In which of the following tables x and y vary inversely:
(i)

x 4 3 12 1

y 6 8 2 24

(ii)

x 5 20 10 4

y 20 5 10 25

(iii)

x 4 3 6 1

y 9 12 8 36

(iv)

x 9 24 15 3

y 8 3 4 25

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Solution

$(i) Since x and y var y inversely, we have : y = \frac{k}{x} \Rightarrow x y = k ∴ The product of x and y is constant . In all cases, the product x y is constant (i . e ., 24) . Thus, in this case, x and y var y inversely . (ii) In all ca s es, the product x y is constant for any two pairs of values for x and y . Here, x y = 100 for all cases Thus, in this case, x and y var y inversely . (iii) If x and y var y inversely, the product x y should be constant . Here, in one case, product = 6 \times 8 = 48 and in the rest, product = 36 Thus, in this case, x and y do not var y inversely . (iv) If x and y var y inversely, the produc t x y should be constant . Here, product is different for all cases . Thus, in this case, x and y do not var y inversely .$

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Similar questions

&Complete the following tables given that $x$ varies directly as $y$

(i)
$\begin{matrix} x & 2.5 & . . . . & . . . . & 15 y & 5 & 8 & 12 & . . . . \end{matrix}$

(ii)
$\begin{matrix} x & 5 & . . . . & 10 & 35 & 25 & . . . y & 8 & 12 & . . . & . . . . & . . . . & 32 \end{matrix}$

(iii)
$\begin{matrix} x & 6 & 8 & 10 & . . . . & 20 y & 15 & 20 & . . . & 40 & . . . . \end{matrix}$

(iv)
$\begin{matrix} x & 4 & 9 & . . . & . . . . & 3 & . . . . y & 16 & . . . & 48 & 36 & . . . . & 4 \end{matrix}$

(v)
$\begin{matrix} x & 3 & 5 & 7 & 9 y & . . . & 20 & 28 & . . . . \end{matrix}$

x	6	10	14	16
y	9	15	21	24

x	5	9	15	3	45
y	18	10	6	30	2

x	9	3	6	36
y	4	12	9	1

x	12	16	...	8	...
y	...	6	4	...	0.25

x	16	32	8	128
y	4	...	...	0.25

x	9	...	81	243
y	27	9	...	1

x	3	5	8	11	26
y	9	15	24	33	78

x	2.5	4	7.5	10	14
y	10	16	30	40	42

x	5	7	9	15	18	25
y	15	21	27	60	72	75

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