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16 Dec 2022, 07:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:28) correct
47%
(02:23)
wrong
based on 265
sessions
History
Date
Time
Result
Not Attempted Yet
12 Days of Christmas GMAT Competition with Lots of Fun
Mary sells both cookies and cakes at her bakery. She sells each cookie for $1.5 and each cake for $2.5. In the month of May, she sold a total of 580 cookies and cakes. If the cost of making one cookie and one cake is $0.7 and $1.2 respectively, did Mary make a profit of more than $600?
(1) She sold 30 more cakes than cookies.
(2) She sold more cakes than cookies.
Mary sells both cookies and cakes at her bakery. She sells each cookie for $1.5 and each cake for $2.5. In the month of May, she sold a total of 580 cookies and cakes. If the cost of making one cookie and one cake is $0.7 and $1.2 respectively, did Mary make a profit of more than $600?
(1) She sold 30 more cakes than cookies.
(2) She sold more cakes than cookies.
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17 Dec 2022, 06:25
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Bunuel
GMATWhiz Official Explanation:
Step 1: Analyse Question Stem:
- • We are given:
• We are given Co+Ca=580….(i)
• We are asked if 0.8Co+1.3Ca>600 or not
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: She sold 30 more cakes than cookies
- • According to this statement, Ca-Co=30….(ii)
• Which we can solve along with equation (i) to get the values of Ca and Co
• And eventually get to know if 0.8Co+1.3Ca>600 or not
• Thus, statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: She sold more cakes than cookies
- • If she sold an equal amount of both cookies and cakes, she would have sold 580/2=290 of each
• According to this statement, Ca>Co
• The least profit would be when Ca=291 and Co=289
- o Profit_min=0.8Co+1.3Ca
=0.8×289+1.3×291
=231.2+378.3
=$609.5
• Minimum possible profit is $609.5 which is greater than 600
• Thus, statement 2 alone is also sufficient
Hence the right answer is Option D.
Attachment:
2022-12-17_17-23-16.png [ 32.28 KiB | Viewed 3385 times ]
General Discussion
Archit3110
16 Dec 2022, 07:19
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SP of Cookie $ 1.5
SP of cake $ 2.5
let cookies sold be x and cakes be y
x+y= 580
Cost of cookie $0.7 & that of cake $ 1.2
Proft = SP-CP
target
x * ( 1.5-0.7)+ y*(2.5-1.2) >600
x * ( 0.8) + y * 1.3>600
#1
She sold 30 more cakes than cookies.
x+30= y
x+y= 580
we get x = 275 , y = 305
275 * ( 0.8 ) + 305 * 1.3 = 616.5
sufficient
#2
She sold more cakes than cookies
y>x
it has to be integer value as cookies / cake cannot be fraction
least value of x is 1 and y is max 579
1* 0.8 + 579 *1.3 ; 753.7
max value of x 289 ; y least 291
289 * 0.8 + 291 * 1.3 ; 609.5
sufficient
OPTION D is correct
SP of cake $ 2.5
let cookies sold be x and cakes be y
x+y= 580
Cost of cookie $0.7 & that of cake $ 1.2
Proft = SP-CP
target
x * ( 1.5-0.7)+ y*(2.5-1.2) >600
x * ( 0.8) + y * 1.3>600
#1
She sold 30 more cakes than cookies.
x+30= y
x+y= 580
we get x = 275 , y = 305
275 * ( 0.8 ) + 305 * 1.3 = 616.5
sufficient
#2
She sold more cakes than cookies
y>x
it has to be integer value as cookies / cake cannot be fraction
least value of x is 1 and y is max 579
1* 0.8 + 579 *1.3 ; 753.7
max value of x 289 ; y least 291
289 * 0.8 + 291 * 1.3 ; 609.5
sufficient
OPTION D is correct
Bunuel
