Assertion (A) & Reason (B) MCQ

MCQ 11 Mark

Assertion (A): In an AP., the sum of 4th term from the beginning and sum of 4th term from the end is equal to sum of 7th term from the beginning and 7th term from the end.
Reason (R): Sum of first n terms of a AP., $\mathrm{S}_{\mathrm{n}} \frac{n}{2}(a+l)$, where $\mathrm{a}=$ first term, $\mathrm{l}=$ last term.

A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
✓
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 but $R$ is true.

Answer

Correct option: B.

Both $A$ and $R$ are true but $R$ is not the correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A.
Explanation: We know that in an A.P., the sum of terms equidistant from beginning and end is constant.
$\therefore$ Assertion is true.
Also, Reason is true but Assertion is not the correct explanation of Assertion.

View full question & answer→

MCQ 21 Mark

Consider the following data

$x_{i}$	4	8	11	17	20	24	32
$f_{i}$	3	5	9	5	4	3	1

Assertion (A): The variance of the data is 45.8.
Reason (R): The standard deviation of the data is 6.77.

A
Both $A$ and $R$ are true and $R$ is the correct explanation of A .
✓
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 but $R$ is true.

Answer

Correct option: B.

Both $A$ and $R$ are true but $R$ is not the correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A .
Explanation: Assertion: Presenting the data in tabular form, we get

$x_{i}$	$f_{i}$	$f_{i}x_{i}$	$x _{ i }-\bar{x}$	$( x _{ i }-\bar{x})^2$	$f _{ i }\left( x _{ i }-\bar{x}\right)^2$
4	3	12	-10	100	300
8	5	40	-6	36	180
11	9	99	-3	9	81
17	5	85	3	9	45
20	4	80	6	36	144
24	3	72	10	100	300
32	1	32	18	324	324
	30	420			1374

$\mathrm{N}=30, \sum_{i=1}^{7} f_{i} x_{i}=420, \sum_{i=1}^{7} f_{i}\left(x_{i}-\bar{x}\right)^{2}=1374$
Therefore, $\bar{x}=\frac{\sum_{i=1}^{7} f_{i} x_{i}}{N}=\frac{1}{30} \times 420=14$
$\therefore$ Variance $\left(\sigma^{2}\right)=\frac{1}{N} \sum_{i=1}^{7} f_{i}\left(x_{i}-\bar{x}\right)^{2}$
$\frac{1}{30} \times 1374=458$
Reason: Standard deviation $(\sigma)=\sqrt{45.8}=6.77$

View full question & answer→

Applied Maths Assertion (A) & Reason (B) MCQ

Test yourself on this topic