# Chapter-Test Measures of Central Tendency ML Aggarwal ICSE Class-10 Solutions

Chapter-Test Measures of Central Tendency ML Aggarwal ICSE Class-10 Solutions Chapter-21 . We Provide Step by Step Answer of Exe-21.1, Exe-21.2, Exe-21.3, Exe-21.4, Exe-21.5, with MCQs and Chapter-Test of **Measures of Central Tendency** Questions / Problems related for ICSE Class-10 APC Understanding Mathematics . Visit official Website **CISCE ** for detail information about ICSE Board Class-10.

Board | ICSE |

Publications | Avichal Publishing Company (APC) |

Subject | Maths |

Class | 10th |

Chapter-21 | Measures of Central Tendency |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe – 21.1 (Mean), Exe – 21.2 (Median) , Exe – 21.3 (Mode) , Exe – 21.4 (Histogram), Exe – 21.5 (Ogive), Exe –21.6, with MCQ Questions and Chapter-Test |

Academic Session | 2021-2022 |

**How to Solve Measures of Central Tendency Problems/Questions / Exerci****se of ICSE Class-10 Mathematics**

Before viewing Answer of Chapter-21 **Measures of Central Tendency **of ML Aggarwal Solutions. Read the Chapter Carefully with formula and then solve all example of Exe-21.1, Exe-21.2, Exe-21.3, Exe-21.4, Exe-21.5, Exe-21.6 given in your text book .

For more practice on **Measures of Central Tendency** related problems /Questions / Exercise try to solve **Measures of Central Tendency** exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE) **Graphical Representations of Statistical Data, Measure of Central Tendency (Mean)** and also Concise Selina Publications ICSE Mathematics **Graphical Representation (Histograms and Ogives), ****Measures of Central Tendency (Mean , Median, Quartiles and Mode)**. Get the Formula of **Measures of Central Tendency** for ICSE Class 10 Maths to understand the topic more clearly in effective way.

**Chapter- Test , Measures of Central Tendency ML Aggarwal Solutions**

## Chapter Test MCQ Measures of Central Tendency ML Aggarwal ICSE Class-10 Solutions

**Page 521-522**

**Question 1 :**The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.

**Answer 1 :**

Total number = 18 × 20 = 360

By adding 3 to first 10 numbers,

The new sum will be = 360 + 3 × 10 = 360 + 30 = 390

New Mean = (390/20) = 19.5

**Question 2 :**The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.

**Answer 2 :**In first case,

Average height of 30 students = 150 cm

Total height = 150 × 30 = 4500 cm

Difference in copying the number = 165 – 135 = 30 cm

Correct sum = 4500 + 30 = 4530 cm

Correct mean = (4530/30) = 151 cm

**Question 3**

There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.

**Answer 3**

Total students of a class = 50

No. of boys = 40

No. of girls = 50 – 40 = 10

Average weight of 50 students = 44 kg

Total weight = 44 × 50 = 2200 kg

Average weight of 10 girls = 40 kg

.’. Total weight of girls = 40 × 10 = 400 kg

Then the total weight of 40 boys = 2200 – 400 = 1800kg

Average weight of boys = 188/40 = 45kg

Page 522

**Question 4**

The heights of 50 children were measured (correct to the nearest cm) giving the following results :

**Answer 4**

Calculate the mean height for this distribution correct to one place of decimal.

Mean = Ʃfx/Ʃf

= 3459/50

= 69.18 = 69.2

**Question 5 Chapter-Test Measures of Central Tendency ML **

Find the value of p if the mean of the following distribution is 18.

**Answer 5**

Mean = Ʃf_{i} x_{i} / Ʃf_{i}

18 = (399+5p^{2}+100p)/( 23+5p) [Given mean = 18]

18(23+5p) = 399+5p^{2}+100p

414 + 90p = 399+5p^{2}+100p

5p^{2}+100p-90p+399-414 = 0

5p^{2}+10p-15 = 0

Dividing by 5, we get

p^{2}+2p-3 = 0

(p-1)(p+3) = 0

p-1 = 0 or p+3 = 0

p = 1 or p = -3

p cannot be negative.

So p = 1

Hence the value of p is 1.

**Question 6**

Find the mean age in years from the frequency distribution given below:

**Answer 6**

Arranging the classes in proper form

**Question 7**

The mean of the following frequency distribution is 62.8. Find the value of p.

**Answer 7**

Mean = 62.8

Hence p = 10

**Question 8**

The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.

**Answer 8**

Mean = 188,

No. of families = 100

Given no. of families = 100

So 35+f_{1}+f_{2 }= 100

f_{1}+f_{2 }= 100-35 = 65

f_{1 }= 65-f_{2} ..(i)

Mean = Ʃf_{i} x_{i} / Ʃf_{i}

188 = (6150+190f_{1}+210f_{2})/100 [Given mean = 188]

188(100) = 6150+190f_{1}+210f_{2}

18800 = 6150+190f_{1}+210f_{2}

18800-6150 = 190f_{1}+210f_{2}

12650 = 190f_{1}+210f_{2} ..(ii)

Substitute (i) in (ii)

12650 = 190(65-f_{2})+210f_{2}

12650 = 12350-190f_{2}+210f_{2}

12650-12350 = -190f_{2}+210f_{2}

300 = 20f_{2}

f_{2} = 300/20 = 15

Put f_{2} in (i)

f_{1 }= 65-15

f_{1 }= 50

Hence the value of f_{1} and f_{2} is 50 and 15 respectively_{.}

**Question 9**

The median of the following numbers, arranged in ascending order is 25. Find x, 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46

**Answer 9**

Here, n = 10, which is even

But median is given = 25

So, x + 3 = 25

x = 25 – 3

= 22

**Question 10**

If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.

**Answer 10**

Arranging in ascending order, 3, 4, 5, x, 8, 9, 11,

Here n = 7 which is odd.

∴ Median = (n + 1)/2 th term = (7 + 1)/2 = 4th term = x

but median = 6

Hence ∴ x = 6

**Question 11**

The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12

Find

(i) the median

(ii) lower quartile

(iii) upper quartile

**Answer 11**

Arranging the given data in ascending order:

1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23

Here n = 16 which is even.

(i) So median = ½ ( n/2 ^{th} term + ((n/2)+1)^{th} term)

= ½ (16/2 ^{th} term + ((16/2)+1)^{th} term)

= ½ (8 ^{th} term + (8+1)^{th} term)

= ½ (8 ^{th} term + 9^{th} term)

= ½ (12+13)

= ½ ×25

= 12.5

(ii) Lower quartile, Q_{1} = (n/4)^{ th} term

= (16)/4

= 4 ^{th }term

= 6

(iii)Upper quartile, Q_{3} = (3n/4)^{ th} term

**= (**3×16/4)^{ th} term

= (3×4)^{th} term

**= **12 ^{th }term

= 18

**Question 12**

Calculate the mean, the median and the mode of the following distribution :

**Answer 12**

**Question 13**

The daily wages of 30 employees in an establishment are distributed as follows :

Estimate the modal daily wages for this distribution by a graphical method.

**Answer 13**

Taking daily wages on x-axis and No. of employees on the y-axis

and draw a histogram as shown. Join AB and CD intersecting each other at M.

From M draw ML perpendicular to x-axis, L is the mode

∴ Mode = Rs 23

**Question 14 Chapter-Test Measures of Central Tendency ML **

Draw a cumulative frequency curve for the following data :

Hence determine:

(i) the median

(ii) the pass marks if 85% of the students pass.

(iii) the marks which 45% of the students exceed.

**Answer 14**

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