There is a number consisting of two digits. T | vedclass

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

Question

(B) Let the $2$-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.Given that the units digit is twice the tens digit,we

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Let the $2$-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.Given that the units digit is twice the tens digit,we have $y = 2x$.The sum of the digits is $x + y$. According to the problem,if $2$ is subtracted from this sum,the result is $1/6$ of the number:$x + y - 2 = \frac{1}{6}(10x + y)$Substitute $y = 2x$ into the equation:$x + 2x - 2 = \frac{1}{6}(10x + 2x)$$3x - 2 = \frac{1}{6}(12x)$$3x - 2 = 2x$$x = 2$Since $y = 2x$,we have $y = 2(2) = 4$.Therefore,the number is $10(2) + 4 = 24$.

There is a number consisting of two digits. The digit in the units place is twice that in the tens place,and if $2$ is subtracted from the sum of the digits,the difference is equal to $1/6^{\text{th}}$ of the number. The number is:

Similar Questions

$42 \%$ of a number is $357$. What is $63 \%$ of that number?

$\frac{(0.188+0.077)^{2}+(0.188-0.077)^{2}}{0.188 \times 0.188+0.077 \times 0.077} = ?$

$A$ number exceeds $20 \%$ of itself by $40$. The number is:

$\sqrt{2601} - \sqrt{1156} + \sqrt{484} = ?$

$\left[ {10 + \left\{ {4 \times \left( {\overline {\frac{2}{3} + \frac{1}{4}} \times \sqrt {\frac{{144}}{{121}}} + 23} \right) \div 12 + 5} \right\} - 3} \right] = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module