Directions for questions: Answer the following questions based on the information given below.
A manufacturing company’s make various alloys, which is be used for certain specific purposes. It is considering a choice of 5 new alloys (A1, A2, A3, A4, and A5), which can be used in different proportions in the formulations. The table below gives the composition of these alloys. The cost per unit of each of these alloys is A1: 150, A2: 50, A3: 200, A4: 500, A5: 100
Composition
Alloys
Cobalt
Iron
Aluminum
Carbon
A1
50
30
10
10
A2
80
20
0
0
A3
10
30
50
10
A4
5
50
40
5
A5
45
50
0
5
The company is planning to make a new alloy required for making ships. This new alloy must contain at least 30% each of cobalt and iron, no more than 25% aluminum and at least 5% carbon. Which one of the following combinations of equally mixed alloys is feasible?
Directions for questions: Answer the following questions based on the information given below.
A manufacturing company’s make various alloys, which is be used for certain specific purposes. It is considering a choice of 5 new alloys (A1, A2, A3, A4, and A5), which can be used in different proportions in the formulations. The table below gives the composition of these alloys. The cost per unit of each of these alloys is A1: 150, A2: 50, A3: 200, A4: 500, A5: 100
Composition |
||||
Alloys |
Cobalt |
Iron |
Aluminum |
Carbon |
A1 |
50 |
30 |
10 |
10 |
A2 |
80 |
20 |
0 |
0 |
A3 |
10 |
30 |
50 |
10 |
A4 |
5 |
50 |
40 |
5 |
A5 |
45 |
50 |
0 |
5 |
The company is planning to make a new alloy required for making ships. This new alloy must contain at least 30% each of cobalt and iron, no more than 25% aluminum and at least 5% carbon. Which one of the following combinations of equally mixed alloys is feasible?
Option 1 -
A1 and A2
Option 2 -
A4 and A5
Option 3 -
A2 and A5
Option 4 -
A1 and A5
-
1 Answer
-
Correct Option - 4
Detailed Solution:As the alloys are fixed in equal amounts, so we can take the average of the constituent percentage of the metal used. A1nly option A1 and A5 satisfies all the conditions.
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From the 1st statement: B2 is now as old as B3 was in the past. Hence B2 is younger to B3 or B2 < B3. Also sometime in the past B1 was twice as old as B4. So B1 is elder to B4 or B1 > B4. B3 will be as old as B5 in future, hence B3 < B5. The second statement suggests: B1 > B6. B1 was as old as B7 in the past. Hence B1 > B7. B4 will be as old as B6 in future. Hence B6 > B4. B6 will be as old as B7 now in future. Hence B7 > B6. B7 was as old as B2, when B1 was as old as B7. Hence B1 = B2. Combining both the results, we get: and B5 > B3 > B2 = B1 > B7 > B6 > B4 (Note by B1 = B2, it is meant that they are of similar age group, not necessarily the same).
B5 is the eldest brother.
According to the information provided following arrangement will be obtained:
(K→ Kerala ; G → Goa ; D → Delhi)
Order according to height | Student | Belong | |||||||||
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| K | G | D |
× | × | × | × | × | × | × | √ | Arun | × | × | √ |
× | × | √ | × | × | × | × | × | Ajit | √ | × | × |
√ | × | × | × | × | × | × | × | Rohan | √ | × | × |
× | × | × | × | × | √ | × | × | Gajraj | × | √ | × |
× | √ | × | × | × | × | × | × | Ronit | × | × | √ |
× | × | × | × | √ | × | × | × | Shubham | √ | × | × |
× | × | × | × | × | × | √ | × | Sweety | × | √ | × |
× | × | × | √ | × | × | × | × | Lovely | × | √ | × |
Looking at the table given in the first question of the set, we get that Talwar lives on the sixth floor.
The data available is not enough to determine where Rubi belongs to. The given data is insufficient to answer the above question.
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