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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% (02:27) correct
47%
(02:24)
wrong
based on 239
sessions
History
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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 3087 times ]
General Discussion
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
