Question

Plzzz solve the problem..... ​

Answer 1

Question:-

• In a two-digit number, the unit digit is $\dfrac{4}{3}$ of the number at tens place. The two-digit number is 4 more than 8 times the units number. What is the two-digit number?

Answer:-

• $\boxed{\red{\bf Original \: Number = 68}}$

Step-by-step explanation:

Let the digit at tens place be x. And, the digit at unit's place be y.

★ $\boxed{\bf Original \: Number = 10x+y}$

Now,

It is given that, the unit digit is $\frac{4}{3}$ of the number at tens place.

Equation :

$\sf \implies y = \dfrac{4x}{3} \: \: \dots (i)$

Also,

The two-digit number is 4 more than 8 times the units number.

Equation :-

$\sf \implies 10x + y = 8y+ 4 \\\\\implies \sf 10x + \frac{4x}{3} = 8\left(\frac{4x}{3}\right) + 4$

[ Using equation (i) ]

$\implies \sf \frac{30x + 4x}{3} - \frac{32x}{3} = 4 \\\\\implies \sf \frac{34x - 32x}{3} =4\\\\\implies \sf \frac{2x}{3} = 4 \\\\\implies \sf 2x = 12 \\\\\implies \sf x = \frac{12}{2} \\\\\boxed{\red{\bf \therefore \: x = 6}}$

Substituting the value of x in (i),

$\sf \implies y = \frac{4x}{3}\\\\\implies \sf y = \frac{4*6}{3} \\\\\implies \sf y = \frac{24}{3}\\\\\boxed{\red{\bf \therefore \: y = 8}}$

Now, Original number = 10x + y

                                     = 10 * 6 + 8

                                     = 60 + 8

                                     = 68

• Hence, the required number is 68.

Answer 2

Answer:

Question:-

In a two-digit number, the unit digit is $\dfrac{4}{3}$ of the number at tens place. The two-digit number is 4 more than 8 times the units number. What is the two-digit number?

Answer:-

$\boxed{\red{\bf Original \: Number = 68}}$

Step-by-step explanation:

Let the digit at tens place be x. And, the digit at unit's place be y.

★ $\boxed{\bf Original \: Number = 10x+y}$

Now,

It is given that, the unit digit is $\frac{4}{3}$ of the number at tens place.

Equation :

$\sf \implies y = \dfrac{4x}{3} \: \: \dots (i)$

Also,

The two-digit number is 4 more than 8 times the units number.

Equation :-

$\sf \implies 10x + y = 8y+ 4 \\\\\implies \sf 10x + \frac{4x}{3} = 8\left(\frac{4x}{3}\right) + 4$

[ Using equation (i) ]

$\implies \sf \frac{30x + 4x}{3} - \frac{32x}{3} = 4 \\\\\implies \sf \frac{34x - 32x}{3} =4\\\\\implies \sf \frac{2x}{3} = 4 \\\\\implies \sf 2x = 12 \\\\\implies \sf x = \frac{12}{2} \\\\\boxed{\red{\bf \therefore \: x = 6}}$

Substituting the value of x in (i),

$\sf \implies y = \frac{4x}{3}\\\\\implies \sf y = \frac{4*6}{3} \\\\\implies \sf y = \frac{24}{3}\\\\\boxed{\red{\bf \therefore \: y = 8}}$

Now, Original number = 10x + y

                                     = 10 * 6 + 8

                                     = 60 + 8

                                     = 68

• Hence, the required number is 68.