MATHS Assertion (A) & Reason (B) MCQ

3. A is true; R is false

Solution:

Mean $\text{(x)}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{350}{25}=14$

$\therefore$ Mean deviation about mean

$\frac{\sum\text{f}_\text{i}|\text{x}_\text{t}-\bar{\text{x}}|}{\sum\text{f}_\text{i}}=\frac{158}{25}=6.32$

Reason 

Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{4000}{80}=50$

$\therefore$ Mean deviation about mean

$=\frac{\sum\text{f}_\text{i}|\text{x}_\text{i}-\bar{\text{x}}|}{\sum\text{f}_\text{i}}$

$=\frac{1280}{80}=16$

3. A is true; R is false

Solution:

Assertion Let the number of boys and girls be x and y.

$\therefore$ 52x + 42y = 50(x + y)

 ⇒ 2x = 8y

⇒ x = 4y

$\therefore$ Total number of students in the class = x + y = 5y

$\therefore$ Required percentage of boys

 $=\frac{4\text{y}}{5\text{y}}\times100\%=80\%$

Reason Let the number of boys be x and number of girls be y.

$\therefore$ 53(x + y) = 55y + 50x

⇒ 3x = 2y

$\Rightarrow\text{x}=\frac{2\text{y}}{3}$

$\therefore$ Total number of students

$=\text{x}+\text{y}=\frac{\text{2y}}{3}+\text{y}=\frac{5}{3}\text{y}$

Hence, required percentage

$=\frac{\text{y}}{\frac{5\text{y}}{3}}\times100\%$

$=\frac{3}{5}\times100\%=60\%$

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion Mean of 1², 2², 3²,...,n² is

$\frac{1^2+2^2+3^2+...+\text{n}^2}{\text{n}}=\frac{\sum\text{n}^2}{\text{n}}$

$\therefore\frac{46\text{n}}{11}\frac{\text{n}\text{(n+1}(2\text{n}+1)}{6\text{n}}$

$\Rightarrow 2\text{m} + 33\text{n}+11-276\text{n}=0$

$ \text{(n-11)} (22\text{n} -1)=0$

$\Rightarrow\text{n}=1\text{ and}\text{ n}\neq\frac{1}{22}$

Reason $\because\sigma_\text{x}^2=4 \text{ and } \sigma_\text{y}^2=5$

Also, $\bar{\text{x}}=2 \text{ and }\bar{\text{y}}=4$

Now, $\frac{\sum\text{x}_\text{i}}{5}=2\Rightarrow\sum\text{x}_\text{i}=10 \frac{\sum\text{y}_\text{i}}{5}=4$

$\Rightarrow\sum\text{y}_\text{i}=20$

Since, $\sigma_\text{x}^2=\frac{1}{5}(\sum\text{x}_\text{i}^2)-(\bar{\text{x}})^2$

$\Rightarrow\sum\text{x}_\text{i}^2=40$

Similarly, $\sum\text{y}_\text{i}^2=105$

$\because\sigma_\text{z}^2 = \frac{1}{10}(\sum\text{x}_\text{i}^2+\sum\text{y}_\text{i}^2)-\Big(\frac{\bar{\text{x}}+\bar{\text{y}}}{2}\Big)^2$

$=\frac{1}{10}(40+105)-9$

$=\frac{55}{10}=\frac{11}{2}$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Let x be the mean of  x₁, x₂..., x_n. Then, variance is given by

$=\sigma_1^2=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}-\bar{\text{x}})^2$

If ais added to each observation, the new observations will be

$\text{y}_\text{i} =\text{x}_\text{i} +\text{a}...(i)$

Let the mean of the new observations be j. Then,

$\bar{\text{y}}=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}\text{y}_\text{i}=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}+\text{a})$

$=\frac{1}{\text{n}}\Bigg[\sum\limits_\text{i=1}^\text{n}\text{x}_\text{i}+\sum\limits_\text{i=1}^\text{n}\text{a}\Bigg]$

$=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}\text{x}_\text{i}+\frac{\text{na}}{\text{n}}=\bar{\text{x}}+\text{a}$

$=\text{i.e.}\ \bar{\text{y}}=\bar{\text{x}}+\text{a}\ ...(ii)$

Thus, the variance of the new observations is

$=\sigma_2^2=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{y}_\text{i}-\bar{\text{y}})^2$

$=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}+\text{a}-\bar{\text{x}}-\text{a})^2$

 [using Eqs. (i) and (ii) ]

$=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}-\bar{\text{x}})^2=\sigma_1^2$

Thus, the variance of the new observations is same as that of the original observations. Reason We may note that adding (or subtracting) a positive number to (or from) each observation of a group does not affect the variance.

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Mean $\bar{\text{(x})}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{760}{40}=19$

Reason Variance $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{\sum\text{f}_\text{i}}\Big(\frac{\sum\text{f}_\text{i}\text{x}\text{i}}{\sum\text{f}_\text{i}}\Big)^2$

$=\frac{16176}{40}-\Big(\frac{760}{40}\Big)^2$

$=404.4-(19)^2$

$=404.4-361=43.4$

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion Since, 44kg is replaced by 46kg and 27kg is replaced by 25kg, then the given series becomes 31, 35, 25, 29, 32, 43, 37, 41, 34, 28, 36, 46, 45, 42, 30. On arranging this series in ascending order, we get 25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46. Total number of students are 15, therefore middle term is 8th whose corresponding value is 35. Reason On arranging the terms in increasing order of magnitude 40, 42, 45, 47, 50, 51, 54, 55, 57.

Number of terms N = 9

$\therefore$ Median $=\Big(\frac{9+1}{2}\Big)$ the term = 5th term = 50kg

MD from median $=\frac{43}{9}=4.78\text{kg}$

2. A is true, R is true; R is not a correct explanation of A.

Solution:

$\text{N}=30\sum\limits_\text{i=1}^7​​​​\text{f}_{\text{i}}\text{x}_\text{i}=420$

$=\sum\limits_\text{i=1}^7​​​​\text{f}_\text{i}(\text{x}_{\text{i}}-\bar{\text{x}})^2=1374$

$\therefore , \bar{\text{x}}=\frac{\sum\limits_\text{i=1}^7​​​​\text{f}_\text{i}\text{x}_{\text{i}}}{\text{N}}=\frac{1}{30}\times420=14$

$\therefore\text{ariance}(\sigma^2)==\sum\limits_\text{i=1}^7​​​​\text{f}_\text{i}(\text{x}_{\text{i}}-\bar{\text{x}})^2$

$=\frac{1}{30}\times1374=45.8$

Reason Standard deviation

$=(\sigma)\times\sqrt{45.8}=6.77$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion In the calculation of variance, we find that the units of individual observations x_i and the unit of their mean $\bar{\text{x}}$ are different from that of variance, since variance involves the sum of squares of $\text{(x}_\text{i} - \bar{\text{x}}).$ For this reason, the proper measure of dispersion about the mean of a set of observations is expressed as positive square-root of the variance and is called standard deviation. Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.

4. A is false; R is true.

Solution:

Assertion Given, $\sum\text{(x-5})=3$

$\therefore\sum\text{x}-\sum5=3$

$\Rightarrow\sum\text{x}-5\times18=3\ [\because\text{n}=18]$

$\Rightarrow\sum\text{x}=3+90\Rightarrow\sum\text{x}=93$

Now, $\sum(\text{x}-5)^2=43$

$\Rightarrow\sum(\text{x}^2+25-10\text{x})=43$

$\Rightarrow\sum\text{x}^2+\sum25-10\sum\text{x}=43$

$\Rightarrow\sum\text{x}^2+25\times18-10\times93=43$

$\Rightarrow\sum\text{x}^2=43+930-450$

$\Rightarrow\sum\text{x}^2=973-450$

$\Rightarrow\sum\text{x}^2=523$

Now, mean $=\frac{\sum\text{x}}{\text{n}}=\frac{93}{18}=5.16$

Reason SD $(\sigma)=\sqrt{\frac{\sum\text{x}^2}{\text{n}}-\Big(\frac{\sum\text{x}}{\text{n}}}\Big)^2$

$=\sqrt{\frac{523}{18}-\Big(\frac{93}{18}\Big)}^2$

$=\sqrt{\frac{523\times18-93\times93}{18\times18}}$

$=\frac{1}{18}\sqrt{9414-8649}$

$=\frac{1}{18}\sqrt{765}=\frac{27.66}{18}=1.54$

1. A is true, R is true; R is acorrect explanation of A

Solution:

Assertion The deviation of an observation x from a fixed value ‘a’ is the difference (x - a). In order to find the dispersion of values of x from a central value a, we find the deviations about a. An absolute measure of dispersion is the mean of these deviations.

Reason To find the mean, we must obtain the sum of the deviations. But, we know that a measure of central tendency lies between the maximum and the minimum values of the set of observations.

Therefore, some of the deviations will be negative and some positive. Thus, the sum of deviations may vanish.Moreover, the sum of the deviations from mean $\bar{\text{(x)}}$ is zero. Also,Mean of deviations $=\frac{\text{Sum of deviations }}{\text{Number of observations }}\frac{0}{\text{n}}=0$

Thus, finding the mean of deviations about mean is not of any use for us, as far as the measure of dispersion is concerned.

4. A is false; R is true.

Solution:

Assertion Sum of n even natural numbers = n(n + 1)

mean $\bar{\text{(x)}}=\frac{\text{n(n+1)}}{\text{n}}=\text{n}+1$

Variance $=\Big[\frac{1}{\text{n}}\sum(\text{x}_\text{i})^2\Big]-\bar{\text{(x)}}^2$

$=\frac{1}{\text{n}}\big[2^2+4^2+...+(2\text{n})^2\big]-(\text{n}+1)^2$

$=\frac{1}{\text{n}}\big[2^2+1^2+...+\text{n}^2\big]-(\text{n}+1)^2$

$=\frac{4}{\text{n}}\frac{\text{n(n+1}(2\text{n+1)}}{6}-(\text{n}+1)^2$

$=\frac{(\text{n+1)}[2(2\text{n+1)}-3(\text{n+1)]}}{3}$

$=\frac{(\text{n+1)}(\text{n}-1)}{3}$

$=\frac{\text{n}^2-1}{3}$

3. A is true; R is false

Solution:

$\bar{\text{x}}=\frac{\text{ Sum of terms}}{\text{Number of terms }}=\frac{\sum\text{x}_\text{i}}{\text{n}}$

$=\frac{4+7+8+9+10++12+13+17}{8}=10$

$\therefore$ Mean deviation about mean $=\frac{\sum|\text{x}_\text{i}-\bar{\text{x}}|}{\text{n}}=\frac{24}{8}=3$

Reason Mean of the given series

$\bar{\text{x}}=\frac{\text{ Sum of terms}}{\text{Number of terms }}=\frac{\sum\text{x}_\text{i}}{\text{n}}$

$\frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10}=50$

$\therefore $ Mean deviation about mean

$=\frac{\sum|\text{x}_\text{i}-\bar{\text{x}|}}{\text{n}}$

$=\frac{84}{10}=8.4$

4. A is false; R is true.

Solution:

Mean $\bar{\text{(x)}}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$

$\bar{\text{(x)}}=64\frac{0}{100}=64$

Reason Standard deviatio $(\sigma)$

$=\sqrt{\frac{\sum\text{f}_\text{i}\text{d}_\text{i}^2}{\sum\text{f}_\text{i}}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2$

$=\sqrt{\frac{286}{100}}-\Big(\frac{0}{100}\Big)^2$

$=\sqrt{286}=1.69$