in a cricket match bye runs were double of the no-balls.the remaining score was greater than twelve times the number of the bye runs. if the total was 276, how were the runs obtained?

Question

about cricket

Answer 1

 

No balls----------x 
byes = 2x
total runs = 276 
remaining runs = (276-3x)=12*2x+6 
 

276-3x=24x+6 
27x=270 
x=10 
No ball
s = 10
Byes = 20
Runs = 246

 

https://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.741217.html
 

Answer 2

A no-ball gives 1 run to the batting side. Since a bye is double this, the total of byes and no-balls=3N where N is the number of no-balls. The score remaining is 276-3N which is greater than 12 times the number of byes=24N, so 276-3N>24N and 27N<276 or, dividing through by 3, 9N<92 and N<92/9, N<10 2/9. This implies N≤10, since fractional scores are not possible. It also implies the number of byes is ≤20.

If N=10 then the batsmen scored at least 6 runs by striking the ball thrown by the bowler, because 276-30=246, and 246-240=6.

If N=9 then we have 276-27=249 which is 249-216=33 runs normally.

If N=8 then we have 276-24=252 which is 252-192=60 runs normally.

So the number of runs scored normally=276-27N=3(92-9N).