Chapter 14
71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers · 68 exercises
Problem 1
Consider the data on \(x\) taking the values \(0,2,4,8, \ldots, 2^{n}\) with frequencies \({ }^{n} \mathrm{C}_{0}{\underline{\phantom{xx}}}^{n} \mathrm{C}_{1},{ }^{n} \mathrm{C}_{2}, \ldots,{ }^{n} \mathrm{C}_{\mathrm{n}}\) respectively. If the mean of this data is \(\frac{728}{2^{n}}\), then \(n\) is equal to
6 step solution
Problem 2
The minimum value of \(2^{\sin x}+2^{\cos x}\) is: \([\) Sep. 04, \(\mathbf{2 0 2 0}\) (II)](a) \(\begin{array}{llll}2^{-1+\frac{1}{\sqrt{2}}} & \text { (b) } 2^{-1+\sqrt{2}} & \text { (c) } 2^{1-\sqrt{2}} & \text { (d) } 2^{1-\frac{1}{\sqrt{2}}}\end{array}\)
5 step solution
Problem 4
The mean and the median of the following ten numbers in increasing order \(10,22,26,29,34, x, 42,67,70, y\) are 42 and 35 respectively, then \(\frac{y}{x}\) is equal to: [April. 09, 2019 (II)] \(x\) (a) \(9 / 4\) (b) \(7 / 2\) (c) \(8 / 3\) (d) \(7 / 3\)
6 step solution
Problem 5
The mean of a set of 30 observations is 75 . If each other observation is multiplied by a non-zero number \(\lambda\) and then each of them is decreased by 25 , their mean remains the same. The \(\lambda\) is equal to [Online April 15, 2018] (a) \(\frac{10}{3}\) (b) \(\frac{4}{3}\) (c) \(\frac{1}{3}\) (d) \(\frac{2}{3}\)
6 step solution
Problem 6
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is : [Online April 8, 2017] (a) 25 (b) 30 (c) 35 (d) 40
4 step solution
Problem 7
The mean of the data set comprising of 16 observations is 16\. If one of the observation valued 16 is deleted and three new observations valued 3,4 and 5 are added to the data, then the mean of the resultant data, is: (a) \(15.8\) (b) \(14.0\) (c) \(16.8\) (d) \(16.0\)
5 step solution
Problem 8
Let the sum of the first three terms of an A. P, be 39 and the sum of its last four terms be 178 . If the first term of this A.P. is 10 , then the median of the A.P. is : [Online April 10, 2015] (a) 28 (b) \(26.5\) (c) \(29.5\) (d) 31
6 step solution
Problem 9
Let the sum of the first three terms of an A. P, be 39 and the sum of its last four terms be 178 . If the first term of this A.P. is 10 , then the median of the A.P. is : [Online April 10, 2015] (a) 28 (b) \(26.5\) (c) \(29.5\) (d) 31
4 step solution
Problem 10
In a set of \(2 \mathrm{n}\) distinct observations, each of the observations below the median of all the observations is increased by 5 and each of the remaining observations is decreased by \(3 .\) Then the mean of the new set of observations: \(\quad[\) Online April 9, 2014] (a) increases by 1 (b) decreases by 1 (c) decreases by 2 (d) increases by 2
4 step solution
Problem 11
If the median and the range of four numbers \(\\{x, y, 2 x+y, x-y\\}\), where
\(0
6 step solution
Problem 12
The mean of a data set consisting of 20 observations is 40 . If one observation 53 was wrongly recorded as 33 , then the correct mean will be : \(\quad\) [Online April9, 2013] (a) 41 (b) 49 (c) \(40.5\) (d) \(42.5\)
3 step solution
Problem 13
The median of 100 observations grouped in classes of equal width is 25 . If the median class interval is \(20-30\) and the number of observations less than 20 is 45 , then the frequency of median class is [Online May 19, 2012] (a) 10 (b) 20 (c) 15 (d) 12
6 step solution
Problem 15
The average marks of boys in class is 52 and that of girls is 42\. The average marks of boys and girls combined is 50 . The percentage of boys in the class is [2007] (a) 80 (b) 60 (c) 40 (d) 20
5 step solution
Problem 16
Let \(x_{1}, x_{2}, \ldots \ldots \ldots \ldots \ldots, x_{n}\) be \(\mathrm{n}\) observations such that \(\sum x_{i}^{2}=400\) and \(\sum x_{i}=80 .\) Then the possible value of \(n\) among the following is (a) 15 (b) 18 (c) 9 (d) 12
6 step solution
Problem 17
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately [2005] (a) \(22.0\) (b) \(20.5\) (c) \(25.5\) (d) \(24.0\)
5 step solution
Problem 18
The median of a set of 9 distinct observations is \(20.5\). If each of the largest 4 observations of the set is increased by 2, then the median of the new set (a) remains the same as that of the original set (b) is increased by 2 (c) is decreased by 2 (d) is two times the original median.
4 step solution
Problem 19
In a class of 100 students there are 70 boys whose average marks in a subject are 75 . If the average marks of the complete class is 72 , then what is the average of the girls? [2002] (a) 73 (b) 65 (c) 68 (d) 74
5 step solution
Problem 20
If \(\sum_{i=1}^{n}\left(x_{i}-a\right)=n\) and \(\sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a,(n, a>1)\), then the standard deviation of \(n\) observations \(x_{1}, x_{2}, \ldots, x_{n}\) is: [Sep. 06, 2020 (I)] (a) \(a-1\) (b) \(n \sqrt{a-1}\) (c) \(\sqrt{n(a-1)}\) (d) \(\sqrt{a-1}\)
7 step solution
Problem 21
The mean and variance of 7 observations are 8 and 16 , respectively. If five observations are \(2,4,10,12,14\), then the absolute difference of the remaining two observations is: (a) 1 (b) 4 (c) 2 (d) 3
6 step solution
Problem 22
If the mean and the standard deviation of the data \(3,5,7, a\), \(b\) are 5 and 2 respectively, then \(a\) and \(b\) are the roots of the equation: [Sep. 05, 2020 (II)] (a) \(x^{2}-10 x+18=0\) (b) \(2 x^{2}-20 x+19=0\) (c) \(x^{2}-10 x+19=0\) (d) \(x^{2}-20 x+18=0\)
7 step solution
Problem 23
The mean and variance of 8 observations are 10 and \(13.5\), respectively. If 6 of these observations are \(5,7,10,12,14\), 15 , then the absolute difference of the remaining two observations is : (a) 9 (b) 5 (c) 3 (d) 7
8 step solution
Problem 25
For the frequency distribution :where \(0
5 step solution
Problem 26
Let \(x_{i}(1 \leq i \leq 10)\) be ten observations of a random variable \(X\). If \(\sum_{i=1}^{10}\left(x_{i}-p\right)=3\) and \(\sum_{i=1}^{10}\left(x_{i}-p\right)^{2}=9\) where \(0 \neq p \in \mathbf{R}\), then the standard deviation of these observations is: \(\quad\) [Sep. 03, 2020 (II)] (a) \(\sqrt{\frac{3}{5}}\) (b) \(\frac{4}{5}\) (c) \(\frac{9}{10}\) (d) \(\frac{7}{10}\)
7 step solution
Problem 27
Let \(X=\\{x \in \mathbf{N}: 1 \leq x \leq 17\\}\) and \(Y=\\{a x+b: x \in X\) and \(a, b \in \mathbf{R}, a>0\\} .\) If mean and variance of elements of \(Y\) are 17 and 216 respectively then \(a+b\) is equal to : [Sep. \(\mathbf{0 2}, \mathbf{2 0 2 0}(\mathbf{I})]\) (a) 7 (b) \(-7\) (c) \(-27\) (d) 9
7 step solution
Problem 28
If the variance of the terms in an increasing A.P., \(b_{1}, b_{2}, b_{3}, \ldots ., b_{11}\) is 90 , then the common difference of this A.P. is [NA Sep. 02, 2020 (II)]
8 step solution
Problem 29
Let the observations \(x_{i}(1 \leq i \leq 10)\) satisfy the equations, \(\sum_{i=1}^{10}\left(x_{i}-5\right)=10\) and \(\sum_{i=1}^{10}\left(x_{i}-5\right)^{2}=40 .\) If \(\mu\) and \(\lambda\) are the mean and the variance of the observations, \(x_{1}-3, x_{2}-3\), \(\ldots, x_{10}-3\), then the ordered pair \((\mu, \lambda)\) is equal to: [Jan. 9, 2020 (I)] (a) \((3,3)\) (b) \((6,3)\) (c) \((6,6)\) (d) \((3,6)\)
4 step solution
Problem 30
The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by \(p\) and then reduced by \(q\), where \(p \neq 0\) and \(q \neq 0 .\) If the new mean and new s.d. become half of their original values, then \(q\) is equal to: [Jan. 8, 2020 (I)] (a) \(-5\) (b) 10 (c) \(-20\) (d) \(-10\)
6 step solution
Problem 31
The mean and variance of 20 observations are found to be 10 and 4 , respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11 . Then the correct variance is: [Jan. 8, 2020 (II)] (a) \(3.99\) (b) \(4.01\) (c) \(4.02\) (d) \(3.98\)
6 step solution
Problem 32
If the variance of the first \(n\) natural numbers is 10 and the variance of the first \(m\) even natural numbers is 16 , then \(m+n\) is equal to \(-\) [NA Jan. 7, 2020 (I)]
6 step solution
Problem 33
If the mean and variance of eight numbers \(3,7,9,12,13\), \(20, x\) and \(y\) be 10 and 25 respectively, then \(x \cdot y\) is equal to [NA Jan. 7, 2020 (II)]
5 step solution
Problem 34
If the data \(x_{1}, x_{2}, \ldots \ldots, x_{10}\) is such that the mean of first four of these is 11 , the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is : \(\quad\) [April 12, 2019 (I)] (a) \(2 \sqrt{2}\) (b) 2 (c) 4 (d) \(\sqrt{2}\)
5 step solution
Problem 35
If both the mean and the standard deviation of 50 observations \(x_{1}, x_{2}, \ldots ., x_{50}\) are equal to 16, then the mean of \(\left(x_{1}-4\right)^{2},\left(x_{2}-4\right)^{2}, \ldots,\left(x_{50}-4\right)^{2}\) is: \(\quad\) [April 10, \(\mathbf{2 0 1 9}\) (II)] (a) 400 (b) 380 (c) 525 (d) 480
4 step solution
Problem 36
If the standard deviation of the numbers \(-1,0,1, \mathrm{k}\) is \(\sqrt{5}\) where \(\mathrm{k}>0\), then \(\mathrm{k}\) is equal to: \(\quad\) [April09, \(\mathbf{2 0 1 9}\) (I)] (a) \(2 \sqrt{6}\) (b) \(2 \sqrt{\frac{10}{3}}\) (c) \(4 \sqrt{\frac{5}{3}}\) (d) \(\sqrt{6}\)
7 step solution
Problem 37
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are \(2,4,10,12,14\), then the product of the remaining two observations is : \([\) April \(08,2019(\mathrm{I})]\) (a) 45 (b) 49 (c) 48 (d) 40
9 step solution
Problem 38
A student scores the following marks in five tests: 45,54 , \(41,57,43\). His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is : [April. \(\mathbf{0 8}, \mathbf{2 0 1 9}\) (II)] (a) \(\frac{10}{\sqrt{3}}\) (b) \(\frac{100}{3}\) (c) \(\frac{10}{3}\) (d) \(\frac{100}{\sqrt{3}}\)
6 step solution
Problem 39
If the sum of the deviations of 50 observations from 30 is 50 , then the mean of these observations is : [Jan. 12, 2019 (I)] (a) 30 (b) 51 (c) 50 (d) 31
5 step solution
Problem 40
The mean and the variance of five observations are 4 and \(5.20\), respectively. If three of the observations are 3,4 and 4; then the absolute value of the difference of the other two observations, is: \(\quad\) [Jan. 12, 2019 (II)] (a) 7 (b) 5 (c) 1 (d) 3
11 step solution
Problem 41
The outcome of each of 30 items was observed; 10 items gave an outcome \(\frac{1}{2}-\mathrm{d}\) each, 10 items gave outcome \(\frac{1}{2}\) each and the remaining 10 items gave outcome \(\frac{1}{2}+\mathrm{d}\) each. If the variance of this outcome data is \(\frac{4}{3}\) then \(|\mathrm{d}|\) equals : [Jan. 11, 2019 (I)] (a) \(\frac{2}{3}\) (b) 2 (c) \(\frac{\sqrt{5}}{2}\) (d) \(\sqrt{2}\)
7 step solution
Problem 42
A data consists of \(\mathrm{n}\) observations: \(x_{1}, x_{2}, \ldots, x_{n} .\) If \(\sum_{i=1}^{\mathrm{n}}\left(x_{i}+1\right)^{2}=9 \mathrm{n}\) and \(\sum_{i=1}^{\mathrm{n}}\left(x_{i}-1\right)^{2}=5 \mathrm{n}\) then the standard deviation of this data is: [Jan. 09, 2019 (II)] (a) 2 (b) \(\sqrt{5}\) (c) 5 (d) \(\sqrt{7}\)
6 step solution
Problem 43
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1,3 and 8 , then a ratio of other two observations is: [Jan. 10, 2019 (I)] (a) \(10: 3\) (b) \(4: 9\) (c) \(5: 8\) (d) \(6: 7\)
3 step solution
Problem 44
If mean and standard deviation of 5 observations \(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\) are 10 and 3, respectively, then the variance of 6 observations \(x_{1}, x_{2}, \ldots, x_{5}\) and \(-50\) is equal to: [Jan. 10, 2019 (II)] (a) \(509.5\) (b) \(586.5\) (c) \(582.5\) (d) \(507.5\)
6 step solution
Problem 45
5 students of a class have an average height \(150 \mathrm{~cm}\) and variance \(18 \mathrm{~cm}^{2}\). A new student, whose height is \(156 \mathrm{~cm}\), joined them. The variance (in \(\mathrm{cm}^{2}\) ) of the height of these six students is: [Jan. 9, 2019 (I)] (a) 16 (b) 22 (c) 20 (d) 18
6 step solution
Problem 46
The mean and the standard deviation (s.d.) of five observations are 9 and 0 , respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10 , then their s.d. is? \(\quad\) [Online April 16, 2018] (a) 0 (b) 4 (c) 2 (d) 1
7 step solution
Problem 47
If the mean of the data \(: 7,8,9,7,8,7, \lambda, 8\) is 8, then the variance of this data is \(\quad\) [Online April 15, 2018] (a) \(\frac{9}{8}\) (b) 2 (c) \(\frac{7}{8}\) (d) 1
4 step solution
Problem 48
If \(\sum_{\mathrm{i}=1}^{9}\left(\mathrm{x}_{\mathrm{i}}-5\right)=9\) and \(\sum_{\mathrm{i}=1}^{9}\left(\mathrm{x}_{\mathrm{i}}-5\right)^{2}=45\), then the standard deviation of the 9 items \(x_{1}, x_{2}, \ldots, x_{9}\) is: [2018] (a) 4 (b) 2 (c) 3 (d) 9
4 step solution
Problem 49
The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3,4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is : \(\quad\) [Online April 9, 2017] (a) \(8.25\) (b) \(8.50\) (c) \(8.00\) (d) \(9.00\)
5 step solution
Problem 50
If the standard deviation of the numbers \(2,3, \mathrm{a}\) and 11 is 3.5, then which of the following is true? (a) \(3 \mathrm{a}^{2}-34 \mathrm{a}+91=0\) (b) \(3 a^{2}-23 a+44=0\) (c) \(3 \mathrm{a}^{2}-26 \mathrm{a}+55=0\) (d) \(3 \mathrm{a}^{2}-32 \mathrm{a}+84=0\)
6 step solution
Problem 51
The mean of 5 observations is 5 and their variance is 124 . If three of the observations are 1,2 and 6 ; then the mean deviation from the mean of the data is : [Online April 10, 2016] (a) \(2.5\) (b) \(2.6\) (c) \(2.8\) (d) \(2.4\)
5 step solution
Problem 52
If the mean deviation of the numbers \(1,1+\mathrm{d}, \ldots, 1+100 \mathrm{~d}\) from their mean is 255 , then a value of \(\mathrm{d}\) is : [Online April 9, 2016] (a) \(10.1\) (b) \(5.05\) (c) \(20.2\) (d) 10
7 step solution
Problem 53
The variance of first 50 even natural numbers is [2014] (a) 437 (b) \(\frac{437}{4}\) (c) \(\frac{833}{4}\) (d) 833
7 step solution