At a stationery shop, Nora spent $84 on sketchbooks and colored marker

Question

At a stationery shop, Nora spent $84 on sketchbooks and colored markers. Each sketchbook cost $12, and each colored marker cost $8. If Nora bought at least one sketchbook and at least one colored marker, how many sketchbooks did she buy?

(1) Nora spent more than $24 on sketchbooks.
(2) Nora spent more than $48 on colored markers.

M47-160

Bunuel · Accepted Answer

GMAT Club Official Solution:

At a stationery shop, Nora spent $84 on sketchbooks and colored markers. Each sketchbook cost $12, and each colored marker cost $8. If Nora bought at least one sketchbook and at least one colored marker, how many sketchbooks did she buy?

Let S be the number of sketchbooks and M be the number of colored markers. The total cost was $84, so:

12S + 8M = 84

3S + 2M = 21

The question asks for S.

(1) Nora spent more than $24 on sketchbooks.

Since each sketchbook cost $12:

12S > 24

S > 2

This is not enough.

If S = 3, then:

3 * 3 + 2M = 21

M = 6

But if S = 5, then:

3 * 5 + 2M = 21

M = 3

Different values of S are possible.

Not sufficient.

(2) Nora spent more than $48 on colored markers.

Since each colored marker cost $8:

8M > 48

M > 6

From 3S + 2M = 21, since 3S and 21 are multiples of 3, 2M must also be a multiple of 3. Therefore, M must be a multiple of 3. Since M > 6, the first possible value is M = 9. The next possible multiple of 3 is M = 12, but then 2M = 24, which is already greater than 21. So M must be 9. Then:

3S + 2 * 9 = 21

S = 1

Sufficient.

Answer: B.

ankushsambare · Answer

Let sketchbooks = x, markers = y
Given:
12x + 8y = 84
x ≥ 1, y ≥ 1
Divide by 4:
3x + 2y = 21
Now find possible integer solutions:
x = 1 → 3 + 2y = 21 → y = 9
x = 3 → 9 + 2y = 21 → y = 6
x = 5 → 15 + 2y = 21 → y = 3
x = 7 → 21 + 2y = 21 → y = 0 (not allowed)
So possible values are:
(1, 9), (3, 6), (5, 3)
x is not fixed, so we use statements.
Statement (1): spent more than $24 on sketchbooks
Sketchbooks cost = 12x
12x > 24 → x > 2 → x = 3 or 5 possible
Still more than one value, so not sufficient.
Statement (2): spent more than $48 on markers
Markers cost = 8y
8y > 48 → y > 6 → only y = 9 works
From equation, y = 9 gives x = 1
So x is uniquely determined.
 Final Answer: B

SS990 · Answer

S  
  C  
 Total 
 
  cost in $  
 12 
  8

number 
  s 
  c

money spent 
 12s 
 8c 
  84

Given : above info. And S and C are atleast 1.

Intial Finding: 12s+8c=84 (3s+2c=21 or s+2c/3=7).. note c has to be a multiple of 3 and s has to be odd given 2c is even and addition will lead to an odd 21 only if s is odd. so given there is atleast 1 of s and c.. c can only take values 3,6,9 and s will be resp 5,3,1.

(fun fact: notice how "s" is reducing in number exactly equal to coefficient of c in 3s+2c=21, the reverse is also true)

To find: what the value of s?

1) money spent on s is 12s>24 so s>2 , so s can be 3 or 5 so thsi is not sufficient

2) Money spent on c is 8c>48, so c>6 , so c can only be 9, wherein s is 1 Hence sufficient.

Answer is B

Edskore · Answer

Answer is B. Here's the full setup.

Let s = number of sketchbooks, c = number of colored markers.
12s + 8c = 84, and both s >= 1 and c >= 1, both integers.

Divide by 4: 3s + 2c = 21. Rearranging: c = (21 - 3s)/2. For c to be a positive integer, (21 - 3s) must be even and positive. Since 21 is odd and 3s must be odd for the difference to be even, s must be odd.

Possible values: s=1 (c=9), s=3 (c=6), s=5 (c=3). s=7 gives c=0, which violates the constraint. So exactly three scenarios exist.

Statement 1: spent more than $24 on sketchbooks, meaning 12s > 24, so s > 2. That rules out s=1, but both s=3 and s=5 are still possible. Not sufficient.

Statement 2: spent more than $48 on colored markers, meaning 8c > 48, so c > 6. Only c=9 qualifies (c=6 doesn't, since 6 is not strictly greater than 6). So s=1 is the only solution. Sufficient alone.

Answer: B

The classic DS trap here is testing Statement 1 and convincing yourself it "almost" works — s=3 and s=5 are both consistent, but that means two possible answers, which is insufficient. Statement 2 cuts it cleanly to one case. When I prep DS questions like this, I always list out all possible integer solutions first before touching the statements.

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At a stationery shop, Nora spent $84 on sketchbooks and colored marker

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