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New answer posted
7 months agoContributor-Level 10
(1) Roll number 1 and 8 have MA, M.Com and BBA as common.
(2) Roll number 3 and 6 have MA, MSC, B.Com as common.
(3) Roll number 9 and 10 have all the four common options.
(4) Roll number 2 and 5 have MSC, M.Com and MBA as common.
New answer posted
7 months agoContributor-Level 10
(1) Roll number 3 and 6
(2) Roll number 1, 4 and 8
(3) Roll number 9, 10
(4) Roll number 7
New answer posted
7 months agoContributor-Level 10
In the first question, we derived the number of females in class secondary section as 288 and the total number of students in the section as 640.
New answer posted
7 months agoContributor-Level 10
Non-vegetarian Females in Class 12 = 0.25 * 32 = 8
Vegetarian Males in class 12 = 48 – 8 = 40
Non-vegetarian Males in class 12 = Non-vegetarian in class 12 – Female non-vegetarian in class 12
= 32 – 8 = 24
Required difference = 40 – 24 = 16
New answer posted
7 months agoContributor-Level 10
From the table given in the question,
Total student = 800
Students in Secondary = 0.8 * 800 = 640
Students in Class 11 = (800 – 640)/2 = 80
Student in Class 12 = 80
Females in Class 11 = 0.55 * 80 = 44
Females in Class 12 = 0.6 * 80 = 48
Females in Secondary = 0.475 * 800 – 44 – 48 = 288
Non-vegetarian in Class 11 = 0.5 * 80 = 40
Non-vegetarian in Secondary = 0.55 * 640 = 352
Non-vegetarian in Class 12 = 800 * 0.53 – 40 – 352 = 32
The percentage of non-vegetarian in class 12 = 32 * 100/80 = 40%
New answer posted
7 months agoContributor-Level 10
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Cities | Max | Min | Avg | Max | Min | Avg | Max | Min | Avg | Max | Min | Avg | Max | Min | Avg | Max | Min | Avg |
Ajmer | 12.0 | 4.0 | 5.6 | 14.0 | 7.0 | 11.2 | 19.0 | 10.0 | 14.5 | 25.0 | 10.0 | 17.5 | 29.0 | 12.0 | 23.9 | 31.0 | 14.0 | 25.9 |
Ayodhya | 6.0 | 1.0 | 2.0 | 8.0 | 3.0 | 6.0 | 10.0 | 5.0 | 7.5 | 14.0 | 9.0 | 11.5 | 16.0 | 11.0 | 14.5 | 19.0 | 14.0 | 17.5 |
Mathura | 20.0 | 12.0 | 13.6 | 24.0 | 13.0 | 19.6 | 29.0 | 17.0 | 23.0 | 36.0 | 21.0 | 28.5 | 38.0 | 24.0 | 33.8 | 41.0 | 27.0 | 36.8 |
Kashi | 13.0 | 7.0 | 8.2 | 16.0 | 10.0 | 13.6 | 19.0 | 14.0 | 16.5 | 21.0 | 16.0 | 18.5 | 24.0 | 18.0 | 22.2 | 26.0 | 21.0 | 24.5 |
New answer posted
7 months agoContributor-Level 10
As not two pipelines connecting same place are carrying same amount, and at least one of the pipelines should carry same as that of the place, either H-I or G-H must carry 300. If G-H carries 300, then H-I must carry 200, which is not possible.
H-I must carry 300 and G-H must carry 400. D-G should carry 400 more than the requirement at G.
G's requirement must equal that of G-H
Requirement at G = 400 D-G = 800
The flow in El must be equal to the requirement at E, and each of them must be equal to 100. Therefore the requirement at I = 100 + 300 = 400.
Considering F, the flow in FB must be equal to the requirement at F, i.e., 100, and the flo
New answer posted
7 months agoContributor-Level 10
Son | Color of shirt | Likes |
A | yellow | writing |
K | white/blue/green | singing |
S | red | traveling |
R | blue/white | reading |
N | green/white | playing |
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