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New answer posted
4 months agoContributor-Level 10
Number of refineries = 6
Number of depots = 7
Number of districts = 9
Therefore, number of possible ways to send petrol from any refinery to any district is 6 * 7 * 9 = 378.
New question posted
4 months agoNew answer posted
4 months agoContributor-Level 10
Total number of excellent students = 120
Number of female excellent students = 70
Required fraction
New answer posted
4 months agoContributor-Level 10
Number of female good students = 8
Number of males average students = 48
Required ratio = 8 : 48 = 1 : 6
New answer posted
4 months agoContributor-Level 10
From the above table, it is clear that number of female good students = 8
New answer posted
4 months agoContributor-Level 10
From (II), 40% of total number of students = 96
Total number of students = (96/40)*100=240
Number of boys = 240 – 96 = 144
From (I), number of average male students =144/3=48
From (III), number of excellent students = 240/2= 120
Number of excellent female = 120 – 50 = 70
From these calculations, we get the following table.
| Performance | Total | ||
Average | Good | Excellent | ||
Male | 48 | 46 | 50 | 144 |
Female | 18 | 8 | 70 | 96 |
Total | 66 | 54 | 120 | 240 |
From the above table, it is clear that number of male good students = 46
New answer posted
4 months agoContributor-Level 10
The answer can be obtained directly by increasing the answer in question 11 by 50%, as all the Children increase the cost by 50% and hence the total cost will increase by 50% only.
New answer posted
4 months agoContributor-Level 10
Note:
C1 Chosen 1 (Doing First 2 works)
C2 Chosen 2 (Doing remaining 3 works)
GM Going for Morning walk
EX Exercise
CH Completing their Homework
DM Drinking Milk
AF Assisting in Family work
ToS Total Cost
C1 | C2 | GM | EX | CH | DM | AF | ToS |
1 | 2 | 35 | 48 | 55 | 21 | 35 | 194 |
1 | 3 | 35 | 48 | 52 | 26 | 34 | 195 |
1 | 4 | 35 | 48 | 38 | 37 | 51 | 209 |
1 | 5 | 35 | 48 | 44 | 42 | 47 | 216 |
2 | 1 | 31 | 42 | 28 | 55 | 26 | 182 |
3 | 1 | 44 | 57 | 28 | 55 | 26 | 210 |
4 | 1 | 23 | 35 | 28 | 55 | 26 | 167 |
5 | 1 | 36 | 37 | 28 | 55 | 26 | 182 |
2 | 3 | 31 | 42 | 52 | 26 | 34 | 185 |
2 | 4 | 31 | 42 | 38 | 37 | 51 | 199 |
2 | 5 | 31 | 42 | 44 | 42 | 47 | 206 |
3 | 2 | 44 | 57 | 55 | 21 | 35 | 212 |
4 | 2 | 23 | 35 | 55 | 21 | 35 | 169 |
5 | 2 | 36 | 37 | 55 | 21 | 35 | 184 |
3 | 4 | 44 | 57 | 38 | 37 | 51 | 227 |
3 | 5 | 44 | 57 | 44 | 42 | 47 | 234 |
4 | 3 | 23 | 35 | 52 | 26 | 34 | 170 |
5 | 3 | 36 | 37 | 52 | 26 | 34 | 185 |
4 | 5 | 23 | 35 | 44 | 42 | 47 | 191 |
5 | 4 | 36 | 37 | 38 | 37 | 51 | 199 |
Choosing Child 4 and Child 1 will be the best option for John.
New answer posted
4 months agoContributor-Level 10
For this we need to maximize the cost each of the task, then we will find the maximum cost.
Maximum Cost = $36 (Child 5 doing Going for Morning walk) + $57 (Child 3 doing Exercise) + $55 (Child 2 doing home work) + $55 (Child 1 drink Milk) + $51 (Child 4 family work) = $254
New answer posted
4 months agoContributor-Level 10
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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