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Liters of 10% acid to be added to 16 liters of 25% acid to achieve a
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25 Jul 2022, 00:54
25 Jul 2022, 00:54
Expert Reply
4
Bookmarks
00:00
Question Stats:
67% (01:58) correct
32% (01:57) wrong based on 64 sessions
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Quantity A |
Quantity B |
Liters of 10% acid to be added to 16 liters of 25% acid to achieve a 20% solution |
Liters of 20% acid to be added to 10 liters of 30% acid to obtain a 25% solution |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
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GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
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GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
Re: Liters of 10% acid to be added to 16 liters of 25% acid to achieve a
[#permalink]
27 Jul 2022, 15:31
27 Jul 2022, 15:31
1
Expert Reply
1
Bookmarks
OE
Quote:
A mixture problem. Let A = the liters to be added in the first problem. The equation for Column A is: A(0.10) + 16(0.25) = (A+16)(0.20). Multiply through by 100 to simplify the decimal, and solve, for A = 8. For the second problem, let B be the number of liters to be added. B(0.20) + 10(0.30) = (B+10)(0.25). Solving, B = 10.
Re: Liters of 10% acid to be added to 16 liters of 25% acid to achieve a
[#permalink]
02 Jan 2026, 15:42
02 Jan 2026, 15:42
1
Quantity A
Liters of 10% acid to be added to 16 liters of 25% acid to achieve a 20% solution
Quantity B
Liters of 20% acid to be added to 10 liters of 30% acid to obtain a 25% solution
Quantity A = x Liters
(0.1)*X + (0.25)(16) = (x+16)(0.2)
0.1 X + 4 = 0.2X + 3.2
0.8 = 0.1X
X = 8
Quantity B = y Liters
(0.2)*y + (0.3)(10) = (y+10)(0.25)
0.2y + 3 = 0.25y + 2.5
0.5 = 0.05y
y = 10
Therefore, y > x
Quantity B > Quantity A
Liters of 10% acid to be added to 16 liters of 25% acid to achieve a 20% solution
Quantity B
Liters of 20% acid to be added to 10 liters of 30% acid to obtain a 25% solution
Quantity A = x Liters
(0.1)*X + (0.25)(16) = (x+16)(0.2)
0.1 X + 4 = 0.2X + 3.2
0.8 = 0.1X
X = 8
Quantity B = y Liters
(0.2)*y + (0.3)(10) = (y+10)(0.25)
0.2y + 3 = 0.25y + 2.5
0.5 = 0.05y
y = 10
Therefore, y > x
Quantity B > Quantity A
