Question

The price of an article is reduced by 10%. By how much percent this value be increased to restore it to its former value?
please answer fast....

Answer 1

Answer:

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Answer 2

Answer:

• The decreased value must be increased by 11.12% to get the original value.

Step-by-step explanation:

Given that:

• The price of an article is reduced by 10%.

To Find:

• By how much percent should the decreased value be increased to get the original value?

Let us assume:

• The original value be ₹100.

Finding Decreased Value:

$\rm{\longmapsto\left\{100-\left(\dfrac{10}{100}\times 100\right)\right\}}$

Cutting off the zeros,

$\rm{\longmapsto\left\{100-\left(\dfrac{1\!\!\!\not{0}}{1\!\!\!\not{0}\!\!\!\not{0}}\times 10\!\!\!\not{0}\right)\right\}}$

$\rm{\longmapsto\left\{100-10\right\}}$

$\rm{\longmapsto90}$

Hence,

• Decreased value = ₹90

Now, let us assume:

• Percentage by which the decreased value should be increased to get original price be x.

Then, according to the question:

$\rm{\longmapsto90+\left(\dfrac{x}{100}\times 90\right)=100}$

Cutting off the zeros,

$\rm{\longmapsto90+\left(\dfrac{x}{10\!\!\!\not{0}}\times 9\!\!\!\not{0}\right)=100}$

$\rm{\longmapsto90+\left(\dfrac{x}{10}\times 9\right)=100}$

Multiplying,

$\rm{\longmapsto90+\dfrac{9x}{10}=100}$

Transposing 90 from LHS to RHS and changing its sign,

$\rm{\longmapsto\dfrac{9x}{10}=100-90}$

Subtracting,

$\rm{\longmapsto\dfrac{9x}{10}=10}$

Transposing 10 from LHS to RHS and changing its sign,

$\rm{\longmapsto9x=10\times10}$

Multiplying,

$\rm{\longmapsto9x=100}$

Transposing 9 from LHS to RHS and changing its sign,

$\rm{\longmapsto x=\dfrac{100}{9}}$

$\rm{\longmapsto\boxed{\rm{x=11.12}}\;\;\;(approx.)}$

Hence,

• The decreased value must be increased by 11.12% to get the original value.