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Difficulty:
75%
(hard)
Question Stats:
67% (03:09) correct
33%
(03:03)
wrong
based on 138
sessions
History
Date
Time
Result
Not Attempted Yet
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
_________________
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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.
\( \mathbf{D} \)
SKaur3
Joined: 12 Aug 2023
Last visit: 08 Jan 2026
Posts: 99
Location: India
Schools: Booth '26 CBS '26 Kellogg '26 Darden '26 (A$$$$)
GMAT Focus 1: 675 Q87 V83 DI81 
GPA: 8.5
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.
Posts: 1,382
Let customers who bought
Juice = j
Cappuccino = c
Bagel = G
Brownie = b
Total customers be T
T = j+c - (1)
j = 1.5c - (2)
Customers who did not buy a cappuccino, are the customers who bought juice
b = 1/3 * j - (3)
G = T * 3/4 - (4)
G = 300+3b - (5)
Substituting 4 in 5
3T/4 = 300+3b
Substituting 1 and 3 in above
3(j+c) = 1200+12*j/3
3j+3c = 1200+4j
3c = 1200+j
Substituting 2 in above
3c = 1200+1.5c
c = 800
j = 1.5c = 1.5*800 = 1200
t = j+c = 800+1200 = 2000
Juice = j
Cappuccino = c
Bagel = G
Brownie = b
Total customers be T
T = j+c - (1)
j = 1.5c - (2)
Customers who did not buy a cappuccino, are the customers who bought juice
b = 1/3 * j - (3)
G = T * 3/4 - (4)
G = 300+3b - (5)
Substituting 4 in 5
3T/4 = 300+3b
Substituting 1 and 3 in above
3(j+c) = 1200+12*j/3
3j+3c = 1200+4j
3c = 1200+j
Substituting 2 in above
3c = 1200+1.5c
c = 800
j = 1.5c = 1.5*800 = 1200
t = j+c = 800+1200 = 2000
Bunuel
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Posts: 303
Let's break this problem down step by step:
Step 1: Define Variables
Let:
- x = number of cappuccinos sold
- y = number of juices sold
Since the café sold 50 percent more juice than cappuccinos, we set up the equation:
y = x + (1/2)x = (3/2)x
Total customers served = x + y:
x + y = x + (3/2)x = (5/2)x
Step 2: Brownie Equation
The problem states that one-third of the customers who did not buy a cappuccino also bought a brownie.
- Customers who did not buy a cappuccino = y
- Customers who bought a brownie = (1/3)y
So, the number of brownies sold is:
(1/3)y
Step 3: Bagel Equation
Three-fourths of the total customers bought a bagel, meaning:
(3/4) (x + y)
And we are given that the number of bagels sold was 300 more than three times the number of brownies:
(3/4) (x + y) = 3 * (1/3)y + 300
(3/4) (x + y) = y + 300
Step 4: Solve for x
Substituting y = (3/2)x:
(3/4) (x + (3/2)x) = (3/2)x + 300
(3/4) * (5/2)x = (3/2)x + 300
(15/8)x = (3/2)x + 300
Multiply everything by 8 to clear fractions:
15x = 12x + 2400
3x = 2400
x = 800
Step 5: Find Total Customers
x + y = x + (3/2)x = (5/2)x = (5/2) * 800 = 2000
Final Answer:
D. 2000
Step 1: Define Variables
Let:
- x = number of cappuccinos sold
- y = number of juices sold
Since the café sold 50 percent more juice than cappuccinos, we set up the equation:
y = x + (1/2)x = (3/2)x
Total customers served = x + y:
x + y = x + (3/2)x = (5/2)x
Step 2: Brownie Equation
The problem states that one-third of the customers who did not buy a cappuccino also bought a brownie.
- Customers who did not buy a cappuccino = y
- Customers who bought a brownie = (1/3)y
So, the number of brownies sold is:
(1/3)y
Step 3: Bagel Equation
Three-fourths of the total customers bought a bagel, meaning:
(3/4) (x + y)
And we are given that the number of bagels sold was 300 more than three times the number of brownies:
(3/4) (x + y) = 3 * (1/3)y + 300
(3/4) (x + y) = y + 300
Step 4: Solve for x
Substituting y = (3/2)x:
(3/4) (x + (3/2)x) = (3/2)x + 300
(3/4) * (5/2)x = (3/2)x + 300
(15/8)x = (3/2)x + 300
Multiply everything by 8 to clear fractions:
15x = 12x + 2400
3x = 2400
x = 800
Step 5: Find Total Customers
x + y = x + (3/2)x = (5/2)x = (5/2) * 800 = 2000
Final Answer:
D. 2000
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