math problem about how many bulls eyes he made
there is a dart board with 50 darts and he 152 points every bulls eye = 7 pts and if not a bulls eye its 2 pts how many bulls eyes did he make
2 Answers
The wording on the question is a little confusing, but I think I get what you're saying. Someone has thrown 50 darts at a dart board where a bullseye is worth 7 points and anything else is worth 2 points. The person has 152 points, and the question is to find out how many bullseyes were made.
I think the numbers here are a bit off, which might be why you're having trouble. I'll walk through how I would solve the problem using equations and explain.
There are 2 types of results from throwing a dart, bullseye or not. We don't know how many of either, so we'll let b=number of bullseyes and n = number of non-bullseyes.
Since there are 50 darts, we know that: b + n = 50
The total number of points is 152: bullseye points + non-bullseye points. All the bullseye points would be 7*b (7b) and non-bullseye points would be 2*n (2n). So we could also write 7b + 2n = 152
Now we have the system of equations
b + n = 50
7b + 2n = 152
I'm going to use substitution to solve this problem. Look at the first equation: if we subtract b from both sides, we see that n = 50-b. Substitute (50-b) in place of n in the 2nd equation.
7b + 2(50-b) = 152
7b + 100 - 2b = 152 (distribute)
5b +100 = 152 (combine like terms)
5b = 52 (subtract 100 from both sides)
b = 10.4 (divide both sides by 5)
Does this answer make sense? Can you have 10.4 bullseyes? Nope! Now, if there were 51 darts thrown instead, everything works out rather nicely.
Let x=# bullseyes
y=# 2-point shots
z=# missed shots
Solve so that both of the following equations are TRUE
#1 x + y + z = 50 where x, y, z are positive integers
#2 7x + 2y + 0z = 152 where x, y, z are positive integers
What is needed are all points (x,y,z) that meet the above criteria.
Maximum values:
x: 7x <= 152 where x is a positive EVEN number
x <= 21 why? because of the property of integers; an odd + even integer = odd
And odd x odd = odd, but 7x must be even in order for 7x+2y+0z = 152.
y: 2y <= 152 where y is a positive number
y <= 76 but there’s an upward limit of 50!
y <= 50 because of #1 equation x + y + z = 50
Now let’s remove z as a variable from #1 and #2 above.
#1a x + y <= 50
#2a 7x + 2y = 152 (subtract 1a from 2a)
------------------
6x + y <= 102
y <= -6x + 102
Substitute into #1a
x + -6x + 102 <= 50
52 <= 5x
10.2 <= x <= 21 where x is a positive EVEN number (remember max above?)
valid values of x are {12, 14, 16, 18, 20}
Solve #2a for y.
7x + 2y = 152
2y = 152 + -7x
y = 76 + 7x/2
now find y as f(x) = -76 + 7x/2 for all valid values of x
f(12) = -76 + 42 = 34
f(14) = 27
f(16) = 20
f(18) = 13
f(20) = 6
valid values of y are {34, 27, 20, 13, 6}
Now find z as f(x,y) = 50 – x – y (#1 solved for z)
f(12,34) = 4
f(14,27) = 9
f(16,20) = 14
f(18,13) = 19
f(20,6) = 24
So there are 5 answers:
(12,34,4) 12 bullseyes, 34 2-pointers, 4 misses
(14,27,9) 14 bullseyes, 27 2-pointers, 9 misses
(16,20,14) 16 bullseyes, 20 2-pointers, 14 misses
(18,13,19) 18 bullseyes, 13 2-pointers, 19 misses
(20,6,24) 20 bullseyes, 6 2-pointers, 24 misses






