CBSE Class 10 Mathematics Statistics MCQs Set 18

Answer: (d) 25

Explanation:
1. Total books = 15 × 12 = 180. If each of the two students got a books and the other 13 averaged m, then 2a + 13m = 180.
2. Testing options: a = 25 gives 13m = 130, so m = 10 (an integer). a = 11, 15, 20 give 158, 150, 140, none divisible by 13.

Teacher's Note:
1. Set up the total and test the options.
2. The remaining average must be a whole number.

Answer: (c) x̄ + a

Explanation:
1. Adding a constant a to each observation adds a to the mean.
2. New mean = x̄ + a.

Teacher's Note:
1. Mean shifts by the constant added.
2. Multiplying would scale the mean instead.

Answer: (b) 81

Explanation:
1. The sum of the first n odd numbers is n², so their mean is n.
2. n = \( \frac{n^2}{81} \) gives n = 81.

Teacher's Note:
1. Sum of the first n odd natural numbers = n².
2. Divide by n to get the mean.

Answer: (b) B

Explanation:
1. Coefficient of range = \( \frac{\text{max} - \text{min}}{\text{max} + \text{min}} \).
2. A: 45/85 = 0.53; B: 30/114 = 0.26; C: 88/108 = 0.81; D: 78/120 = 0.65.
3. B has the smallest coefficient of range, so it is the most consistent.

Teacher's Note:
1. A smaller coefficient of range means less variation.
2. Compare the four values.

Answer: (d) D

Explanation:
1. Coefficient of range = (max − min)/(max + min).
2. A: 1480/1720 = 0.86; B: 1206/1654 = 0.73; C: 1075/1445 = 0.74; D: 304/1304 = 0.23.
3. D has the smallest value, so it is the most consistent.

Teacher's Note:
1. Consistency is better when the coefficient of range is smaller.
2. Use the largest and smallest value of each company.

Answer: (b) 4

Explanation:
1. Counting: 5 occurs 4 times, 4 occurs 4 times and 3 occurs 4 times. For 3 to be the unique mode, x = 3 (3 then occurs 5 times).
2. Sorted: 3,3,3,3,3,4,4,4,4,5,5,5,5 (13 values). The 7th value is 4, so the median is 4.

Teacher's Note:
1. Find x from the condition on the mode.
2. With 13 observations, the median is the 7th value.

Answer: (c) 5 : 2

Explanation:
1. Let mean = 2k and median = 3k. Mode = 3 × median − 2 × mean = 9k − 4k = 5k.
2. Mode : mean = 5 : 2.

Teacher's Note:
1. Empirical relation: Mode = 3 Median − 2 Mean.
2. Substitute using a common multiple k.

Answer: (c) 0

Explanation:
1. The sum of the deviations of observations from their mean is always zero.

Teacher's Note:
1. Deviations from the mean add up to zero.
2. This holds for any data set.

Answer: (d) None of these

Explanation:
1. The new mean is \( \frac{1}{k} \).
2. Whether this is greater than, less than or equal to 1 depends on k, so none of the first three statements is always true.

Teacher's Note:
1. Dividing every observation by k divides the mean by k.
2. The comparison with 1 depends on k.

Answer: (c) 32

Explanation:
1. The largest listed value is 18 and the smallest is 12.
2. If x were the smallest, 18 − x = 20 gives a negative x, which is not allowed. So x is the largest: x − 12 = 20, x = 32.

Teacher's Note:
1. Range = maximum − minimum.
2. Check which of the two cases is possible.

Answer: (b) ab − c

Explanation:
1. Multiplying by b multiplies the mean by b, giving ab.
2. Subtracting c from each product lowers the mean by c: ab − c.

Teacher's Note:
1. Mean follows the same operations as the observations.
2. Apply multiplication first, then subtraction.

Answer: (c) 5

Explanation:
1. The mode is the value with the greatest frequency.
2. The highest frequency is 12, corresponding to x = 5.

Teacher's Note:
1. Mode is the most frequent value.
2. Do not confuse the frequency with the value.

Answer: (b) 13.5

Explanation:
1. Sum of 12 observations = 12 × 15 = 180.
2. Remaining sum = 180 − 45 = 135 for 10 observations, so the mean = 13.5.

Teacher's Note:
1. Mean = sum ÷ number of observations.
2. Update both the sum and the count.

Answer: (a) 20

Explanation:
1. Using Mode = 3 Median − 2 Mean: 12 = 3 Median − 48.
2. 3 Median = 60, so the median is 20.

Teacher's Note:
1. The empirical relation links mean, median and mode.
2. Solve for the median.

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

Explanation:
1. Their mean is \( \frac{a}{2} \).
2. Each observation deviates from the mean by \( \frac{b}{2} \), so the mean deviation is \( \frac{b}{2} \).

Teacher's Note:
1. Mean deviation is the average of absolute deviations from the mean.
2. The two values are symmetric about a/2.

Answer: (c) −1

Explanation:
1. Sum = 10x + 14, so the mean = \( \frac{10x + 14}{4} \).
2. Setting this equal to x + 2: 10x + 14 = 4x + 8, so x = −1.

Teacher's Note:
1. Add the four expressions and divide by 4.
2. Solve the linear equation.

Answer: (b) 12

Explanation:
1. The maximum of the other numbers is 35 and the minimum is 14.
2. If x is the smallest, 35 − x = 23 gives x = 12. (If x were the largest, x = 37.)
3. The least possible value is 12.

Teacher's Note:
1. Range = max − min.
2. Consider whether x is the minimum or the maximum.

Answer: (c) 24

Explanation:
1. N = 70, so N/2 = 35. Cumulative frequencies: 12, 25, 50, 70, so the median class is 20-30.
2. Median = \( 20 + \frac{35 - 25}{25} \times 10 \) = 24.

Teacher's Note:
1. Median = l + ((N/2 − cf)/f) × h.
2. Locate the class where the cumulative frequency first exceeds N/2.

Answer: (d) −25

Explanation:
1. Sum of five numbers = −25, so the other three add up to −25 − 50 = −75.
2. Their average = −75 ÷ 3 = −25.

Teacher's Note:
1. Use sum = mean × count.
2. Subtract the known pair.

Answer: (a) 7

Explanation:
1. Arranged in order: x = 3 (5), 5 (4), 6 (2), 7 (6), 8 (6), 12 (7); total = 30.
2. Cumulative: 5, 9, 11, 17, 23, 30. The 15th and 16th values both fall in x = 7, so the median is 7.

Teacher's Note:
1. Arrange the values in ascending order first.
2. For an even count, average the two middle values.