After being dropped a certain ball always bounces back to 2/5 of the height of its previuos bounce. After the first bounce it reaches a height of 125 inches. How high (in inches) will it reach after its fourth?
Created: 5 years ago |
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According to the question,
⇒(x + y + z)/3 = 45
⇒x + y + z = 135.....(i)
⇒x = (y + z)/2 + 9
⇒2x−y−z = 18.....(ii)
x + y + z = 135
2x−y−z = 18
−−−−−−−−−−−− 3x = 153
x = 51
From (i)
y + z = 135 - 51 = 84.....(iii)
Also,
⇒(y + z)/2 = y + 2
⇒y + z = 2y + 4
⇒z−y = 4
+ y + z = 84
−y + z = 4
−−−−−−−−−
2z = 88
z = 44
Required difference = 51 - 44 = 7
⇒(x + y + z)/3 = 45
⇒x + y + z = 135.....(i)
⇒x = (y + z)/2 + 9
⇒2x−y−z = 18.....(ii)
x + y + z = 135
2x−y−z = 18
−−−−−−−−−−−− 3x = 153
x = 51
From (i)
y + z = 135 - 51 = 84.....(iii)
Also,
⇒(y + z)/2 = y + 2
⇒y + z = 2y + 4
⇒z−y = 4
+ y + z = 84
−y + z = 4
−−−−−−−−−
2z = 88
z = 44
Required difference = 51 - 44 = 7
Md Azizur Rahman
Founder of Satt
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