Question

In examination, there are 100 question divided into a
three groups A, B and c such that each group
contains at least one question. Each question in
gooup A a carries 1 mark each question in group B
carries 2 Marks and each in C is carries 3 marks.
it is known that the question in group A together
carry at least 60% of the total marks.

so if group B contains 23 questions, then how many
question are in group c
(a)1
(b)2
( c)3
(d) cannot be determined

If group c contains 8 questions and group B carries
at least 20% of the total marks , which of the
following best describes the number of question in
grouр В?
(a) 11 or 12
(b) 13 or 14
(c) 12 or 3
(D) 19 or 15​

Answer 1

Answer:

• the questions in group c cannot be determined.
• the group B contains 11 or12 questions

Step-by-step explanation:

• because bth the groups made more than 100 marks.
• because group b contains 20% of the total marks.

Answer 2

Step-by-step explanation:

• So let the questions in A be p, B be q and C be r.                              So we get p + q + r = 100.

• Now questions A has 1 mark, B has 2 marks and C has 3 marks. So total marks will be p + 2q + 3z.
• Now p + q + r = 100
• According to the question q = 23, so we get
•          So p + r = 100 – 23
•             Or p + r = 77
•               Or r = 77 – p
•     Now total marks will be p + 2q + 3r = p + 2(23) + 3r
•                                                                = p + 46 + 3r
•         Now r = 77 – p, substituting we get
• Total marks = p + 46 + 3r = p + 46 + 3(77 – p)
•                                            = p + 46 + 231 – 3p
•                                            = 277 – 2p
• According to the question A needs to carry 60% of the total marks.
• So p >= 60/100 x (277 – 2p)
• Or p >= 3/5 (277 – 2p)
• 5p >= 831 – 6p
•    Or 11 p >= 831
•     Or p >= 75.54
• So minimum value of p is 76
• So now q = 23, therefore 100 – 76 = 24
•                                             24 – 23 = 1
• So r is 1
• Now for the second part we have
• So r = 8 questions
• So p + q = 92 (100 – 8 = 92)
•     Or q = 92 – p
•     Now total marks will be p + 2q + 3r = p + 2q + 3(8)
•                                                                = p + 2q + 24
•         Now q = 92 – p, substituting we get
• Total marks = p + 2q + 24 = p + 2(92 – p) + 24
•                                            = p + 184 + 24 – 2p
•                                            = 208 – p
• According to the question A needs to carry 60% of the total marks.
• So p >= 60/100 x (208 – p)
• Or p >= 3/5 (208 – p)
• 5p >= 624 – 3p
•    Or 8p >= 624
•     Or p >= 78
• Therefore p will be 78,79 and so on.
• So maximum value of q is 14.
• Now substituting 14 for q we get total marks in q will be 28, so that the value is 20% of the total marks

Reference link will be

https://brainly.in/question/27772870