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

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

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)

Posted from my mobile device

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)

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.

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.

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.

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

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

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

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)

Posted from my mobile device

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.

chetan2u VeritasKarishma
shouldn't the answer be C? (5,2,5) and (4,4,4) both are deductable from statement 2.

Yes, you are correct.

But the question is - Is any of the type of the three stamps greater than 5.
Be it (5,5,2) or (4,4,4), the answer is negative, that is none of them is greater than 5 in numbers.

Oh yes. I missed that. Thanks.

Posted from my mobile device

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 the number of 30 cent stamps, 35 cent stamps and 40 cent stamps be x, y and z. The confusion in the statement II occurs when we assume that x, y and z can take 0 value too, that is it is a possibility that there are none of a particular type of stamp.
I would believe that the question wants us to take that there is at least one of each as it mentions three specific types of stamps when there can be many other types.
However, it may be preferable and would generally be the case in official questions that this is mentioned explicitly.

\(30x+35y+40z=420\)

(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
\(y=z\)
\(30x+35y+40z=420……..30x+75y=420…..2x+5y=28\)
y has to be even.
y=z=2, x=9…..>5, yes
y=z=4, x=4……>5, no
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.
\(x=z\)
\(30x+35y+40z=420……..70x+35y=420…..2x+y=12\)
y has to be even.
y=2, x=z=5…..>5, no
y=4, x=z=4……>5, no
If we assume that y can be 0, then x and z will be greater than 5, but the question does not suggest so.
Sufficient


B

You can also play around with numbers to get the answer easily. First note that the number of 35 cents stamps must be even since the total is $4.20 and both other amounts are multiples of 10 (30 and 40).
So, assuming that she did buy at least 1 stamp of each type (which would normally be mentioned), she must have bought 2/4/6/8/10 stamps of 35 cents.

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

If she bought 2 stamps each of 35c and 40c, she would have bought 9 stamps of 30c.
If she bought 4 stamps each of 35c and 40c, she would have bought 4 stamps of 30c.
Not sufficient.

(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.

Assuming she bought 2 stamps (smallest number) of 35 cents, she would have bought 5 stamps of 30c and 40c.
Can she buy 6 stamps of 35 cents? Then she would have bought 3 stamps of 30c and 40c but that is not possible because no. of 35c stamps cannot be more than number of 40c stamps.
So she could not have bought more than 5 stamps of any value.
Sufficient

Answer (B)