country A and B joined together to defend the attack by country A, each soldier of B fired 8 bullets while each soldier of C fired 4 bullets if total number of bullets fired is less by 100 compared to eight times that of the total number of soldiers of B and C, then how many soldiers from C were there on the front of this joint defence?
Answers
Answered by
3
Answer:
8b + 4c = 8 (b+c) - 100
8b + 4c = 8b + 8c -100
-4c = -100
c = 25
Answered by
4
Solution :-
Let us assume that,
- Country A has total soldiers = a
- Country B has total soldiers = b
- Country C has total soldiers = c
given that,
- Each soldier of B fired = 8 bullets
- Each soldier of C fired = 4 bullets
then,
→ Total bullets fired by (B + C) = 8b + 4c .
now, given that,
- Total bullets fired by (B + C) is 100 less than 8 times of the total number of soldiers of B and C .
therefore,
→ 8b + 4c = 8(b + c) - 100
→ 8b + 4c = 8b + 8c - 100
→ 8b - 8b + 8c - 4c = 100
→ 4c = 100
→ c = 25 soldiers. (Ans.)
Hence, 25 soldiers from C were there on the front of this joint defence .
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