The digits of a two-digit number differ by 3. If digits are interchanged and the resulting number is added to the original number, we get 121. Find the original number.
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let the digit at ones place be x and tens place be y
so the number will be 10y+x
x-y=3 ...(i)
by eq (i)... x=3+y ... (ii)
if the digits are interchanged, the number becomes 10x+y
so, (10x+y)+(10y+x) = 121
11x + 11y =121
takung 11 common, x+y=11 ...(iii)
putting value of x from eq (ii) in eq (iii)...
(3+y)+y=11
2y=11-3
2y=8
y=4
putting value of y in eq (ii)
x=3+y = 3+4 = 7
the no. is 10y+x = 10(4)+7 = 47
so the number is 47.
hope it helps... :) pls mark me as brainliest if it helps....
so the number will be 10y+x
x-y=3 ...(i)
by eq (i)... x=3+y ... (ii)
if the digits are interchanged, the number becomes 10x+y
so, (10x+y)+(10y+x) = 121
11x + 11y =121
takung 11 common, x+y=11 ...(iii)
putting value of x from eq (ii) in eq (iii)...
(3+y)+y=11
2y=11-3
2y=8
y=4
putting value of y in eq (ii)
x=3+y = 3+4 = 7
the no. is 10y+x = 10(4)+7 = 47
so the number is 47.
hope it helps... :) pls mark me as brainliest if it helps....
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