THE DIGIT IN THE TEN'S PLACE OF A TWO DIGIT NUMBER IS THREE TIMES THAT IN ONE'S PLACE.IF THE DIGITS ARE REVERSED,THE NEW NUMBER WILL BE 36 LESS THAN THE ORIGINAL NUMBER ,FIND THE NUMBER?
Answers
Answer:
Let the digit in ones place be a
Thus the digit in tens place will be 3a
given in question
=((3a x 10) + a) - ((a x 10)+3a) = 36
=(30a + a) - (10a + 3a) = 36
=31a - 13a = 36
=18a = 36
=a = 36/18
=a = 2
Therefore the number = ((3a x 10) +a)= (( 3 x 2 x 10) +2)
=((6 x 10) +2) = 60 + 2= 62
number is 62
Step-by-step explanation:
2 digit number can be written as
62 = ((6 x 10) +2)
hope i helped you
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Question :
The digits in the tens place of a two digit number is three times that in the one's place. If the digits are reversed the new number is 36 less than the original number. What is the original number?
Solution :
Let ten's digit be M and one's digit be N.
Original number (digit) = 10M + N
According to question,
=> M = 3N _________ (eq 1)
Revered number = 10N + M
So,
=> 10N + M = 10M + N - 36
=> 10N - N + M - 10M = - 36
=> 9N - 9M = - 36
=> N - M = - 4
=> N - (3N) = - 4 [From (eq 1)]
=> N - 3N = - 4
=> - 2N = - 4
=> N = 2 (one's digit)
Put value of N in (eq 1)
=> M = 3(2)
=> M = 6 (ten's digit)
Original number = 10M + N
=> 10(6) + 2
=> 60 + 2
=> 62
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Original number = 62.
____________ [ ANSWER ]
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☆ VERIFICATION :
From above calculations we have M = 6 and N = 2
Put value of M and N in (eq 1)
=> M = 3N
=> 6 = 3(2)
=> 6 = 6
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