CAT 2021 Question Paper | Quant Slot 3

CAT 2021 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 1 : If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

Video Explanation

Explanatory Answer

Let the weight of Silver in the initial Alloy be Ag kgs
and the weight of Copper in the initial Alloy be Cu kgs
So, the total weight of the initial Alloy = (Ag + Cu) kgs
This Alloy is mixed with 3kgs of Pure Silver to form a new mixture.
Total weight of the mixture = (Ag + Cu) + 3 kgs
Weight of Silver in the mixture = (Ag + 3) kgs
Since this mixture contains 90% Silver,
\( \frac { \text { Weight of Silver } } { \text { Total weight of the Mixture } } = \frac { 90 } { 100 } \)
\( \frac { A g + 3 } { A g + C u + 3 } = \frac { 90 } { 100 } \) — (1)
The initial Alloy is mixed with 2kgs of another alloy having 90% Silver by weight.
Total weight of the mixture = (Ag + Cu) + 2 kgs
Weight of Silver in the mixture = (Ag + 90% of 2) kgs = (Ag + 1.8) kgs
Since this mixture contains 84% Silver,
\( \frac { \text { Weight of Silver } } { \text { Total weight of the Mixture } } = \frac { 84 } { 100 } \)
\( \frac { A g + 1.8 } { A g + C u + 2 } = \frac { 84 } { 100 } \) — (2)
Simplifying (1)
\( \frac { A g + 3 } { A g + C u + 3 } = \frac { 90 } { 100 } \)
\( \frac { A g + 3 } { A g + C u + 3 } = \frac { 9 } { 10 } \)
10(Ag + 3) = 9(Ag + Cu + 3)
Ag + 3 = 9 Cu
Ag = 9 Cu - 3
Simplifying (1)
\( \frac { A g + 1.8 } { A g + C u + 2 } = \frac { 84 } { 100 } \)
\( \frac { A g + 1.8 } { A g + C u + 2 } = \frac { 21 } { 25 } \)
25(Ag + 1.8) = 21(Ag + Cu + 2)
4 Ag + 3 = 21 Cu
Ag = \( \frac { 21 C u - 3 } { 4 } \)
9 Cu - 3 = \( \frac { 21 C u - 3 } { 4 } \)
4(9 Cu - 3) = 21 Cu - 3
36 Cu - 12 = 21 Cu - 3
15 Cu = 9
Cu = \( \frac { 9 } { 15 } \)= \( \frac {3} {5} \)= 0.6
Ag = 9 Cu - 3 = 9(0.6) - 3 = 5.4 - 3 = 2.4
The weight of the initial Alloy = Ag + Cu = 2.4 + 0.6 = 3 kgs

The answer is '3'

Choice A is the correct answer.