Ratios & Proportion
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Q.

A noodles merchant buys two varieties of noodles the price of the first being twice that of the second. He sells the mixture at Rs 17.50 per kilogram thereby making a profit of $25%$. If the ratio of the amounts of the first noodles and the second noodles in the mixture is $2:3$, then the respective costs of each noodles are

 A.

Rs 20, Rs 10

 B.

Rs 24, Rs 12

 C.

Rs 16, Rs 8

 D.

Rs 26, Rs 13

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Solution:
Option(A) is correct

Let the price of one noodles = $k$

\(\Rightarrow\) the price of other noodle = \(\dfrac{k}{2}\)

Price of 1 kg = \(\dfrac{2k}{5}+\dfrac{3}{5}\times \dfrac{k}{2}=\dfrac{7k}{10}\)

But CP = \(\dfrac{17.50\times 100}{125}=14\)

\(\Rightarrow \dfrac{7k}{10}=14\)

\(\Rightarrow k=20\)

So price of the noodles's are 20 and 10.


(2) Comment(s)


Asmit Anand
 ()

Let price of 2nd noodle be x and price of 1st noodle be 2x

Given, profit on selling 1 kg of mixture = Rs 25

sp of 1kg of mixture = Rs 17.50

Therefore Sp of 1kg of mixture is given by-

sp = (100 + profit)/ 100 * cp

17.50 = (125/100) x cp

on solving above eq-

cp = Rs 14 for 1 kg of mixture

Now let amount of 1st type of noodle be t1 kg and amount of 2nd type of noodle be t2 kg

and let total amount of mixture be y kg

hence

cp of total mixture is given by

14 * y = t2 * x + t1 * (2x)

y = (t2 * x )/14 + (t1 * 2x)/14

t1 + t2 = (t2 * x)/14 + (t1 * x)/7

dividing both sides by t2, we get -

t1/t2 + 1 = x/14 + (t1/7*t2 )*x ------- eq-1

given- total amount 1st noodle /total amount of 2nd noodle = t1/t2 =2/3

thus putting the value of t/t2 in eq - 1

we have,

2/3 +1 = x/14 + 2/21 * x

on solving above eq-

x = 10

therfore,

price of 1st noodle is Rs 10 and 2nd noodle is 2x = 2* 10 = Rs 20



Kruthi M
 ()

PLs explain more... Wat is CP ... Y is it equal to (7k/10)??