Problem Solving

Hello could someone help me to solve this problem?

A mixture of two candies, A and B, costs
$11.625 per kilogram. Candy A costs $10.2
per kilogram while candy B costs $14 per
kilogram. What is the ratio A:B in which the
two candies are mixed?
A) 3:5
B) 5:3
C) 3:8
D) 5:8
E) 3:4

thanks

Notice that on the number line, the overall average of 11.625 is closer to A (10.2) than it is to B (14)

10.2----11.625----------14

Therefore, we must have more A than B. If A > B, only option that works is B

Vincelauret wrote:Hello could someone help me to solve this problem?

A mixture of two candies, A and B, costs
$11.625 per kilogram. Candy A costs $10.2
per kilogram while candy B costs $14 per
kilogram. What is the ratio A:B in which the
two candies are mixed?
A) 3:5
B) 5:3
C) 3:8
D) 5:8
E) 3:4

Cost of A = 10.2.
Cost of B = 14.
Cost of the MIXTURE of A and B = 11.625.

The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 prices on a number line, with the prices for A and B on the ends and the price for the mixture in the middle.
A 10.2--------------11.625--------------14 B

Step 2: Calculate the distances between the prices.
A 10.2----1.425-----11.625----2.375-----14 B

Step 3: Determine the ratio in the mixture.
The required rate of A to B is equal to the RECIPROCAL of the distances in red.
A:B = 2.375/1.425 = 5/3.

The correct answer is B.

As David notes above, since the cost of the mixture (11.625) is closer to A's cost (10.2) than to B's cost (14), the mixture must contain more A than B.
Only B offers a ratio with more A than B.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html.

Thanks to all of you, I especially like the answer of David, this simple idea will certainly help me for further problems 😄