The sum of the digits of three digit number is 20 middle NUMBER is equal to one half four times of sum of its extreme digits if the digits are reserved the new number is 198 more than the old number. find original numbers.
pls answer correctly, don't be greedy for points
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Answers
Solution :-
Let us assume that, the original three digit number is 100x + 10y + z .
So,
→ New number after digits are reversed = 100z + 10y + x .
A/q,
→ (100z + 10y + x) - (100x + 10y + z) = 198
→ 100z - z + 10y - 10y + x - 100x = 198
→ 99z - 99x = 198
→ 99(z - x) = 198
→ (z - x) = 2 ----------- Eqn.(1)
also,
→ x + y + z = 20 ------- Eqn.(2)
and,
→ y = (1/4)(x + z)
→ 4y = x + z ------------ Eqn.(3)
putting value of Eqn.(3) in Eqn.(2),
→ (x + z) + y = 20
→ 4y + y = 20
→ 5y = 20
→ y = 4
putting value of y in Eqn.(3) now,
→ x + z = 4 * 4
→ x + z = 16 --------- Eqn.(4)
adding Eqn.(4) and Eqn.(1) now,
→ (x + z) + (z - x) = 16 + 2
→ x - x + z + z = 18
→ 2z = 18
→ z = 9
putting value of z in Eqn.(1) now,
→ 9 - x = 2
→ x = 9 - 2
→ x = 7
therefore,
→ Original number = 100x + 10y + z = 100 * 7 + 10 * 4 + z = 700 + 40 + 9 = 749 (Ans.)
Hence, Original number is equal to 749 .
Verification :-
→ 9 + 4 + 7 = 20
→ 947 - 749 = 198
→ 4 * 4 = 9 + 7
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Answer:
the correct answer -96.5 ans.
but I've still confusion on 198