In Fig. 60, the values of x and y are
(a) x = 20, y = 130
(c) x = 20, y = 140
(b) x = 40, y = 140
(d) x = 15, y = 140

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Solution

∠ACB + ∠ACD = 180°
⇒ 40° + y° = 180°
⇒ y° = 140°
⇒ y = 140
Now, ∠ACD = ∠ABC + ∠BAC [Exterior angle property]
⇒ 3x° + 4x° = y°
⇒ 7x° = 140°
⇒ x = 20
Hence, the correct answer is option (c).

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The values of $x$ and $y$ are
(a) $x = 130, y = 120$
(b) $x = 120, y = 130$
(c) $x = 120, y = 120$
(d) $x = 130, y = 130$

x^{2} : y^{2} = 1

(y - x) (x + y) = 40 % of 60 - 120 % of 20

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∠ A D B = ∠ A D C = 90^{\circ} .

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Class-interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	Total
Frequency	$17$	$x$	$32$	$y$	$19$	$120$

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