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MCQ Questions for Class 12 Informatics Practices – Python Pandas Part 1

Python Pandas Class 12 MCQ Questions Part 1

Question . 1.

To create an empty series object, you can use:

(a) p(d)Series(empty)
(b) p(d)Series(np.NaN)
(c) p(d)Series( )
(d) All of these
Answer:
(c) p(d)Series( )

Question 2.

To specify datatype int 16 for a series object,you can write:

(a) p(d)Series(data = array, dtype = int16)
(b) p(d)Series(data = array, dtype = numpy.int 16)
(c) p(d)Series(data= array.dtype = pandas.int 16)
(d) All of the above
Answer:
(b) p(d)Series(data = array, dtype = numpy.int16)

Question 3.

To get the number of dimensions of a series object, attribute is displaye(d)

(a) index
(b) size
(c) itemsize
(d) ndim
Answer:
(d) ndim

Question 4.

To get the size of the datatype of the items in series object, you can display attribute.

(a) index
(b) size
(c) itemsize
(d) ndim
Answer:
(c) itemsize

Question 5.

To get the number of elements in a series object, attribute may be use(d)

(a) index
(b) size
(c) itemsize
(d) ndim
Answer:
(b) size

Question 6.

To get the number of bytes of the series data, attribute is displaye(d)

(a) hasnans
(b) nbytes
(c) ndim
(d) dtype
Answer:
(b) nbytes

Question 7.

To check if the series object contains NaN values, attribute is displaye(d)

(a) hasnans
(b) n bytes
(c) n dim
(d) dtype
Answer:
(a) hasnans

Question 8.

To display third element of a series object S, you will write

(a) S(:3)
(b) S[2]
(c) S[3]
(d) S[:2]
Answer:
(b) 5(2]

Question 9.

To display first three elements of a series object S, you may write

(a) S(:3]
(b) 5(3]
(c) 5 (3rd]
(d) All of these
Answer:
(a) S(:3]

Question 10.

To display last five rows of a series object S, you may write

(a) head()
(b) head(5)
(c) tail()
(d) tail(5)
Answer:
(c) tail(), (d) tail(5)

Question 11.

Missing data in Pandas object is represented through:

(a) Null
(b) None
(c) Missing
(d) NaN
Answer:
(d) NaN

Question 12.

Given a Pandas series called SeQuestion uences, the command which will display the first 4 rows is

(a) print(SeQuestion uences.head(4))
(b) print (SeQuestion uences. Head(4))
(c) print(SeQuestion uences.heads(4)
(d) print(SeQuestion uences.Heads(4))
Answer:
(a) print(SeQuestion uences.head(4))

Question 13.

If a dataframe is created using a 2D dictionary, then the indexes/row labels are formed from

(a) dictionary’s values
(b) inner dictionary’s keys
(c) outer dictionary’s keys
(d) None of these
Answer:
(b) inner dictionary’s keys

Question 14.

If a dataframe is created using a 2D dictionary, then the column labels are formed from

(a) dictionary’s values
(b) inner dictionary’s keys
(c) outer dictionary’s keys
(d) None of these
Answer:
(c) outer dictionary’s keys

Question 15.

The axis 0 identifies a dataframe’s

(a) rows
(b) columns
(c) values
(d) datatype
Answer:
(a) rows

Question 16.

The axis 1 identifies a dataframe’s

(a) rows
(b) columns
(c) values
(d) datatype
Answer:
(b) columns

Question 17.

To get the number of elements in a dataframe attribute may be use(d)

(a) size
(b) shape
(c) values
(d) ndim
Answer:
(a) size

Question 18.

To get NumPy representation of a dataframe ……… attribute may be use(d)

(a) size
(b) shape
(c) values
(d) ndim
Answer:
(c) values

Question 19.

To get a number representing number of axes in a dataframe, ……….. attribute may be use(d)

(a) size
(b) shape
(c) values
(d) ndim
Answer:
(d) ndim

Question 20.

To get the transpose of a dataframe Dl, you can write

(a) D1.T
(b) D1. Transpose
(c) D1.Swap
(d) All of these
Answer:
(a) D1.T

Question 21.

To extract row/column from a dataframe,……….. function may be use(d)

(a) row( )
(b) column( )
(c) loc( )
(d) All of these
Answer:
(c) loc( )

Question 22.

To display the 3rd, 4th and 5th columns from the 6th to 9th rows of a dataframe DF,you can write

(a) DF.loc[6:9, 3:5]
(b) DF.loc[6:10, 3:6]
(c) DF.iloc(6:10, 3:6]
(d) DF.iloc[6:9, 3:5]
Answer:
(c) DF.iloc[6:10,3:6]

Question 23.

To change the 5th column’s value at 3rd row as 35 in dataframe DF, you can write

(a) DF(4, 6] = 35
(b) DF(3, 5] = 35
(c) DF.iat(4, 6] = 35
(d) DF.iat(3, 5] = 35 .
Answer:
(d) DF.iat(3, 5] = 35

Question 24.

Which among the following options can be used to create a dataframe in Pandas?

(a) A scalar value
(b) An ndarray
(c) A python diet
(d) All of these
Answer:
(d) All of these

Question 25.

Identify the correct statement:

(a) The standard marker for missing data in Pandas is NaN
(b) Series act in a way similar to that of an array
(c) Both (a) and (b)
(d) None of the above
Answer:
(c) Both (a) and (b)

Question 26.

To delete a column from a dataframe, you may use statement.

(a) remove
(b) del
(c) drop
(d) cancel
Answer:
(b) del

Question 27.

To delete a rowfrom a DataFrame.you may use……… statement.

(a) remove
(b) del
(c) drop
(d) cancel
Answer:
(c) drop

Question 28

……… is a popular data-science library of Python.

(a) numpy
(b) pandas
(c) Both (a) and (b)
(d) None of these
Answer:
(b) pandas

Question 29.

A ………… is a Pandas data structure that represents a ID array like object.

(a) dataframe
(b) vector
(c) series
(d) All of these
Answer:
(c) series

Question 30.

A is a Pandas data structure that represents ……… a 2D array like object.

(a) dataframe
(b) vector
(c) series
(d) All of these
Answer:
(a) dataframe

Question 31.

You can use numpy …… for missing dat(a)

(a) none
(b) no
(c) NaN
(d) None of these
Answer:
(c) NaN

Question 32.

To specify datatype for a series object …… argument is use(d)

(a) ctype
(b) atype
(c) ntype
(d) dtype
Answer:
(d) dtype

Question 33.

The function on series object returns total ……. elements in it including NaNs.

(a) gen( )
(b) len( )
(c) gen( )
(d) cen( )
Answer:
(b) len()

Question 34.

The function on series object returns only the count of non-NaN values in it.

(a) acount( )
(b) allcount( )
(c) count( )
(d) dcount( )
Answer:
(c) count( )

Question 35.

Series is mutable.

(a) non- value
(b) text
(c) numeric
(d) value
Answer:
(d) value

Question 36.

Series is not ……… mutable.

(a) size
(b) variable
(c) shape
(d) value
Answer:
(a) size

Question 37.

Dataframe is …….. mutable as well as ………..mutable.

(a) value, size
(b) size, value
(c) size, size
(d) value, value
Answer:
(b) size, value

Question 38.

In a dataframe, Axis = 1 represents the ……… elements.

(a) row
(b) record
(c) column
(d) None of these
Answer:
(c) column

Question 39.

To access values using row labels you can use DF

(a) loc
(b) voc
(c) doc
(d) toe
Answer:
(a) loc

Question 40.

To access individual value, you can use DF …….. using row/column index labels.

(a) in
(b) or
(c) more
(d) at
Answer:
(d) at

Question 41.

To access individual value, you can use DF ……. using row/column integer position.

(a) lot
(b) doc
(c) toe
(d) iat
Answer:
(d) iat

Question 42.

The rename() function requires argument ……… to make changes in the original dataframe.

(a) outplace()
(b) atplace
(c) inplace
(d) ofplace
Answer:
(c) inplace

Question 43.

Which of the following are modules/libraries in Python?

(a) NumPy
(b) Pandas
(c) Matplotlib
(d) All of these
Answer:
(d) All of these

Question 44.

NumPy stands for ……….

(a) Number Python
(b) Numerical Python
(c) Numbers in Python
(d) None of these
Answer:
(b) Numerical Python

Question 45.

Which of the following libraries allows to manipulate, transform and visualize data easily and efficiently?

(a) Pandas
(b) NumPy
(c) Matplotlib
(d) All of these
Answer:
(d) All of these

Question 46.

PANDAS stands for……….

(a) Panel Data Analysis
(b) Panel Data Analyst
(c) Panel Data
(d) Panel Dashboard
Answer:
(c) Panel Data

Question 47.

……… is an important library used for analysing dat(a)

(a) Math
(b) Random
(c) Pandas
(d) None of these
Answer:
(c) Pandas

Question 48.

Important data structure of pandas is/are

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

Question 49.

Which of the following library in Python is used for plotting graphs and visualization?

(a) Pandas
(b) NumPy
(c) Matplotlib
(d) None of these
Answer:
(c) Matplotlib

Question 50.

Pandas series can have data types.

(a) float
(b) integer
(c) string
(d) All of these
Answer:
(d) All of these

Question 51

…….. is used when data is in Tabular Format.

(a) NumPy
(b) Pandas
(c) Matplotlib
(d) All of these
Answer:
(b) Pandas

Question 52.

Which of the following command is used to install pandas?

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

Question 53.

A is a collection of data values and operations that can be applied to that dat(a)

(a) data structure
(b) dataframe
(c) table
(d) None of these
Answer:
(a) data structure

Question 54.

A is a one-dimensional array.

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

Question 55.

Which of the following statement is wrong?

(a) We can create Series from Dictionary in Python,
(b) Keys of dictionary become index of the series.
(c) Order of indexes created from Keys may not be in the same order as typed in dictionary.
(d) All are correct
Answer:
(d) All are correct

Question 56.

A Series by default have numeric data labels starting from

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

Question 57.

The data label associated with a particular value of Series is called its

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

Question 58.

Which of the following module is to be imported to create Series?

(a) NumPy
(b) Pandas
(c) Matplotlib
(d) None of these
Answer:
(b) Pandas

Question 59.

Which of the following function/method help to create Series?

(a) series( )
(b) 5eries( )
(c) createSeries( )
(d) None of these
Answer:
(b) Series( )

Question 60.

Write the output of the following:
>>> import pandas as pd
>>> seriesl = p(d)Series(10,20,30)
>>> print(seriesl)

(a) Output:
0 10
1 20
2 30
dtype: int64

(b) Output:
10
20
30
dtype: int64

Answer:
(a) Output:
0 10
1 20
2 30
dtype: int64

Question 61.

When you print/display any series then the left most column is showing value.

(a) index
(b) data
(c) value
(d) None of these
Answer:
(a) index

Question 62.

How many values will be there in arrayl, if given code is not returning any error?>>> series4 = p(d)Series(arrayl, index = [“Jan”, “Feb”, “Mar”, “Apr”])

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

Question 63.

Which of the following statement will create an empty series named “SI”?

(a) 51 = p(d)Series(None)
(b) 51 = p(d)5eries()
(c) Both of the above
(d) None of these
Answer:
(c) Both of the above

Question 64.

How many elements will be there in the series named “SI”? »> SI = p(d)Series(range(5)) >» print(Sl)

(a) 5
(b) 4
(c) 6
(d) None of these
Answer:
(a) 5

Question 65.

When we create a series from dictionary then the keys of dictionary become

(a) index of the series
(b) value of the series
(c) caption of the series
(d) None of these
Answer:
(a) index of the series

Question 66.

Write the output of the following:>>> Sl=p(d)Series(14, index = [‘a’, ‘b’, ‘c’]) >>> print(Sl)

(a) a 14
b 14
c 14
dtype: int64

(b) a 14
dtype: int64

(d) None of these

Answer:
(a) a 14
b 14
c 14
dtype: int64

Question 67.

Write the output of the following: >>>Sl=p(d)Series([14,7, index = [‘a’, ‘b’, ‘c’]) >>> print(Sl)

(a) a 14
b 7
c 7

(b) a 14
b 7
dtype: int64

(d) None of these
Answer:
(c) Error

Question 68.

Write the output of the following:
>>> Sl=p(d)Series([14,7,9] .index = range(l, 8,3)) >>> print(Sl)

(a) 14 1
7 4
9 7
dtype: int64

(b) 1 14
47
7 9
dtype: int64

(d) None of these
Answer:
(b) 1 14
47
79
dtype: int64

Question 69.

Which of the following code will generate the following output?
Jan 31
Feb 28
Mar 31 dtype: int64

(a) import pandas as pd
51 = p(d)Series(data = [31,28,31], index=[“Jan”,”Feb”,,,Mar”]) print(SI)

(b) import pandas as pd
S1 = p(d)Series((31,28,31], index=[“Jan”,”Feb”,”Mar”]) print(SI)

(d) None of the above

Answer:
(c) Both of the above

Question 70.

Write the output of the following: import pandas as pd SI = p(d)Series(data = range(31, 2, -6), index = [x for x in “aeiou”]) print(Sl)

(a) a 31
e 25
i 19
o 13
u 7
dtype: int64

(b) a 31
e 25
i 19
dtype: int64

(d) None of these
Answer:
(a) a 31
e 25
i 19
o 13
u 7
dtype: int64

Question 71.

What type of error is returned by following code? import pandas as pd SI = p(d)Series(data = (31,2, -6), index = [7,9,3,2]) print(Sl)

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

Question 72.

Write the output of the following: import pandas as pd SI = p(d)Series(data – 2*(31,2,-6)) print(Sl)

(a) 0 31
12
2-6
dtype: int64

(b) 0 31
12
2-6
3 31
4 2
dtype: int64

(d) 0 31
12
2-6
3 31
dtype: int64

Answer:
(d) 0 31
12
2-6
3 31
dtype: int64

Question 73.

We can imagine a Pandas series as a …….. spreadsheet.

(a) column
(b) cell
(c) table
(d) None of these
Answer:
(a) column

Question 74.

We can assign user-defined labels to the index of the series

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

Question 75.

Write the output of the following: import pandas as pd series2 = p(d)Series([“Kavi7Shyam7Ravi”], index=[3,5 print(series2 > “S”)

(a) 3 False
5 False
1 False
dtype: bool

(b) 3 False
5 False
1 False
dtype: bool

(d) None of these

Answer:
(b) 3 False
5 False
1 False
dtype: bool

MCQ Questions for Class 12 Informatics Practices with Answers

MCQ Questions for Class 12 Informatics Practices – Python Pandas Part 1 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 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 »

MCQ Questions for Class 10 Maths Chapter 10 Circles

MCQ Questions / Prasanna

Circles Class 10 MCQ Questions with Answers

Question 1

If the radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is the tangent to the other circle is:

(A) 3 cm
(B) 6 cm
(C) 9 cm
(D) 1cm
Answer:
(B) 6 cm

Explanation:
Let C₁; C₂ be two coni oniric circles with their centre C.

Chord AB of circle C² touches C₁ at P AB is tangent at P and PC is radius
CP ⊥ AB
Given, ∠P = 90°, CP = 4 cm and CA = 5 cm
∴ IN is tangent angle ∆PAC,
AP² = AC² – PC²
= 5² – 4²
= 25 – 16
= 9
AP = 3 cm
∴ Perpendicular from centre to chord bisects the chord.
∴ AB = 2AP
= 2 × 3
= 6 cm.

Question 2.

In the given figure, if ∠AOB = 125°, then ∠COD is equal to:

(A) 62.5°
(B) 45°
(C) 35°
(D) 55°

Answer:
(D) 55°

Explanation:
Since, quadrilateral circumscribing a circle subtends supplementary angles at the centre of the circle.
∴∠AOB + ∠COD = 180°
125° + ∠COD = 180°
∠COD = 180° – 125° = 55°

Question 3.

In the given figure, AB is a chord of the circle and AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to:

(A) 65°
(B) 60°
(C) 50°
(D) 40°

Answer:
(C) 50°

Explanation:
Since, the angle between chord and tangent is equal to the angle subtended by the same chord in alternate segment of circle. ⇒ ∠BAT = 50°.

Question 4.

From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is:

(A) 60 cm²
(B) 65 cm²
(C) 30 cm²
(D) 32.5 cm²

Answer:
(A) 60 cm²

Explanation:
PQ is tangent and QO is radius at contact point Question
∴ PQO = OP² – OQ²
= 132 – 52
= 169 – 25 = 144
PQ = 12 cm
∆OPQ = ∆OPR [SSS congruence]
∴ Area of ∆OPQ = area of ∆OPR
[Since, congruent figures are equal in areas]
Area of quadrilateral QORP = 2 area of ∆OPR
= 2 × \(\frac{1}{2}\) base × height
= RP × OR
= 12 × 5
= 60 cm²

Question 5.

At one end A of diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the cirde. The length of the chord CD parallel to XY and at a distance 8 cm from A is:

(A) 4cm
(B) 5cm
(C) 6cm
(D) 8cm
Answer:
(D) 8cm

Explanation:
XAY is tangent and AO is radius at contact point A of circle.

AO = 5 cm
∠OAY = 90°
∴ CD is another chord at distance (perpendicular) of 8 cm from A and CMD || XAY meets AB at M.
join OD.
OD = 5 cm
OM = 8 – 5 = 3 cm
∠OMD = ∠OAY = 90°
Now, in right angled ∆OMD
MD2 = OD2 – MO2
= 52 – 32
= 25 – 9
= 16
⇒MD = 4 cmWe know thaat perpendiculars from centre O of circle bisect the chorde.
CD = 2MD
= 2 × 4
= 8 cm.
Hence, length of chord, CD = 8 cm.

Question 6.

In the given figure, AT is a tangent to the circle with centre ‘O’ such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to:

(A) 4 cm
(B) 2 cm
(C) 2√3 cm
(D) 4√3 cm

Answer:
(C) 2√3 cm

Explanation:
Join OA. OA is radius and AT is tangen at contact point A.
∴∠OAT = 90°,
Given that, OT = 4 cm
Now, \(\frac{A T}{4}\) = \(\vec{a}\) = cos 30°
⇒ AT = 4 × \(\frac{\sqrt{3}}{2}\) = 2√3 cm.

Question 7.

In the given figure, ‘O’ is the centre of circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then POQ is equal to:

(A) 100°
(B) 80°
(C) 90°
(D) 75°

Answer:
(A) 100°

Explanation:
OP is radius and PR is tangent at P.
so, ∠OPR = 90°
∠OPQ + 50 = 90°
∠OPQ = 90 – 50°
∠OPQ = 40°
In OPQ, OP = OQ
[Radii of same circle]
∴∠Q = ∠OPQ = 40°
[ Angle opposite to equal sides are equal]
But, ∠POQ = 180° – ∠P – ∠QO= 180° – 40° – 40°
= 180 – 80° = 100°
⇒ ∠POQ = 100° .

Question 8.

In the given figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to:

(A) 25°
(B) 30°
(C) 40°
(D) 50°

Answer:
(A) 25°

Explanation:
In ∆OAB, we have OA = OB [Radii of same circle ]
∴∠OAB = ∠OBA [Angle opposite to equal sides are equal]
As OA and PA are radius and tangent respectively at contact point A.
So, ∠OAP = 90°
Similariy, ∠OBP = 90°
Now, in quadrilateral PAOB,
∠P + ∠A + ∠O +∠B = 360°
⇒ 50° + 90 +∠ O + 90 = 360°
⇒ ∠O = 360° – 90° – 90° – 50°
⇒ ∠O = 130°
Again, in ∆OAB,
∠O + ∠OAB +∠ OBA = 180°
⇒ 130° +∠OAB + ∠OAB = 180° [∴ ∠OBA = ∠OAB]
⇒ 2 ∠OAB = 180° – 130° = 50v
⇒ ∠OAB = 25°
Hence,∠OAB = 25°

Assertion and Reason Based MCQs

Question 1.

Assertion (A): The length of the tangent drawn from a point 8 cm away from the centre of circle of radius 6 cm is 2 √7 cm.
Reason (R): If the angle between two radii of a circle is 130°, then the angle between the tangents at the end points of radii at their point of intersection is 50°.

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

Explanation:
In case of assertion:
Length of the tangent =\(\sqrt{d^{2}-r^{2}}\)
= \(\sqrt{(8)^{2}-(6)^{2}}\)
= \(\sqrt{64-36}\)
= \(\sqrt{28}\) = 2√7 cm
∴ Assertion is correct.
In case of reason:
Since, sum of the angles between radii and between intersection point of tangent is 180°. Angle at the point of intersection of tangents = 180° -130° = 50°.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 2.

Assertion (A): If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents at A and B is also 60°.
Reason (R): The length of the tangent from an external point P on a circle with centre O is always less than OP.

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

Explanation:
In case of assertion:
Chord AB subtends ∠60° at O

∴ ∠OAP = 90°
Similarly, ∠OBP = 90°
In quadrilateral OAPB,
∠O + ∠P + ∠OAP + ∠OBP = 360°
⇒ 60 + ∠P + 90° +90° = 360°
⇒ ∠P = 360° – 240°
⇒ ∠P = 120°
∴ Assertion is correct.
In case of reason:
PT and OT are the tangent and radius, respectivly at contact point T.

So, OTP = 90°
⇒ ∆OPT is right angled triangle.
Again, in ∆OPT
∴ ∠T > ∠O
∴ ∠OP > ∠PT
[ side opposite to greater angle is larger]
∴ Reason is correct.
Hence, assertion is incorrect but reason is correct.

Question 3.

Assertion (A): The tangent to the circumcircle of an isosceles AABC at A, in which AB = AC, is parallel to BC.
Reason (R): PQ is a tangent drawn from an external point P to a circle with centre O. QOR is the diameter of the circle. If ∠POR = 120°, then the measure of ∠OPQ is 60°.

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

Explanation:
In case of assertion:
given that, ∆ABC, inscribed in a circle in which AB = AC

PAQ is tangent at A.
AB is chord.
∴∠ PAB = ∠C
As, we know that , ∠PAB formed by chord AB with tangente segment.
In ∆ABC,
AB = AC [ Given]
∴ ∠B = ∠C [∴Angles opposite to equai sides are equal]……(ii)
From (i) and (ii), ∠B = ∠PAB
These are alternate interior angles.
Therefore, PAQ || BC.
∴ Assertion is correct.
In case of reason:

Civen that,
PQ is a tangent, ∠POR = 120°
∠POR = ∠OQP + ∠OPQ [ Exterior angle sum property]
∠OPQ = 120 – 90° = 30°
∠OPQ = 30°
∴ Reason is correct.
Hence, assertion is correct but reason is incorrect.

Question 4.

Assertion (A): PQ is a tangent to a circle with centre O at point P. If AOPQ is an isosceles triangle, then ∠OQP = 45°.
Reason (R): If two tangents inclined at 60° are drawn to a circle of radius 3 cm, then the length of each tangent is 3√3 cm.

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

Explanation:
In case of assertion:
As we know that ∠OPQ = 90° ( Angle between tangent and radius)
Let ∠PQO be x°, then∠QOP = x°
Since∠ OPQ is an isosceles triangle.
(given)(OP = OQ)

In ∆OPQ,
∠OPQ + ∠PQO + ∠QOP = 180
(property of the sum of angles of a triangle)
∴ 90° + x° + x° = 180°
⇒ 2x° = 180° – 90° = 90°
⇒ x = \(\frac{90}{2}\) = 45.
Hence, OQP is 45

∴ Assertion is correct.
In case of reason:

In ∆PAO,
tan 30 = \(\frac{A O}{P A}\) (Using trigonometry)
\(\frac{1}{\sqrt{3}}\) = \(\frac{3}{P A}\)
PA = 3 √3 cm.
∴ Reason is correct.
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:
There is a circular filed of radius 5 m. Kanabh, Chikoo and Shubhi are playing with ball, in which Kanabh and Chikoo are standing on the boundary of the circle. The distance between Kanabh and Chikoo is 8 m. From Shubhi point S, two tangents are drawn as shown in the figure. Give the answer of the following questions.

Question 1.

What is the relation between the lengths of SK and SC?

(A) SK ≠ SC
(B) SK = SC
(C) SK > SC
(D) SK < SC.
Answer:
(B) SK = SC

Explanation:
We know that the lengths of tangents drawn from an external point to a circle are equal. So SK and SC are tangents to a circle with centre O.
∴SK = SC

Question 2.

The length (distance) of OR is:

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

Explanation:
In question 1, we have proved SK = SC
Then ASKC is an isosceles triangle and SO is the angle bisector of ZKSC. So, OS X KC.
∴ OS bisects KC, gives KR = RC = 4 cm
Now, OR = \(\sqrt{O K^{2}-K R^{2}}\)
(By using Pythagoras theorem)
= \(\sqrt{5^{2}-4^{2}}=\) = \(\sqrt{25-16}\)
= \(\sqrt{25-16}\)
= \(\sqrt{9}\)
= 3 m

Question 3.

The sum of angles SKR and OKR is:

(A) 45°
(B) 30°
(C) 90°
(D) none of these
Answer:
(C) 90°
∠SKR + ∠OKR = ∠OKR
= 90 (Radius is ⊥ to tangent)

Question 4.

The distance between Kanabh and Shubhi is:

(A) \(\frac{10}{3} \mathrm{~m}\)
(B) \(\frac{13}{3} \mathrm{~m}\)
(C) \(\frac{16}{3} \mathrm{~m}\)
(D) \(\frac{20}{3} \mathrm{~m}\)
Answer:
(D) \(\frac{20}{3} \mathrm{~m}\)

Explanation:
∆SER and ∆RKO,
∠RKO = ∠KSR
∠SRK = ∠ORK
∆KSR ~ ∆OKR . (By AA Similarity)
\(\frac{S K}{K O}\) = \(\frac{R K}{R O}\)
Then \(\frac{S K}{5}\) = \(\frac{4}{3}\)
(RO = 3 m, proved in Question 2)
⇒ 3SK = 20
⇒ Sk = \(\frac{20}{3}\)
Hence, the distance between Kanabh and Shubhi is \(\frac{20}{3}\)

Question 5.

What is the mathematical concept related to this question ?

(A) Constructions
(B) Area
(C) Circle
(D) none of these
Answer:
(C) Circle

Explanation:
The mathematical concept (Circle) is related to this question.

II. Read the following text and answer the following question on the basis of the same:
ABCD is a playground. Inside the playground a circular track is present such that it touches AB at point P, BC at Q, CD at R and DA at S.

Question 1.

If DR = 5 m, then DS is equal to:

(A) 6 m
(B) 11 m
(C) 5 m
(D) 18 m
Answer:
(C) 5 m

Explanation:
∴ DR =5 m (given)
DR = DS (Length of tangents are equal)
DS = 5 m.

Question 2.

The length of AS is:

(A) 18 m
(B) 13 m
(C) 14 m
(D) 12 m
Answer:
(A) 18 m

Explanation:
We have AD = 23 m.
ands DS = 5 m (Proved in Question 1)
∴ AS = AD – DS
= (23 – 5)m = 18 m.

Question 3.

The length of PB is:

(A) 12 m
(B) 11 m
(C) 13 m
(D) 20 m
Answer:
(B) 11 m

Explanation:
We have,
AB = 29 m
But AS = AP (lengths of tangents are equal)
and AS = 18 m(proved in Question 2)
∴ AP = 18 m
NOw, PB = AB – AP
= (29 – 18) m
= 11 m

Question 4.

What is the angle of OQB?

(A) 60°
(B) 30°
(C) 45°’
(D) 90°
Answer:
(D) 90°

Explanation:
∠OQB = 90°
(Radius is ’ to tangent)

Question 5.

What is the diameter of given circle?

(A) 22 m
(B) 33 m
(C) 20 m
(D) 30 m
Answer:
(A) 22 m

Explanation:
PB = 11 m (proved in Question 3) PB = BQ (lengths oftangents are equal)
But PB = BQ
∴BQ = 11 m
or = OQ = QB = 11m
Hence, diameter = 2r = 2 × 11 = 22 m.

III. Read the following text and answer the following question on the basis of the same:
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride. She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.

Question 1.

In the given figure find ∠ROQuestion

(A) 60°
(B) 100°
(C) 150°
(D) 90°
Answer:
(C) 150°

Explanation:
∠ORP = 90° & ∠OQP [∵ radius of circle is perpendicular to tangent]
∴ ∠ROQ + ∠ORP + ∠OQP + ∠QPR = 360° ∠ROQ + 90° + 90° + 30° = 360° ∠ROQ + 210° = 360° ∠ROQ = 360°-210°
∠ROQ = 150°

Question 2.

Find ∠RQP.

(A) 75°
(B) 60°
(C) 30°
(D) 90°
Answer:
(A) 75°

Explanation:
∠OQR = ∠ORQ
∠RPQ = 150
and ∠ROQ + ∠OQR + ∠ORQ = 180
150 + 2∠ORQ =180
2∠ORQ = 30
∠ORQ = 15
∠OQR = ORQ = 1
∠RQP = ∠OQP – ∠OQR= 90 – 15

Question 3.

Find ∠RSQuestion

(A) 60°
(B) 75
(C) 100°
(D) 30
Answer:
(B) 75

Question 4.

Find ∠ORP

(A) 90°
(B) 70
(C) 100°
(D) 60
Answer:
(A) 90°

Explanation:
∠ORP = 90°
Because, radius of circle is perpendicular to tangent.

IV Read the following text and answer the following question on the basis of the same:
Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff. The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a AABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.

Question 1.

Find the length of AD.

(A) 7
(B) 8
(C) 5
(D) 9
Answer:
(A) 7

Explanation:
Given, AB = 12 cm
BC = 8 cm
CA = 10 cm

In ∆ABC
x + y =12 cm -(i)
x + z = 10 cm …(ii)
y + z = 8 cm -(Mi)
Adding (i), (ii) and (iii)
2(x + y + z) =30 cm
x + y + z =15 cm …(iv)
from eQuestion (iii) and (iv)
x = 7 cm
AD = x = 7 cm

Question 2.

Find the Length of BE.

(A) 8
(B) 5
(C) 2
(D) 9
Answer:
(B) 5

Explanation:
We have
x + y + z = 15 cm
and x + z = 10 cm
By solving above equations
we get y = 5 cm
i.e., BE = y = 5 cm

Question 3.

Find the length of CF.

(A) 9
(B) 5
(C) 2
(D) 3
Answer:
(D) 3

Explanation:
We have
x y + z = 15 cm
and x + y =12 cm
By solving above equations,
we get z = 3 cm
i.e., CF = z = 3 cm

Question 4.

If radius of the circle is 4 cm, find the area of ∆OAB.

(A) 20
(B) 36
(C) 24
(D) 48
Answer:
(C) 24

Explanation:
Area of ∆OAB = \(\frac{1}{2}\) × AB × OD
= \(\frac{1}{2}\) × 12 × 4
= 24 cm²

MCQ Questions for Class 10 Maths with Answers

MCQ Questions for Class 10 Maths Chapter 10 Circles Read More »

MCQ Questions

MCQ Questions for Class 12 Informatics Practices – Python Pandas Part 1

Python Pandas Class 12 MCQ Questions Part 1

To create an empty series object, you can use:

To specify datatype int 16 for a series object,you can write:

To get the number of dimensions of a series object, attribute is displaye(d)

To get the size of the datatype of the items in series object, you can display attribute.

To get the number of elements in a series object, attribute may be use(d)

To get the number of bytes of the series data, attribute is displaye(d)

To check if the series object contains NaN values, attribute is displaye(d)

To display first three elements of a series object S, you may write

To display last five rows of a series object S, you may write

Missing data in Pandas object is represented through:

Given a Pandas series called SeQuestion uences, the command which will display the first 4 rows is

If a dataframe is created using a 2D dictionary, then the indexes/row labels are formed from

If a dataframe is created using a 2D dictionary, then the column labels are formed from

The axis 0 identifies a dataframe’s

The axis 1 identifies a dataframe’s

To get the number of elements in a dataframe attribute may be use(d)

To get NumPy representation of a dataframe ……… attribute may be use(d)

To get a number representing number of axes in a dataframe, ……….. attribute may be use(d)

To get the transpose of a dataframe Dl, you can write

To extract row/column from a dataframe,……….. function may be use(d)

To display the 3rd, 4th and 5th columns from the 6th to 9th rows of a dataframe DF,you can write

To change the 5th column’s value at 3rd row as 35 in dataframe DF, you can write

Which among the following options can be used to create a dataframe in Pandas?

Identify the correct statement:

To delete a column from a dataframe, you may use statement.

To delete a rowfrom a DataFrame.you may use……… statement.

……… is a popular data-science library of Python.

A ………… is a Pandas data structure that represents a ID array like object.

A is a Pandas data structure that represents ……… a 2D array like object.

You can use numpy …… for missing dat(a)

To specify datatype for a series object …… argument is use(d)

The function on series object returns total ……. elements in it including NaNs.

The function on series object returns only the count of non-NaN values in it.

Series is mutable.

Series is not ……… mutable.

Dataframe is …….. mutable as well as ………..mutable.

In a dataframe, Axis = 1 represents the ……… elements.

To access values using row labels you can use DF

To access individual value, you can use DF …….. using row/column index labels.

To access individual value, you can use DF ……. using row/column integer position.

The rename() function requires argument ……… to make changes in the original dataframe.

Which of the following are modules/libraries in Python?

NumPy stands for ……….

Which of the following libraries allows to manipulate, transform and visualize data easily and efficiently?

PANDAS stands for……….

……… is an important library used for analysing dat(a)

Important data structure of pandas is/are

Which of the following library in Python is used for plotting graphs and visualization?

Pandas series can have data types.

…….. is used when data is in Tabular Format.

Which of the following command is used to install pandas?

A is a collection of data values and operations that can be applied to that dat(a)

A is a one-dimensional array.

Which of the following statement is wrong?

A Series by default have numeric data labels starting from

The data label associated with a particular value of Series is called its

Which of the following module is to be imported to create Series?

Which of the following function/method help to create Series?

Write the output of the following: >>> import pandas as pd >>> seriesl = p(d)Series(10,20,30) >>> print(seriesl)

When you print/display any series then the left most column is showing value.

How many values will be there in arrayl, if given code is not returning any error?>>> series4 = p(d)Series(arrayl, index = [“Jan”, “Feb”, “Mar”, “Apr”])

Which of the following statement will create an empty series named “SI”?

How many elements will be there in the series named “SI”? »> SI = p(d)Series(range(5)) >» print(Sl)

When we create a series from dictionary then the keys of dictionary become

Write the output of the following:>>> Sl=p(d)Series(14, index = [‘a’, ‘b’, ‘c’]) >>> print(Sl)

Write the output of the following: >>>Sl=p(d)Series([14,7, index = [‘a’, ‘b’, ‘c’]) >>> print(Sl)

Write the output of the following: >>> Sl=p(d)Series([14,7,9] .index = range(l, 8,3)) >>> print(Sl)

Which of the following code will generate the following output? Jan 31 Feb 28 Mar 31 dtype: int64

Write the output of the following: import pandas as pd SI = p(d)Series(data = range(31, 2, -6), index = [x for x in “aeiou”]) print(Sl)

What type of error is returned by following code? import pandas as pd SI = p(d)Series(data = (31,2, -6), index = [7,9,3,2]) print(Sl)

Write the output of the following: import pandas as pd SI = p(d)Series(data – 2*(31,2,-6)) print(Sl)

We can imagine a Pandas series as a …….. spreadsheet.

We can assign user-defined labels to the index of the series

Write the output of the following: import pandas as pd series2 = p(d)Series([“Kavi7Shyam7Ravi”], index=[3,5 print(series2 > “S”)

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
>>> seriesl = p(d)Series(10,20,30)
>>> print(seriesl)

Write the output of the following:
>>> Sl=p(d)Series([14,7,9] .index = range(l, 8,3)) >>> print(Sl)

Which of the following code will generate the following output?
Jan 31
Feb 28
Mar 31 dtype: int64

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\(\frac{3}{4}\).