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MCQ Questions for Class 12 Informatics Practices – Python Pandas Part 2

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Python Pandas Class 12 MCQ Questions Part 2

Question 1.

Which of the following statement is correct for importing pandas in python?

(a) import pandas
(b) import pandas as pd
(c) import pandas as pds
(d) All of these
Answer:
(d) All of these

Question 2.

What type of error is returned by following statement? import pandas as pnd pn(d)Series([l,2,3,4], index = [aVb’,’c’])

(a) SyntaxError
(b) IndexError
(c) ValueError
(d) None of these
Answer:
(c) ValueError

Question 3.

Which attribute is used to give user defined labels in Series?

(a) Index
(b) Data
(c) Values
(d) None of these
Answer:
(a) Index

Question 4.

Fill in the blank to get the ouput as 3 import pandas as pnd Sl=pn(d)Series([l,2,3,4], index = [‘a’,’b’,’c’,’d’]) print(Sl[ ])

(a) ‘c’
(b) 2
(c) c
(d) All of these
Answer:
(d) All of these

Question 5.

Write the statement to get NewDelhi as output using positional index, import pandas as pd SI=p(d)Series([‘NewDelhi’,’WashingtonDC’,’London’,’Paris’], index=[‘lndia’, ‘USA’, ‘UK’, ‘France’])

(a) print(S1(0])
(b) print(S1(‘lndia’])
(c) Both of the above
(d) print(Sl.lndia)
Answer:
(a) print(S1(0])

Question 6.

We can access elements in Series by using …… index and index.

(a) Numeric, Labelled
(b) Positional, Naming
(c) Positional, Labelled
(d) None of these
Answer:
(c) Positional, Labelled

Question 7.

Write the output of the following: import pandas as pd
SI = p(d)Series([‘NewDelhi’,’WashingtonDC’, ‘London’, ’Paris’], index=[’lndia’, ‘USA’, ‘UK’, ‘France’]) print(Sl[‘lndia’,’UK’])

(a) India NewDelhi UK London dtype: object
(b) India NewDelhi UK Washington dtype: object
(c) Error
(d) None of these
Answer:
(c) Error

Question 8.

Which of the following statement will print Series ‘SI’ in reverse order?

(a) print(S1(:: 1]
(b) print(S1(:: -1]
(c) print(S1(-1:: 1]
(d) print(S1.reverse())
Answer:
(b) print(S1(:: -1]

Question 9.

How many values will be modified by last statement of given code? import pandas as pd
SI=p(d)Series([‘NewDelhi’,’WashingtonDC’,’London’,’Paris’], index=[‘A’, ‘B’, ‘C’, ‘D’]) S1[A’: C’] = ‘ND’

(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 10.

How many values will be modified by last statement of given code? import pandas as pd
SI = p(d)Series([’NewDelhi’, ’WashingtonDC’, ‘London’, ‘Paris’], index=[‘A’,’B’,’C’,’D’]) Sl[l: 3] = ‘ND’

(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 11.

Which of the following property/attribute assign name to the Series?

(a) name
(b) index.name
(c) size
(d) Series.name
Answer:
(a) name

Question 12.

Sl.vatues will return all the values of Series ‘SI’ in

(a) Tuple
(b) Dictionary
(c) List
(d) String
Answer:
(c) List

Question 13.

Which of the following property/attribute return total number of values in Series‘SI’?

(a) Size
(b) Values
(c) lindex
(d) None of these
Answer:
(a) Size

Question 14.

Which of the following attributes returns True if there is no value in Series?

(a) Index
(b) Size
(c) Empty
(d) Values
Answer:
(c) Empty

Question 15.

Which of the following attributes returns all the values of Series?

(a) Size
(b) Index
(c) Name
(d) Values
Answer:
(d) Values

Question 16.

Write the output of the following code: import pandas as pd Sl=p(d)Series() print(p(d)Series().empty)

(a) True
(b) False
(c) Error
(d) None of these
Answer:
(a) True

Question 17.

Write the output of the following code : import pandas as pd Sl=p(d)Series([l,2,3,4])
S2=p(d)Series([7,8]) S3=S1+S2 print(S3.size) (a) 2 (c) 6

Answer:
(a) 2
(b) 4
(c) 6
(d) Error
Answer:
(b) 4

Question 18.

Which of the following statement shows first five values of Series ‘SI’?

(a) S1.head( )
(b) S1.head( 5 )
(c) Both of the above
(d) None of these
Answer:
(c) Both of the above

Question 19.

Write the output of the following: import pandas as pd Sl=p(d)Series([l,2,3,4])
S2=p(d)Series([7,8]) print((Sl+S2).count()) .

(a) 6
(b) 4
(c) 2
(d) 0
Answer:
(c) 2

Question 20.

Which of the following returns number of non-NaN values of Series?

(a) count
(b) size
(c) index
(d) values
Answer:
(a) count

Question 21.

Write the output of the following: import pandas as pd Sl=p(d)Series([l,2,3,4])
S2=p(d)Series([7,8,9,10]) S2.index=[‘a’,’b’,’c’,’d’] print{{Sl+S2).count( ))

(a) 8
(b) 4
(c) 0
(d) 6
Answer:
(c) 0

Question 22.

We can perform on two series in Pandas.

(a) addition
(b) subtraction
(c) multiplication
(d) All of these
Answer:
(d) All of these

Question 23.

Which of the following method is used to add two series?

(a) sum( )
(b) addition( )
(c) add( )
(d) None of these
Answer:
(c) add( )

Question 24.

Which of the following statement will display the difference of two Series ‘A’ and ‘B’?

(a) >>>A-B
(b) >>>(a)sub(B)
(c) Both of the above
(d) None of these
Answer:
(c) Both of the above

Question 25.

Which of the following fills the missing values in Series?

(a) fill-value
(b) fill-value
(c) fill_value
(d) fill_value( )
Answer:
(c) fill_value

Question 26.

Which of the following function is used for basic mathematical operations in Series?

(a) add( )
(b) mul( )
(c) div( )
(d) All of these
Answer:
(d) All of these

Question 27.

Which of the following statement is replacing missing values of Series A and Series B by 100?

(a) >>> (a)add(B, fill_value = 100)
(b) >>>(a)add(B, fill_value : 100)
(c) >>> (a)add(B, fill-value = 100)
(d) >>> (a)add(B, fill-value : 100)
Answer:
(a) >>> A,add(B, fill_value = 100)

Question 28.

Mathematical operations on two Series object is done by matching

(a) indexes
(b) values
(c) Both of the above
(d) None of these
Answer:
(a) indexes

Question 29.

Which of the following statement will display values more ‘ than 40 from Series ‘SI’?

(a) >>>S1
(b) >> SI > 40
(c) >>>S1(S1 > 40]
(d) None of these
Answer:
(c) >>> S1(S1 > 40]

Question 30.

Which of the following statement will return 10 values from the bottom/end of the Series ‘SI’?

(a) S1.tail()
(b) S1.tail(10)
(c) S1.head(10)
(d) S1(10)
Answer:
(b) S1.tail(10)

Question 31.

Which of the following are valid operations on Series ‘SI’?

(a) >>>S1 + 2
(b) >>>S1**2
(c) >>> S1*2
(d) All of these
Answer:
(d) All of these

Question 32.

When an operation is carried out on every value of Series object is called

(a) Scalar operation
(b) Vector operation
(c) Both of the above
(d) None of these
Answer:
(b) Vector operation

Question 33.

Which of the following statement will modify the first three values of Series ‘SI’?

(a) S1[0,1, 2] = 100
(b) S1[0 : 3] = 100
(c) S1[: 3] = 100
(d) All of these
Answer:
(d) All of these

Question 34.

We can have duplicate indexes in Series?

(a) True
(b) False
(c) On
(d) Off
Answer:
(a) True

Question 35.

What is the index value of 31 in given Series ‘SI’? import pandas as pd
Sl=p(d)Series([l,2,31,4], index = [’a’.’bVcVd’])

(a) ‘c’
(b) 2
(c) Both of the above
(d) None of these
Answer:
(c) Both of these

Question 36.

S2 is ……..in given code.
S2 = Sl[2 : 5] #S1 is a Series object

(a) list
(b) tuple
(c) series
(d) None of these
Answer:
(c) series

Question 37.

Following two statements will provide the same output:
>>>L1 “ 2 #L1 is a list
>>>S1 “ 2 #S1 is a Series

(a) True
(b) False
(c) On
(d) Off
Answer:
(b) False

Question 38.

Best way to import the pandas module in your program:

(a) import pandas
(b) import pandas as p
(c) from pandas import
(d) All of these
Answer:
(d) All of these

Question 39.

Dataframe in pandas is:

(a) 1 dimensional array
(b) 2 dimensional array
(c) 3 dimensional array
(d) None of these
Answer:
(b) 2 dimensional array

Question 40.

For what purpose a Pandas is used?

(a) To create a GUI programming
(b) To create a database
(c) To create a High level array
(d) All of the above
Answer:
(c) To create a High level array

Question 41.

In data science, which of the python library are more popular?

(a) Numpy
(b) Pandas
(c) Open CV
(d) Django
Answer:
(b) Pandas

Question 42.

Minimum number of argument we require to pass in pandas series?

(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 43.

Series in Pandas is:

(a) 1 dimensional array
(b) 2 dimensional array
(c) 3 dimensional array
(d) None of these
Answer:(a) 1 dimensional array

Question 44.

Way to install the pandas library?

(a) install pandas
(b) pandas install python
(c) python install pandas
(d) None of these
Answer:
(d) None of these

Question 45.

We can analyse the data in pandas with:

(a) series
(b) dataframe
(c) Both of the above
(d) None of these
Answer:
(c) Both of the above

Question 46.

What we pass in Dataframe in pandas?

(a) Integer
(b) String
(c) Pandas series
(d) All of these
Answer:
(c) Pandas series

Question 47.

Choose the correct function to rename city columns to location using rename() function.

(a) df.rename(columns={‘City’:’Location’})
(b) df.rename(columns={‘City’=’Location’})
(c) df.rename(‘City’=’Location’)
(d) df.rename(df.columns(‘City’,’Location’))
Answer:
(a) df.rename(columns={,City’:’Location’})

Question 48.

Which of the following statement(s) is/are correct with respect to df.columns properties to rename columns? All columns must be specified . Columns must be in the form of a list Old column names not required Columns can be specified with columns number

(a) Only 1 is correct
(b) 1, 2 and 3 are correct
(c) 1 and 3 are correct
(d) All of them are correct
Answer:
(b) 1, 2 and 3 are correct

Question 49.

df.index properties ca be used to:

(a) rename rows
(b) rename columns
(c) rename rows and columns both
(d) None of the above
Answer:
(a) rename rows

Question 50.

To display 2 rows from the top in the dataframe, which of the following statement is correct?

(a) df.head( )=2
(b) df.head(n=2)
(c) df.head(range(2))
(d) All of these
Answer:
(b) df.head(n=2)

Question 51.

A function is used to display top rows (n) from …….. dataframe.

(a) Tail( )
(b) Head( )
(c) Top( )
(d) Header( )
Answer:
(b) Head( )

Question 52.

A function is used to display bottom rows (n) from dataframe.

(a) Tail( )
(b) Head( )
(c) Top( )
(d) Header( )
Answer:
(a) Tail( )

Question 53.

…… refers to passing True and False value as an index in Dataframes.

(a) Ignore_head
(b) Ignore_index
(c) Ignore_tail
(d) Ignore_header
Answer:
(b) ignore_index

Question 54.

To pass index label keyword is used to in p(d)dataframe() function.

(a) index
(b) value
(c) size
(d) empty
Answer:
(a) index

Question 55.

By default you can display……..no. of top/bottom …… rows using head()/tail() function.

(a) four
(b) six
(c) five
(d) seven
Answer:
(c) five

Question 56.

Pandas is an opensource Library?

(a) Ruby
(b) Javascript
(c) Java
(d) Python
Answer:
(d) Python
Pandas is an opensource Python Library providing high- performance data manipulation and analysis tool using its powerful data structures.

Question 57.

Pandas key data structure is called:

(a) Keyframe
(b) Dataframe
(c) Statistics
(d) Econometrics
Answer:
(b) Dataframe
Pandas is built on the Numpy package and its key data structure is called the Dataframe.

Question 58.

In pandas, Index values must be:

(a) uniQuestion ue
(b) hashable
(c) Both a and b
(d) None of these
Answer:
(c) Both a and b
Index values must be uniQuestion ue and hashable, same length as dat(a) Default np.arrange(n) if no index is passe(d)

Question 59.

Which of the following is correct features of Dataframe?

(a) Potentially columns are of different types
(b) Can perform Arithmetic operations on rows and columns
(c) Labeled axes (rows and columns)
(d) All of the above

Question 60.

A panel is a container of dat(a)

(a) ID
(b) 2D
(c) 3D
(d) Infinite
Answer:
(c) 3D
A panel is a 3D container of dat(a) The term Panel data is derived from econometrics and is partially responsible for the name pandas : pan(el)-da(ta)-s.

Question 61.

Which of the following is true?

(a) If data is an ndarray, index must be the same length as dat(a)
(b) Series is a one-dimensional labeled array capable of holdihg any data type.
(c) Both a and b
(d) None of the above
Answer:
(c) Both a and b

Question 62.

Which of the following takes a diet of diets or a diet of arraylike sequences and returns a Dataframe?

(a) Dataframe.fromjtems
(b) Dataframe.from_records
(c) Dataframe.from_dict
(d) All of the above
Answer:
(a) Dataframe.fromjtems
Dataframe.from_dict operates like the DataFrame constructor except for the orient parameter which is ‘columns’ by default.

Question 63.

Which of the following makes use of pandas and returns data in a series or dataframe?

(a) panda SDMX
(b) freedapi
(c) OutPy
(d) Inpy
Answer:
(b) freedapi
freedapi module requires a FRED API key that you can obtain for free on the FRED website.

Question 64.

What will be output for the following code? import pandas as pd import numpy as np s – p(d)Series(np.random.randn(4)) print s.ndim

(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Returns the number of dimensions of the object. By definition, a Series is a ID data structure, so it returns 1.

Question 65.

Which of the following indexing capabilities is used as a concise means of selecting data from a pandas object?

(a) In
(b) ix
(c) ipy
(d) iy
Answer:
(b) ix
ix and reindex are 100% eQuestion uivalent.

Question 66.

Which of the following thing can be data in Pandas?

(a) a python diet
(b) an ndarray
(c) a scalar value
(d) All of the mentioned
Answer:
(d) all of the mentioned
The passed index is a list of axis labels.

Question 67.

Point out the correct statement.

(a) If data is a list, if index is passed the values in data corresponding to the labels in the index will be pulled out
(b) NaN is the standard missing data marker used in pandas
(c) Series acts very similarly to a array
(d) None of the mentioned
Answer:
(b) NaN is the standard missing data marker used in pandas If data is a diet, if index is passed the values in data corresponding to the labels in the index will be pulled out.

Question 68.

The result of an operation between unaligned Series will have the of the indexes involve(d)

(a) intersection
(b) union
(c) total
(d) All of the mentioned
Answer:
(b) union
If a label is not found in one Series or the other, the result will be marked as missing NaN.

Question 69.

Which of the following input can be accepted by Dataframe?

(a) Structured ndarray
(b) Series
(c) Dataframe
(d) All of the mentioned
Answer:
(d) All of the mentioned
Dataframe is a 2-dimensional labeled data structure with columns of potentially different types advertisement.

Question 70.

Point out the wrong statement.

(a) A Dataframe is like a fixed-size diet in that you can get and set values by index label
(b) Series can be be passed into most NumPy methods expecting an ndarray
(c) A key difference between Series and ndarray is that operations between Series automatically align the data based on label
(d) None of the mentioned
Answer:
(a) A Dataframe is like a fixed-size diet in that you can get and set values by index label
A Series is like a fixed-size diet in that you can get and set values by index label.

Question 71.

Which of the following takes a diet of diets or a diet of arraylike sequences and returns a Dataframe?

(a) DataFrame.fromtems
(b) DataFrame.from records
(c) DataFrame.from dict
(d) All of the mentioned
Answer:
(a) DataFrame.fromtems
DataFrame.fromdict operates like the DataFrame constructor except for the orient parameter which is ‘columns’ by default.

Question 72.

Series is a one-dimensional labeled array capable of holding any data type.

(a) True
(b) False
(c) Error
(d) None of these
Answer:
(a) True
The axis labels are collectively referred to as the index.

Question 73.

Which of the following works analogously to the form of the diet constructor?

(a) DataFrame.fromJtems
(b) DataFrame.from_records
(c) DataFrame.from_dict
(d) All of the mentioned
Answer:
(a) DataFrame.fromJtems
DataFrame.from records takes a list of tuples or an ndarray with structured dtype.

Question 74.

Which of the following operation works with the same syntax as the analogous diet operations?

(a) Getting columns
(b) Setting columns
(c) Deleting columns
(d) All of the mentioned
Answer:
(d) All of the mentioned
You can treat a Dataframe semantically like a diet of like- indexed Series objects.

Question 75.

If data is an array; index must be the same length as dat(a)

(a) True
(b) False
(c) Error
(d) None of these
Answer:
(a) True
If no index is passed, one will be created having values [0 len(data) -1],

MCQ Questions for Class 12 Informatics Practices with Answers

MCQ Questions for Class 12 Informatics Practices – Python Pandas Part 2 Read More »

MCQ Questions for Class 10 Maths Chapter 14 Statistics

MCQ Questions / Prasanna

Statistics Class 10 MCQ Questions with Answers

Question 1.

In the formula \(\bar{x}=a+\frac{\sum x_{i} d_{i}}{\sum f_{i}}\) for finding thc mean of grouped data d;s are the deviations from a of:

(A) lower limits of the classes
(B) upper limits of the classes
(C) mid-points of the classes
(D) frequencies of the class marks
Answer:
(B) upper limits of the classes

Explanation:
In the given formula, a is assumed mean from class marks (aq) and d_i = x₁ – a Therefore, d{ is the deviation of class mark (mid-value) from the assumed mean ‘a’.

Question 2.

While computing mean of grouped data, we assume that the frequencies are :

(A) evenly distributed over all the classes
(B) centred at the class marks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes
Answer:
(B) centred at the class marks of the classes

Explanation:
In grouping the data from ungrouped data, all the observations between lower and upper limits of class marks are taken in one group then mid-value or class mark is taken for further calculation. Therefore, frequencies or observations must be centred at the class marks of the classes.

Question 3.

If x_i‘s are the mid-points of the class intervals of grouped data f_i s are the corresponding frequencies
and \( \bar{x}\) is the mean, then \( \sum\left(f_{i} x_{i}-\bar{x}\right)\) is equal to :

(A) 0
(B) -1
(C) 1
(D) 2
Answer:
(A) 0

From equation (i) and (ii), we have

Question 4.

In the formula \( \bar{x}\) = a + h \( \frac{\Sigma f_{i} l l_{i}}{\sum f_{i}}\) , for finding the mean of grouped frequency distribution

(A) \(\frac{x_{i}+a}{h}\)
(B) h (x¹ – a)
(C) \(\frac{x_{i}-a}{i}\)
(D) \( \frac{a-x_{i}}{h}\)
Answer:
(C) \(\frac{x_{i}-a}{i}\)
Explanation:
\( \bar{x}\) = a + h\(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)
u₁ = \(\frac{x_{i}-a}{h}\)

Question 5.

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its :

(A) mean
(B) median
(C) mode
(D) All of these
Answer:
(B) median

Explanation:
The point of intersection of the less than type and of the more than type cumulative frequency curves gives the median on abscissa as on x-axis we take the upper or lower limits, respectively and on y-axis we take cumulative frequency.

Question 6.

For the following distribution :

Class	0-5	5 – 10	10 – 15	15 – 20	20 – 25
Frequency	10	15	12	20	9

the sum of lower limits of median class and modal class is :

(A) 15
(B) 25
(C) 30
(D) 35
Answer:
(B) 25

Explanation:

Class	Frequency	Cumulative frequency
0-5	10	10
5 – 10	15	25
10 – 15	12	37
15 – 20	20	57
20 – 25	9	66

The modal class is the class having the maximum frrquency 20 belongs to class (15 – 20).
Hrer, n = 66
So, \( \frac{n}{2}\) = \(\frac{66}{2}\) = 33
33 lies in the class 10 – 15.
Therefore, 10 – 15 is the median class.
So, sum of lower limits of (15 – 20) and (10 -15) is (15 – 10) = 25

Question 7.

Consider the following frequency distribution :

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

the upper limit of the median class is :

(A) 7
(B) 17.5
(C) 18
(D) 18.5
Answer:
(B) 17.5

Explanation:

Class	Frequency	Cumulative frequency
0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

The median of 57 (odd) observations = \(\frac{(57+1)}{2}\) = \(\frac{58}{2}\) = 29th term
29th term lies in class 11.5 – 17.5.
So, upper, limit is 17. 5

Question 8.

For the following distribution :

Marks	Number of students
Below 10	3
Below 20	12
Below 30	27
Below 40	57
Below 50	75
Below 60	80

the modal class is :

(A) 10-20
(B) 20-30
(C) 30-10
(D) 50-60 .
Answer:
(C) 30-10

Explanation:

Marks	Number of students	f_i
0-10	3 – 0=3	3
10-20	12 – 3=9	9
20-30	27 – 12=15	15
30-40	57 – 27=30	30
40-50	75 – 57=18	18
50-60	80 – 75=5	5

Modal class has maximum frequency (30) in class 30 – 40.

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of of A Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): Consider the following data:

Class	Frequency
65-85	4
85 -105	5
105 -125	13
125 -145	20
145 -165	14
165-185	7
185 – 205	4

The difference of the upper limit of the median class and the lower limit of the modal class is 20.
Reason (R): The median class and modal class of grouped data always be different.

Answer:
(C) A is true but R is false
Explanation:
In case of assertion :

Class	Frequency	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	7	63
185-205	4	67

Hence, n= 67 (odd)
so, Median = \(\frac{67+1}{2}\) = 34
34 lies in class 125 – 145 and upper limit is 145.
Now, the maximum frequency is 20 and it lies in class 125 – 145 (modal class).
Lower limit of modal class = 125
Hence, the reuired difference 145 – 125 = 20.
∴ Assertion is correct.
In case of reason :
The median andmodal class may be same. If modal class is median class which is not always possible as the number of trequencies may be maximum in any class.
So, given dtatement is not true.
∴ Reason is incorrect.
Hence, assertion is correct but reason is incorrect.

Question 2.

Assertion (A): Consider the data:

Class	4-7	8-11	12-15	16-19
Frequency	5	4	9	10

The mean of the above data is 12.93.
Reason (R): The following table gives the number of pages written by Sarika for completing her own book for 30 days:

Class	16-18	19-21	22-24	25-27	28-30
Frequency	1	3	4	9	13

The mean of the above date is 26

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
In case of assertion :
Class marks of these classes are same ,so no need to convert given date continuous.

Class	Class marks(x_i)	marks(x_i) d_i = (x_i – a)	Frequency (f_i)	f_id_i
4-7	5.5	-4	5	-20
8-11	9.5 = a	0	4	0
12-15	13.5	+4	9	36
16-19	17.5	+8	10	80
			Σf_i = 28	Σf_id_i = 96

a = Assumed mean, di = Deviation fro mean
Mean, \( \bar{x}\) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\) = 9.5 + \(\frac{96}{28}\) = 9.5 + 3.43
∴\( \bar{x}\) = 12.93
∴ Hence, the mean = 12.93
∴Assertion is correct.
In case of reason:
No need to change the class intervals of continuous intervals as class marks of continuous and discontinuous classes are same. di is deviation from assumed mean.

Class interval	Midvalue	d_i =(x_i – a)	Number of days	f_id_i
16-18	17	-6	1	-6
19-21	20	-3	3	-9
22-24	a = 23	0	4	0
25-27	26	3	9	27
28-30	29	6	13	78
			Σf_i = 30	Σf_id_i = 90

a = Assumed mean = 23
Mean, \( \bar{x}\) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\) = 23 + \(\frac{90}{30}\) = 23 + 3 = 26
∴\( \bar{x}\) = 26
Hence, the mean of pages written per day is 26.
∴Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 3.

Assertion (A): If the median of a series exceeds the mean by 3, then mode exceeds mean by 10.
Reason (R): If made = 12.4 and mean = 10.5, then the median is 11.13.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
Given, median = mean + 3
Since, Mode = 3 Median – 2 Mean
= 3 (Mean + 3) – 2 Mean
⇒ Mode = Mean + 9
Hence, mode exceed mean by 9.
∴ Assertion is correct.
In case of reason:
Median = \(\frac{1}{3}\) Mode + \(\frac{2}{3}\) Mean
= \(\frac{1}{3}\) (12 . 4) + \(\frac{2}{3}\) (10.5)
\(\frac{12.4}{3}\) + \(\frac{21}{3}\)
= \(\frac{12.4+21}{3}\) = \(\frac{33.4}{3}\)
= \(\frac{33.4}{3}\) = 11.13
∴ Reason is correct.
Hence, Assertion is incorrect but reason is correct.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following question on the basis of the same:
COVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.

The following tables shows the age distribution of case admitted during a day in two different hospitals

Question 1.

The averge age for which maximum cases occurred is

(A) 32.24
(B) 34.36
(C) 36.82
(D) 42.24
Answer.
(C) 36.82
Explanation:
Since, highest frequency is 23. So, modal class is 35 – 45
Now, Mode = l +\(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \) × h
Here, l =35, h = 10, f_i = 23, f0 = 21, f2 = 14,
⇒ Mode = 35 + \(\frac{23-21}{46-21-14}\) × 10
= 35 + \(\frac{2}{11}\) × 10
= 35 + \(\frac{20}{11}\)
= 35 + 1.81
= h 36.818 = 36.82

Question 2.

The upper limit of modal class is

(A) 15
(B) 25
(C) 35
(D) 45
Answer:
(D) 45

Question 3.

The mean of the given data is

(A) 26.2
(B) 32.4
(C) 33.5
(D) 35.4
Answer:
(D) 35.4

Explanation:

Age (in years)	Class marks – (x_i)	frequency (f_i)	Deviation d_i= (x_i – a)	f_id_i
5-15	10	6	-3	-15
15-25	20	11	6	66
25-35	30	21	16	336
35-45	40	23	26	598
45-55	50	14 → a	1	18
55-65	60	5	46	46
		Σf_i = n = 80		Σf_id_i = 1,716

Now, MEan (\( \bar{x}\)) a + \( \bar{x}\)
= 14 + \( \bar{x}\)
= 14 + 21.45
= 35. 45
Refer to table 2

Question 4.

The mode of the given data is

(A) 41.4
(B) 48.2
(C) 55.3
(D) 64.6
Answer:
(A) 41.4
Question 5.

The median of the given data is

(A) 32.7
(B) 40.2
(C) 42.3
(D) 48.6
Answer:
(B) 40.2

Explanation:

Age (in years)	frequency (f_i) (No. of cases)	C.f.
5-15	8	8
15-25	16	24
25-35	10	34
35-45	42 (frequency)	76 (Nearest to \(\frac{n}{2}\)))
45-55	24	100
55-65	12	112
	Σf_i = n = 12

Now, \(\frac{n}{2}\) = \(\frac{112}{2}\) =56
l = 35 (lower mlimit of median class)
Cf = 34 (Preceding to median class)
Here, Madian = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 35 + \(\left(\frac{56-34}{42}\right)\) × 10
= 35 + \(\left(\frac{22}{42}\right)\) × 10
= 35 + \(\left(\frac{11}{21}\right)\) × 10
= 35 + \(\frac{110}{21}\)
= 40.25

II. Read the following text and answer the following question on the basis of the same:
Electricity Energy ConsumptionElectricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).

Tariff : LT – Residential	Bill Number : 384756′
Type of Supply : Single Passes	Connected lead : 3 kW
Mater Reading : 31-11-13 Date	Mater Reading : 65700
Previous Reading : 31-10-13 Date	Previous Mater : 65500 Reading
	Units consumed : 289

A survey is conducted for 56 families of a Colony A. The following tables gives the weekly consumption of electricity of these families.

Weekly consumption (in units)	0-10	10-20	20-30	30 – 40	40-50	50-60
No. of families	16	12	18	6	4	0

Weekly consumption (in units)	0-10	10-20	20-30	30-40	40-50	50-60
No. of families	0	5	10	20	40	5

Refer to data received from Colony A

Question 1.

The median weekly consumption is

(A) 12 units
(B) 16 units
(C) 20 units
(D) None of these
Answer:
(C) 20 units

Explanation:

Weekly consumption (in units)	frequency (f_i) (No. of families)	C.f.
0 – 10	8	16
10-20(Medianclass)	12 (frequency)	28(Nearest to \(\frac{n}{2}\))
20-30	18	46
30-40	6	52
40-50	4	56
50-60	0	56
	Σf_i = n = 56

Now, \(\frac{n}{2}\) = \(\frac{56}{2}\) = 28
l = 10, Cf = 16 ,f = 12, h = 10
Here,
Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 10 + \(\left(\frac{28-16}{12}\right)\) × 10
= 10 + \(\left(\frac{12}{12}\right)\) × 10
= 10 + 10
= 20
Hence, median weekiy consumption = 20 units.

Question 2.

The mean weekly consumption is

(A) 19.64 units
(B) 22.5 units
(C) 26 units
(D) None of these
Answer:
(A) 19.64 units

Question 3.

The modal class of the above data is I

(A) 0 – 10
(B) 10 – 20
(C) 20 – 30
(D) 30-40
Answer:
(C) 20 – 30
Refer to data received from Colony B

Question 4.

The modal weekly consumption is

(A) 38.2 units
(B) 43.6 units
(C) 26 units
(D) 32 units
Answer:
(B) 43.6 units

Question 5.

The mean weekly consumption is

(A) 15.65 units
(B) 32.8 units
(C) 38.75 units
(D) 48 units
Answer:
(C) 38.75 units

III. Read the following text and answer the following question on the basis of the same:
The weights (in kg) of 50 wrestlers are recorded in the following table :

Question 1

What is the upper limit of modal class.

(A) 120
(B) 130
(C) 100
(D) 150
Answer:
(B) 130

Explanation:
Modal Class = 120 – 130
Upper limit = 130

Question 2.

What is the mode class frequency of the given data

(A) 21
(B) 50
(C) 25
(D) 80
Answer:
(A) 21

Explanation:
Mode class frequency of the given data is 21.

Question 3.

How many wrestlers weights have more than 120 kg weight?

(A) 32
(B) 50
(C) 16
(D) 21
Answer:
(A) 32

Explanation:
No. of wrestlers with more than 120 kg weight = 21 + 8 + 3 = 32

Question 4.

What is the class mark for class 130 – 140?

(A) 120
(B) 130
(C) 135
(d) 150
Answer:
(C) 135

Explanation:
For class mark of 130 -140,
= \(\frac{130+140}{2}\)
= \(\frac{270}{2}\) = 135

Question 5.

Which method is more suitable to find the mean of the above data ?

(A) Direct method
(B) Assumed mean method
(C) Step-Deviation method
(D) None of these
Answer:
(C) Step-Deviation method
IV Read the following text and answer the following question on the basis of the same:
The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows.

Question 1.

What is the modal class of the given data?

(A) 85 – 100
(B) 100 – 115
(C) 115 – 130
(D) 130 – 145
Answer:
(A) 85 – 100

Explanation:
Modal class is the class with highest frequency i.e., 85 – 100

Question 2.

What is the value of class interval for the given data set?

(A) 10
(B) 15
(C) 5
(D) 20
Answer:
(B) 15

Explanation:

The value of class interval = 100 – 85 = 15
again =115-100 = 15
and = 130-115 = 145 – 130

Question 3.

What is the median class of the given data?

(A) 85-100
(B) 100-115
(C) 115 – 130
(D) 130 – 145
Answer:
(B) 100-115

Explanation:
N = Number of observations = 33
Median of 33 observations = 17.5^th observation, which in class 100 – 115

Question 4.

What is the median of bowling speed?

(A) 109.17 km/hr (Approx)
(B) 109.71 km/hr (Approx)
(C) 107.17 km/hr (Approx)
(D) 109.19 km/hr (Approx)
Answer:
(A) 109.17 km/hr (Approx)Explanation:
Median = l + \(\frac{\left(\frac{n}{2}-c . f\right)}{f}\) × h
l = 100, f = 9, c.f. = , h = 100 – 85 = 15
Median = l + \(\frac{\left(\frac{n}{2}-c . f\right)}{f}\) × h
= 100 + \(\frac{\left(\frac{33}{2}-11\right)}{9}\) × 15
= \(\frac{100+(16.5-11)}{9 \times 15}\)
= 100 + \(\frac{5.5 \times 15}{9}\)
= 100 + \(\frac{82.5}{9}\)
= 100 + 9.166
= 109.17 km/h (Approx)
Hence, the median bowling sppeed is 109. 17 km/h (Approx)

Question 5.

What is the sum of lower limit of modal class and upper limit of median class?

(A) 100
(B) 200
(C) 300
(D) 400
Answer:
(B) 200

Explanation:
Lower limit of modal class = 85 and upper limit of median class =115 sum = 85 + 115 200V. Read the following text and answer the following question on the basis of the same:
100 m RACE A stopwatch was used to find the time that it took a group of students to run 100 m.

Question 1.

Estimate the mean time taken by a student to finish the race.

(A) 54
(B) 63
(C) 43
(D) 50
Answer:
(C) 43

Explanation:

Time (in sec)	x	f	cf	fx
0 – 20	10	8	8	80
20 – 40	30	10	18	300
40 – 60	50	13	31	650
60 – 80	70	6	37	420
80 – 100	90	3	40	270
Total		40		1720

Mean = \(\frac{1720}{40}\) = 43

Question 2.

What will be the upper limit of the modal class?

(A) 20
(B) 40
(C) 60
(D) 80
Answer:
(C) 60

Explanation:
Modal class = 40-60
Upper limit = 60

Question 3.

The construction of cumulative frequency table is useful in determining the

(A) Mean
(B) Median
(C) Mode
(D) All of the above
Answer:
(B) Median

Explanation:
The construction of c.f. table is useful in determining the median.

Question 4.

The sum of lower limits of median class and modal class is

(A) 60
(B) 100
(C) 80
(D) 140
Answer:
(C) 80

Explanation:
Median class = 40-60
Modal class = 40-60
Therefore, the sum of the lower limits of median and modal class = 40 + 40 = 80

Question 5.

How many students finished the race within 1 minute?

(A) 18
(B) 37
(C) 31
(D) 8
Answer:
(C) 31

Explanation:
Number of students who finished the race within 1 minute = 8 + 10 + 13 = 31

MCQ Questions for Class 10 Maths with Answers

MCQ Questions for Class 10 Maths Chapter 14 Statistics Read More »

MCQ Questions for Class 10 Maths Chapter 15 Probability

MCQ Questions / Prasanna

Probability Class 10 MCQ Questions with Answers

Question 1.

If an event that cannot occur, then its probability is:

(A) 1
(B) \(\frac{3}{4}\)
(C) \(\frac{1}{2}\)
(D) 0
Answer:
(D) 0

Explanation:
An event that cannot occur has 0 probability, such an event is called impossible event.

Question 2.

Which of the following cannot be the probability of an event ?

(A) \(\frac{1}{3}\)
(B) 0.1
(C) 3%
(D) \(\frac{17}{16} \)
Answer:
(D) \(\frac{17}{16}\)

Explanation:
Probability of any event cannot be more than one or negative as \(\frac{17}{16}\) > 1.

Question 3.

An event is very unlikely to happen. Its probability is closest to:

(A) 0.0001
(B) 0.001
(C) 0.01
(D) 0.1
Answer:
(A) 0.0001

Explanation:
The probability of the event, which is very unlikely to happen, will be very close to zero. So it’s probability is 0.0001 which is minimum amongst the given values.

Question 4.

If the probability of an event is p, then the probability of its complementary event will be:

(A) p – 1
(B) p
(C) 1 – p
(D) \(1-\frac{1}{p}\)
Answer:
(C) 1 – p

Explanation:
Probability of an event + Probability of its complementary event = 1
∴ p + Probability of complement = 1
Probability of complement = 1 – p

Question 5.

The probability expressed as a percentage of a particular occurrence can never be:

(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number
Answer:
(B) less than 0

Explanation:
Probability lies between 0 and 1 and when it is converted into percentage it will be between 0 and 100. So, cannot be negative.

Question 6.

If P(A) denotes the probability of an event A, then:

(A) P(A) < 0 (B) P(A) > 1
(C) 0 ≤ P(A) ≤ 1
(D) -1 ≤ P(A) ≤ 1
Answer:
(C) 0 ≤ P(A) ≤ 1

Explanation:
As the probability of an event lies between 0 and 1.

Question 7.

If a card is selected from a deck of 52 cards, then the probability of its being a red face card is :

(A) \(\frac{3}{26}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{2}{13}\)
(D) \(\frac{1}{2}\)
Answer:
(A) \(\frac{3}{26}\)

Explanation:
In a deck of 52 cards, there are 26 red cards.
Number of red face cards = 3 of hearts + 3 of diamonds = 6
So, probability of having a red face card
= (A) \(\frac{6}{52}\) = (A) \(\frac{3}{26}\)

Question 8.

The probability that a non-leap year selected at random will contain 53 Sundays is:

(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{3}{7}\)
(D) \(\frac{5}{7}\)
Answer:
(A) \(\frac{1}{7}\)
Number of weeks = \(\frac{365}{7}\) = 52\(\frac{1}{7}\) = 52 weeks
Number of days left = 1
For example, it may be any of 7 days which from Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday; so,
T(E) = 7
F(E) = 1 (Sunday)
p(F) = \(\frac{F(E)}{T(E)}\) = \(\frac{1}{7}\)

Asertion and Reased Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, the probability of each is \(\frac{1}{2}\).
Reason (R): When we toss a coin, there are two possible outcomes: head or tail. Therefore, the probability of each outcome is i.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
From 1 to 100 numbers, there are 50 even and 50 odd numbers.
Total number of outcomes T(E) = 100 Number of outcomes favourable for event E (even numbers) = F (E) = 50
so, P (E) = \(\frac{50}{100}\) = \(\frac{1}{2}\)
Similarly, the probability of getting odd numbers =\(\frac{1}{2}\).
Hence the probability of getting odd and even each = \(\frac{1}{2}\).
Hence, the given statement is true.
∴ Assertion is correct.
In case of reason:
Since, there are two outcomes equal in all manners. So, probability of both head and tail is equal to \(\frac{1}{2}\). each.
Hence, the given statement is true.
∴ Reason is correct:
Hence, both assertion and reason are correct but reason is not correct explanation for assertion.

Question 2.

Assertion (A): If P(F) = 0.20, then the probability of ‘not E’ is 0.80.
Reason (R): If two dice are thrown together, then the probability of getting a doublet is \(\frac{5}{6}\).

Answer:
(C) A is true but R is false
Explanation:
In case of assertion:
P(E) = 0.20
P(not E) = 1 – P(E)
∴Assertion is correct.
In case of reason:
Total number of possible outcomes = 6² = 36
E : (doublets are (1,1), (2,2), (3, 3), (4,4), (5,5), (6,6)
No. of favourable outcomes to E = 6
P(a doublet)
= \(\frac{Number of outcomes favourable to E}{Total number of outcomes}\).
=\(\frac{6}{36}\). = \(\frac{1}{6}\).
∴Reason is incorrect:
Hence, assertion is correct but reason is incorrect.

Question 3.

Assertion (A): The probability that a number selected at random from the number 1, 2, 3, , 15
is a multiple of 4, is \(\frac{1}{3}\)..
Reason (R): Two different coins are tossed simultaneously. The probability of getting at least one head is\(\frac{3}{4}\).

Answer:
(D) A is false and R is True

Explanation:
In case of reason:
n(S) = 15
n(A) = 3
p(A) = \(\frac{n(A)}{n(S) }\) = \(\frac{3}{15}\) = \(\frac{1}{5}\)
S = HH, HT, TH,TT
A = HH, HT, TH
n(S) = 4
n(A) = 3
p(A) = \(\frac{n(A)}{n(S) }\) = \(\frac{3}{4 }\)
∴Reason is incorrect:
Hence, assertion is correct but reason is incorrect.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same: On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card.

Question 1.

Find the probability of getting a king of red colour.

(A) \(\frac{1}{26}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Answer:
(A) \(\frac{1}{26}\)

Explanation:
No. of cards of a king of red colour = 52
Probability of getting a king of red colour
= \(\frac{No. of king of red colour}{Total number of cards }\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 2.

Find the probability of getting a face card.

(A) \(\frac{1}{26}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{2}{13}\)
(D) \(\frac{3}{13}\)
Answer:
(D) \(\frac{3}{13}\)

Question 3.

Find the probability of getting a jack of hearts.

(A) \(\frac{1}{26}\)
(B) \(\frac{1}{52}\)
(C) \(\frac{3}{52}\)
(D) \(\frac{3}{26}\)
Answer:
(B) \(\frac{1}{52}\)

Question 4.

Find the probability of getting a red face card.

(A) \(\frac{3}{13}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Answer:
(A) \(\frac{3}{13}\)

Explanation:
No. of face card = 13
Total no of cards = 52
Probability of getting a face card
= \(\frac{ No. of face cards }{Total no. of cards}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

Question 5.

Find the probability of getting a spade.

(A) \(\frac{1}{26}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)

Explanation:
No. of face card = 13
Total no of cards = 52
Probability of getting a face card
= \(\frac{ No. of face cards }{Total no. of cards}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

II. Read the following text and answer the following questions on the basis of the same:
Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice.

Question 1.

Ravi got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?

(A) \(\frac{1}{26}\)
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) 0
Answer:
(B) \(\frac{5}{36}\)

Explanation:
The outcomes when two dice are thrown together are:
= (1, 1), (1, 2), (1, 3), (1,4), (1, 5), (1,6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total outcomes = 36
No. of outcomes when the sum is 8 = 5
Probability = \(\frac{5}{36}\)

Question 2.

Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13?

(A) 1
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) 0
Answer:
(D) 0

Explanation:
No. of outcomes when the sum is 13 = 0
Total outcomes =36
Probability =\(\frac{0}{35}\) = 0

Question 3.

Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12?

(A) 1
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) 0
Answer:
(A) 1
Explanation:
No. of outcomes when the sum is less than or equal to 12 = 36
Total outcomes = 36
Probability = \(\frac{36}{36}\) = 1

Question 4.

Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7?

(A) \(\frac{5}{9}\)
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{6}\)
(D)0
Answer:
(C) \(\frac{1}{6}\)

Question 5.

Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8 ?

(A) 1
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) \(\frac{5}{18}\)
Answer:
(D) \(\frac{5}{18}\)

MCQ Questions for Class 10 Maths with Answers

MCQ Questions for Class 10 Maths Chapter 15 Probability Read More »

MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes

MCQ Questions / Prasanna

Surface Areas and Volumes Class 10 MCQ Questions with Answers

Question 1.

A cylindrical pencil sharpened at one edge is the combination of:

(A) a cone and a cylinder
(B) frustum of a cone and a cylinder
(C) a hemisphere and a cylinder
(D) two cylinders
Answer:
(A) a cone and a cylinder

Explanation:
The sharpened part of the pencil is cone and unsharpened part is cylinder.

Question 2.

A surahi is the combination of:

(A) a sphere and a cylinder
(B) a hemisphere and a cylinder
(C) two hemispheres
(D) a cylinder and a cone
Answer:
(A) a sphere and a cylinder

Explanation:
A surahi is the combination of a sphere and a cylinder.

Question 3.

A plumbline (Sahul) is the combination of:

(A) a cone and a cylinder
(B) a hemisphere and a cone
(C) frustum of a cone and a cylinder
(D) sphere and cylinder
Answer:
(B) a hemisphere and a cone

Explanation:
Plumbline is an instrument used to check the vertically of an object. It is a combination of a hemisphere and a cone.

Question 4.

The shape of a gilli, in the gilli-danda game (see in Figure) is a combination of:

(A) two cylinders
(B) a cone and a cylinder
(C) two cones and a cylinder
(D) two cylinders and a cone
Answer:
(C) two cones and a cylinder

Explanation:
The shape of gilli, in the gilli- danda game is a combination of two cones and a cylinder.

Question 5.

A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is:

(A) 142296
(B) 142396
(C) 142496
(D) 142596
Answer:
(A) 142296

Explanation:
Let the spherical marble has radius r. Diameter of the marble = 0.5 cm
⇒ r = \(\frac{0.5}{2}\) cm = 0.25 cm
Length of side of l = 22 cm
Let n marbles can fill the cube.
⇒Volume of n marbles = (1 – \(\frac{1}{8}\))
part of volume of cube
⇒ n. \(\frac{4}{3}\) πr³ = \(\frac{7}{8}\) × l³
n = \(\frac{7 l^{3}}{8}\)× \(\frac{3}{4 \pi r^{3}}\)
⇒ = \(\frac{7 \times 3 \times 22 \times 22 \times 22 \times 7}{8 \times 4 \times 22 \times 0.25 \times 0.25 \times 0.25}\)
⇒ π n = 7 × 3 × 22 × 22 × 2 × 7
= 42 × 484 × 7
n = 142296
So, cube can accommodate up 142296 marbles.

Question 6.

A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each to its ends. The length of entire capsule is 2 cm. The capacity of the capsule is :

(A) 0.36 cm³
(B) 0.35 cm³
(C) 0.34 cm³
(D) 0.33 cm³
Answer:
(A) 0.36 cm³

Explanation:
Capsule consists of 2 hemispheres and a cylinder.
⇒ r = \(\frac{0.5}{2}\) cm = 0.25 cm
⇒ r = 0. 25 cm
Total length of capsule = r + h + r
⇒ 2 cm = 2r + h
⇒ 2 = 2 × 0.25 + h
⇒ h = 2 – 0.5 = 1.5 cm
Volume of capsule = Volume of two
hemispheres + Volume of cylinder
= 2 × \(\left(\frac{4}{3} \pi r^{3} \times \frac{1}{2}\right)\) + πr²h
= \(\frac{4}{3}\) πr² + πr²h
= πr² \(\left(\frac{4}{3} r+h\right)\)
= \(=\frac{22}{7}\) × 0. 25 × 0. 25 \(\left(\frac{4}{3} \times 0.25+\frac{15}{10}\right)\)
= \(\frac{22}{7}\) × 0. 25 × 0. 25 \(\left(\frac{1}{3}+\frac{3}{2}\right)\)
= \(\frac{22}{7}\) × \(\frac{25}{100}\) × \(\frac{25}{100}\) × \(\frac{11}{6}\) = \(\frac{121}{336}\)
⇒Volume of capsule = 0. 3601 cm³ = 0. 36 cm³

Question 7.

If two solid hemispheres of same base radius V are joined together along their bases, then curved surface area of this new solid is :

(A) 4 πr²
(B) b πr²
(C) 3 π²
(D) 8 πr²
Answer:
(A) 4 πr²

Explanation:
When two hemispheres of equal radii are joined base to base, new solid becomes sphere and curved surface area of sphere is im².

Question 8.

A right circular cylinder of radius r cm and height h cm (where h > 2r) just encloses of sphere of diameter :

(A) r cm
(B) 2r cm
(C) h cm
(D) 2h cm
Answer:
(B) 2r cm

Explanation:
As the cylinder just enclosed the sphere so the radius or diameter of cylinder and sphere are equal, i.e., 2r and height h > 2r.

Question 9.

A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively, is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is:

(A) 12 cm
(B) 14 cm
(C) 15 cm
(D) 18 cm
Answer:
(B) 14 cm

Explanation:
During recasting a shape into another shape it’s volume does not change. For Spherical shell
r₁ = \(\frac{4}{2}\) = 2 cm
r₂ = \(\frac{8}{2}\) = 4 cm
For Cone r = \(\frac{8}{2}\) = 4 cm
h = π
During recasting voleme remains same so, Volume of cone = Volume of hollow spherical shell
⇒ \(\frac{1}{3}\)πr² h = \(\frac{4}{3} \pi r_{2}^{3}-\frac{4}{3} \pi r_{1}^{3}\)
⇒ \(\frac{1}{3}\) = \(\frac{4}{3} \pi\left(r_{2}^{3}-r_{1}^{3}\right)\)
⇒ r² h = 4\(\left(r_{2}^{3}-r_{1}^{3}\right)\)
⇒ 4 × 4h = 4[(4)3 – (2)3]
⇒ 4h = 64 – 8
⇒ h = \(\frac{56}{4}\)
⇒ h = 14 cm

Question 10.

A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid sphere. The radius of the sphere is:

(A) 21 cm
(B) 23 cm
(C) 25 cm
(D) 19 cm
Answer:
(A) 21 cm

Explanation:
Solid cuboid of iron is moulded into solid sphere. For hence, volume of cuboid and sphere are equal.
For sphere
r = π
cuboid
l = 49 cm
b = 33 cm
h = 24 cm
⇒Volume of sphere (solid) = Volu me of cuboid
⇒ \(\frac{4}{3 \pi r^{3}}\) = l x b x h
πr³ = \(\frac{(l \times b \times h \times 3)}{(4 \times \pi)}\)
= \(\frac{(49 \times 33 \times 24 \times 3 \times 7)}{(4 \times 22)}\)
πr³ = 7 × 7 × 7× 3 × 3 × 3
r = 21 cm

Question 11.

A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that 1/8 space is covered by the mortar. Then the number of bricks used to construct the wall is:

(A) 11100
(B) 11200
(C) 11000
(D) 11300
Answer:
(B) 11200

Explanation:
The volume of the wall covered mortar = \(\frac{1}{8}\) part So, the volume covered by bricks of wall = \(\left(1-\frac{1}{8}\right)\) volume of wall
= \(\frac{7}{8}\)
volume of wall
Bricks (Cuboid) Wall (Cuboid)
l₁= 22.5 cm l – 270 cm
b₁ = 11.25 cm b = 300 cm
h1 = 8.75 cm 350 cm
Bricks (Cuboid) Wall (Cuboid)
l₁= 22.5 cm l – 270 cm
b₁ = 11.25 cm b = 300 cm
h1 = 8.75 cm 350 cm

Bricks (Cuboid)	Wall (Cuboid)
l₁ = 22.5 cm	l – 270 cm
b₁ = 11.25 cm	b = 300 cm
h1 = 8.75 cm	h = 350 cm

Let n be the number of bricks.
According to the question, we have
Volume of n bricks = \(\frac{7}{8}\) Volume of wall (Cuboid)
⇒ n × l₁× b₁ × h₁ = \(\frac{7}{8}\) × l × b × h
⇒ n = \(\frac{(7 \times l \times b \times h)}{\left(8 \times l_{1} \times b_{1} \times h_{1}\right)}\) = \(\frac{(7 \times 270 \times 300 \times 350)}{(8 \times 22.5 \times 11.25 \times 8.75)}\)
⇒ n = \(\frac{(7 \times 270 \times 300 \times 350 \times 100 \times 10 \times 100)}{(8 \times 225 \times 1125 \times 875)}\)
⇒ n = 2 × 4 × 350 × 4 = 32 × 350 = 11,200 bricks.

Question 12.

Twelve solid sphere of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is:

(A) 4 cm
(B) 3 cm
(C) 2 cm
(D) 6 cm
Answer:
(C) 2 cm

Explanation:
Solid cylinder is recasted into 12 spheres. So, the volume of 12 spheres will be equal to the volume of the cylinder.
12 For spheres
R = π
For cylinder
r = \(\frac{2}{2}\) = 1 cm
h – 16 cm
⇒ Volume , diameter = 2R = 2 × 1 = 2 cm
⇒ \(\frac{4}{3 \pi r^{3}}\) = πr²h
⇒ R³ = \(\frac{\left(3 r^{2} h\right)}{(4 \times 12)}\)
= \(\frac{(3 \times 1 \times 1 \times 16)}{(4 \times 12)}\) = 1
⇒ R = 1 cm
Hence, diameter = 2R = 2 × 1 = 2 cm.

Question 13.

During conversion of a solid from one shape to another, the volume of new shape will:

(A) increase
(B) decrease
(C) remains unaltered
(D) be doubled
Answer:
(C) remains unaltered

Explanation:
During reshaping a solid, the volume of new solid will be equal to old one or remains unaltered.

Question 14.

A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is:

(A) 3.5
(B) 7
(C) \(\frac{80}{7}\)
(D) 5
Answer:
(A) 3.5

Explanation:
Circumference = 22 cm
2πr = 22
2 × \(\frac{22}{7}\) × r = 22
r = 3.5 cm

Question 15.

The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is:

(A) 3
(B) 5
(C) 4
(D) 6
Answer:
(B) 5

Explanation:
No. of solid spheres = \(\frac{\text { Volume of cylinder }}{\text { Volume of sphere }}\)
= \(\frac{\pi R^{2} h}{\frac{4}{3} \pi r^{3}}\)
= \(\frac{\pi(2)^{2} \times 45 \times 3}{4 \times \pi \times(3)^{3}}\)
= 5.

Assertion and Reason Based MCQs

Question 1.

Assertion (A): In a right circular cone, the cross-section made by a plane parallel to the base is a circle.
Reason (R): If the volume and the surface area of a solid hemisphere are numerically equal, then the diameter of hemisphere is 9 units.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
In a right circular cone, if any cut is made parallel to its base, we get a circle.
∴ Assertion is correct.
In case of reason:
Let radius of sphere be r.
Given, volume of hemisphere = Surface area of hemisphere
or, \(\vec{a}\) πr³ = 3πr²
or, r = \(\vec{a}\) units
⇒ Diameter = \(\vec{a}\) × 2 = 9 units
∴ Reason is correct :
Hence, both assertion and reason are correct but reason is the correct explanation for assertion.

Question 2.

Assertion (A): If the volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is 4 : 3.
Reason (R): If the surface areas of two spheres are in the ratio 16 : 9, then the ratio of their volumes is 64 : 27.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
\(\frac{V_{1}}{V_{2}}\) = \(\frac{64}{27}\)
\(\frac{\frac{4}{3} \pi r_{1}{ }^{3}}{\frac{4}{3} \pi r_{2}{ }^{3}}\) = \(\frac{64}{27}\)
\(\left(\frac{r_{1}}{r_{2}}\right)^{3}\) = \(\frac{64}{27}\)
\(\frac{r_{1}}{r_{2}}\) = \(\frac{4}{3}\)
Now the ratio of their surface areas,
\(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\) = \(\left(\frac{r_{1}}{r_{2}}\right)^{2}\) = \(\left(\frac{4}{3}\right)^{2}\) = \(\frac{16}{9}\)
∴ Assertion is correct.
In case of reason:
Given,
\(\frac{A_{1}}{A_{2}}\) = \(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\) = \(\frac{16}{9}\)
\(\left(\frac{r_{1}}{r_{2}}\right)^{2}\) = \(\frac{16}{9}\)
\(\frac{r_{1}}{r_{2}}\) = \(\sqrt{\frac{16}{9}}\) = \(\frac{4}{3}\)
Now, volumes of two spheres,
\(\frac{V_{1}}{V_{2}}\) = \(\frac{\frac{4}{3} \pi r_{1}^{2}}{\frac{4}{3} \pi r_{2}^{2}}\)
= \(\left(\frac{r_{1}}{r_{2}}\right)^{3}\) = \(\left(\frac{4}{3}\right)^{3}\) = \(\frac{64}{27}\)
= 64 : 27
∴ Reason is correct:
Hence, assertion is incorrect but reason is correct.

Question 3.

Assertion (A): The volume of a right circular cylinder of base radius 7 cm and height 10 cm is 1540 cm3
Reason (R): If the curved surface area of a cylinder is 264m2 and its volume is 924m3, then the ratio of its height to its diameter is 4 : 7.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
Here r = 7 cm,
h = 10 cm,
Volume of cylinder = πrh
= \(\frac{22}{7}\) × (7)² × 10
= 1540 cm³
∴ Assertion is correct.
In case of reason:
Curved Surface area of cylinder = 2πrh
Volume of cylinder = 2π2rh
\(\frac{\pi r^{2} h}{2 \pi r h}\) =\(\frac{924}{264}\)
\(\frac{y^{*}}{2}\) = \(\frac{7}{2}\)
r = 7 m
2πrh = 264
or, 2 × \(\frac{22}{7}\) × 7 × h = 264
h = 6 m
∴ \(\frac{h}{2 r}\) = \(\frac{6}{14}\) = \(\frac{3}{7}\)
Hwnce, h : r = 3 : 7
∴ Reason is correct:
Hence, assertion is incorrect but reason is correct.

Question 4.

Assertion (A): The number of solid spheres of diameter 6 cm can be made by melting a solid metallic cylinder of height 45 cm and diameter 4 cm, is 5.
Reason (R): If three solid metallic spherical balls of radius 3 cm, 4 cm and 5 cm are melted into a single spherical ball, then its radius is 6 cm.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
Let the number of spheres be n.
Radius of sphere r₁ = 3 cm, radius of cylinder r₂ = 2 cm
Volume of spheres = Volume of cylinder
n × \(\frac{4}{3} \pi r^{3}\) πr³ = \(\pi r_{1}^{2} h\)
or, n × \(\frac{4}{3}\) x \(\frac{22}{7}\) × (3)³ = \(\frac{22}{7}\) × (2)² × 45
or, 36 n = 180
or, n = \(\frac{180}{36}\) = 5
Thus, the number of solid spheres = 5 r. Assertion is correct.
In case of reason:
Let the radius of spherical ball be R.
Volume of spherical ball = Volume of three
\(\frac{4}{3} \pi R^{3}\) = \(\frac{4}{3} \pi\)[(3)³ + (4)³ + (5)³]
or, R³ = 27 + 64 + 125
or, R³ = 216
or, R = 6 cm
∴ Reason is correct:
Hence, both assertion and reason are correct but reason is not the correct explantion for assertion.

Question 5.

Assertion (A): If 12 solid spheres of the same size are made by melting a solid metallic cone of base radius 1 cm and height of 48 cm, then radius of each sphere is 2 cm.
Reason (R): If three cubes of iron whose edges are 3 cm, 4 cm and 5 cm respectively are melted and formed into a single cube, then the edge of the single cube is 6 cm.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
No. of spheres = 12
Radius of cone, r = 1 cm
Height of the cone = 48 cm
∴ Volume of 12 spheres = Volume of cone
Let the raddius of sphere be R cm
12 × \(\frac{4}{3} \pi R^{3}\) = \(\frac{1}{3} \pi r^{2} h\)
or, 12 × \(\frac{4}{3} \pi R^{3}\) = \(\frac{1}{3} \pi\) × (1)² × 48
16R³ = 16
R³ = 1
or, R = 1 cm
∴ Assertion is incorrect.
In case of reason:
Let the edge of single cube be x cm Volume of single cube = Volume of three cubes
x³ = (3)³ + (4)³ + (5)³
= 27 + 64 + 125
= 216
= 6 cm
∴ Reason is correct:
Hence, both assertion and reason are correct but reason is not the correct explantion for assertion.

Question 6.

Assertion (A): If a solid sphere of radius r is melted and recast into the shape of a solid cone of height r, then the radius of the base of a cone is 2r.
Reason (R): If a metallic sphere of total volume n is melted and recast into the shape of a right circular cylinder of radius 0.5 cm, then the height of cylinder is 8 cm.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
Volume of sphere = Volume of cone Let the radius of cone be R cm.
∴ \(\frac{4}{3} \pi r^{3}\)
= \(\frac{1}{3} \pi R^{2} \) × r
or, 4r3 = R2r
or, R2 = 4r2
or, R = 2r
∴ Assertion is correct.
In case of reason:
Volume of cylinder = Volume of sphere,
πr²h = π
where r and h are radius of base and height of cylinder
(0.5)² h = 1
\(\left(\frac{1}{2}\right)\) h = 1
h = 4 cm.
∴ Reason is incorrect:
Hence, assertion is correct but reason is incorrect.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following question on the basis of the same:
Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move. Some students of a school reached for adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below. After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it was time to take shelter. Each group of four students was given a canvas of area 551 m2. Each group had to make a conical tent to accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1 m2, the students put the tents. The radius of the tent is 7 m.

Question 1.

The volume of cylindrical cup is

(A) 295.75 cm3
(B) 7415.5 cm3
(C) 384.88 cm3
(D) 404.25 cm3
Answer:
(D) 404.25 cm3

Explanation:
diameter = 7 cm
radius = 3.5 cm
height = 10.5 cm
Volume of cylindrical cup
= πr²h
\(\frac{22}{7}\) × 3.5 × 3.5 × 10. 5
= 404 . 25 cm²

Question 2.

The volume of hemispherical cup is

(A) 179.67 cm³
(B) 89.83 cm³
(C) 172.25 cm³
(D) 210.60 cm³
Answer:
(B) 89.83 cm³

Question 3.

Which container had more juice and by how much?

(A) Hemispherical cup, 195 cm3
(B) Cylindrical glass, 207 end
(C) Hemispherical cup, 280.85 cm3
(D) Cylindrical glass, 314.42 cm3
Answer:
(D) Cylindrical glass, 314.42 cm3

Question 4.

The height of the conical tent prepared to accommodate four students is

(A) 18 m
(B) 10 m
(C) 24 m
(D) 14 m
Answer:
(C) 24 m

Explanation:
Radius = 7m
Area of conical tent = 551 m² – 1 m²
= 550 m²
πrl = 551
\(\frac{22}{7}\) x 7\(\sqrt{r^{2}+h^{2}}\) = 550
\(\frac{22}{7}\) x 7\(\sqrt{7^{2}+h^{2}}\) =550
\(\sqrt{7^{2}+h^{2}}\) = \(\frac{550}{22}\)
\(\sqrt{7^{2}+h^{2}}\) = \(\frac{50}{2}\)
\(\sqrt{7^{2}+h^{2}}\) = 25
72 + h2 = (25)2
h2 = 625 – 49
h2 = 576
h = √576
= 24 m

Question 5.

How much space on the ground is occupied by each student in the conical tent

(A) 54 m²
(B) 38.5 m²
(C) 86 m²
(D) 24 m²
Answer:
(B) 38.5 m²

Explanation:
Area of Base of conical tent = πr²
\(\frac{22}{7}\) ×7 × 7
= 154 m²
Area of occupied by each
student = \(\frac{1}{4}\) × 154 m²
= 38.5 m²

II. Read the following text and answer the following question on the basis of the same:
A The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus was a simple hemispherical brick structure built over the relics of the Buddha. It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure
\(\left(\text { Take } \pi=\frac{22}{7}\right)\)

Question 1.

Calculate the volume of the hemispherical dome if the height of the dome is 21 m:

(A) 19404 sq. m
(B) 2000 sq. m
(C) 15000 sq. m
(D) 19000 sq.. m
Answer:
(A) 19404 sq.. m

Explanation:
height of hemispherical dome = Radius of hemispherical dome = 21 m.
Volume of dome = \(\frac{2}{3}\)πr³
= \(\frac{2}{3}\) x \(\frac{22}{7}\) ×21 × 21 × 21
= 19,404 m³

Question 2.

The formula to find the Volume of Sphere is:

(A)\(\frac{2}{3}\){7}πr³
(B) \(\frac{4}{3}\)πr³
(C) 4πr²
(D) 2πr²
Answer:
(B) \(\frac{4}{3}\)πr³

Question 3.

The cloth require to cover the hemispherical dome if the radius of its base is 14m is:

(A) 1222 sq. m
(B) 1232 sq. m
(C) 1200 sq. m
(D) 1400 sq. m
Answer:
(B) 1232 sq. m

Question 4.

The total surface area of the combined figure i.e. hemispherical dome with radius 14 m and cuboidal shaped top with dimensions 8m × 6m × 4mis

(A) 1200 sq. m
(B) 1232 sq. m
(C) 1392 sq. m
(D) 1932 sq. m
Answer:
(C) 1392 sq. m

Explanation:
Total surface Area of Combined figure
= 2πr² + 2(lb + bh + hl) – lb
= 2 x \(\frac{22}{7}\) × 14 × 4 + 2( 8 × 6 + 6 × 4 + 4 x 8) – 8 × 6lm²
= [1232 + 208 – 48] m²
= 1392 m²

Question 5.

The volume of the cuboidal shaped top is with dimensions mentioned in question 4.

(A) 182.45 m³
(B) 282.45 m³
(C) 292 m³
(D) 192 m³
Answer:
(D) 192 m³

Explanation:
Volume of the cuboidal shaped 1 top
= l × b × h
= 8m × 6m × 4m
= 192 m³.

III. Read the following text and answer the following question on the basis of the same:
On a Sunday, your Parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’s cube and strawberry ice-cream for you.
Observe the figures and answer the questions:

Question 1.

The length of the diagonal if each edge measures 6 cm is

(A) 3√3
(B) 3√6
(C) √l2
(D) 6√3
Answer:
(D) 6√3

Question 2.

Volume of the solid figure if the length of the edge is 7 cm is:

(A) 256 cm³
(B) 196 cm³
(C) 343 cm³
(D) 434 cm³
Answer:
(C) 343 cm³

Question 3.

What is the curved surface area of hemisphere (ice cream) if the base radius is 7 cm ?

(A) 309 cm²
(B) 308 cm²
(C) 803 cm²
(D) 903 cm²
Answer:
(B) 308 cm²

Question 4.

Slant height of a cone if the radius is 7 cm and the height is 24 cm……..

(A) 26 cm
(B) 25 cm
(C) 52 cm
(D) 62 cm
Answer:
(B) 25 cm

Question 5.

The total surface area of cone with hemispherical ice cream is

(A) 858 cm²
(B) 885 cm²
(C) 588 cm²
(D) 855 cm²
Answer:
(A) 858 cm²

IV. Read the following text and answer the following question on the basis of the same:
A carpenter made a wooden pen stand. It is in the shape of cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. (See Figure).

Question 1.

What is the volume of cuboid?

(A) 525 cm³
(B) 225 cm³
(C) 552 cm³
(C) 255 cm³
Answer:
(A) 525 cm³

Explanation:
For cuboid
l = 15 cm, b = 10 cm and h = 3.5 cm
Volume of the cuboid = l × b × h
= 15 × 10 × 3.5
= 525 cm³

Question 2.

What is the volume of a conical depression ?

(A) \(\frac{11}{3}\) cm³
(B) \(\frac{11}{30}\) cm³
(C) \(\frac{3}{11}\) cm³
(D) \(\frac{30}{11}\) cm³
Answer:
(B) \(\frac{11}{30}\) cm³

Explanation:
For conical depression:
r = 0.5 cm,
h = 1.4 cm
Volume of conical depression
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0. 5 × 0.5 × 1.4
= \(\frac{11}{30}\) cm³

Question 3.

What is the total volume of conical depressions?

(A) 1.74 cm³
(B) 1.44 cm³
(C) 1.47 cm³
(D) 1.77 cm³
Answer:
(C) 1.47 cm³

Explanation:
Volume of four conical depressions
= 4 × \(\vec{a}\) = 1.47 cm³

Question 4.

What is the volume of wood in the entire stand?

(A) 522.35 cm³
(B) 532.53 cm³
(C) 523.35 cm³
(D) 523.53 cm³
Answer:
(D) 523.53 cm³

Explanation:
Volume of the wood in the entire stand = Volume of cuboid – Volume of 4 conical depressions
= 525 – 1.47
= 523.53 cm3

Question 5.

The given problem is based on which mathematical concept?

(A) Triangle
(B) Surface Area and Volumes
(C) Height and Distances
(D) None of these
Answer:
(B) Surface Area and Volumes

MCQ Questions for Class 10 Maths with Answers

MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes Read More »

MCQ Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Uncategorized / Prasanna

Areas Related to Circles Class 10 MCQ Questions with Answers

Question 1.

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then:

(A) R₁ + R₂ = R
(B) \(R_{1}^{2}+R_{2}^{2}=R^{2}\)
(C) R1 + R2 < R
(D) \(R_{1}^{2}+R_{2}^{2}=R^{2}\)
Answer:
(B) \(R_{1}^{2}+R_{2}^{2}=R^{2}\)

Explanation:
According to the given condition, Area of circle =Area of first circle + Area of second circle
πR² = π \(R_{1}{ }^{2}\) + π\(R_{2}{ }^{2}\)
R² =\(R_{1}{ }^{2}\) + \(R_{2}{ }^{2}\).

Question 2.

If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then:

(A) R₁ + R₂ = R
(B) R₁ + R₂ > R
(C) R₁ + R₂ < R
(D) Nothing definite can be said about the relation among R₁, R₂ and R
Answer:
(A) R₁ + R₂ = R

Explanation:
According to question, Circumference of circle
= Circumference of first circle + Circumference of second circle
2πR = 2πR₁ + 2πR₂
R = R₁ + R₂

Question 3.

If the circumference of a circle and the perimeter of a square are equal, then:

(A) Area of the circle = Area of the square
(B) Area of the circle > Area of the square
(C) Area of the circle < Area of the square
(D) Nothing definite can be said about the relation between the areas of the circle and square.
Answer:
(B) Area of the circle > Area of the square

Explanation:
According to question, Circumference of a circle = Perimeter of square Let ‘r’ and ‘a’ be the radius of circle and side of square.
2πr = 4 a
\(\frac{22}{7}\)r = 2a
11r = 7a
r = \(\frac{7 a}{11}\) ….(i)
A₁ = πR²
From equation (i), we have
A₁ = π\(\left(\frac{7 a}{11}\right)^{2}\)
= \(\frac{22}{7}\left(\frac{49 a^{2}}{121}\right)\)
= \(\frac{14 a^{2}}{11}\)
A₂ = a²
From equation (ii) and (iii), we haveA1 = \(\frac{14}{11}\)A2
A₁ > A₂
Area equation is greater than the area of square.

Question 4.

Area of the largest triangle that can be inscribed in a semi-circle of radius ‘r’ units is:

(A) r² sq. units
(B) \(\frac{1}{2}\) r² sq. units
(C) 2r² sq units
(D) √2r² sq. units
Answer:
(A) r² sq. units

Explanation:
Take a point C on the circumference of the semi-circle and join it by the end points of diameter AB.

∠C = 90 [Angle in a semi – circle is right angle]
So ABC = \(\frac{1}{2}\) × AB × CD
= \(\frac{1}{2}\) × 2r × r = r² sQuestion units

Question 5.

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:

(A) 22 : 7
(B) 14 : 11
(C) 7 : 22
(D) 11 : 14
Answer:
(B) 14 : 11

Explanation:
Let the radius of circle be ‘r’ and side of square be ‘a’.
According to given question,
Perimeter of circle = Perimeter of square
2πr = 4a
∴ a = \(\frac{\pi r}{2}\)
So, \(\frac{ Area of circle}{Area of square}\) = \(\frac{\pi r^{2}}{\left(\frac{\pi r}{2}\right)^{2}}\) [From equation (i)]
Solving equation (i), we get result as \(\frac{14}{11}\).

Question 6.

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be:

(A) 10 m
(B) 15 m
(C) 20 m
(D) 24 m
Answer:
(A) 10 m

Explanation:
Area of first circular park whose diameter is 16 m,
= π\(\left(\frac{16}{2}\right)^{2}\)
= π\((8)^{2}\)
= 64π m²
Area of second circle park whose dimeter is 12 m,
= π\(\left(\frac{12}{2}\right)^{2}\)
= π(6)²
= 36π m²
According to question,
Area of single circlar park = Area of first circular park + Area of second circular park
πr² = 64π + 36π
πr² = 100π
r =10 m

Question 7.

The area of the circle that can be inscribed in a square of side 6 cm is:

(A) 36π cm²
(B) 18π cm²
(C) 12π cm²
(D) 9π cm²
Answer:
(D) 9π cm²

Explanation:

Given, side of square = 6 cm
Diameter of a circle, (d) = Side of square = 6 cm
Radius of a circle (r) = \(\frac{d}{2}\) = 6 = 3 cm.
Area of circle = πr²
π(3)² = 9π cm²

Question 8.

The area of the square that can be inscribed in a circle of radius 8 cm is:

(A) 256 cm²
(B) 128 cm²
(C) 64 cm²
(D) 64 cm²
Answer:
(B) 128 cm²

Explanation:
Given, radius of circle, r =OC = 8 cm
Diameter of the circle
= AC = 2 × OC
= 2 × 8 = 16 cm
Which is equal to the diagonal of a square.

Let side of square be ‘a’a.
Using Pythagoras theorem,
AB² + BC² = AC2
a² + a² = 16²
2a² = 256
a² = 128 cm ²

Assertion and Reason Based MCQs

Question 1.

Assertion (A): The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is 50 cm.
Reason (R): If the perimeter and the area of a circle are numerically equal, then the radius of the circle is 2 units.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In the case of assertion:
Let r₁ = 24 cm and r₂ = 7 cm Area of first circle = π\(r_{1}^{2}\) = π(24)² = 576π cm²
Area of second circle = π\(r_{2}^{2}\) = π(7)² = 49π cm² cm²
According to question,
Area of circle = Area of first circle + Area of second circle
πR² = 576π + 49π
[where, R be radius of circle]
R² =625 = 25 cm
Diameter of a circle = 2R = 2 × 25 = 50 cm.
∴ Assertion is correct.
In case of reason:
Let the radius of the circle be r.
Circumference of circle = 2πr
Area of circle = πr²
Given that, the circumference of the circle and the area of the circle are equal. This implies,
2πr = πr²
r = 2
Therefore, the radius of the circle is 2 units.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not correct explanation for assertion.

Question 2.

Assertion (A): In covering a distance s meter, a circular wheel of radius r meter makes \(\frac{s}{2 \pi r}\) revolution.
Reason (R): The distance travelled by a circular wheel of diameter d cm in one revolution is 2πd cm.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
The distance covered in one revolution is 2πr, that is, its circumference.
∴ Assertion is correct In case of reason:
Because the distance travelled by the wheel in one revolution is equal to its circumference Circumference of wheel = π × diameter ,
= π × d
= πd
Hence the given answer in the question is incorrect.
∴Reason is incorrect.
Hence, assertion is correct and reason is incorrect.

Question 3.

Assertion (A): If circumferences of two circles are equal, then their areas will be equal.
Reason (R): If the areas of two circles are equal, then their circumferences are equal.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
If circumferences of two circles are equal, then their corresponding radii are equal. So, their areas will be equal.
∴ Assertion is correct.
In case of reason:
If areas of two circles are equal, then their corresponding radii are equal. So, their circumference will be equal.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following question on the basis of the same:
A horse is tied to a peg at one comer of a square shaped grass field of sides 15 m by means of a 5 m long rope (see the given figure)

Question 1.

What is the area of the grass field?

(A) 225 m²
(B) 225 m
(C) 255 m²
(D) 15 m
Answer:
(A) 225 m²

Explanation:
Area of square = (side)²
= 15 × 15
= 225 m²

Question 2.

The area of that part of the field in which the horse can graze.

(A) 19.625 m²
(B) 19.265 m²
(C) 19 m²
(D) 78.5 m²
Answer:
(A) 19.625 m²

Explanation:
From the figure, it can be observe that the horse can graze a sector of 90° in a circle of 5 m radius.
Area that can be grazed by horse = Area of sector
= \(\frac{90^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{1}{4}\) × 3 . 14 × 5 × 5
= 19 . 625 m²

Question 3.

The grazing area if the rope were 10 m long instead of 5 m.

(A) 7.85 m²
(B) 785 m²
(C) 225 m²
(D) 78.5 m²
Answer:
(D) 78.5 m²

Explanation:
Area that can be grazed by the horse when length of rope is 10 m long
= \(\frac{90^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{1}{4}\) × 3 . 14 × 10 × 10
= 78.5 m²

Question 4.

The increase in the grazing area if the rope were 10 m long instead of 5 m.

(A) 58.758 m²
(B) 58.875 m²
(C) 58 m²
(D) 78.5 m²
Answer:
(B) 58.875 m²

Explanation:
Increase in grazing area
= (78.5 – 19.625) m²
= 58.875 m²

Question 5.

The given problem is based on which concept?

(A) Coordinate geometry
(B) Area related to circles
(C) Circle
(D) None of these
Answer:
(B) Area related to circles

II. Read the following text and answer the following question on the basis of the same:
In a workshop, brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the given figure.

Question 1.

What is the radius of the circle?

(A) \(\frac{35}{2} \mathrm{~mm}\)
(B) \(\frac{5}{2} \mathrm{~mm}\)
(C) 35 mm
(D) 10 mm
Answer:
(A) \(\frac{35}{2} \mathrm{~mm}\)

Explanation:
Radius of circle = \(\frac{\text { Diameter }}{2}\)
= \(\frac{35}{2} \mathrm{~mm}\)

Question 2.

What is the circumference of the brooch?

(A) 100 mm
(B) 110 mm
(C) 50 mm
(D) 10 mm
Answer:
(B) 110 mm

Explanation:
Circumference of brooch = 2πr
= 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\)
= 110 mm

Question 3.

What is the total length of silver wire required?

(A) 528 mm
(B) 825 mm
(C) 285 mm
(D) 852 mm
Answer:
(C) 285 mm

Explanation:
Length of wire required
= 110 + 5 x 35
= 110 + 175
= 285 mm.

Question 4.

What is the area of the each sector of the brooch?

(A) \(\frac{385}{2} \mathrm{~mm}^{2}\)
(B) \(\frac{358}{4} \mathrm{~mm}^{2}\)
(C) \(\frac{585}{4} \mathrm{~mm}^{2}\)
(D) \(\frac{385}{4} \mathrm{~mm}^{2}\)
Answer:
(D) \(\frac{385}{4} \mathrm{~mm}^{2}\)

Explanation:
It can be observed from the figure that an angle of each 10 sectors of the circle is subtending at the centre of the circle.
∴Area of each sector = \(\frac{36}{360^{\circ}}\) × πr²
= \(\frac{1}{10}\) × \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\)
= \(\frac{385}{4}\) mm2

Question 5.

The given problem is based on which mathematical concept ?

(A) Areas Related to circles
(B) Circles
(C) Construction
(D) none of these
Answer:
(A) Areas Related to circles

III. Read the following text and answer the following question on the basis of the same:
AREAS RELATED TO CIRCLES Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam, your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below.
Observe these carefully.

Design I: This design is made with a circle of radius 32 cm leaving equilateral triangle ABC in the middle as shown in the given figure.
Design II: This Pookalam is made with 9 circular design each of radius 7 cm.
Refer Design I:

Question 1.

The side of equilateral triangle is

(A) 12√3 cm
(B) 32√3 cm
(C) 48 cm
(D) 64 cm
Answer:
(B) 32√3 cm

Question 2.

The altitude of the equilateral triangle is

(A) 8 cm
(B) 12 cm
(C) 48 cm
(D) 52 cm
Answer:
(C) 48 cm

Refer Design II:

Question 3.

The area of square is

(A) 1264 cm²
(B) 1764 cm²
(C) 1830 cm²
(D) 1944 cm²
Answer:
(B) 1764 cm²

Explanation:
radius = 7 cm
diameter = 2 x 7 cm = 14 cm
side of square = 14 cm + 14 cm + 14 cm
= 42 cm.
Area of square = side²
= (42 cm)²
1764 cm²

Question 4.

Area of each circular design is

(A) 124 cm²
(B) 132 cm²
(C) 144 cm²
(D) 154 cm²
Answer:
(D) 154 cm²

Explanation:
radius = 7 cm
Area of each circular design = πr²
= \(\frac{22}{7}\) × 7 × 7
= 154 cm²

Question 5.

Area of the remaining portion of the square ABCD

(A) 378 cm²
(B) 260 cm²
(C) 340 cm²
(D) 278 cm²
Answer:
(A) 378 cm²

Explanation:
Area of 9 circular design = 9 × πr²
= 9 × \(\frac{22}{7}\) × 7 × 7
= 1386 cm²
Area of square = 1764 cm²
Area of remaining portion of square – Area of 9 circular design
= 1764 cm ² – 1386 cm²
= 378 cm²

IV. Read the following text and answer the following question on the basis of the same:
A Brooch A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.

Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which divide the circle into 8 equal parts.
Design B: Brooch b is made two colours i.e. Gold and silver. Outer part is made with Gold. The circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere.
Refer to Design A

Question 1.

The total length of silver wire required is

(A) 180 mm
(B) 200 mm
(C) 250 mm
(D) 280 mm
Answer:
(B) 200 mm

Explanation:
Diameter = 28 mm
radius = 14 mm
Total length of wire = length of 4 diameter + circumference of circle.
= 4 × 28 + 2πr²
= 112 + 2 × \(\frac{22}{7}\) × 14
= 112 + 88
= 200 mm

Question 2.

The area of each sector of the brooch is

(A) 44 mm2
(B) 52 mm2
(C) 77 mm2
(D) 68 mm2
Answer:
(C) 77 mm2

Explanation:
Area of each sector of Brooch
= \(\frac{1}{8}\) × Area of Brooch
= \(\frac{1}{8}\) × πr²
= \(\frac{1}{8}\) x \(\frac{22}{7}\) × 14 × 14
= 77 mm²

Refer to Design B

Question 3.

The circumference of outer part (golden) is

(A) 48.49 mm
(B) 82.2 mm
(C) 72.50 mm
(D) 62.86 mm
Answer:
(D) 62.86 mm

Question 4.

The difference of areas of golden and silver parts is

(A) 18π
(B) 44π
(C) 51π
(D) 64π
Answer:
(C) 51π

Question 5.

A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm ?

(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(C) 4

Explanation:
Circumference of silver part of Brooch
= 44 cm
2πr = 44 mm
2 × \(\frac{22}{7}\) × r = 44
r = 7 mm.
radius of whole Brooch
= 7 mm + 8 mm
= 10 mm.
Circumference of outer edge
= 2πr
= 2 × \(\frac{22}{7}\) × 10
= \(\frac{440}{7}\) mm
let the number of revolutions = n
Now, According to question,
n . 2πr = 80π
n . \(\frac{440}{7}\) = 80π
n . \(\frac{440}{7}\) = 80 x \(\frac{22}{7}\)
n = 4

MCQ Questions for Class 10 Maths with Answers

MCQ Questions for Class 10 Maths Chapter 12 Areas Related to Circles Read More »

MCQ Questions for Class 10 Maths Chapter 11 Constructions

MCQ Questions / Prasanna

Constructions Class 10 MCQ Questions with Answers

Question 1.

To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

(A) 8
(B) 10
(C) 11
(D) 12
Answer:
(D) 12

Explanation:
Minimum number of the points marked = sum of ratios = 5 + 7 = 12.

Question 2.

To divide a line segment AB in ratio 4 : 7, a ray AX is drawn first such that ∠ BAX is an acute angle and then points A1, A2, A3, … are located at equal distances on the ray AX and the point B is joined to :

(A) A₁₂
(B) A₁₁
(C) A₁₀
(D) A₉
Answer:
(B) A₁₁

Explanation:
We have to divide the line segment into 7 + 4 = 11 equal parts and 11th part will bem joined to B, here A₁₂ will never appear.

Question 3.

To divide a line segment AB in the ratio 5 : 6 draw a ray AX such that∠BAX is an acute angle, then draw a ray BY parallel to AX, and the points, A₁, A₂, A₃,… and B₁, B₂, B₃, … are located at. Equal distances on ray AX and BY, respectively. Then the points joined are

(A) A₅ and B₆
(B) A₆ and B₅
(C) A₄ and B₅
(D) A₅ and B₄
Answer:
(A) A₅ and B₆

Explanation:
In the figure; segment AB of given length is divided into 2 parts of ratio 5 : 6 in following steps:
(i) Draw a line-segment AB of given length.
(ii) Draw an acute angle BAX as shown in figure either upside or down side.
(iii) Draw angle ∠ABY = ∠BAX on other side of AX, that is, down side.
(iv) Divide AX into 5 equal parts by using compass.
(v) Divide BX into same distance in 6 equal parts as AX was divided.
(vi) Now, join A5 and B6 which meet AB at P. P divides AB in ratio AP : PB = 5 : 6.

Question 4.

To draw a pair of tangents to a circle which are in¬clined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:

(A) 135°
(B) 90°
(C) 60°
(D) 120°
Answer:
(D) 120°

Explanation:

We know that tangent and radius at contact point are perpendicular to each other. So, ∠P and ∠Q in quadrilateral TPOQ formed by tangents and radii will be of 90° each.
So, the sum of
∠T+ ∠O = 180°
as T = 60° [Given]
∴∠O = 180° – 60°
= 120°

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the below questions:
A school conducted Annual Sports Day on a triangular playground. On the ground, parallel lines have been drawn with chalk powder at a distance of 1 m. 7 flower pots have been placed at a distance of from each other along DM as shown in the figure.

Now answer the following questions:

Question 1.

PD₃ is parallel to:

(A) PD
(B) PE
(C) ED₇
(D) None of these.
Answer:
(C) ED₇

Explanation:
We have,
PD(C) ED₃ || ED₇

Question 2.

If ∠PD3D = 82°, then the measure of ∠ED7D is:

(A) 98°
(B) 82°
(C) 90°
(D) 45°
Answer:
(B) 82°

Explanation:
We have,
PD(C) ED₃ || ED₇
Then, ∠ED₇D = ∠PD₃D
(Corresponding angles)
∴∠ED₇D = 82°.

Question 3.

The ratio in which P divides DE, is:

(A) 3 : 4
(B) 7 : 3
(C) 3 : 7
(D) 2 : 5
Answer:
(A) 3 : 4

Explanation:
P divides DE in the ratio 3 : 4.

Question 4.

The ratio of DE to DP will be:

(A) 2:5
(B) 3 : 4
(C) 3 : 7
(D) 7 : 3
Answer:
(D) 7 : 3

Explanation:
DE = 7 m
[∵ DD₁ = D₁D₂ = D₂D₃ = D₃D₄
= D₄D₅= D₅D₆ = D₆D₇]
and DP = 3 m
[∴ DD₁ = D₁D₂ = D₂ D₃]
∴ \(\frac{D E}{D P}\) = \(\frac{7 \mathrm{~m}}{3 \mathrm{~m}}\) = \(\frac{7}{3}\)
Hence, the ratio of DE to DP is 7 : 3.

Question 5.

The total distance used for putting 7 flower pots is:

(A) 6 m
(B) 7 m
(C) 5 m
(D) 8 m.
Answer:
(B) 7 m

Explanation:
Since, 7 flower pots have been placed at a distance of 1 m from each other, then total distance = 7 m.

MCQ Questions for Class 10 Maths with Answers

MCQ Questions for Class 10 Maths Chapter 11 Constructions Read More »

Author name: Prasanna

MCQ Questions for Class 12 Informatics Practices – Python Pandas Part 2

Python Pandas Class 12 MCQ Questions Part 2

Which of the following statement is correct for importing pandas in python?

What type of error is returned by following statement? import pandas as pnd pn(d)Series([l,2,3,4], index = [aVb’,’c’])

Which attribute is used to give user defined labels in Series?

Fill in the blank to get the ouput as 3 import pandas as pnd Sl=pn(d)Series([l,2,3,4], index = [‘a’,’b’,’c’,’d’]) print(Sl[ ])

Write the statement to get NewDelhi as output using positional index, import pandas as pd SI=p(d)Series([‘NewDelhi’,’WashingtonDC’,’London’,’Paris’], index=[‘lndia’, ‘USA’, ‘UK’, ‘France’])

We can access elements in Series by using …… index and index.

Write the output of the following: import pandas as pd SI = p(d)Series([‘NewDelhi’,’WashingtonDC’, ‘London’, ’Paris’], index=[’lndia’, ‘USA’, ‘UK’, ‘France’]) print(Sl[‘lndia’,’UK’])

Which of the following statement will print Series ‘SI’ in reverse order?

How many values will be modified by last statement of given code? import pandas as pd SI=p(d)Series([‘NewDelhi’,’WashingtonDC’,’London’,’Paris’], index=[‘A’, ‘B’, ‘C’, ‘D’]) S1[A’: C’] = ‘ND’

How many values will be modified by last statement of given code? import pandas as pd SI = p(d)Series([’NewDelhi’, ’WashingtonDC’, ‘London’, ‘Paris’], index=[‘A’,’B’,’C’,’D’]) Sl[l: 3] = ‘ND’

Which of the following property/attribute assign name to the Series?

Sl.vatues will return all the values of Series ‘SI’ in

Which of the following property/attribute return total number of values in Series‘SI’?

Which of the following attributes returns True if there is no value in Series?

Which of the following attributes returns all the values of Series?

Write the output of the following code: import pandas as pd Sl=p(d)Series() print(p(d)Series().empty)

Write the output of the following code : import pandas as pd Sl=p(d)Series([l,2,3,4]) S2=p(d)Series([7,8]) S3=S1+S2 print(S3.size) (a) 2 (c) 6

Which of the following statement shows first five values of Series ‘SI’?

Write the output of the following: import pandas as pd Sl=p(d)Series([l,2,3,4]) S2=p(d)Series([7,8]) print((Sl+S2).count()) .

Which of the following returns number of non-NaN values of Series?

Write the output of the following: import pandas as pd Sl=p(d)Series([l,2,3,4]) S2=p(d)Series([7,8,9,10]) S2.index=[‘a’,’b’,’c’,’d’] print{{Sl+S2).count( ))

We can perform on two series in Pandas.

Which of the following method is used to add two series?

Which of the following statement will display the difference of two Series ‘A’ and ‘B’?

Which of the following fills the missing values in Series?

Which of the following function is used for basic mathematical operations in Series?

Which of the following statement is replacing missing values of Series A and Series B by 100?

Mathematical operations on two Series object is done by matching

Which of the following statement will display values more ‘ than 40 from Series ‘SI’?

Which of the following statement will return 10 values from the bottom/end of the Series ‘SI’?

Which of the following are valid operations on Series ‘SI’?

When an operation is carried out on every value of Series object is called

Which of the following statement will modify the first three values of Series ‘SI’?

We can have duplicate indexes in Series?

What is the index value of 31 in given Series ‘SI’? import pandas as pd Sl=p(d)Series([l,2,31,4], index = [’a’.’bVcVd’])

S2 is ……..in given code. S2 = Sl[2 : 5] #S1 is a Series object

Following two statements will provide the same output: >>>L1 “ 2 #L1 is a list >>>S1 “ 2 #S1 is a Series

Best way to import the pandas module in your program:

Dataframe in pandas is:

For what purpose a Pandas is used?

In data science, which of the python library are more popular?

Series in Pandas is:

Way to install the pandas library?

We can analyse the data in pandas with:

What we pass in Dataframe in pandas?

Choose the correct function to rename city columns to location using rename() function.

Which of the following statement(s) is/are correct with respect to df.columns properties to rename columns? All columns must be specified . Columns must be in the form of a list Old column names not required Columns can be specified with columns number

df.index properties ca be used to:

To display 2 rows from the top in the dataframe, which of the following statement is correct?

A function is used to display top rows (n) from …….. dataframe.

A function is used to display bottom rows (n) from dataframe.

…… refers to passing True and False value as an index in Dataframes.

To pass index label keyword is used to in p(d)dataframe() function.

By default you can display……..no. of top/bottom …… rows using head()/tail() function.

Pandas is an opensource Library?

Pandas key data structure is called:

In pandas, Index values must be:

Which of the following is correct features of Dataframe?

A panel is a container of dat(a)

Which of the following is true?

Which of the following takes a diet of diets or a diet of arraylike sequences and returns a Dataframe?

Which of the following makes use of pandas and returns data in a series or dataframe?

What will be output for the following code? import pandas as pd import numpy as np s – p(d)Series(np.random.randn(4)) print s.ndim

Which of the following indexing capabilities is used as a concise means of selecting data from a pandas object?

Which of the following thing can be data in Pandas?

Point out the correct statement.

The result of an operation between unaligned Series will have the of the indexes involve(d)

Which of the following input can be accepted by Dataframe?

Point out the wrong statement.

Which of the following takes a diet of diets or a diet of arraylike sequences and returns a Dataframe?

Series is a one-dimensional labeled array capable of holding any data type.

Which of the following works analogously to the form of the diet constructor?

Which of the following operation works with the same syntax as the analogous diet operations?

If data is an array; index must be the same length as dat(a)

MCQ Questions for Class 12 Informatics Practices with Answers

MCQ Questions for Class 10 Maths Chapter 14 Statistics

Statistics Class 10 MCQ Questions with Answers

Write the output of the following: import pandas as pd
SI = p(d)Series([‘NewDelhi’,’WashingtonDC’, ‘London’, ’Paris’], index=[’lndia’, ‘USA’, ‘UK’, ‘France’]) print(Sl[‘lndia’,’UK’])

How many values will be modified by last statement of given code? import pandas as pd
SI=p(d)Series([‘NewDelhi’,’WashingtonDC’,’London’,’Paris’], index=[‘A’, ‘B’, ‘C’, ‘D’]) S1[A’: C’] = ‘ND’

How many values will be modified by last statement of given code? import pandas as pd
SI = p(d)Series([’NewDelhi’, ’WashingtonDC’, ‘London’, ‘Paris’], index=[‘A’,’B’,’C’,’D’]) Sl[l: 3] = ‘ND’

Write the output of the following code : import pandas as pd Sl=p(d)Series([l,2,3,4])
S2=p(d)Series([7,8]) S3=S1+S2 print(S3.size) (a) 2 (c) 6

Write the output of the following: import pandas as pd Sl=p(d)Series([l,2,3,4])
S2=p(d)Series([7,8]) print((Sl+S2).count()) .

Write the output of the following: import pandas as pd Sl=p(d)Series([l,2,3,4])
S2=p(d)Series([7,8,9,10]) S2.index=[‘a’,’b’,’c’,’d’] print{{Sl+S2).count( ))

What is the index value of 31 in given Series ‘SI’? import pandas as pd
Sl=p(d)Series([l,2,31,4], index = [’a’.’bVcVd’])

S2 is ……..in given code.
S2 = Sl[2 : 5] #S1 is a Series object

Following two statements will provide the same output:
>>>L1 “ 2 #L1 is a list
>>>S1 “ 2 #S1 is a Series

If x_i‘s are the mid-points of the class intervals of grouped data f_i s are the corresponding frequencies
and \( \bar{x}\) is the mean, then \( \sum\left(f_{i} x_{i}-\bar{x}\right)\) is equal to :

The difference of the upper limit of the median class and the lower limit of the modal class is 20.
Reason (R): The median class and modal class of grouped data always be different.

The mean of the above data is 12.93.
Reason (R): The following table gives the number of pages written by Sarika for completing her own book for 30 days:

Assertion (A): If the median of a series exceeds the mean by 3, then mode exceeds mean by 10.
Reason (R): If made = 12.4 and mean = 10.5, then the median is 11.13.

Assertion (A): If P(F) = 0.20, then the probability of ‘not E’ is 0.80.
Reason (R): If two dice are thrown together, then the probability of getting a doublet is \(\frac{5}{6}\).

Assertion (A): The probability that a number selected at random from the number 1, 2, 3, , 15
is a multiple of 4, is \(\frac{1}{3}\)..
Reason (R): Two different coins are tossed simultaneously. The probability of getting at least one head is\(\frac{3}{4}\).