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Marks Obtained by 40 Students in a Short Assessment is Given Below, Where a and B Are Two Missing Data. If the Mean of the Distribution is 7.2, Find a and B. - ICSE Class 10 - Mathematics

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Question

Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.

Marks 5 6 7 8 9
Number of Students 6 a 16 13 b

If the mean of the distribution is 7.2, find a and b.

Solution 1

Marks (x) Number of students (f) fx
5 6 30
6 a 6a
7 16 112
8 13 104
9 b 9b
  `sumf = 34 + a + b` `sumfx = 246 + 6a + 9b`

It is given that the number of students is 40.

∴ 35 + a + b = 40

⇒ a + b – 5 = 0      ....(1)

`Mean = (sum fx)/(sum f)`

`=> (246 + 6a + 9b)/(35 + a + b) = 7.2`

`=> 246 + 6a + 9b = 7.2(35 + a + b)`

`=> 246 + 6a + 9b = 252 + 7.2a  + 7.2b`

`=> 0 = 252 - 246 + 7.2a - 6a + 7.2b - 9b`

`=> 6 + 1.2a - 1.8b = 0`

`=> 10 + 2a - 3b = 0`    .... (2)

Solving equations (1) and (2), we have

5a - 5 = 0

`=> a = 1`

From (1), we have b = 4

Hence, the values of a and b are 1 and 4 respectively.

Solution 2

Mean=`(∑ fx)/(∑f)` 

⇒`7.2=(6xx5+axx6+16xx7+13xx8+bxx9)/(6+a+16+13+b)`  

⇒`7.2=(246+6a+9b)/(35+a+b)` 

⇒`1.2-1.8b=-6`..............(i)  

Total number of students= `6+a+16+13+b ` 

⇒`40=35+a+b` 

⇒`a+b=5`..............(ii) 

Multiple equation (ii) by 1.8 and add it to equation (i)

1.8a + 1.8b =  9
1.2a - 1.8b = -6
3a = 3

⇒ a = 1 

Substituting = 1 in equation (ii) we get,

1+b =5 

⇒ b = 4

  Is there an error in this question or solution?

APPEARS IN

 Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 24: Measure of Central Tendency(Mean, Median, Quartiles and Mode)
24E | Q: 21
 2011-2012 (March) (with solutions)
Question 3.3 | 4.00 marks
Solution Marks Obtained by 40 Students in a Short Assessment is Given Below, Where a and B Are Two Missing Data. If the Mean of the Distribution is 7.2, Find a and B. Concept: Measures of Central Tendency - Mean, Median, Mode for Raw and Arrayed Data.
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