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Standard IX

Mathematics

Direct Proportionality

Question 5 1 ...

Question

Question 51

If x and y are in inverse proportion, then (x + 1) and (y + 1) are also in inverse proportion.

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Solution

False
If x and y are inverse proportion, then xy =k (constant)
e.g. Let x = 2 and y = 3
$∴ x y = 2 \times 3 = 6$ , Now, x + 1 = 2 + 1 = 3 and y + 1 = 3 + 1 = 4
Then, (x + 1) (y + 1) = $3 \times 4 = 12$ [not in inverse proportion]
Hence, (x + 1) and (y + 1) cannot be in inverse proportion.

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Similar questions

Question 51

If x and y are in inverse proportion, then (x + 1) and (y + 1) are also in inverse proportion.

Question 10

Both x and y are in direct proportion, then $\frac{1}{x}$ and $\frac{1}{y}$ are
(a) In indirect proportion
(b) In inverse proportion
(c) Neither in direct in inverse proportion
(d) Sometimes in direct and sometimes in inverse proportion

Question 74

Find the values of x and y if a and b are in inverse proportion

Q. Find inverse proportionaly constant if

x

and

y

are in inverse proportion.

$x$	$9$	$6$	$3$	$18$
$y$	$2$	$3$	$6$	$1$

Q. If

x

y

are in inverse proportion then find proportionality constant at

x = 3, y = 2

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Direct Proportionality

Standard IX Mathematics

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Question 51 If x and y are in inverse proportion, then (x + 1) and (y + 1) are also in inverse proportion.

False If x and y are inverse proportion, then xy =k (constant) e.g. Let x = 2 and y = 3 ∴xy=2×3=6, Now, x + 1 = 2 + 1 = 3 and y + 1 = 3 + 1 = 4 Then, (x + 1) (y + 1) = 3×4=12 [not in inverse proportion] Hence, (x + 1) and (y + 1) cannot be in inverse proportion.

Question 51

If x and y are in inverse proportion, then (x + 1) and (y + 1) are also in inverse proportion.

False
If x and y are inverse proportion, then xy =k (constant)
e.g. Let x = 2 and y = 3
$∴ x y = 2 \times 3 = 6$ , Now, x + 1 = 2 + 1 = 3 and y + 1 = 3 + 1 = 4
Then, (x + 1) (y + 1) = $3 \times 4 = 12$ [not in inverse proportion]
Hence, (x + 1) and (y + 1) cannot be in inverse proportion.