Please solve the fourth question
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Let the digit at tens place be x and the digit at units place be y.
Therefore the decimal expansion is 10x + y. ------ (*)
Given that A two-digit number is 3 more than 4 times the sum of its digits.
= > 10x + y = 4(x + y) + 3
= > 10x + y = 4x + 4y + 3
= > 10x + y - 4x - 4x = 3
= > 6x - 3y = 3
= > 2x - y = 1 ------ (1)
Given that if 18 is added to the number, its digits are reversed.
= > 10x + y + 18 = 10y + x
= > 10y + x - 10x - y = 18
= > 9x - 9y = -18
= > x - y = -2 ------- (2)
On solving (1) & (2), we get
2x - y = 1
x - y = -2
-----------------
x = 3
Substitute x = 3 in (1), we get
= > x - y = -2
= > 3 - y = -2
= > -y = -2 - 3
= > y = 5.
Substitute x = 3 and y = 5 in (*), we get
= > 10x + y
= > 10(3) + 5
= > 35.
Therefore the original number = 35.
Hope this helps!
Therefore the decimal expansion is 10x + y. ------ (*)
Given that A two-digit number is 3 more than 4 times the sum of its digits.
= > 10x + y = 4(x + y) + 3
= > 10x + y = 4x + 4y + 3
= > 10x + y - 4x - 4x = 3
= > 6x - 3y = 3
= > 2x - y = 1 ------ (1)
Given that if 18 is added to the number, its digits are reversed.
= > 10x + y + 18 = 10y + x
= > 10y + x - 10x - y = 18
= > 9x - 9y = -18
= > x - y = -2 ------- (2)
On solving (1) & (2), we get
2x - y = 1
x - y = -2
-----------------
x = 3
Substitute x = 3 in (1), we get
= > x - y = -2
= > 3 - y = -2
= > -y = -2 - 3
= > y = 5.
Substitute x = 3 and y = 5 in (*), we get
= > 10x + y
= > 10(3) + 5
= > 35.
Therefore the original number = 35.
Hope this helps!
siddhartharao77:
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1
I hope you got your answer
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