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A number, p, varies directly as q and partly inversely as q
{}^(2). Given that p = 11 and q = 7 and p = 25.16 when q = 5, calculate the value of p when q = 7​​

1 Answer

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Since it varies directly as q and partly inversely as q^2 it is represented has this below

p = aq + b/q^2

The constants of proportionality here are a and b

11 = 2a + b/4 ----- i x 5

25.16 = 5a + b/25 ------ ii x 2

Multiply equation i by 5 and equation ii by 2

55 = 10a + 5b/4 ----- iii

50.32 = 10a + 2b/25 ----- iv

Subtract equation iv from equation iii

b = 4

Put the value of b in equation iv

50.32 = 10a + 8/25

Multiply through by 25

1258 = 250a + 8

a = 5

p =5q + 4/q^2

Therefore using this equation put into the equation the value of q which is 7

p = 5(7) + 4/(7)^2

p = 35 +4/49

p = 35.082

hope this helped.

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