Statistics Assertion–Reason PYQs Class 10 CBSE (With Answers)

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Mean = $\frac{\text{sum of observations}}{\text{number of observations}}$.
The first ten natural numbers are 1, 2, …, 10.
Their sum = $\frac{\text{10 × 11}}{\text{2}}$ = 55.
Mean = $\frac{\text{55}}{\text{10}}$ = 5.5.

Reason: True.
For n natural numbers 1, 2, …, n, their sum is $\frac{\text{n(n + 1)}}{\text{2}}$.

Link: R gives the sum needed to calculate the mean in A. Dividing that sum by 10 gives 5.5, so R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
Here n is a positive integer.
Mean = $\frac{\text{sum of observations}}{\text{number of observations}}$.
= $\frac{\text{n(n + 1)}}{\text{2n}}$
= $\frac{\text{n + 1}}{\text{2}}$, not $\frac{\text{n − 1}}{\text{2}}$.

Reason: True.
The sum 1 + 2 + … + n = $\frac{\text{n(n + 1)}}{\text{2}}$.

Link: R is the correct sum formula. Dividing it by n disproves A. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: False.
In the source, N denotes the total frequency; we write this as n.
$\frac{\text{n}}{\text{2}}$ helps locate the median class; it is not the median value.
For example, the data 10, 20, 30 have n = 3 and median 20, whereas $\frac{\text{n}}{\text{2}}$ = 1.5.

Reason: True in the sense of ordered position.
The median is the central value of ordered data.
With repeated values, the counts strictly below and above the median need not be equal.

Link: R describes the median’s position. It does not identify the median value with half the total frequency. A is false and R is true. Answer: D.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Total frequency n = 4 + 5 + 13 + 20 + 14 + 7 + 4 = 67.
$\frac{\text{n}}{\text{2}}$ = 33.5.
The cumulative frequencies are 4, 9, 22, 42, 56, 63, 67.
The first cumulative frequency greater than 33.5 is 42, so the median class is 125–145.
The greatest frequency is 20, so the modal class is also 125–145.
Required difference = 145 − 125 = 20.

Reason: False.
This distribution itself is a counterexample: the median class and modal class are both 125–145.

Link: A is correct, while R makes a false general claim. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True, using the empirical relationship intended in this question.
Mode = 3 Median − 2 Mean.
= 3 × 170 − 2 × 169
= 510 − 338
= 172.

Reason: True in the empirical relationship used here.
Mode = 3 Median − 2 Mean.
This is an empirical relationship, not an exact identity for every dataset.

Link: R supplies the relationship used to obtain 172. Under the question’s intended empirical relationship, R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True, using the empirical relationship intended in this question.
Mode = 3 Median − 2 Mean.
Mode − Median = 2(Median − Mean).
Therefore, the magnitude of the first difference is twice the second.
Required difference = $\frac{\text{24}}{\text{2}}$ = 12.

Reason: False.
The correct empirical relationship is Mode = 3 Median − 2 Mean.
R interchanges Mean and Median.

Link: A follows from the correct empirical relationship, but R states an incorrect one. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True, using the empirical relationship intended in this question.
Mode = 3 Median − 2 Mean.
3 Median = Mode + 2 Mean.
Median = $\frac{\text{60 + 2 × 66}}{\text{3}}$
= $\frac{\text{192}}{\text{3}}$ = 64.

Reason: False.
Rearranging the empirical relationship gives 2 Mean = 3 Median − Mode.
R incorrectly adds Mode instead of subtracting it.
The empirical relationship is also not an exact rule for any dataset.

Link: A uses the correct rearrangement. R gives a different, incorrect relationship. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True, using the empirical relationship intended in this question.
Mode = 3 Median − 2 Mean.
3 Median = Mode + 2 Mean.
Median = $\frac{\text{12k + 2 × 15k}}{\text{3}}$
= $\frac{\text{42k}}{\text{3}}$ = 14k.

Reason: False.
The correct rearrangement is Mean = $\frac{\text{3 Median − Mode}}{\text{2}}$.
R gives incorrect coefficients.

Link: A follows from the correct empirical relationship, whereas R misstates it. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True, using the empirical relationship intended in this question.
Mode = 3 Median − 2 Mean.
Median = $\frac{\text{Mode + 2 Mean}}{\text{3}}$.
= $\frac{\text{50 + 2 × 56}}{\text{3}}$
= $\frac{\text{162}}{\text{3}}$ = 54.

Reason: False.
The correct numerator is Mode + 2 Mean.
R has a minus sign in place of the plus sign.

Link: A gives the result of the correct formula. R states an incorrect rearrangement. A is true and R is false. Answer: C.