Question

a number consists of two digits the digit in the tens place exceeds the digit in the units place by 4 the sum of the digits is 1 by 7 of the number the number is

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Answer 1

Question:

A number consists of two digits. The digit in the tens place exceeds the digit in the unit place by 4 . The sum of the digit is (1/7)th of the number. Then find the number.

Solution:

Let the tens digit of the required number be x and the unit digit of the number be y.

Thus,

The number will be (10x+y).

According to the question,

The digit in the tens place exceeds the digit in unit place by 4.

ie;

=> x = y + 4

=> y = x - 4 -----------(1)

Also;

It is said that, the sum of the digits is (1/7)th of the number.

ie;

=> x + y = 1/7 of (10x + y)

=> x + y = (1/7)•(10x + y)

=> x + y = (10x + y)/7

=> 7(x + y) = 10x + y

Now;

Putting y=x-4 in the above equation,

We get;

=> 7(x + y) = 10x + y

=> 7(x + x - 4) = 10x + x - 4

=> 7(2x - 4) = 11x - 4

=> 14x - 28 = 11x - 4

=> 14x - 11x = 28 - 4

=> 3x = 24

=> x = 24/3

=> x = 8

Now,

Putting x=8 in eq-(1) , we get;

=> y = x - 4

=> y = 8 - 4

=> y = 4

Hence;

The tens digit of the required number is 8 and the unit digit of the required number is 4.

Also;

The required number = 10x + y

= 10•8 + 4

= 80 + 4

= 84

Hence,

The required number is 84.

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Number = 84

$\bold{\underline{\underline{Step\:by\:-\:step\:explanation:}}}$

Given :

• The tens place exceeds the digit in the units place by 4 in a two digit number.
• The sum of the digits is $\bold{\dfrac{1}{7}}$of the number

To find :

• The Numbers

Solution :

Let the digit in the tens place be x.

Let the digit in the units place be y.

Original Number = 10x + y

$\bold{\underline{\underline{As\:per\:the\:first\:condition:}}}$

• The tens place exceeds the digit in the units place by 4 in a two digit number.

Constituting it mathematically,

$\rightarrow$$\bold{x=y+4}$

$\rightarrow$$\bold{x-y=4}$ ---> (1)

$\bold{\underline{\underline{As\:per\:the\:second\:condition:}}}$

• The sum of the digits is $\bold{\dfrac{1}{7}}$of the number

Constituting the second condition mathematically,

$\rightarrow$$\bold{x+y={\dfrac{1}{7}\times\:10x+y}}$

$\rightarrow$$\bold{x+y={\dfrac{10x+y}{7}}}$

Cross multiplying,

$\rightarrow$$\bold{7(x+y)=10x+y}$

$\rightarrow$$\bold{7x+7y=10x+y}$

$\rightarrow$$\bold{7y-y=10x-7x}$

$\rightarrow$$\bold{6y=3x}$

$\rightarrow$$\bold{3x=6y}$

$\rightarrow$$\bold{3x-6y=0}$ ---> (2)

Multiply equation 1 by 3,

$\rightarrow$$\bold{x-y=4}$ ---> (1)

$\rightarrow$3x - 3y = 12 ----> (3)

Solve equation 3 and equation 2 simultaneously by elimination method.

Subtract equation 3 from equation 2,

3x - 3y = 12

3x - 6y = 0

-------------

3y = 12

$\rightarrow$$\bold{y={\dfrac{12}{3}}}$

$\rightarrow$$\bold{y=4}$

Substitute y = 4 in equation 3,

3x - 3y = 12 ----> (3)

$\rightarrow$$\bold{3x-3(4)=12}$

$\rightarrow$$\bold{3x-12=12}$

$\rightarrow$$\bold{3x=12+12}$

$\rightarrow$$\bold{3x=24}$

$\rightarrow$$\bold{x={\dfrac{24}{3}}}$

$\rightarrow$$\bold{x=8}$

$\bold{\red{\boxed{\sf{Tens\:digit\:=\:x\:=\:8}}}}$

$\bold{\red{\boxed{\sf{Units\:digit\:=\:y\:=\:4}}}}$

$\bold{\red{\boxed{\sf{Original\:Number\:=\:10x+y\:=\:10(8)\:+4\:=\:80\:+4\:=\:84}}}}$