A civil engineering laboratory prepares a composite soil mixture using three material grad… — GRE General Test

A civil engineering laboratory prepares a composite soil mixture using three material grades: Grade $X$ , Grade $Y$ , and Grade $Z$ . Initially, the ratio of Grade $X$ to Grade $Y$ by weight is $1 : 2$ , and the ratio of Grade $Y$ to Grade $Z$ by weight is $3 : 4$ . A technician adds $30\text{ kg}$ of Grade $X$ to the batch, causing Grade $X$ to constitute exactly $\frac{1}{4}$ of the total weight of the new mixture. Which of the following statements about the batch must be true? Select all such statements.

The initial total weight of the mixture before Grade $X$ was added was $306\text{ kg}$ .Cevap
The weight of Grade $Z$ in the mixture is $144\text{ kg}$ .Cevap
In the final mixture, the ratio of Grade $Y$ to Grade $Z$ by weight is $3 : 4$ .Cevap
D
The initial weight of Grade $X$ in the mixture was $72\text{ kg}$ .
E
Grade $Y$ accounts for $35\%$ of the total weight of the final mixture.

Cevap

The correct statements are those indicating that the initial total weight of the mixture was 306 kg, the weight of Grade Z in the mixture is 144 kg, and the ratio of Grade Y to Grade Z in the final mixture is 3 : 4.

The unified ratio of X : Y : Z is 3 : 6 : 8. Setting up the fraction of Grade X after adding 30 kg gives (3k + 30) / (17k + 30) = 1/4, which yields k = 18. This gives an initial total weight of 17 * 18 = 306 kg, a Grade Z weight of 8 * 18 = 144 kg, and an unchanged Grade Y to Grade Z ratio of 108 : 144 = 3 : 4. Therefore, the statements asserting the initial total weight as 306 kg, the Grade Z weight as 144 kg, and the final Y to Z ratio as 3 : 4 are all correct.

Adım Adım Çözüm

Unify the two given ratio relationships into a single 3-part ratio.

Grade X : Grade Y = 1 : 2 = 3 : 6, and Grade Y : Grade Z = 3 : 4 = 6 : 8. Unified ratio Grade X : Grade Y : Grade Z = 3 : 6 : 8.

Matching the term for Grade Y across both ratios enables setting up unified algebraic variable parts.

Define initial quantities using a common multiplier k.

Initial weight of X = 3k, Y = 6k, Z = 8k. Initial total weight = 3k + 6k + 8k = 17k.

Expressing quantities in terms of k allows setting up an equation after adding material.

Formulate and solve the equation based on the addition of Grade X.

(3k + 30) / (17k + 30) = 1/4 => 4(3k + 30) = 17k + 30 => 12k + 120 = 17k + 30 => 5k = 90 => k = 18.

Setting the new weight of X over the new total weight equal to 1/4 determines the exact value of k.

Evaluate the individual component weights and statements.

Initial total weight = 17(18) = 306 kg. Initial X = 3(18) = 54 kg. Weight of Y = 6(18) = 108 kg. Weight of Z = 8(18) = 144 kg. Final total weight = 306 + 30 = 336 kg. Final Y : Z ratio = 108 : 144 = 3 : 4.

Determining all numerical values allows verifying which statements are true.

Anahtar Kavram

Combining three-variable ratio streams into a unified ratio and solving linear rate/proportion equations upon single-component addition.

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