MATHEMATICS
JAMB 2015 - Question 5
Mathematics 2015 JAMB Past Questions - Question 5: x varies directly as y² and x=4,when y=6.find the value of y when x =16
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Correct Answer
A
Explanation
xᾳ y² ,x=ky²4=k(6)² =4<= 36 k=k=1/9 , x = ky² = x 1/9 y² = 16 = y²/9 y = sqr root of 9 X 16 = y =12To solve this problem, we can use the direct variation formula, which states that when two variables, x and y, vary directly, their relationship can be expressed as x = ky², where k is the constant of variation.Given that x varies directly as y² and x = 4 when y = 6, we can find the value of k using the given values:4 = k * 6²4 = 36kk = 4/36k = 1/9Now that we have the value of k, we can use it to find the value of y when x = 16:16 = (1/9)y²y² = 16 * 9y² = 144y = ±√144y = ±12So, when x = 16, the value of y can be either 12 or -12.
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