Directions for questions: Answer the questions on the basis of the information given below.
Mr. John has 5 children & he gives different pocket money to all of his 5 children to do 5 particular tasks viz. going for morning walk, Exercise, Completing their home work, Drinking Milk and Assisting in family work. A particular child can do only one task at a time and each task can be assigned to one child only. (The values are in $)
Children
1
2
3
4
5
Going for
Morning Walk
35
31
44
23
36
Exercise
48
42
57
35
37
Completing their Homework
28
55
52
38
44
Drinking Milk
55
21
26
37
42
Assisting in
Family work
26
35
34
51
47
If no two children can work simultaneously so that each task is done one after than the other, then what will be the least amount a parent has to give to all the tasks done?
Directions for questions: Answer the questions on the basis of the information given below.
Mr. John has 5 children & he gives different pocket money to all of his 5 children to do 5 particular tasks viz. going for morning walk, Exercise, Completing their home work, Drinking Milk and Assisting in family work. A particular child can do only one task at a time and each task can be assigned to one child only. (The values are in $)
Children |
1 |
2 |
3 |
4 |
5 |
Going for Morning Walk |
35 |
31 |
44 |
23 |
36 |
Exercise |
48 |
42 |
57 |
35 |
37 |
Completing their Homework |
28 |
55 |
52 |
38 |
44 |
Drinking Milk |
55 |
21 |
26 |
37 |
42 |
Assisting in Family work |
26 |
35 |
34 |
51 |
47 |
If no two children can work simultaneously so that each task is done one after than the other, then what will be the least amount a parent has to give to all the tasks done?
Option 1 -
$150
Option 2 -
$147
Option 3 -
$143
Option 4 -
Cannot be determined
-
1 Answer
-
Correct Option - 3
Detailed Solution:For this we need to minimize the cost for each of the task, then we will find the minimum cost.
Minimum cost = $23 (Child 4 doing Going for Morning walk) + $37 (Child 5 doing Exercise) + $28 (Child 1 doing home work) + $21 (Child 2 drink Milk) + $34 (Child 3 family work) = $143
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Please find the answer below
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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