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03 Mar 2026, 20:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
86% (03:05) correct
14%
(02:43)
wrong
based on 86
sessions
History
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Not Attempted Yet
Jars A and B each have a mix of red and blue stones. The ratio of number red stones to blue stones in jar A is 1:4 and the ratio of the same in jar B is 3:2. When the contents of the two jars are mixed, the ratio of red stones to blue stones turns out to be 1:3. If jar A had 112 blue stones, what was the total number of all stones in the two jars combined?
A. 84
B. 120
C. 140
D. 144
E. 160
A. 84
B. 120
C. 140
D. 144
E. 160
GraemeGmatPanda
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04 Mar 2026, 03:37
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We first define variables for the counts of red and blue stones in jars A and B
\(R_A = \text{number of red stones in jar A}\)
\(B_A = \text{number of blue stones in jar A}\)
\(R_B = \text{number of red stones in jar B}\)
\(B_B = \text{number of blue stones in jar B}\)
We translate the given number of blue stones in jar A
\(B_A = 112\)
We translate the ratio of red to blue stones in jar A
\(\frac{R_A}{B_A} = \frac{1}{4}\)
We translate the ratio of red to blue stones in jar B
\(\frac{R_B}{B_B} = \frac{3}{2}\)
We translate the ratio of red to blue stones after mixing the contents of both jars
\(\frac{R_A + R_B}{B_A + B_B} = \frac{1}{3}\)
We calculate the number of red stones in jar A using the ratio and known blue stones
\(\frac{112}{4} = \frac{100}{4} + \frac{12}{4} = 25 + 3 = 28\)
We substitute \(R_A\) and express \(R_B\) in terms of \(B_B\) into the mixture equation
\(3(28 + \frac{3}{2}B_B) = 112 + B_B\)
We eliminate the fraction by multiplying both sides of the equation by 2
\(2\times3(28 + \frac{3}{2}B_B) = 2\times(112 + B_B)\)
We distribute and simplify to combine like terms
\(6(28 + \frac{3}{2}B_B) = 224 + 2B_B\)
\(168 + 9B_B = 224 + 2B_B\)
We subtract \(2B_B\) from both sides to isolate terms involving \(B_B\)
\(168 + 9B_B - 2B_B = 224 + 2B_B - 2B_B\)
We combine like terms on the left side
\(168 + 7B_B = 224\)
We subtract 168 from both sides to solve for \(7B_B\)
\(7B_B = 224 - 168 = 56\)
We divide both sides by 7 to find \(B_B\)
\(\frac{56}{7} = 8\)
We simplify \(\frac{3}{2}B_B\) to find \(R_B\)
\(\frac{3}{2}\times8 = \frac{3\times8}{2} = \frac{24}{2} = 12\)
We sum all red and blue stones from both jars to find the total
\(28 + 12 + 112 + 8 = (28 + 12) + (112 + 8) = 40 + 120 = 160\)
Answer E
Hope this helps!
ʕ•ᴥ•ʔ
\(R_A = \text{number of red stones in jar A}\)
\(B_A = \text{number of blue stones in jar A}\)
\(R_B = \text{number of red stones in jar B}\)
\(B_B = \text{number of blue stones in jar B}\)
We translate the given number of blue stones in jar A
\(B_A = 112\)
We translate the ratio of red to blue stones in jar A
\(\frac{R_A}{B_A} = \frac{1}{4}\)
We translate the ratio of red to blue stones in jar B
\(\frac{R_B}{B_B} = \frac{3}{2}\)
We translate the ratio of red to blue stones after mixing the contents of both jars
\(\frac{R_A + R_B}{B_A + B_B} = \frac{1}{3}\)
We calculate the number of red stones in jar A using the ratio and known blue stones
\(\frac{112}{4} = \frac{100}{4} + \frac{12}{4} = 25 + 3 = 28\)
We substitute \(R_A\) and express \(R_B\) in terms of \(B_B\) into the mixture equation
\(3(28 + \frac{3}{2}B_B) = 112 + B_B\)
We eliminate the fraction by multiplying both sides of the equation by 2
\(2\times3(28 + \frac{3}{2}B_B) = 2\times(112 + B_B)\)
We distribute and simplify to combine like terms
\(6(28 + \frac{3}{2}B_B) = 224 + 2B_B\)
\(168 + 9B_B = 224 + 2B_B\)
We subtract \(2B_B\) from both sides to isolate terms involving \(B_B\)
\(168 + 9B_B - 2B_B = 224 + 2B_B - 2B_B\)
We combine like terms on the left side
\(168 + 7B_B = 224\)
We subtract 168 from both sides to solve for \(7B_B\)
\(7B_B = 224 - 168 = 56\)
We divide both sides by 7 to find \(B_B\)
\(\frac{56}{7} = 8\)
We simplify \(\frac{3}{2}B_B\) to find \(R_B\)
\(\frac{3}{2}\times8 = \frac{3\times8}{2} = \frac{24}{2} = 12\)
We sum all red and blue stones from both jars to find the total
\(28 + 12 + 112 + 8 = (28 + 12) + (112 + 8) = 40 + 120 = 160\)
Answer E
Hope this helps!
ʕ•ᴥ•ʔ
04 Mar 2026, 06:05
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ExpertsGlobal5
Let’s use alligations and mixtures to solve this problem
Jar A blue stone : Total = 4:5
Jar B blue stone : Total = 2:5
Combined jar blue stone : Total = 3:4
(4/5). (2/5)
(3/4)
Multiply by 20, we get
16. 8.
.15
7. 1.
The jar A and Jar B are mixed in the ratio 7:1 = 7k : 1k
Jar A contains 112 blue = 4x , thus, x = 28
Jar A total = 5x = 5*28 = 140.
If 140 denotes 7k, then K =20.
Total number of stones = 7k +1k = 8k = 8*20 = 160
Option E
