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IPMAT Indore 2019 (MCQ) PYQs

IPMAT Indore 2019

Arithmetic
>
Mixture & Alligation

Hard

An alloy PP has copper and zinc in the proportion of 5:25:2 (by weight), while another alloy QQ has the same metals in the proportion of 3:43:4 (by weight). If these two alloys are mixed in the proportion of a:ba:b (by weight), a new alloy RR is formed, which has equal contents of copper and zinc. Then, the proportion of copper and zinc in the alloy SS, formed by mixing the two alloys PP and QQ in the proportion of b:ab:a (by weight) is

Correct Option: 3
Understanding the given information:
Alloy PP: Copper : Zinc = 5:25:2 (by weight)
Alloy QQ: Copper : Zinc = 3:43:4 (by weight)
Alloy RR: Made by mixing PP and QQ in ratio a:ba:b, has equal copper and zinc
Alloy SS: Made by mixing PP and QQ in ratio b:ab:a
Finding the fraction of copper in each alloy:
In Alloy PP: Total parts = 5+2=75 + 2 = 7
Copper fraction = 57\dfrac{5}{7}
In Alloy QQ: Total parts = 3+4=73 + 4 = 7
Copper fraction = 37\dfrac{3}{7}
Setting up equation for Alloy RR:
When mixing PP and QQ in ratio a:ba:b, copper fraction in mixture:
a×57+b×37a+b\dfrac{a \times \dfrac{5}{7} + b \times \dfrac{3}{7}}{a + b}
Since RR has equal copper and zinc, copper fraction = 12\dfrac{1}{2}
a×57+b×37a+b=12\dfrac{a \times \dfrac{5}{7} + b \times \dfrac{3}{7}}{a + b} = \dfrac{1}{2}
Solving for ratio a:ba:b:
Cross multiplying:
a×57+b×37=12(a+b)a \times \dfrac{5}{7} + b \times \dfrac{3}{7} = \dfrac{1}{2}(a + b)
Multiplying by 1414:
10a+6b=7a+7b10a + 6b = 7a + 7b
3a=b3a = b
Therefore: a:b=1:3a:b = 1:3
Finding composition of Alloy SS:
Alloy SS mixes PP and QQ in ratio b:a=3:1b:a = 3:1
Copper fraction in SS:
3×57+1×373+1=157+374=1874=1828=914\dfrac{3 \times \dfrac{5}{7} + 1 \times \dfrac{3}{7}}{3 + 1} = \dfrac{\dfrac{15}{7} + \dfrac{3}{7}}{4} = \dfrac{\dfrac{18}{7}}{4} = \dfrac{18}{28} = \dfrac{9}{14}
Zinc fraction in SS:
1914=5141 - \dfrac{9}{14} = \dfrac{5}{14}
Copper : Zinc = 914:514=9:5\dfrac{9}{14} : \dfrac{5}{14} = 9:5
The proportion of copper and zinc in alloy SS is 9:59:5

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