a number consist of two digits such that the digit in ten's place is less by 2 than the digit in the unit place. three times the number added to 6/7 times the number obtained by reversing the digits equal 108. the sum of digits of number is:
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Answered by
34
let the digit in unit's place be x
digit in ten's place = x - 2
∴no. = 10(x-2) + x=10x-20+x=11x - 20
∴3(11x - 20) + {6/7×(11x - 2)}=108
or, 33x - 60 +(66x - 12)/7 = 108
or,231x- 420 +66x - 12 = 108 × 7
or,297x - 432 = 756
or, 297x = 756+432=1188
or, x=1188/297
or, x=4
∴sum of the digit of the no. = x-2+x = 2x-2 = 8 - 2 = 6
digit in ten's place = x - 2
∴no. = 10(x-2) + x=10x-20+x=11x - 20
∴3(11x - 20) + {6/7×(11x - 2)}=108
or, 33x - 60 +(66x - 12)/7 = 108
or,231x- 420 +66x - 12 = 108 × 7
or,297x - 432 = 756
or, 297x = 756+432=1188
or, x=1188/297
or, x=4
∴sum of the digit of the no. = x-2+x = 2x-2 = 8 - 2 = 6
Answered by
7
Step-by-step explanation:
33x-60 +6/7*10x+x-2
231x+60x+6x-420-12=756
297x=1188
x=4
x-2=2
4+2=6
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