GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 28Live Online Class with Top GRE Expert Mike Butville
08:00 PM EDT
-10:00 PM EDT
Top scores are possible when you enroll in a powerful GRE course, taught live online + 6 months access to TTP GRE OnDemand video courses included! Class starts Mon/Wed Sept. 28, 2026 - Dec. 9, 2026, 8:00pm-10:00pm EST
Liters of 10% acid to be added to 16 liters of 25% acid to achieve a
[#permalink]
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
Hide Show timer Statistics
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)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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
