 is a valid solution that fits this statement, and gives us a "no" answer (none of the values are greater than 5.)

Therefore, this statement is insufficient.

Statement 2 The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.

This says that x = z, and y<=z.

The second case we tested above fits the bill: x = y = z = 4. In this case, we get an answer of "no." Can we also get a "yes" to show that this is insufficient?

To do that, we should simplify the equation:

3x + 3.5y + 4z = 42

3x + 3.5y + 4x = 42

7x + 3.5y = 42

I multiplied by 2 to get rid of the decimal:

14x + 7y = 84

Then, divide by 7:

2x + y = 12

The constraint from this statement is that y is no bigger than z. Since x and z are equal, we know that y can't be any bigger than x. Is there a solution that fits, where one of the values is greater than 5?

x = 6 and y = 0 works (and that implies that z = 6 as well.)

Therefore, this statement is also insufficient.

Statements 1 + 2 together

Interestingly, we already found a case that works with both statements and gives a "yes" answer: x = y = z = 4.

Can we find a case that works with both statements and gives a "no" answer?

Take the info from both statements: x = y, x = z. That is, all three values must be equal.

Plug that into the equation:

3x + 3.5x + 4x = 42
10.5x = 42
21x = 84
x = 4

In other words, x = y = z = 4 is the only solution that still works. So, both statements together are sufficient and the correct answer is C.
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 28 Mar 2020, 05:58
Ans C

Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?


(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.

Make basic equation ... and procedd evaluating choices ...

30a+35b+40c = 420

St1: you get (a,b,c) as (9,2,2) and (4,4,4) Insufficient
St2: you get (a,b,c) as (6,0,6) , (5,2,5) and (4,4,4) .. Insufficient

Combining both only a=b=c satisfies ... Hence sufficent
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 28 Mar 2020, 07:43
1
Bunuel wrote:
Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?


(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.

(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.


Let stamps for 30 cents be 'a'
Let stamps for 35 cents be 'b'
Let stamps for 40 cents be 'c'

Hence, according to stem -

30a + 35b + 40c = 420 cents -- (1)

As per statement 1,

b = c -- (2)

Using (1) and (2), we get

30a + 35b + 40b = 420
30a + 70b = 420

Now here we can have the following options -

a = 4 and b = c = 4 ,OR
a = 9 and b = c = 2

So in first case values are below 5 but in the second case value of a is above 5.

Hence Statement 1 alone is Insufficient

As per statement 2,

a = b -- (3) ,and
b < c -- (4)

Using (1) and (3), we get

30a + 35a + 40c = 420
65a + 40c = 420

Now here we can have the only the following case

c = 4 and b = a = 4

which is sufficient to determine if alice bought more than 5 stamps of any of the three values

Hence Statement 2 alone is Sufficient

Hence B
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 29 Mar 2020, 13:10
This seems to be a divisibility question.

We want to know if a, b, c are all <= 5 so this is a YES/NO DS question.

let a = # of 30 cent stamps, b = # 35 cent stamps, c = # of 40 cent stamps

so 30a + 35b + 40c = 420

1) # of 35 cent stamps is equal to # of 45 cent stamps. so b = c.

35+40 = 75. There are FIVE multiples of 75 that are less than or equal to 420 are 75, 150, 225, 300, 375.

now we can quickly see what values of a 30 cent stamp work with these combinations of the multiples of 75 we listed:
420 - 375 = 45, so c has to be less than 2 stamps (60 cents) so this is sufficient as our max possibility here is

b = c = 4 -->300
a = 4 ----->120

all values are <5 so:
SUFFICIENT

2) # of 30 cent stamps = # of 40 cent stamps, and # of 35 cent stamps is not more than # of 40 cent stamps
so we have equations a = c and b <=c

we know a = c and so 30 + 40 = 70, there are 6 multiples of 70 less than or equal to 420:
70, 140, 210, 280, 350, 420

so we can have a = c = 6 here, while b = 0 which still satisfies b <= c so we get NO since a and c are > 5

but we can ALSO have a = c = 5 (total of 350), while b = 2 (value of 70) which adds to 420 so we get YES since all are <=5. INSUFFICIENT

So the answer is A
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 29 Mar 2020, 18:05
30x + 35y + 40z = 420
—> 6x + 7y + 8z = 84
x >5?
y >5?
z >5?

(Statement1): y = z
If y= z = 2, then 6x = 84 —30 = 54
—> x= 9 (yes), y = z= 2 (No)
Insufficient

(Statement2): x= z, y <= z.
If x=z =6, then 14*6 + 7y = 84
—> y = 0(No), x=z=6 (Yes)
Insufficient

Taken together 1&2,
x =y = z —> 21*4 = 84
—> x= y = z = 4 (Always NO)
Sufficient

Answer(C)

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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40  [#permalink]

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New post 29 Mar 2020, 19:35
Let the number of 30cents stamps bought be \(a\), the number of 35cents stamps bought be \(b\), and the number 40cents stamps bought be \(c\).

Then \(30a + 35b + 40c = 420\) -----(1)
Is \(a>5\), \(b>5\), or \(c>5\)?

Statement 1: The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
This means that \(b = c\)
Hence \(30a = 420 - 75b\)
\(a = 14 - 5b/2\)
when \(b=2\), \(a=9 \) Yes
when \(b=4\), \(a = 4\) No
Statement 1 is insufficient.

Statement 2: The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought
\(a = c\) and \(b ≤ c\)
\(70a + 35b = 420\)
\(2a + b = 12\)
\(a = 6 - b/2\)
When \(b=2, a = 5,\) condition that \(b≤c\) satisfied. Yes
When \(b=4, a = 4,\) condition that \(b≤c\) satisfied. No
Statement 2 is insufficient.

1+2
\(a=b=c \)and naturally the condition that\( b≤c\) is satisfied.
\(105a = 420\)
\(a = 4.\) No.
Both statements combined are sufficient.

The answer is C.
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Re: Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40   [#permalink] 29 Mar 2020, 19:35

Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40

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