The sum of the digit in the units place and twice that in the ten's place of a two digit number is 15. When 41 is substracted from the number, the digits becomes equal. Find the number.
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Answered by
2
Answer:
Answer is 28
Step-by-step explanation:
Let the unit digital be A and tent's digit be B
A+B=10……1
A=4B
A-4B=0……..2
From “1” And “2”
5B=10
B=10/5=2
From “1”:
A+2=10
A=10–2=8
Number:
28
Answered by
25
ATQ
X+2Y=15
10Y+X-41= 10Z+Z=11Z
THEREFORE THE DIGIT IS 41 MORE THAN A MULTIPLE OF 11 AND SOME OF IT'S UNIT DIGIT AND TWICE THE TEN'S PLACE DIGIT IS =15
SO FROM THIS I CAN DEDUCE THAT
2(4+Z) +1(1+Z) =15
9+3Z=15
Z=6/3=2
SO THE NO. IS
41 +11*Z
41+22=63
Hope it Helps
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