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Heya Frnds.............☺
Plz give proper solution
Class X board..........^_^
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The speed of the peacock and snake is equal; they cover an equal distance in the same time limit.
Let the distance covered each of them be ‘x’ m.
Here, PS and RS represent the distance covered by the peacock and the snake respectively.
Triangle PQR forms a right angled triangle,
⇒ PQ²+ QS²= PS²
⇒ 9² + (27 – x)²= x²
⇒ 81 + (729 – 54x + x²) = x²
⇒ 54x = 810
⇒ x = 15
QS = (27 –x) m = (27 – 15) m = 12 m
Hence, the snake is caught at a distance of 12 m from the hole.
OR
LET THE ONE NO BE X AND ANOTHER BE
(X-4)
AND, 1/(X-4) - 1/X =4/21
=>[-X+4+X]/(X-4)X = 4/21
=>4/(X²-4X) = 4/21 [MULTIPLY BOTH SIDE BY 1/4]
=>WE GET,
=>21=X²-4X
=>X²-4X-21=0
=>X²-7X+3X-21=0
=>X(X-7)+3(X-7)=0
=>(X+3)(X-7)=0
=>X=7,& - 3
I HOPE YOU UNDERSTAND VERY WELL DEAR,
REGARDED BY :- ROHIT KUMAR GUPTA
THANKS :-)
The speed of the peacock and snake is equal; they cover an equal distance in the same time limit.
Let the distance covered each of them be ‘x’ m.
Here, PS and RS represent the distance covered by the peacock and the snake respectively.
Triangle PQR forms a right angled triangle,
⇒ PQ²+ QS²= PS²
⇒ 9² + (27 – x)²= x²
⇒ 81 + (729 – 54x + x²) = x²
⇒ 54x = 810
⇒ x = 15
QS = (27 –x) m = (27 – 15) m = 12 m
Hence, the snake is caught at a distance of 12 m from the hole.
OR
LET THE ONE NO BE X AND ANOTHER BE
(X-4)
AND, 1/(X-4) - 1/X =4/21
=>[-X+4+X]/(X-4)X = 4/21
=>4/(X²-4X) = 4/21 [MULTIPLY BOTH SIDE BY 1/4]
=>WE GET,
=>21=X²-4X
=>X²-4X-21=0
=>X²-7X+3X-21=0
=>X(X-7)+3(X-7)=0
=>(X+3)(X-7)=0
=>X=7,& - 3
I HOPE YOU UNDERSTAND VERY WELL DEAR,