Question

The digits in the tens place of 2 digit number is three times that in unit place.if the digits are reversed the new number will 36 less than the original number. Find the number

Answer 1

Given: The digit at tens place is 3 times of digit at ones place.

Let, ten's digit be x and ones be y then x = 3y.

Given: If digits are reversed then new number will be 36 less than orginal one.

Let Original number be 10x + y.

→ 10y + x = 10x + y - 36

→ 36 = 9x - 9y

→ 4 = x - y

Since, x = 3y therefore

→ 4 = 3y - y → 4 = 2y

→ y = 2, hence x = 3(2) = 6

Hence the number is 62. [10(6) + 2 = 62]

Answer 2

• Let ten's digit be M

and

• One's digit be N.

• Original number = 10M + N

» The digits in the tens place of 2 digit number is three times that in unit place.

M = 3N _____ (eq 1)

So, now..

• Ten's digit = N

• One's digit = 3M

» If the digits are reversed the new number will 36 less than the original number.

• Revered number = 10N + M

A.T.Q.

→ 10N + M = 10M + N - 36

→ 10N - N + M - 10M = - 36

→ 9N - 9M = - 36

→ N - M = - 4 _______ (eq 2)

→ N - (3N) = - 4 [From (eq 1)]

→ - 2N = - 4

→ N = 2

Put value of N in (eq 1)

→ M = 3(2)

→ M = 6

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Original number = 10M + N

From above calculations we have M = 6 and N = 2

Put them above;

=> 10(6) + 2

=> 62

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Number = 62

_______ [ ANSWER ]

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