Frequency Distribution MCQs Class 9 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-15. In this article you will learn how to solve Multiple Choice Questions on Pythagoras Theorem very easily**.** Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics.

**Frequency Distribution MCQs Class 9 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-15**

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-15 | Frequency Distribution |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of MCQs |

Academic Session | 2024-2025 |

**Multiple Choice Questions**

**( Frequency Distribution MCQs Class 9 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-15 )**

**Que-1: If in a grouped frequency distribution, the classes are 20-30, 30-40, 40-50, ……., then the observation 40 is included in the class**

(a) 30-40 (b) 40-50 (c) 50-60 (d) 20-30

**Sol: **(b) 40-50

**Reason: **In a grouped frequency distribution, the observation that lies on the boundary between two classes is typically included in the upper class. This depends on whether the classes are closed or open at the boundary. If the intervals are continuous, the class 40-50 would include 40.

Thus, the observation 40 would be included in the class:

(b) 40-50

**Que-2: If in a grouped frequency distribution, the class intervals are 20-29, 30-39, 40-49, ………, then the observation 29.4 is included in the class :**

(a) 20-29 (b) 30-39 (c) 40-49 (d) 29-30

**Sol: **(a) 20-29

**Reason: **In a grouped frequency distribution, the observation that lies on the boundary between two classes is typically included in the upper class. This depends on whether the classes are closed or open at the boundary. If the intervals are continuous, the class 20-29 would include 29.4.

Thus, the observation 29.5 would be included in the class:

(a) 20-29

**Que-3: The classes of a frequency distribution are 30-34, 35-39, …….., 50-54. The lower boundary of the class 35-39 is :**

(a) 30.5 (b) 35 (c) 34.5 (d) 39

**Sol: **(c) 34.5

**Reason: **Here, the classes are 30-34 and 35-39. To find the lower boundary of the class 35-39:

Upper limit is 34

Lower limit is 35

Lower boundary = [Upper limit of the previous class + Lower limit of the current class]/2

Lower boundary =(34+35)/2

Lower boundary = 69/2 = 34.5

**Que-4: In a grouped frequency distribution, the class intervals are 40-49, 50-59, …….., 80-89. The upper boundary of the class (50-59) is :**

(a) 59 (b) 59.5 (c) 49.5 (d) 60

**Sol: **(b) 59.5

**Reason: **Here, the classes are 50-59 and 60-69. To find the lower boundary of the class 50-59 :

Upper limit is 59

Lower limit is 60

Lower boundary = [Upper limit of the previous class + Lower limit of the current class]/2

Lower boundary =(59+60)/2

Lower boundary = 119/2 = 59.5

**Que-5: If 0-9, 10-19, 20-29, …….., are the classes of a grouped frequency distribution, then the width of each class is**

(a) 9 (b) 9.5 (c) 10 (d) 9 or 10

**Sol: **(c) 10

**Reason: **The width of the class is also called class interval

For class 0-9:

Class width = Upper limit – Lower limit

= 9 – 0 = 9

Thus, the width of each class is 10.

**Que-6: If 10,15,20,……. are respectively the mid-values of the classes of a grouped frequency distribution, then the class whole mid-value is 25 is :**

(a) 20-27 (b) 22.5-27.5 (c) 21.5-28.5 (d) 21-28

**Sol: **(b) 22.5-27.5

**Reason: **Lower limit = 22.5

Upper limit = 27.5

Mid value = (Lower limit + Upper limit)/2

= (22.5+27.5)/2

= 50/2 = 25.

**Que-7: The width of each class interval of a grouped frequency distribution is 8. The lower boundary of the class whole mid-value is 10 is**

(a) 2 (b) 4.5 (c) 5.5 (d) 6

**Sol: **(d) 6

**Reason: **Mid-value of class is 10

The width of the class interval is 8

Lower Limit is ‘L’

Upper limit is ‘U’

Mid value = (L+U)/2

10 = (L+U)/2

L+U = 20 ……….. (1)

U-L = 8 …………… (2)

Add the equation (1) and (2)

(L+U) + (U-L) = 20+8

2U = 28

U = 14

Now substitute U in L+U is

L + 14 = 20

L = 20 – 14

L = 6.

**Que-8: The width of each class of a frequency distribution is 5. The upper boundary of the class whose class mark is 25 is :**

(a) 22 (b) 22.5 (c) 27 (d) 27.5

**Sol: **(d) 27.5

**Reason: **Class marks = 25

Class width = 5

Class mark = (L+U)/2

25 = (L+U)/2

L+U = 50 …………(1)

U-L = 5 ………….. (2)

Add the equation (1) and (2)

(L+U) + (U-L) = 50+5

2U = 55

U = 27.5.

**Que-9: Class marks of distribution are : 47, 52, 57, 62, 67, 72, 77, 82. Class size of the given distribution is :**

(a) 5 (b) 6 (c) 7 (d) 2

(a) 5 (b) 6 (c) 7 (d) 2

**Sol: **(a) 5

**Reason: **Given class marks: 47, 52, 57, 62, 67, 72, 77, 82.

we can subtract two consecutive class marks. For example:

52−47 = 5

Similarly :

57−52 = 5, 62−57 = 5, and so on

Thus, the class size (width) of the given distribution is: 5

**Que-10: Mid value for the class intervals 19.5-29.5 is :**

(a) 14.5 (b) 24.5 (c) 15.5 (d) 25.5

**Sol: **(b) 24.5

**Reason: **Lower limit = 19.5

Upper limit = 29.5

Mid value = (Lower limit + Upper limit)/2

= (19.5+29.5)/2

= 49/2 = 24.5.

–: Frequency Distribution MCQs Class 9 RS Aggarwal Goyal Brothers ICSE Maths Solutions :–

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