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18 May 2026, 19:00
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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.
The OA will be automatically revealed on Friday 22nd of May 2026 07:00:13 PM Pacific Time Zone
(1) Nora spent more than $24 on sketchbooks.
(2) Nora spent more than $48 on colored markers.
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The OA will be automatically revealed on Friday 22nd of May 2026 07:00:13 PM Pacific Time Zone
ankushsambare
Joined: 13 Jun 2022
Last visit: 19 May 2026
Posts: 171
Own Kudos:
Given Kudos: 107
Location: India
Schools: ISB '27 Wharton '26 INSEAD Aug '26
GPA: 2.4
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18 May 2026, 20:43
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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
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
19 May 2026, 03:26
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| 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
