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01 Jun 2026, 04:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
62% (02:27) correct
38%
(02:20)
wrong
based on 138
sessions
History
Date
Time
Result
Not Attempted Yet
Bo profits $6.06 every time he sells a plate of hummus and $0.79 every time he sells a pack of pita. Yesterday, Bo only sold plates of hummus and packs of pita, and he sold a total of 214 plates and packs. Was Bo's profit more than $700 yesterday?
(1) Bo sold more packs than plates yesterday.
(2) Bo sold fewer than 64 plates yesterday.
(1) Bo sold more packs than plates yesterday.
(2) Bo sold fewer than 64 plates yesterday.
01 Jun 2026, 05:29
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GMAT shortcut:
With fewer than 64 hummus plates, even the maximum possible profit is only about $533, which is nowhere near $700. Therefore Statement (2) alone is sufficient.
Detailed
We need to know whether profit exceeded $700.
Each hummus earns about $6 profit.
There are 214 total items.
(2) Fewer than 64 plates
Maximum possible profit occurs if plates = 63.
Profit < 63(6.06) + 151(1)
= 381.78 + 151
= 532.78
Still far below 700.
Answer to the question is definitely No.
Sufficient.
(1) More packs than plates
Since total = 214:
Plates could be 100
Profit = 100(6.06) + 114(0.79) = 696.06
Not more than 700.
Plates could be 106
Profit = 106(6.06) + 108(0.79) = 727.44
More than 700.
Both Yes and No are possible.
Not sufficient.
Answer: B
With fewer than 64 hummus plates, even the maximum possible profit is only about $533, which is nowhere near $700. Therefore Statement (2) alone is sufficient.
Detailed
We need to know whether profit exceeded $700.
Each hummus earns about $6 profit.
There are 214 total items.
(2) Fewer than 64 plates
Maximum possible profit occurs if plates = 63.
Profit < 63(6.06) + 151(1)
= 381.78 + 151
= 532.78
Still far below 700.
Answer to the question is definitely No.
Sufficient.
(1) More packs than plates
Since total = 214:
Plates could be 100
Profit = 100(6.06) + 114(0.79) = 696.06
Not more than 700.
Plates could be 106
Profit = 106(6.06) + 108(0.79) = 727.44
More than 700.
Both Yes and No are possible.
Not sufficient.
Answer: B
01 Jun 2026, 06:42
Kudos
Bookmarks
say total hummus plates - m
cost price for each = a $
selling price for each = p $
profit for 1 plate = p - a
6.06 = p - a
p = 6.06 + a
we got selling price in terms of cost price.
he sells m plates.
revenue = m * (6.06+a)
profit= revenue- cost = m(6.06+a) - ma
m (6.06+ a-a)
profit = 6.06m
we know total he sold 214 of plates and packs. since plates we assumed as m, so packs will be 214-m
cost for each = b
sp for each = q
profit = q-b
0.79 = q-b
q= 0.79 + b
now total revenue for 214-m packs = (0.79 + b) (214-m)
profit = (0.79 + b) (214-m) - (214-m)b
= (214-m) (0.79)
now need to find that,
total profit > 700
profit from plates + profit from packs > 700
60.0m + (214-m)(0.79) > 700
solving this we get,
m>100.74
so we need to find m is greater than 100 or less than?
s-1
214-m > m
214> 2m
107>m
m< 107
this is yes or no. not a definitive ans. reject.
s-2
m < 64
this is definitely less than 100.
so this is sufficient.
choice B alone is sufficient.
cost price for each = a $
selling price for each = p $
profit for 1 plate = p - a
6.06 = p - a
p = 6.06 + a
we got selling price in terms of cost price.
he sells m plates.
revenue = m * (6.06+a)
profit= revenue- cost = m(6.06+a) - ma
m (6.06+ a-a)
profit = 6.06m
we know total he sold 214 of plates and packs. since plates we assumed as m, so packs will be 214-m
cost for each = b
sp for each = q
profit = q-b
0.79 = q-b
q= 0.79 + b
now total revenue for 214-m packs = (0.79 + b) (214-m)
profit = (0.79 + b) (214-m) - (214-m)b
= (214-m) (0.79)
now need to find that,
total profit > 700
profit from plates + profit from packs > 700
60.0m + (214-m)(0.79) > 700
solving this we get,
m>100.74
so we need to find m is greater than 100 or less than?
s-1
214-m > m
214> 2m
107>m
m< 107
this is yes or no. not a definitive ans. reject.
s-2
m < 64
this is definitely less than 100.
so this is sufficient.
choice B alone is sufficient.
Bunuel
