Question

the digit at once place of a two digit number is 4 times the digit at 10th place the number obtained by reversing the digit exceeds the given number by 54 find the given number

Answer 1

Hi there !!
Here's your answer

Given,
the digit at once place of a two digit number is 4 times the digit at tenths place

So,
Let the digit in the tens place be x
Digit in ones place = 4x

The original number is
10(x) + 4x = 10 + 4x
= 14x ___(i)

By interchanging the digits,
we have,
digit in tens place = 4x
Digit in units place = x
The new number is 10(4x) + 4x = 40x + x
= 41x ____(ii)

Also,
41x(new number) exceeds 14x(original number) by 54

So,
the following balanced equation will be formed

41x = 14x + 54

41x - 14x = 54

27x = 54

x = 54/27

x = 2

Thus,
digit in tens place = x = 2
digit in units place = 4x = 4 × 2 = 8


Therefore,
the given number is 28


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Hope it helps !!

Answer 2

Hi!
Here is your answer !

Let the digit at tens place be x & digit at the ones place be y.

By the given condition.,

y = 4x

4x - y = 0 ..................................(1)

Number : 10x + y
Reversed number : 10y + x

By the 2nd given condition,

10y + x = 10x + y + 54

10y - y + x - 10x = 54

9y - 9x = 54

[Dividing throughout by 9 , we get]

y - x = 6

x - y = -6................................(2)

[Subtracting equation no. (2) from (1)]

4x - y = 0
x - y = -6
_________
3x = 6
x =6/3
x = 2

Now, Substitute x = 2 in equation no.1

4x - y = 0
4(2) - y = 0
8 - y = 0
y = 8

Therefore.,
-----------------------
x = 2. & y = 8.
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Required Number :
10x + y
=10(2) + (8)
=20 + 8
= 28
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●Final Answer :
Thus , the given number is 28.
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◆Hope this Helps ! ◆☺