The sum of the digits of a number consisting of three digits is 20. the middle digit is equal to 1/4th sum of the other two. if the order of the digits be reversed, the number is increased by 198. Find the original number.
Answers
You the sum it up to 20 .....given
We then get y=4
We are then given that the reverse of the digits is equal to the original no. Plus 198 the solve as solved in the pic
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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