John and Joseph ran a 2km race twice. Joseph completed the first round 2minutes earlier than John. In the second round,John increased his speed by 2km/hr and Joseph decreased his speed by 2km/hr. John finished 2 minutes earlier than Joseph. Find their speeds of running the first round.
Answers
Answer:
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Step-by-step explanation:
Let Jonh run with a speed of x km/hr
Let joseph run with a speed of y km/hr
According to the 1
st
condition,
x
2
−
y
2
=
60
2
∴
x
1
−
y
1
=
60
1
∴
xy
y−x
=
60
1
∴60(y−x)=xy ...........(1)
According to the 2
nd
condition,
y−2
2
−
x+2
2
=
60
2
∴
y−2
1
−
x+2
1
=
60
1
∴
(y−2)(x+2)
x+2−y+2
=
60
1
∴
xy+2y−2x−4
x−y+4
=
60
1
∴60(4−y+x)=xy+2y−2x−4
∴240−60(y−x)=xy−4+2(y−x)
∴244=xy+2(y−x)+60(y−x)
∴244=xy+62(y−x)
∴244=60(y−x)+62(y−x) ........[from (1)]
∴244=122(y−x)
∴y−x=2 ............(2)
∴xy=60×2=120 .........(from 1)
Since, (y+x)
2
=(y−x)
2
+4xy
=2
2
+4×120=4+480=484
∴(y+x)
2
=(22)
2
∴y+x=22 ............(3)
Adding equation (2) and (3),
2y=24
∴y=12
Putting y=12 in equation (3),
12+x=22
∴x=10
Jonh ran with a speed of 10 km/hr and joseph ran with speed of 12 km/hr.