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22 Jul 2026, 09:02
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:37) correct
41%
(03:29)
wrong
based on 39
sessions
History
Date
Time
Result
Not Attempted Yet
Before the 50 people celebrating Pedro’s birthday arrived twenty minutes ago, there were no children at Sally’s Burger Shack. After Pedro’s group arrived, there were twice as many adults as children. Another group of 50 people celebrating Raquel’s birthday has just arrived, and now there are 4 adults for every 3 children. No other people have arrived, nor has anybody left yet. If there are the same number of children in the two groups, how many adults are now at Sally’s Burger Shack?
(A) 80
(B) 92
(C) 100
(D) 124
(E) 140
(A) 80
(B) 92
(C) 100
(D) 124
(E) 140
22 Jul 2026, 14:54
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Before the 50 people celebrating Pedro’s birthday arrived twenty minutes ago, there were no children at Sally’s Burger Shack. After Pedro’s group arrived, there were twice as many adults as children. Another group of 50 people celebrating Raquel’s birthday has just arrived, and now there are 4 adults for every 3 children. No other people have arrived, nor has anybody left yet. If there are the same number of children in the two groups, how many adults are now at Sally’s Burger Shack?
Set variables:
Let the number of adults at the shack before the groups arrived be A.
Let the number of children in each group be C.
Translate:
After Pedro’s group arrived, there were twice as many adults as children.
A + (50 - C) = 2C
Another group of 50 people celebrating Raquel’s birthday has just arrived, and now there are 4 adults for every 3 children.
A + 2(50 - C) = 4/3 × 2C
Solve for C by subtracting:
A + 2(50 - C) = 4/3 × 2C
A + (50 - C) = 2C
50 - C = 1/3 × 2C
150 - 3C = 2C
30 = C
Answer the question asked:
A + 2(50 - C) = 4/3 × 2C
A + 2(50 - C) = 4/3 × 60
A + 2(50 - C) = 80
(A) 80
(B) 92
(C) 100
(D) 124
(E) 140
Correct answer: A
Set variables:
Let the number of adults at the shack before the groups arrived be A.
Let the number of children in each group be C.
Translate:
After Pedro’s group arrived, there were twice as many adults as children.
A + (50 - C) = 2C
Another group of 50 people celebrating Raquel’s birthday has just arrived, and now there are 4 adults for every 3 children.
A + 2(50 - C) = 4/3 × 2C
Solve for C by subtracting:
A + 2(50 - C) = 4/3 × 2C
A + (50 - C) = 2C
50 - C = 1/3 × 2C
150 - 3C = 2C
30 = C
Answer the question asked:
A + 2(50 - C) = 4/3 × 2C
A + 2(50 - C) = 4/3 × 60
A + 2(50 - C) = 80
(A) 80
(B) 92
(C) 100
(D) 124
(E) 140
Correct answer: A
22 Jul 2026, 18:37
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Before Pedro’s group arrived
Adults = x, children = 0
In Pedro’s group of 50 people:
Adults = 50-y, children = y
In Raquel’s group of 50 people:
Adults = 50-y, children = y
After Pedro’s group arrived:
Adults = x + 50 - y; children = y
Given: there were twice as many adults as children
So, x + 50 - y = 2y
So, x - 3y = -50
After Raquel’s group arrived:
Adults = x + 50 - y + 50 - y = x + 100 - 2y; children = 2y
Given: there are 4 adults for every 3 children
x + 100 - 2y = 4/3*(2y)
3x + 300 - 6y = 8y
3x - 14y = -300
So, the equations are:
x - 3y = -50
3x - 14y = -300
Solving, we get
y = 30
x=40
Total adults = x + 100 - 2y = 40 +100-60 = 80
Quite a time taking question though. Wonder how we can do it under 2 minutes.
Adults = x, children = 0
In Pedro’s group of 50 people:
Adults = 50-y, children = y
In Raquel’s group of 50 people:
Adults = 50-y, children = y
After Pedro’s group arrived:
Adults = x + 50 - y; children = y
Given: there were twice as many adults as children
So, x + 50 - y = 2y
So, x - 3y = -50
After Raquel’s group arrived:
Adults = x + 50 - y + 50 - y = x + 100 - 2y; children = 2y
Given: there are 4 adults for every 3 children
x + 100 - 2y = 4/3*(2y)
3x + 300 - 6y = 8y
3x - 14y = -300
So, the equations are:
x - 3y = -50
3x - 14y = -300
Solving, we get
y = 30
x=40
Total adults = x + 100 - 2y = 40 +100-60 = 80
Quite a time taking question though. Wonder how we can do it under 2 minutes.
