Question

9. In a class of 100 students, the mean marks obtained in a subject is 30 and in another class of 50 students
the mean marks obtained in the same subject is 60. The mean marks obtained by the students of two
classes taken together is :
a) 40
(c) 48
(d) 50
(b) 45
12 ore What​

Answer 1

Required Answer:-

Mean or Average is calculated by dividing the total sum of all items by the total number of items.

In this question, we have to add all the marks and divide it by total number of students.

Class - (1)

No. of students = 100

Mean marks = 30

Then, Total marks = 100 × 30 = 3000

Class - (2)

No. of students = 50

Mean marks = 60

Total marks = 50 × 60 = 3000

Find the mean marks of both classes:

= Total marks / Total no. of students

= 3000 + 3000 / 100 + 50

= 6000 / 150

= 40

$\therefore$ The mean marks of two classes taken together is 40 (Option A).

Note:-

• Sometimes average is asked instead of mean. Both are synonymous to each other.
• Always add the all total at the last for final answer.

Answer 2

$\: \: \: \: \:{\Large{\bf{\underbrace{Required \; answer}}}}$

${\large{\bold{\rm{\underline{Question}}}}}$

⚕ In a class of 100 students, the mean marks obtained in a subject is 30 and in another class of 50 students the mean marks obtained in the same subject is 60. The mean marks obtained by the students of two classes taken together is :

• (a) 40
• (b) 45
• (c) 48
• (d) 50

${\large{\bold{\rm{\underline{Given \; that}}}}}$

$\; \; \; \; \; \; \; \; \; \;{\sf{Class \: - \: A}}$

✴ There are 100 students.

✴ In exams the mean marks obtained in a subject is 30.

$\; \; \; \; \; \; \; \; \; \;{\sf{Class \: - \: B}}$

✴ There are 50 students.

✴ In exams the mean marks obtained in a subject is 60.

${\large{\bold{\rm{\underline{To \; find}}}}}$

✴ The mean marks obtained by the students of two classes taken together.

${\large{\bold{\rm{\underline{Solution}}}}}$

✴ The mean marks obtained by the students of two classes taken together = 40 ; option (a)

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

$\; \; \; \; \; \; \; \; \; \;{\sf{Class \: - \: A}}$

✴ There are 100 students.

✴ In exams the mean marks obtained in a subject is 30.

✴ Total marks = 100 × 30 = 3000

$\; \; \; \; \; \; \; \; \; \;{\sf{Class \: - \: B}}$

✴ There are 50 students.

✴ In exams the mean marks obtained in a subject is 60.

✴ Total marks = 60 × 50 = 3000

$\; \; \; \; \;{\sf{Let's \: find \: mean \: together}}$

✴ Mean = Total sum of items / Total number of items.

${\sf{\rightarrow \dfrac{Total \: marks}{Total \: no. \: of \: students}}}$

$\: \: \: \:{\sf{\rightarrow \dfrac{3000+3000}{100+50}}}$

$\: \: \: \: \: \:{\sf{\rightarrow \dfrac{6000}{150}}}$

$\: \: \:{\sf{\rightarrow \dfrac{600}{15}}}$

${\sf{\rightarrow 40}}$

${\large{\bold{\rm{\underline{Additional \; knowledge}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

${\sf{Where,}}$

$\; \; \;{\tt{\rightarrow a \: denotes \: side \: or \: height}}$

$\; \; \;{\tt{\rightarrow b \: denotes \: base}}$