can u solve this for me .. I need the reason why the equation formed too
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pariangel1:
let the tens digit be x and ones be (9-x). This way the no. is 10x+(9-x) and the other is 10(9-x)+x..... I hope you can solve Else
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i think you understand
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it is wrong. it will be 18......
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Let the tens digit be x.
The ones digit = (9-x) [as 9 is the sum]
The no. so would be = 10*x + (9- x)
[ ten times x will make it tens digit and add with the ones digit]
Reversing digits the number= 10(9-x) + x
A.T.Q.
9 {10x + (9-x)} = 2 {10(9-x) + x}
or, 9 {9x + 9} = 2 {90-10x+x}
or, 81x + 81 = 2 {90 - 9x}
or, 81x + 81 = 180 - 18x
or, 81x + 18x = 180 - 81
or, 99x = 99
or, x = 99/99
or, x = 1
So tens digit is = 1
ones digit = 9-1 = 8
18 is the number which satisfies the above conditions
The ones digit = (9-x) [as 9 is the sum]
The no. so would be = 10*x + (9- x)
[ ten times x will make it tens digit and add with the ones digit]
Reversing digits the number= 10(9-x) + x
A.T.Q.
9 {10x + (9-x)} = 2 {10(9-x) + x}
or, 9 {9x + 9} = 2 {90-10x+x}
or, 81x + 81 = 2 {90 - 9x}
or, 81x + 81 = 180 - 18x
or, 81x + 18x = 180 - 81
or, 99x = 99
or, x = 99/99
or, x = 1
So tens digit is = 1
ones digit = 9-1 = 8
18 is the number which satisfies the above conditions
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