in a 13 round tv match, two boxer a and b compete for gold medal. the boxer a delivers 8 punches in the first round and his punch delivery increases by 1 in each successive round. the boxer b delivers 4 punches in the first round and his punch delivery increases by 2 punch in each successive round. of 1 punch delivered equals 1 point, then who should win the gold medal
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This is a question of arithmetic progression.
n = 13
We need to find whose sum of terms of AP is greater.
For boxer A,
a = 8
d = 1
S = ½×n[2a+(n-1)d]
= ½×13(16+12×1)
= ½×13×28
= 182
For boxer B
a = 4
d = 2
S = ½×13(8+12×2)
= ½×13×32
= 208
Since number of punches of B are greater than those of A, B will have more points.
Thus, B will get the medal.
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Given : Total Number of Rounds = 13
Let us Calculate Total Number of Punches delivered by Boxer A in 13 Rounds
Given that Boxer A delivers 8 Punches in the 1st Round and his Punch Delivery increase by 1 in each Successive round.
⇒ 1st Round Punches of A = 8
⇒ 2nd Round Punches of A = 9
⇒ 3rd Round Punches of A = 10
We can notice that the Punches of A are in Arithmetic Progression with 1st Term (a) = 8 , Common Difference (d) = 1 and n = 13
We know that Sum of 'n' terms is Given by :
Total Number of Punches of A =
⇒
Number of Points of A = 182
Let us Calculate Total Number of Punches delivered by Boxer B in 13 Rounds
Given that Boxer B delivers 4 Punches in the 1st Round and his Punch Delivery increase by 2 in each Successive round.
⇒ 1st Round Punches of B = 4
⇒ 2nd Round Punches of B = 6
⇒ 3rd Round Punches of B = 8
We can notice that the Punches of B are in Arithmetic Progression with 1st Term (a) = 4 , Common Difference (d) = 2 and n = 13
Total Number of Punches of B =
⇒
Number of Points of B = 208
As, Points of B are Greater than A :
⇒ B is the Winner of the Gold Medal