Question

Q2
Arun was a Class 10 student who lived in
a small town. He always rode his 5-year-
old bicycle everywhere he went. But,
after five years of regular use, looking
after the worn-out bike had become
difficult. He asked his father if he could
buy him a new bicycle. His father told
him that he would only get him a new
bicycle if he scored well in his next Maths
examination. Arun was worried, because
he was a little weak in Maths, but he
agreed to his father's challenge and
began working hard to get high scores in
his upcoming exam.
Arun showed his father his Mathematics
results. His father was pleased. The next
morning, Arun and his father went to the
local sports store. The shopkeeper was
selling a bicycle for Rs. 2,850. The sale
would have earned the shopkeeper a
profit of 14%. However, Arun's father
negotiated with the seller and asked him
if he could reduce the price a bit. If the
shopkeeper reduced his profit to 8%,
then what would the new selling price of
the bicycle be?​

Answer 1

let x be the original price of the bicycle

Thus,

x+14% of x= 2850

x+ .14x= 2850

1.14x= 2850

x= 2850/1.14 = 285000/114= 2500

profit %= p/cp ×100

8 = p/2500 × 100

p= (25×8)

= 200

Sp= CP+ profit = 2500+200= 2700

the new selling price is ₹2,700

Answer 2

➤ Given :-

Selling price of the bicycle :- ₹2850

Profit percentage :- 14%

$\:$

➤ To Find :-

The selling price if it's sold at profit of 8%.

$\:$

★ How to do :-

Here, we are given with the selling price and the profit percentage of a bicycle. The shopkeeper reduces his profit percentage to 8%. We are asked to find the selling price if the shopkeeper sells the bicycle for 8%. So, first we should find the cost price of the bicycle by using the information in first statement. Later, we can find the new selling price of the bicycle by using the cost price and new gain percentage. The appropriate formulas are provided below. So, let's start solving!!

$\:$

➤ Solution :-

Cost price of the bicycle :-

${\tt \leadsto \underline{\boxed{\tt \dfrac{100}{(100 + Profit \bf\%)} \times SP}}}$

Substitute the given values.

${\tt \leadsto \dfrac{100}{(100 + 14)} \times 2850}$

Add the bracket in denominator of the fraction and write the denominator and the whole number in lowest form by cancellation method.

${\tt \leadsto \dfrac{100}{\cancel{114}} \times \cancel{2850} = \dfrac{100}{1} \times 25}$

Multiply the remaining numbers.

${\tt \leadsto \dfrac{2500}{1} = \underline{2500}}$

$\:$

Now, find the new selling price of the bicycle by using the cost price and new gain percentage.

New selling price of bicycle :-

${\tt \leadsto \underline{\boxed{\tt \dfrac{(100 +Profit \bf\%)}{100} \times CP}}}$

Substitute the given values.

${\tt \leadsto \dfrac{(100 + 8)}{100} \times 2500}$

Add the bracket in numerator of the fraction and cancel the zeros in denominator and whole number.

${\tt \leadsto \dfrac{108}{1 \cancel{00}} \times 25 \cancel{00} = \dfrac{108}{1} \times 25}$

Multiply the remaining numbers to get the final answer.

${\tt \leadsto \dfrac{2700}{1} = \pink{\underline{\boxed{\tt Rs \: \: 2700}}}}$

$\Huge\therefore$ The new selling price of the bicycle is ₹2700.

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$\dashrightarrow$ Some related formulas :-

$\small\boxed{\begin{array}{cc}\large\sf\dag \: {\underline{More \: Formulae}} \\ \\ \bigstar \: \sf{Gain = S.P - C.P} \\ \\ \bigstar \:\sf{Loss = C.P - S.P} \\ \\ \bigstar \: \sf{Gain \: \% = \Bigg( \dfrac{Gain}{C.P} \times 100 \Bigg)\%} \\ \\ \bigstar \: \sf{loss \: \% = \Bigg( \dfrac{loss}{C.P} \times 100 \Bigg)\%} \\ \\\bigstar \: \sf{S.P = \dfrac{100-loss\%}{100} \times C.P} \\ \\ \bigstar \: \sf{C.P =\dfrac{100}{100-loss\%} \times S.P}\end{array}}$