Problem Solving

Hi, Can someone help me solve this question without equation? Thanks

A can contains a mixture of two liquids A and B in ratio of 7:5. 9 liters of mixture is drawn off and the can is filled with B. The new ratio is 7:9. How many liters of A was there in the can initially?

A. 25
B. 21
C. 20
D. 10
E. 14

B

ahahkhyati.j wrote:Hi, Can someone help me solve this question without equation? Thanks

A can contains a mixture of two liquids A and B in ratio of 7:5. 9 liters of mixture is drawn off and the can is filled with B. The new ratio is 7:9. How many liters of A was there in the can initially?

A. 25
B. 21
C. 20
D. 10
E. 14

Let B = the pure liquid B, S = the original solution, and M = the mixture.

The following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with fractions or percentages.

Step 1: Convert any ratios to FRACTIONS.
B:
Here, B/total = 1/1.
S:
Since A:B = 7:5, and 7+5=12, B/total = 5/12.
M:
Since A:B = 7:9, and 7+9=16, B/total = 9/16.

Step 2: Put the fractions over a COMMON DENOMINATOR.
W = 1/1 = 48/48.
S = 5/12 = 20/48.
M = 9/16 = 27/48.

Step 3: Plot the 3 numerators on a number line, with the numerators for B and S on the ends and the numerator for M in the middle.
B 48-----------------27------------------20 S

Step 4: Calculate the distances between the numerators.
B 48-------21--------27---------7--------20 S

Step 5: Determine the ratio in the mixture.
The ratio of B to S in the mixture is equal to the RECIPROCAL of the distances in red.
B:S = 7:21 = 1:3 = 9:27.

Since B:S = 9:27, 9 liters of pure B was added to 27 liters of the original solution, implying that the can = 9+27 = 36 liters.
Since in the original solution B/total = 5/12, A/total = 7/12.
Thus:
Before the pure B was added, the amount of A in the can = (7/12)(36) = 21 liters.

The correct answer is B.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583

ahahkhyati.j wrote:A can contains a mixture of two liquids A and B in ratio of 7:5. 9 liters of mixture is drawn off and the can is filled with B. The new ratio is 7:9. How many liters of A was there in the can initially?

A. 25
B. 21
C. 20
D. 10
E. 14

An alternate approach is to PLUG IN THE ANSWERS, which represent the original volume of A.

Since original A:B = 7:5, the original volume of A is likely to be a MULTIPLE OF 7.
Implication:
The correct answer choice is probably B or E.

Since new A:B = 7:9, and 7+9=16, (new A)/(total volume) = 7/16.
Implication:
After 9 liters are removed from the can and replaced with pure B, the remaining volume of A must constitute 7/16 of the total volume.

Answer choice B: 21
Since original A:B = 7:5 = 21:15, original A = 21 liters and original B = 15 liters, for a total of 36 liters.
When 9 of the 36 liters are removed, since 9/36 = 1/4, A's volume and B's volume are each reduced by 1/4.
Thus:
Remaining A = (3/4)(21) = 63/4.
(remaining A)/(total volume) = (63/4)/36 = 7/16.
Success!

The correct answer is B.

Answer choice E: 14
Since original A:B = 7:5 = 14:10, original A = 14 liters and original B = 10 liters, for a total of 24 liters.
When 9 of the 24 liters are removed, since 9/24 = 3/8, A's volume and B's volume are each reduced by 3/8.
Thus:
Remaining A = (5/8)(14) = 35/4.
(remaining A)/(total volume) = (35/4)/24 ≠ 7/16.
Eliminate E.

Thank you... That was really helpful...

If anyone reading this doesn't mind a *little* bit of algebra, I think there's a shorter way.

Suppose our initial ratio is 7x / 5x. That means that B is (5/12) of the mixture, that A is (7/12) of the mixture, and that the total amount in the mixture is 12x.

When we replace 9 liters of the mix with 9 liters of B, that's equivalent to replacing (7/12)*9 liters of A with (7/12)*9 liters of B. Our total remains 12x.

That gives us an equation of

New A = 7/(7 + 9) of New Total, or

7x - (7/12)*9 = (7/16)*12x, or

x = 3

Since the initial amount of A was 7x, our answer is 7*3, or 21.

Actually, even my last explanation made it harder than it needs to be!

Suppose our initial amounts of A and B are called a and b. From our ratios, we know that

Old Amount of A = (7/12) * Total
New Amount of A = (7/16) * Total

Since the total is the same both times, we have

Old A / (7/12) = New A / (7/16), or

Old A = (4/3) * New A

Since New A = Old A - (7/12)*9, we have

x = (4/3) * (x - 63/12), or

x = (4/3)x - 7, or

7 = (1/3)x, and x = 21.

Too easy!