Advertisements
Advertisements
प्रश्न
A two digit number is such that the product of its digit is 14. When 45 is added to the number, then the digit interchange their places. Find the number.
उत्तर
Let the ten's digit be x, then unit's digit = `(14/x)`
Then, the number is `(10x +14/x)`
Where 45 is added to the number, the digits get interchanged.
∴ `10x + (14)/x + 45 = 10 xx (14)/x + x`
⇒ x2 + 5x - 14 = 0
⇒ (x + 7) (x + 2) = 0
⇒ x = 2
and x = -7 (inadmissible)
Hence, the number is `(10x + 14/x)`
= `(10 xx 2 + 14/2)`
= 27.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
a2x2 - 3abx + 2b2 = 0
Solve the following quadratic equations by factorization:
a(x2 + 1) - x(a2 + 1) = 0
Solve the following quadratic equations by factorization:
`1/(x-1)-1/(x+5)=6/7` , x ≠ 1, -5
Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.
Find the value of k for which the following equations have real and equal roots:
\[\left( k + 1 \right) x^2 - 2\left( k - 1 \right)x + 1 = 0\]
If 2 is a root of the equation x2 + ax + 12 = 0 and the quadratic equation x2 + ax + q = 0 has equal roots, then q =
Solve the following equation: 3x2 + 25 x + 42 = 0
Solve the Following Equation : x2- x - a (a + 1) = o
Solve the following equation by factorization
`(x - 3)/(x + 3) + (x + 3)/(x - 3) = 2(1)/(2)`
A two digit number contains the bigger at ten’s place. The product of the digits is 27 and the difference between two digits is 6. Find the number.
