In world cup 2019 a league match was played between india and Australia sum of 1/3 part of india's total score and 1/7 part of score of Australia's total score is 137 runs. If indian team won the match with minimum required run in front of australia then find the score of indian team
Answers
Step-by-step explanation:
Given In world cup 2019 a league match was played between India and Australia sum of 1/3 part of India's total score and 1/7 part of score of Australia's total score is 137 runs. If Indian team won the match with minimum required run in front of Australia then find the score of Indian team
- Let the Australia team be x score and score of India be (x + 1)
- Given 1/3 of Indian + 1/7 of Australian = 137
- So x + 1 / 3 + x/7 = 137
- So 7x + 7 + 3x / 21 = 137
- So 10 x + 7 / 21 = 137
- 10 x + 7 = 137 x 21
- 10 x + 7 = 2877
- 10 x = 2,870
- Or x = 287
- So Australian score is 287 and Indian score will be 287 + 1 = 288
Reference link will be
https://brainly.in/question/15364113
Step-by-step explanation:
Let the runs scored by India be 'x' and the runs scored by Srilanka be 'y'.
One third runs scored by Indian team =
3x .
One seventh runs scored by Indian team =
7y
∴ 3x+ 7y =137
∴ 217x+3y=137
∴7x+3y=2877 ....... (i)
Also, India won this match by minimum required runs, hence the difference between scores of India and Srilanka will be 1.
∴x−y=1 ....... (ii)
Multiplying (ii) by 3 and (i) by 1,
∴7x+3y=2877
∴3x−3y=3
Adding both the equations,
∴10x=2880
∴x=288
Substituting x=288 in (ii)
∴y=287
Runs scored by the Indian team =288