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12 Mar 2025, 00:33
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Difficulty:
75%
(hard)
Question Stats:
66% (03:09) correct
34%
(03:06)
wrong
based on 119
sessions
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Josh, the manager of a bookstore café, is completing his weekly sales summary. Each customer bought either a cappuccino or a juice and the café sold 50% more juice than cappuccino. One third of the customers who did not buy a cappuccino also bought a brownie and 3/4 of the total customers bought a bagel. If the number of bagels sold was 300 more than three times the number of brownies, how many customers did Josh's café serve?
A. 400
B. 800
C. 1200
D. 2000
E. 6000
A. 400
B. 800
C. 1200
D. 2000
E. 6000
12 Mar 2025, 00:44
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Let the number of customers who bought a cappuccino be \( C \), and the number of customers who bought a juice be \( J \).
We are given that the café sold 50% more juice than cappuccino, so:
\( J = 1.5C \)
The total number of customers is:
\( T = C + J \)
\( T = C + 1.5C = 2.5C \)
One third of the customers who did not buy a cappuccino (i.e., those who bought juice) also bought a brownie:
\( B = \frac{1}{3} J \)
\( B = \frac{1}{3} \times 1.5C = 0.5C \)
We are given that 3/4 of the total customers bought a bagel:
\( G = \frac{3}{4} T \)
\( G = \frac{3}{4} \times 2.5C = 1.875C \)
We are given that the number of bagels sold was 300 more than three times the number of brownies:
\( G = 3B + 300 \)
Substituting \( G = 1.875C \) and \( B = 0.5C \):
\( 1.875C = 3(0.5C) + 300 \)
\( C = 800 \)
\( T = 2.5 \times 800 = 2000 \)
The café served \(\mathbf{2000}\) customers.
\( \mathbf{D} \)
We are given that the café sold 50% more juice than cappuccino, so:
\( J = 1.5C \)
The total number of customers is:
\( T = C + J \)
\( T = C + 1.5C = 2.5C \)
One third of the customers who did not buy a cappuccino (i.e., those who bought juice) also bought a brownie:
\( B = \frac{1}{3} J \)
\( B = \frac{1}{3} \times 1.5C = 0.5C \)
We are given that 3/4 of the total customers bought a bagel:
\( G = \frac{3}{4} T \)
\( G = \frac{3}{4} \times 2.5C = 1.875C \)
We are given that the number of bagels sold was 300 more than three times the number of brownies:
\( G = 3B + 300 \)
Substituting \( G = 1.875C \) and \( B = 0.5C \):
\( 1.875C = 3(0.5C) + 300 \)
\( C = 800 \)
\( T = 2.5 \times 800 = 2000 \)
The café served \(\mathbf{2000}\) customers.
\( \mathbf{D} \)
12 Mar 2025, 00:49
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Cappuchino – C, Juice = J , Brownie = B = J/3 , Bagel = G = 3(J+C)/4
Given
J = 1.5 C - I
3B + 300 = G - II
Solving II
J + 300 = 3(J+C)/4
4J + 1200 = 3J + 3C
J + 1200 = 3C
Using I
1.5C = 1200 = J
C = 800
J + C = Total Customers = 800+1200 = 2000.
Given
J = 1.5 C - I
3B + 300 = G - II
Solving II
J + 300 = 3(J+C)/4
4J + 1200 = 3J + 3C
J + 1200 = 3C
Using I
1.5C = 1200 = J
C = 800
J + C = Total Customers = 800+1200 = 2000.
