The sum of a two digit number and the number formed by interchanging its digit is 132. If nine is subtracted from the first number, the new number is 3 more than 6 times of the sum of the digits in the first number. Find the first number.
84
Let the two-digit number be represented as $10a + b$, where 'a' is the tens digit and 'b' is the units digit.
The number formed by interchanging the digits is $10b + a$.
According to the first condition:
$ (10a + b) + (10b + a) = 132 $Combine like terms:
$ 11a + 11b = 132 $Factor out 11:
$ 11(a + b) = 132 $Solve for the sum of the digits ($a + b$):
$ a + b = \frac{132}{11} $ $ a + b = 12 $This tells us the sum of the digits of the number is 12.
The second condition states that if nine is subtracted from the first number ($10a + b$), the result is 3 more than 6 times the sum of its digits ($a + b$).
The equation is:
$ (10a + b) - 9 = 6(a + b) + 3 $We already found that $a + b = 12$. Substitute this value into the equation:
$ (10a + b) - 9 = 6(12) + 3 $ $ (10a + b) - 9 = 72 + 3 $ $ (10a + b) - 9 = 75 $Now, solve for the original number ($10a + b$):
$ 10a + b = 75 + 9 $ $ 10a + b = 84 $The calculated first number is 84.
Both conditions are satisfied.
The first number is 84.
The sum of two numbers m and n is 84 (m > n) and their difference is 6. What is the ratio of the two numbers?
A piece of cloth costs Rs. 35. If the piece were 4 m longer and each meter was to cost Rs. 1 lesser, then the total cost would remain unchanged. How long is the piece of cloth?
A. 10 m
B. 14 m
C. 12 m
D. 8 m
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What is the value of x, 2x/3 + y/ 2 = 4 and x/3 - y/2 = 1?
The values of x and y from the equations x - y = 6 and x/3 + y/2 = 12 are: