Aptitude Discussion

Q. |
Teas worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs 153 per Kg , the price of the third variety per Kg will be? |

✖ A. |
Rs. 147.50 |

✖ B. |
Rs. 785.50 |

✔ C. |
Rs. 175.50 |

✖ D. |
Rs. 258.50 |

**Solution:**

Option(**C**) is correct

Since first and second varieties are mixed in equal proportions.

So, their average price $= \text{Rs. } \left(\dfrac{126+135}{2}\right)$

$= \text{Rs. } 130.50$

So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. $ x $ per kg in the ratio 2 : 2, i.e., 1 : 1.

We have to find $ x $

By the rule of alligation, we have:

Cost of 1 kg Cost of 1 kg

of 1st kind of 2^{nd} kind

(Rs. 130.50) (Rs. $*x$*)

\ /

Mean Price

(Rs. 153)

/ \

$x - 153$ 22.50

$\Rightarrow x-\dfrac{153}{22.50}=1$

$\Rightarrow x-153=22.50$

$\Rightarrow x=$ 175.50 Rs.

**Edit:** For a quicker approach, using the weighted average method, check comment by ** Jaideep.**

**Adithya**

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**SHRINWANTU RAHA**

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1 st kind:

126*1/4=31.50

2nd kind

135*1/4=33.75

Let price of the third is x rs./kg.

2/4kg worth of 1/2x rs.

so, 31.50+33.75+x/2=153

x= 175.50

**Kkpan**

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but if it is done by taking 1st and third ratio..then we will have to consider 1:1 ratio and then it will come as 180

**Murali**

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if we assume third party as $x$,then we have to take $1:1:2$ for all to $153$. so $\dfrac{(135+126+2x)}{4} = 153$. solve it we will get easily $\textbf{175.50}$.

**Jaideep**

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\(\dfrac{126+135+2x}{4} =153\)

this is alternate solution and time saving also.

Not able to understand why 4 came below?? I know 153 is mean value so it must come 3 becoz quantity is there

BECAUSE OF THE RATIO OF DISTRIBUTION

what if 1st two elements aren't mixed in equal proportion?