Question

In a football game, Team A defeated Team B by a score of 51 to 44. The total points scored came from 28 different scoring plays, which were a combination of touchdowns, extra-point kicks, field goals, and safeties, worth 6, 1, 3, and 2 points, respectively. There were four times as many touchdowns as field goals, and the number of extra-point kicks was equal to the number of touchdowns. How many touchdowns, extra-point kicks, field goals, and safeties were scored during the game?

Answer 1

Solution-

Let the number of touchdowns, extra-point kicks, field goals, and safeties are a, b, c and d respectively .

Now according to question

Since total number of scoring plays are 28.S O,

a+b+c+d=28 ..........1

Since these were a combination of touchdowns, extra-point kicks, field goals, and safeties, worth 6, 1, 3, and 2 points, respectively and total points are 51 + 44=95.S o,

6a + b +3c + 2d= 95 ..........2

Since there were four times as many touchdowns as field goals.So,

a=4c .........3

Since the number of extra-point kicks was equal to the number of touchdowns. So,

a=b ..........4

Putting a=b in equation 3, we get

b=4c ...........5

Putting the values of a =4c and b=4c (equation 4 and 5) in equations 1 and 2, we get,

4c+4c+c+d =28

Or 9c+d=28 ...........6

And

6(4c)+(4c)+3c +2d=95

Or 24c+4c+3c+2d=95

Or 31c+2d=95 ..........7

Subtracting 2 times of equation 6 from equation 7, we get

(31c+2d)–2(9c+d)=95 -2(28)

31c +2d - 18c -2d=95 - 56

13c= 39

Or c=39/13

Or c=3 ...............8

Putting the value of c in equations 3, 5 and 6, to get of a, b, and d.

a=4c=4(3)=12 ...........9

b=4c=4(3)=12 ............10

9(3)+d=28

27 +d=28

Or d=28–27

Or d=1 ..........11

From equations 9, 10,8 and 11 ,we get

a=12, b=12 , c=3 and d=1

Hence, the required number of touchdowns, extra-point kicks, field goals, and safeties were scored during the game are 12, 12 , 3 and 1 respectively.