Aptitude Discussion

Q. |
In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? |

✔ A. |
11 : 77 : 7 |

✖ B. |
25 : 45 : 8 |

✖ C. |
27 : 23 : 6 |

✖ D. |
11 : 45 : 7 |

**Solution:**

Option(**A**) is correct

**Step1:**

Mix wheat of first and third kind to get a mixture worth Rs 1.41 per Kg.

C.P of 1 Kg wheat of 1st kind 120p

C.P of 1 Kg wheat of 3rd kind 174p

Mean Price 141p

C.P of 1 Kg C.P of 1 Kg

Wheat of 1st wheat of 3rd

kind (120p) kind (174p)

\ /

Mean Price

(141p)

/ \

33 21

By Alligation rule:

$\dfrac{\text{Quantity of 1st kind of wheat}}{\text{Quantity of 3rd kind of wheat}} = \dfrac{33}{21}=\dfrac{11}{7}$

So they must be mixed in the ratio $11 : 7$

**Step 2: **

Mix wheat of 1st kind and 2nd kind to obtain a mixture worth of Rs. 1.41per Kg

C.P of 1 Kg C.P of 1 Kg

Wheat of 1st wheat of 3rd

kind (120p) kind (144p)

\ /

Mean Price

(141p)

/ \

3 21

By alligation rule:

$\dfrac{\text{Quantity of 1st kind of wheat}}{\text{Quantity of 2nd kind of wheat}}=\dfrac{3}{21}=\dfrac{1}{7}$

So they must be mixed in the ratio 1 : 7

Thus,

\begin{align*}

&\dfrac{\text{Quantity of 2nd kind of wheat}}{\text{Quantity of 3rd kind of wheat}}=\\

&\left(\dfrac{\text{Quantity of 1st kind of wheat}}{\text{Quantity of 3rd kind of wheat}}\right)\times \left(\dfrac{\text{Quantity of 2nd kind of wheat}}{\text{Quantity of 1st kind of wheat}}\right)\\

& \Rightarrow \dfrac{\text{Quantity of 2nd kind of wheat}}{\text{Quantity of 3rd kind of wheat}}=\left(\dfrac{11}{7}\times \dfrac{7}{1}\right) =\left(\dfrac{11}{1}\right)

\end{align*}

Thus, Quantities of wheat of $1^{st} : 2^{nd} : 3^{rd} = 11 : 77 : 7$

**Edit:** For a different and shortcut approach, check **Murugan's** comment.

**Rajesh Kumar**

*()
*

Hi Rajesh,

In order to get avg. 1.41 we need to have values such that 1.41 lies between them. We can not use a combination 1.44 and 1.74 to get 1.41. Hence two possible alligations will be 1.2 and 1.44 , 1.2 and 1.74. And every time while paring to solve alligation we have to take care of it.

**Rajesh Kumar**

*()
*

#rajesh

can any one tell me why we use 120 two times to mix but not 144 ..confused !!

**Anuj Bhawsar**

*()
*

why are we do we add the ratio

c:a:b=7:(11+1):7.

Simple way is to get by substitution and ans comes as 11 : 77 : 7.

**Prince**

*()
*

Alligation has failed this question.

Correct answer: Multiple ratios possible

One possibility =12:7:7

this is the only possibility. 11 : 77 : 7 is wrong

**Murugan**

*()
*

$a:b=1:7$

$a:c=11:7$

$c:a:b=c:a::a:b$

So,

$c:a:b=7:11::7:1$

To find out the value of $a$,

$=(7*1):(7*11)::(7*11):(1*11)$

Therefore,

$=7:77::77:11$

$a:b:c=11:77:7$

**Sachin Dahiya**

*()
*

Approach for this question and the last 2 question is different and also the answer shown as correct is not matching the given information. it should be 12:7:7 #Sachin

**Naresh**

*()
*

$c:a=7:11$

a:b=1:7

So

$c:a:b=7:(11+1):7$

means,

$a:b:c = 12:7:7$

is the correct answer

I agree with Naresh.

the correct answer is 12:7:7

mean average= 12*120 +7*144 + 7*174 /26 equals to 141. Otherwise, also answer is calculated adding the ratio.A: C 11:7

A : B 7:7

12 : 7 : 7

There are multiple answers possible in such ratio questions. But I guess challenge is to follow an approach that matches the given options.

can any one tell me why we use 120 two times .. not 144 confused !!