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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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.