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put this solution on YOUR website!Let Ann's age be A and Bill's age be B
ann's age is seven-tenths of Bill's age.
A = 7/10(B) ....eq1
In 4 years Ann's age will be eight-elevenths of Bill's age.
A + 4 = 8/11(B+4).....eq2
Substitute eq1 in eq2
(7/10(B)) + 4 = 8/11(B+4)
Cross multiply
11(7B + 40) = 80(B + 4)
77B + 440 = 80B + 320
-3B = -120
B=40
Bill's age is 40
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Let the ten digit be T and units digit U
The number will be 10T + U
The Ten's digit of a two-digit number is 3 more than the unit's digit.
T = 3 + U .....eq1
The number is 8 more than six times the sum of the digits.
10T + U = 8 + 6(T + U)
10T + U = 8 + 6T + 6U
10T + U - 6T - 6U = 8
4T - 5U = 8
SUBSTITUTE T = 3 + U
4(3 + U) - 5U = 8
12 + 4U -5U =8
-U = -4
U = 4( UNIT'S DIGIT IS 4)
Since Ten's digit is 3 more than the unit's digit, it 7
Number is 74
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