CBSE Class 11 Mathematics Statistics JEE Previous Year Questions Set 01

Answer: (b) 65

Explanation:
1. Total marks = 100 × 72 = 7200; boys' total = 70 × 75 = 5250.
2. Girls' total = 7200 − 5250 = 1950 for 30 girls.
3. Girls' average = 1950/30 = 65.

Teacher's Note:
1. Convert averages to totals before combining or splitting groups.
2. The class average 72 lies between 65 and 75, closer to 75 because boys are more.

Answer: (d) 78.00

Explanation:
1. Corrected Σx = 170 − 20 + 30 = 180.
2. Corrected Σx² = 2830 − 400 + 900 = 3330.
3. Variance = \( \frac{3330}{15} - \left(\frac{180}{15}\right)^2 = 222 - 144 = 78 \).

Teacher's Note:
1. Update both Σx and Σx² before using the variance formula.
2. Option (a) is just Σx²/n with the old data; always subtract the square of the mean.

Answer: (c) remains the same as that of the original set

Explanation:
1. For 9 ordered observations the median is the 5th value.
2. Only the 6th to 9th values are increased, and they stay above the 5th value, so the 5th value is unchanged.
3. The median remains 20.5.

Teacher's Note:
1. The median depends only on the middle position, not on the size of the extreme values.
2. The mean, by contrast, would increase by 8/9 here.

Answer: (a) only (1) and (2)

Explanation:
1. (1) True: the modal class is the class with the tallest bar of the histogram, and the mode can be read from it.
2. (2) True: if every value is multiplied by k, the median is also multiplied by k, so it depends on scale.
3. (3) False: variance is independent of change of origin but not of change of scale (it gets multiplied by k²).
4. Hence only (1) and (2) are correct.

Teacher's Note:
1. Measures of central tendency change with both origin and scale.
2. Measures of dispersion ignore change of origin but change with scale.

Answer: (a) 2

Explanation:
1. Mean = \( \frac{na + n(-a)}{2n} = 0 \).
2. Variance = \( \frac{na^2 + na^2}{2n} - 0 = a^2 \).
3. S.D. = |a| = 2.

Teacher's Note:
1. Symmetric data about 0 has mean 0, so variance is simply the mean of the squares.
2. The S.D. is the square root of a², which is |a|, not a.

Answer: (a) 18

Explanation:
1. Variance ≥ 0 ⇒ \( \frac{\Sigma x_i^2}{n} \geq \left(\frac{\Sigma x_i}{n}\right)^2 \).
2. \( \frac{400}{n} \geq \frac{6400}{n^2} \) ⇒ 400n ≥ 6400 ⇒ n ≥ 16.
3. Among the options only 18 satisfies n ≥ 16.

Teacher's Note:
1. Variance can never be negative; this gives the inequality mean of squares ≥ square of mean.
2. Equality would hold only when all observations are equal.

Answer: (a) 1

Explanation:
1. A = 100 + (1, 2, …, 100) and B = 150 + (1, 2, …, 100).
2. Both are shifts of the same data 1, 2, …, 100, so \( V_A = V_B \).
3. Ratio = 1.

Teacher's Note:
1. Variance does not change when a constant is added to every observation.
2. Look for a common underlying data set before doing any long calculation.

Answer: (a) 80

Explanation:
1. Let p be the fraction of boys: 52p + 42(1 − p) = 50.
2. 10p = 8 ⇒ p = 0.8, i.e. 80% boys.

Teacher's Note:
1. The combined mean is a weighted average of the group means.
2. Since 50 is much closer to 52, boys must form the larger part of the class.

Answer: (a) a = 3, b = 4

Explanation:
1. a + b + 23 = 30 ⇒ a + b = 7.
2. Σx² = 5(6.8 + 36) = 214 ⇒ a² + b² + 64 + 25 + 100 = 214 ⇒ a² + b² = 25.
3. So ab = (49 − 25)/2 = 12, giving a, b = 3, 4.

Teacher's Note:
1. Use Σx² = n(σ² + x̄²) to turn the variance into an equation in a and b.
2. All four options satisfy a + b = 7, so the variance condition decides the answer.

Answer: (b) 10.1

Explanation:
1. There are 101 numbers; mean = 1 + 50d (middle term).
2. Σ|x − mean| = d × 2(1 + 2 + … + 50) = 2550d.
3. \( \frac{2550d}{101} = 255 \) ⇒ d = 10.1.

Teacher's Note:
1. For an AP, the mean equals the middle term when the number of terms is odd.
2. The deviations are symmetric: 50d, 49d, …, d, 0, d, …, 50d.

Answer: (c) Statement 1 is false, Statement 2 is true.

Explanation:
1. First n even numbers = 2 × (1, 2, …, n).
2. Variance = 2² × \( \frac{n^2 - 1}{12} = \frac{n^2 - 1}{3} \), not \( \frac{n^2 - 1}{4} \); Statement 1 is false.
3. Statement 2 gives the standard sum formulas, which are true.

Teacher's Note:
1. Write the even numbers as 2 times the natural numbers and use the scaling rule for variance.
2. Test Statement 1 with n = 2: data 2, 4 has variance 1, while (n² − 1)/4 gives 3/4.

Answer: (b) \( \frac{11}{2} \)

Explanation:
1. Set 1: Σx = 10, Σx² = 5(4 + 4) = 40. Set 2: Σx = 20, Σx² = 5(5 + 16) = 105.
2. Combined: n = 10, Σx = 30, Σx² = 145.
3. Variance = \( \frac{145}{10} - 3^2 = 14.5 - 9 = \frac{11}{2} \).

Teacher's Note:
1. Combine data sets through Σx and Σx², not by averaging the variances.
2. The combined variance exceeds the average of 4 and 5 because the two means are different.

Answer: (a) 4

Explanation:
1. Median of a, 2a, …, 50a = average of 25th and 26th terms = 25.5a.
2. Σ|ka − 25.5a| = |a| × 2(0.5 + 1.5 + … + 24.5) = |a| × 2 × 312.5 = 625|a|.
3. M.D. = \( \frac{625|a|}{50} = 12.5|a| = 50 \) ⇒ |a| = 4.

Teacher's Note:
1. For an even number of terms, the median is the average of the two middle terms.
2. Pair the deviations symmetrically to add them quickly.

Answer: (b) Statement 1 is true, Statement 2 is false

Explanation:
1. Multiplying data by 2 multiplies the variance by 2² = 4, so Statement 1 is true.
2. The mean is multiplied by 2, giving \( 2\bar{x} \), not \( 4\bar{x} \); Statement 2 is false.

Teacher's Note:
1. The mean scales by k, the variance by k².
2. Mixing these two rules is the trap in this question.

Answer: (c) variance

Explanation:
1. Adding 10 to every mark increases the mean, median and mode by 10.
2. Deviations from the mean are unchanged, so the variance does not change.

Teacher's Note:
1. Measures of dispersion are unaffected by adding a constant to all observations.
2. Measures of central tendency shift by the same constant.

Answer: (a) 41

Explanation:
1. Wrong total = 20 × 40 = 800.
2. Correct total = 800 − 33 + 53 = 820.
3. Correct mean = 820/20 = 41.

Teacher's Note:
1. The total was short by 20, so the mean was short by 20/20 = 1.
2. Fix the total first and then divide by the number of observations.

Answer: (a) 4

Explanation:
1. Missing value = 35 − (6 + 7 + 8 + 10) = 4.
2. Deviations from 7: −1, 0, 1, 3, −3; squares sum = 1 + 0 + 1 + 9 + 9 = 20.
3. Variance = 20/5 = 4.

Teacher's Note:
1. Find the missing observation from the total before computing the variance.
2. Use the deviations from the mean directly when the mean is a whole number.

Answer: (d) 14

Explanation:
1. Since x < 2y, x − y < y; so the order is x − y < y < x < 2x + y.
2. Median = \( \frac{x + y}{2} = 10 \) ⇒ x + y = 20; Range = (2x + y) − (x − y) = x + 2y = 28.
3. So y = 8, x = 12; numbers 4, 8, 12, 32 have mean 56/4 = 14.

Teacher's Note:
1. Use the given inequalities to arrange the numbers in order before finding the median.
2. For four numbers, the median is the average of the 2nd and 3rd values.

Answer: (a) 833

Explanation:
1. The numbers are 2 × (1, 2, …, 50).
2. Variance = 4 × \( \frac{50^2 - 1}{12} = 4 \times \frac{2499}{12} = 833 \).

Teacher's Note:
1. Write even numbers as 2 × natural numbers and use Var(kx) = k² Var(x).
2. Var(1, 2, …, n) = (n² − 1)/12 is the key formula.

Answer: (a) increases by 1

Explanation:
1. n observations increase by 5 and n observations decrease by 3.
2. Change in total = 5n − 3n = 2n.
3. Change in mean = 2n/2n = 1, i.e. the mean increases by 1.

Teacher's Note:
1. Track the change in the total, then divide by the number of observations.
2. Since the 2n observations are distinct, exactly n lie below the median.

Answer: (b) \( \bar{x} + 5 \), M.D.

Explanation:
1. Adding 5 to every observation increases the mean by 5.
2. Each |x_i − x̄| is unchanged, so the mean deviation stays the same.

Teacher's Note:
1. Mean deviation, like variance, is independent of change of origin.
2. The mean always follows a shift of the data.

Answer: (b) Statement-1 and Statement-2 are both true

Explanation:
1. \( d_i = (-1)x_i - a \): variance = (−1)²σ² = σ², so Statement-1 is true.
2. Mean of \( d_i \) = \( -\bar{x} - a \); the most frequent value of \( d_i \) corresponds to the most frequent \( x_i \), so mode = −M − a. Statement-2 is true.

Teacher's Note:
1. The sign of the multiplier does not affect the variance because it is squared.
2. Mean, median and mode all follow the same linear change as the data.

Answer: (c) 14.0

Explanation:
1. Original total = 16 × 16 = 256.
2. New total = 256 − 16 + 3 + 4 + 5 = 252; number of observations = 16 − 1 + 3 = 18.
3. New mean = 252/18 = 14.

Teacher's Note:
1. Update both the total and the count when observations are added or removed.
2. Forgetting to change the count from 16 to 18 is the most common error.

Answer: (c) 3a² − 32a + 84 = 0

Explanation:
1. Σx = 16 + a, Σx² = 134 + a², n = 4.
2. \( \frac{134 + a^2}{4} - \left(\frac{16 + a}{4}\right)^2 = 12.25 \).
3. Multiplying by 16: 536 + 4a² − 256 − 32a − a² = 196 ⇒ 3a² − 32a + 84 = 0.

Teacher's Note:
1. Square the standard deviation first: 3.5² = 12.25.
2. Clearing the denominator 16 early keeps the algebra simple.

Answer: (c) 8.4

Explanation:
1. Let the other two be x, y: x + y = 25 − 9 = 16; x² + y² = 5(124 + 25) − 41 = 704.
2. xy = (256 − 704)/2 = −224, so x, y = \( 8 \pm \sqrt{288} \), one above 5 and one below 5.
3. |x − 5| + |y − 5| = x − y = 2√288 ≈ 33.94.
4. M.D. = \( \frac{4 + 3 + 1 + 33.94}{5} \) ≈ 8.4.

Teacher's Note:
1. Use Σx = n × mean and Σx² = n(σ² + mean²) to find the unknown observations.
2. When one unknown is above the mean and the other below, the sum of their absolute deviations is just their difference.

Answer: (b) 35

Explanation:
1. Old total = 25 × 40 = 1000; new total = 25 × 39 = 975.
2. 975 = 1000 − 60 + x ⇒ x = 35 years.

Teacher's Note:
1. The count stays 25 because one teacher replaces another.
2. The total fell by 25, so the new teacher is 25 years younger than the retired one.

Answer: (c) 9.00

Explanation:
1. Remaining: n = 97, Σx = 400 − 12 = 388, Σx² = 2474 − (9 + 16 + 25) = 2424.
2. Mean = 388/97 = 4.
3. Variance = \( \frac{2424}{97} - 16 \) ≈ 24.99 − 16 ≈ 9.00.

Teacher's Note:
1. When omitting observations, subtract them from Σx and Σx² and reduce n.
2. Here the mean conveniently stays 4, which simplifies the work.

Answer: (c) 2

Explanation:
1. Let \( d_i = x_i - 5 \); variance of x equals variance of d.
2. \( \sigma^2 = \frac{45}{9} - \left(\frac{9}{9}\right)^2 = 5 - 1 = 4 \).
3. σ = 2.

Teacher's Note:
1. Deviations from any assumed mean give the same variance as the original data.
2. This is the short cut method with A = 5.

Answer: (c) \( \frac{4}{3} \)

Explanation:
1. New mean = 75λ − 25.
2. 75λ − 25 = 75 ⇒ λ = \( \frac{100}{75} = \frac{4}{3} \).

Teacher's Note:
1. The mean follows the same linear change as every observation.
2. The number of observations, 30, is not needed.

Answer: (b) 1

Explanation:
1. 54 + λ = 64 ⇒ λ = 10.
2. Deviations from 8: −1, 0, 1, −1, 0, −1, 2, 0; squares sum = 8.
3. Variance = 8/8 = 1.

Teacher's Note:
1. Find the unknown from the mean before computing the variance.
2. With a whole-number mean, the deviation method is quickest.

Answer: (c) 2

Explanation:
1. S.D. 0 means all five observations equal 9.
2. New total = 50, so the changed observation is 50 − 36 = 14: data 9, 9, 9, 9, 14.
3. Variance = \( \frac{4(1) + 16}{5} = 4 \), S.D. = 2.

Teacher's Note:
1. A standard deviation of 0 means all observations are equal to the mean.
2. Find the new value from the new total, then compute the spread about the new mean 10.