If each of the stamps Carla bought cost 20, 25, or 30 cents and she bo

Carla bought at least one of each denomination. At a minimum, she has 75 cents of stamps (30, 20, 25)

(1) If she spent a total of $1.45, we already know 75 cents consists of one of each stamp.

We have 70 cents unaccounted for; she could have purchase a 30 cent stamp and 2 20 cent stamps or 2 25 cent stamps and 1 20 cent stamp. INSUFFICIENT.

(2) We have no idea what these 6 stamps consist of; INSUFFICIENT.

(1&2) The two scenarios mentioned for statement 1 still apply here. INSUFFICIENT.

Answer is E.

GMATinsight Bunuel chetan2u I don't get why we CAN distribute 20 for instance to 25y (see highlighted part above) if y has to be an INTEGER. Can you explain?

BrentGMATPrepNow can you help with ST1? I get this equation 4x+5y+6z=29 and nothing seems to work, no matter what values I plug into it.

Your equation, 4x+5y+6z=29, is correct.
Two possible solutions are:
1) x = 2, y = 3 and z = 1
2) x = 3, y = 1 and z = 2

If each of the stamps Carla bought cost 20, 25, or 30 cents and she bought at least one of each denomination, what is the number of 25-cent stamps that she bought?

(1) She spent a total of $1.45 for stamps.
(2) She bought exactly 6 stamps.



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Let # of 20 cents coins be -> x
Let # of 25 cents coins be -> y
Let # of 30 cents coins be -> z

x+y+z >=3


(1) 20x + 20y +30z =145 (1.45 *100)
Insufficient - 3 Variables 1 equation

(2) x +y +Z = 6
Insufficient - 3 Variables 1 equation


(1) & (2) combined
20x + 20y +30z =145
x +y +Z = 6

3 variables 2 equation - Insufficient

hence Answer is E

KarishmaB is this approach correct ?

Each statement alone are obviously insufficient so I'll jump directly to the part where we combine Statment I and Statement II and then check

let x be the number of 20 cents stamps
let y be the number of 25 cents stamps
let z be the number of 30 cents stamps

so, the total cost of all the stamps combined would be

20x+25y+30z

Statement (I) She spent a total of $1.45 for stamps
Statement (II) She bought exactly 6 stamps

so she spent a total of 145cents[$1.45] on the stamps and bought a total of 6 stamps

note that to get the value of 20x+25y+30z=145

y either has to be 1 or 3. Why?
because only in that case can we have 5 in the unit digit of the total cost of the stamps [145]

if y is 2 or 4 then that would make the value of y stamps either 50cents or 100cents and we can never get a value of units digit of x or z type stamps to be 5 because they cost 20cents and 30cents respectively [multiples of 10]

y obviously can't be 5 because that would either make x=0 or z=0 and it has been mentioned that she did buy at least 1 stamp of each kind

Now that we know that y is either has to be 1 or 3

case I: y=1

if y=1 then value of y stamps will be 25cents and the total value of x and z stamps will be 145-25=120cents

can we get 20x+30z=120?

yes, if x=3 and z=2 then we will have 20x=60 and 30z=60

and the total value of 20x+25y+30z=145 [60+25+60]

case II: y=3

if y=3 then value of y stamps will be 75cents and the total value of x and z stamps will be 145-75=70cents

can we get 20x+30z=70?

yes, if x=2 and z=1 then we will have 20x=40 and 30z=30

and the total value of 20x+25y+30z=145 [40+75+30]

So we have 2 cases in which the total value of the stamps can be 145cents [$1.45] and the total number of stamps can be 6

But, observe that in Case I: the number of y stamps[25cents] = 1 and in Case II: the number of y stamps[25cents] = 3

Hence, we have two different answers for the number of 25cents stamps(y), therefore, both statements together are also not sufficient

Answer Choice E

KarishmaB
How to solve questions like these efficiently?

This tests the integer solutions to linear equations concept.
First thing to do is remove the "at least one of each denomination" constraint. She bought 1 of each type and spent 20+25+30 = 75 cents.
Now say a, b and c are the number of additional stamps.
a = no of 20 cent stamps,
b = no of 25 cent stamps,
c = no of 30 cent stamps

We need the value of b + 1 so we need the value of b.

(1) She spent a total of $1.45 for stamps.

We have accounted for 75 cents so total left to account for is 70 cents.

20a + 25b + 30c = 70
4a + 5b + 6c = 14
These are simple numbers and a, b, c can be 0.
I see that a = 1, b = 2, c = 0 will work
I see that a = 2, b = 0, c = 1 will work
Not sufficient


(2) She bought exactly 6 stamps.

We have accounted for 3 stamps. a + b + c = 3 now.
b can take any value 0/1/2/3
Not sufficient

Using both, in both cases given, a+b+c = 3 so still not sufficient.

Answer (E)

Very late to reply but no, this approach does not work for "integer solutions to linear equations."
It is possible that 1 equation with two variables is enough to get the value of both variables in case there is an integer constraint on the value of the variables. In this question, luckily, both solutions had different values of y and hence the answer was (E).