A competition requires teams to attempt 10 long jumps and 5 high jumps

A competition requires teams to attempt 10 long jumps and 5 high jumps. Each successful long jump is worth the same number of points, and each successful high jump is worth the same number of points. Team A completed all 15 jumps, receiving 20 points. Team B succeeded in only 8 long jumps and 2 high jumps, earning 10 points. Team C missed 2 long jumps and successfully completed all the high jumps. How many points did Team C score?
10L+5H=20
8L+2H=10
8L+5H=?
solve by substitution we get
L=1/2
H=3
therefore, 8(1/2)+5(3)=19
E

Let's x be points earned from each long jump, and y be points earned from each high jump.

Team A: completed all 15 jumps and receiving 20 points -- 10x + 5y = 20 (Equation 1)
Team B: completed 8 long jumps, 2 high jumps -- 8x + 2y = 10 (Equation 2)

Solve the equation: x = 0.5, y = 3

Finding Team C points: completed 8 long jumps, 5 high jumps -- 8x + 5y = 8(0.5) + 5(3) = 19 points. -> Answer E

Total number of jumps in competition = 10 + 5 = 15 where n(L) = 10 and n(H) = 5.

To find = Total points of Team C.

Given that there are three teams – Team A, Team B and Team C. Let L and H be the points per jumps or Long jumps and High jumps respectively.

Team A: \(n(L) = 10\) & \(n(H) = 5\) and Points = 20
 \(10L + 5H = 20\)
 \(2L + H = 4\) Eqn. ①

Team B: \(n(L) = 8\) & \(n(H) = 2\) and Points = 10
 \(8L + 2H = 10\)
 \(4L + H = 5\) Eqn. ②

Team C: \(n(L) = 8\) & \(n(H) = 5\) and Points = ?
 Points = \(8L + 5H\)

Since we have two variables and two equations we can solve for the variables.

\(Eqn. ② - Eqn. gives 4L - 2L + H – H = 5 – 4\)
 \(2L = 1\)
 \(L = ½\)

Solving for H either equation gives \(H = 3\)

Thus, \(8L + 5H = 8 * \frac{1}{2} + 5 * 3\)
 \(8L + 5H = 19\)

Answer (E).

[Q] 8x+5y=?
[1] 10x+5y=20
[2] 8x+2y=10
[3:1-2] 2x+3y=10
[3-2] 6x-y=0…6x=y
[2] 8x+2y=10…8x+2(6x)=10…20x=10…x=0.5
[Q] 8x+5y=…8x+5(6x)=…38x=…38(0.5)=19

Answer (E)

IMO it's E, 19
We can make two equations by the question stem:
10L+5H=20
8L+2H=10
Solving these 2 we will get, H=3, L=0.5
Team C completed 8L and 5H which will total make 4+15=19 points.

A competition requires teams to attempt 10 long jumps and 5 high jumps. Each successful long jump is worth the same number of points, and each successful high jump is worth the same number of points. Team A completed all 15 jumps, receiving 20 points. Team B succeeded in only 8 long jumps and 2 high jumps, earning 10 points. Team C missed 2 long jumps and successfully completed all the high jumps. How many points did Team C score?

(A) 12
(B) 13
(C) 15
(D) 18
(E) 19

We have two equations and two variables, looking to solve for the 2 variables. The first equation is \(10L + 5H = 20\), second equation is \(8L + 2H = 10\). We can multiply the second equation by 2.5 so that we can eliminate the variable \(H\), this gives \(10L = 5\) and \(L = 0.5\), thus \(H = 3\). The answer is \(8L + 5H = 4 + 15 = 19\), E.
It would be good practice to try solving this in your head, the key here is to find a nice variable to eliminate and here we multiply the second equation by 2.5 to match the \(5H\) in the first equation.

given
h+l=15
h=10 and l=5
also
10x+5y=20
8x+2y=10
find
8x+5y
solve for x& y ; x= 0.5 and y=3
so 8x+5y= 19
IMO E


A competition requires teams to attempt 10 long jumps and 5 high jumps. Each successful long jump is worth the same number of points, and each successful high jump is worth the same number of points. Team A completed all 15 jumps, receiving 20 points. Team B succeeded in only 8 long jumps and 2 high jumps, earning 10 points. Team C missed 2 long jumps and successfully completed all the high jumps. How many points did Team C score?

(A) 12
(B) 13
(C) 15
(D) 18
(E) 19