# A two-digit number is obtained by either multiplying the sum of the digits by 8 and adding 1 or by multiplying the difference of the digits by 13 and adding 2. Find the number.

Last updated date: 20th Mar 2023

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Answer

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Hint: Assume the two digits as variables i.e. ‘x’ and ‘y’ and then apply both the conditions given in the problem; you will get the system of equations and solving these equations you will get the values of variables. Put these values in assumed number i.e. 10y+x to get the final answer.

Complete step-by-step answer:

To solve the above example we will assume the digits of the number to be variables,

Therefore we will assume the unit digit of the number as ‘x’ and the tens digit of the number as ‘y’ therefore we can represent the number as,

Two digit number = 10y+x …………………………………. (1)

As we have assumed our number therefore now we can go through the conditions given in the problem,

According to the first condition given in the problem we can write,

Two digit number \[=\left( x+y \right)\times 8+1\]

From equation (1) we can write,

\[\Rightarrow 10y+x=\left( x+y \right)\times 8+1\]

\[\Rightarrow 10y+x=8x+8y+1\]

\[\Rightarrow 10y-8y=8x-x+1\]

\[\Rightarrow 2y=7x+1\]

\[\Rightarrow 2y-7x=1\] …………………………………………….. (2)

Now by using the second condition given in the problem we can write,

Two digit number \[=\left( y-x \right)\times 13+2\]

From equation (1) we can write,

\[\Rightarrow 10y+x=\left( y-x \right)\times 13+2\]

\[\Rightarrow 10y+x=13y-13x+2\]

\[\Rightarrow 13x+x=13y-10y+2\]

\[\Rightarrow 14x=3y+2\]

\[\Rightarrow 3y-14x=-2\] …………………………………………. (3)

Now if we observe equation (2) and equation (3), these equations form a system of equations and they can be solved using elimination method.

Now, if we multiply equation (2) by 2 we will get,

\[2\times \left( 2y-7x \right)=2\times 1\]

\[\Rightarrow 2\times 2y-2\times 7x=2\]

\[\Rightarrow 4y-14x=2\] ………………………………………. (4)

Now to eliminate ‘x’ from the system of equations we will subtract equation (3) from equation (4) therefore we will get,

\[\begin{align}

& 4y-14x=2 \\

& - \\

& 3y-14x=-2 \\

& -\_+\_\_\_\_\_+\_\_\_\_\_\_\_\_\_ \\

& y+0=4 \\

\end{align}\]

\[\therefore y=4\] ………………………………………….. (5)

Now if put the value of equation (5) in equation (2) we will get,

\[\therefore 2\times \left( 4 \right)-7x=1\]

By shifting the term -7x on the right hand side of the equation we will get,

\[\Rightarrow 2\times \left( 4 \right)-1=7x\]

\[\Rightarrow 8-1=7x\]

\[\Rightarrow 7=7x\]

\[\Rightarrow 7x=7\]

\[\Rightarrow x=\dfrac{7}{7}\]

\[\Rightarrow x=1\]……………………………………….. (6)

To find the required number we will substitute the values of equation (5) and equation (6) in equation (1), therefore we will get,

Two digit number = 10y+x

Therefore, Two digit number \[=\text{ }10\times \left( 4 \right)+\left( 1 \right)\]

Therefore, Two digit number = 40 + 1

Therefore, Two digit number = 41

Therefore the required number is 41.

Note: Do remember that we have to represent the number in tens and units to represent the number only i.e. Two digit number = 10y+x but while writing the conditions the digits are independent numbers and if you consider it like number then you will get the wrong answer i.e. use \[\left( x+y \right)\times 8+1\] as the first condition and not as \[\left( 10y+x \right)\times 8+1\]

Complete step-by-step answer:

To solve the above example we will assume the digits of the number to be variables,

Therefore we will assume the unit digit of the number as ‘x’ and the tens digit of the number as ‘y’ therefore we can represent the number as,

Two digit number = 10y+x …………………………………. (1)

As we have assumed our number therefore now we can go through the conditions given in the problem,

According to the first condition given in the problem we can write,

Two digit number \[=\left( x+y \right)\times 8+1\]

From equation (1) we can write,

\[\Rightarrow 10y+x=\left( x+y \right)\times 8+1\]

\[\Rightarrow 10y+x=8x+8y+1\]

\[\Rightarrow 10y-8y=8x-x+1\]

\[\Rightarrow 2y=7x+1\]

\[\Rightarrow 2y-7x=1\] …………………………………………….. (2)

Now by using the second condition given in the problem we can write,

Two digit number \[=\left( y-x \right)\times 13+2\]

From equation (1) we can write,

\[\Rightarrow 10y+x=\left( y-x \right)\times 13+2\]

\[\Rightarrow 10y+x=13y-13x+2\]

\[\Rightarrow 13x+x=13y-10y+2\]

\[\Rightarrow 14x=3y+2\]

\[\Rightarrow 3y-14x=-2\] …………………………………………. (3)

Now if we observe equation (2) and equation (3), these equations form a system of equations and they can be solved using elimination method.

Now, if we multiply equation (2) by 2 we will get,

\[2\times \left( 2y-7x \right)=2\times 1\]

\[\Rightarrow 2\times 2y-2\times 7x=2\]

\[\Rightarrow 4y-14x=2\] ………………………………………. (4)

Now to eliminate ‘x’ from the system of equations we will subtract equation (3) from equation (4) therefore we will get,

\[\begin{align}

& 4y-14x=2 \\

& - \\

& 3y-14x=-2 \\

& -\_+\_\_\_\_\_+\_\_\_\_\_\_\_\_\_ \\

& y+0=4 \\

\end{align}\]

\[\therefore y=4\] ………………………………………….. (5)

Now if put the value of equation (5) in equation (2) we will get,

\[\therefore 2\times \left( 4 \right)-7x=1\]

By shifting the term -7x on the right hand side of the equation we will get,

\[\Rightarrow 2\times \left( 4 \right)-1=7x\]

\[\Rightarrow 8-1=7x\]

\[\Rightarrow 7=7x\]

\[\Rightarrow 7x=7\]

\[\Rightarrow x=\dfrac{7}{7}\]

\[\Rightarrow x=1\]……………………………………….. (6)

To find the required number we will substitute the values of equation (5) and equation (6) in equation (1), therefore we will get,

Two digit number = 10y+x

Therefore, Two digit number \[=\text{ }10\times \left( 4 \right)+\left( 1 \right)\]

Therefore, Two digit number = 40 + 1

Therefore, Two digit number = 41

Therefore the required number is 41.

Note: Do remember that we have to represent the number in tens and units to represent the number only i.e. Two digit number = 10y+x but while writing the conditions the digits are independent numbers and if you consider it like number then you will get the wrong answer i.e. use \[\left( x+y \right)\times 8+1\] as the first condition and not as \[\left( 10y+x \right)\times 8+1\]

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