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02 Jul 2010, 19:58
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
75% (01:47) correct
25%
(02:21)
wrong
based on 130
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History
Date
Time
Result
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How many mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to form an alloy containing 78% nickel?
A. 4
B. 5
C. 6
D. 7
E. 8
A. 4
B. 5
C. 6
D. 7
E. 8
02 Nov 2016, 05:49
Kudos
Bookmarks
Hi mikesteven1000bc,
Questions of this kind can be easily solved by making use of the mixture diagram given below.
Capture.png [ 3.96 KiB | Viewed 5934 times ]
Putting in the values in mixture diagram and subtracting along the diagonals gives us a ratio in which two quantities need to be mixed. This ratio can now be used to find out what specific amounts of two quantities need to be mixed to obtain a solution of a specific concentration.
The only thing that you need to keep in mind is that the values you need to use that is the higher value, lower value and mean value have to be values which are associated with the word 'per' (percents, average, per km, per kg etc.).
Now in the question given we are combining 6 mg of pure nickel (100% nickel) with 'x'mg of a metal containing 45% nickel to form an alloy containing 78%
nickel. The question here an be solved using the mixture diagram by taking the percentages of nickel as the higher value (100%), lower value (45%) and mean value (78%) as shown below.
Untitled 2.png [ 6.33 KiB | Viewed 5921 times ]
On solving the mixture diagram we get a ratio 3 : 2. This means that in order to obtain an alloy containing 78% nickel, we need to combine 3 parts of pure nickel with 2 parts of a metal containing 45% nickel. Since we have 6gm of 100% nickel and we need to maintain the ratio 3 : 2, we need to have 4mg of the metal containing 45% nickel.
Answer : 4mg
The mixture diagram proves useful not only when mixing solutions or combining solids but also to explain the weighted average concept (the word average is also associated with the word per i.e. if the average marks of the class is 80, then it can be understood as 80 marks per student). Say if we have a class A where the average marks is 80 and another class B where the average marks is 70 and the combined average of both class A and B is 74, then we can definitely comment upon which class has the greater number of students. If we represent the average values in the mixture diagram, the ratio of students of Class A and Class B will be 2 : 3. This clearly indicates that class B has the greater number of students.
Hope this helps!
CrackVerbal Academics Team
Questions of this kind can be easily solved by making use of the mixture diagram given below.
Attachment:
Capture.png [ 3.96 KiB | Viewed 5934 times ]
Putting in the values in mixture diagram and subtracting along the diagonals gives us a ratio in which two quantities need to be mixed. This ratio can now be used to find out what specific amounts of two quantities need to be mixed to obtain a solution of a specific concentration.
The only thing that you need to keep in mind is that the values you need to use that is the higher value, lower value and mean value have to be values which are associated with the word 'per' (percents, average, per km, per kg etc.).
Now in the question given we are combining 6 mg of pure nickel (100% nickel) with 'x'mg of a metal containing 45% nickel to form an alloy containing 78%
nickel. The question here an be solved using the mixture diagram by taking the percentages of nickel as the higher value (100%), lower value (45%) and mean value (78%) as shown below.
Attachment:
Untitled 2.png [ 6.33 KiB | Viewed 5921 times ]
On solving the mixture diagram we get a ratio 3 : 2. This means that in order to obtain an alloy containing 78% nickel, we need to combine 3 parts of pure nickel with 2 parts of a metal containing 45% nickel. Since we have 6gm of 100% nickel and we need to maintain the ratio 3 : 2, we need to have 4mg of the metal containing 45% nickel.
Answer : 4mg
The mixture diagram proves useful not only when mixing solutions or combining solids but also to explain the weighted average concept (the word average is also associated with the word per i.e. if the average marks of the class is 80, then it can be understood as 80 marks per student). Say if we have a class A where the average marks is 80 and another class B where the average marks is 70 and the combined average of both class A and B is 74, then we can definitely comment upon which class has the greater number of students. If we represent the average values in the mixture diagram, the ratio of students of Class A and Class B will be 2 : 3. This clearly indicates that class B has the greater number of students.
Hope this helps!
CrackVerbal Academics Team
General Discussion
