The digit at the ten's place is 4 in a two-di | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the digit at the ten's place and $y$ is the digit at the unit's place.Given that

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the digit at the ten's place and $y$ is the digit at the unit's place.Given that the digit at the ten's place is $x = 4$.The product of the digits is $x 	imes y = 4y$.According to the problem,the product of the digits is four times the digit at the ten's place:$x 	imes y = 4 	imes x$Substitute $x = 4$ into the equation:$4 	imes y = 4 	imes 4$$4y = 16$$y = 4$Therefore,the digit at the unit's place is $4$.The number is $10x + y = 10(4) + 4 = 44$.

The digit at the ten's place is $4$ in a two-digit number,and the product of both the digits is four times greater than the digit at the ten's place. Find the number.

Similar Questions

In a two-digit number,the digit at the ten's place is double the digit at the unit's place. If the digits of the number are interchanged,the new number so formed is $27$ less than the original number. Find the original number.

Solve the following pair of linear equations using a graph: $3x + 6y = 3900, x + 3y = 1300$

If $x=a$ and $y=b$ is the solution of the equations $x-y=2$ and $x+y=4,$ then the values of $a$ and $b$ are,respectively:

Solve the following pairs of equations by the cross-multiplication method:
$4x - 19y + 13 = 0$
$13x - 23y + 19 = 0$

In a two-digit number,the digit at the unit's place is $3x$ and the digit at the ten's place is $2x$. Then,that number is $\ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module