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In Mathematics / High School | 2025-07-08

One cylinder has a volume that is $8 cm^3$ less than $\frac{7}{8}$ of the volume of a second cylinder. If the first cylinder's volume is $216 cm^3$, what is the correct equation and value of $x$, the volume of the second cylinder?

A. $\frac{7}{8} x+8=216 ; x=182 cm^3$
B. $\frac{7}{8} x-8=216 ; x=196 cm^3$
C. $\frac{7}{8} x+8=216 ; x=238 cm^3$
D. $\frac{7}{8} x-8=216 ; x=256 cm^3

Asked by Igfennikus

Answer (1)

Set up the equation representing the relationship between the volumes of the two cylinders: 216 = 8 7 ​ x − 8 .
Add 8 to both sides of the equation: 224 = 8 7 ​ x .
Multiply both sides by 7 8 ​ to solve for x : x = 224 × 7 8 ​ .
Calculate the value of x : x = 256 c m 3 ​ .

Explanation

Problem Analysis Let's analyze the problem. We are given that the volume of the first cylinder is 8 c m 3 less than 8 7 ​ of the volume of the second cylinder. We are also given that the volume of the first cylinder is 216 c m 3 . We need to find the correct equation and the volume of the second cylinder, which we'll call x .

Setting up the Equation We can express the relationship between the volumes of the two cylinders as an equation. The volume of the first cylinder is equal to 8 7 ​ of the volume of the second cylinder minus 8. So, we can write the equation as: 216 = 8 7 ​ x − 8

Isolating the Term with x Now, let's solve the equation for x . First, we add 8 to both sides of the equation: 216 + 8 = 8 7 ​ x 224 = 8 7 ​ x

Solving for x Next, we multiply both sides of the equation by 7 8 ​ to isolate x :
x = 224 × 7 8 ​ x = 7 224 × 8 ​ x = 7 1792 ​ x = 256

Finding the Solution Therefore, the volume of the second cylinder is 256 c m 3 . The correct equation is 8 7 ​ x − 8 = 216 , and the value of x is 256.


Examples
Imagine you're comparing the sizes of two water tanks. One tank's capacity is related to the other, but with a known difference and a fractional relationship. This problem helps determine the capacity of the second tank based on the first. Understanding such relationships is useful in various real-world scenarios, such as comparing storage capacities, calculating material requirements, or analyzing proportional changes in quantities. This algebraic approach provides a clear method for solving similar comparative problems.

Answered by GinnyAnswer | 2025-07-08