Multiple Choice Questions on Measures of Central Tendency Class 11 OP Malhotra Exe-20C ISC Maths Solutions Ch-20. In this article you would learn to solve hard mcq questions on Measures of Central Tendency. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Measures of Central Tendency Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-20
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-20 | Measures of Central Tendency |
| Writer | OP Malhotra |
| Exe-20(C) | Multiple Choice Questions. |
Multiple Choice Questions on Measures of Central Tendency
OP Malhotra ISC Class 11 Maths Solutions
Que-1: The number of accidents taking place in a city over a week are as under:
| Day | Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |
|---|---|---|---|---|---|---|---|
| Number of accidents | 25 | 30 | 15 | 14 | 20 | 17 | 12 |
Find the average number of accidents per day.
(a) 18.5 (b) 17 (c) 19 (d) 20
Answer: (c) 19
Sol: Total accidents = 25 + 30 + 15 + 14 + 20 + 17 + 12 = 133
Number of days = 7
Average = Total / Number of days
= 133 / 7 = 19
Que-2: If Σfᵢ = 20, Σfᵢxᵢ = 2p + 20 and mean of distribution is 12, then the value of p is:
(a) 110 (b) 100 (c) 90 (d) 105
Answer: (a) 110
Sol: Mean = Σfᵢxᵢ / Σfᵢ
12 = (2p + 20) / 20
⇒ 12 × 20 = 2p + 20
⇒ 240 = 2p + 20
⇒ 2p = 220
⇒ p = 110
Que-3: If the mean of 5, 8, 10, 15, p, 11, 13, q, 18 is 12, then the value of p + q is:
(a) 30 (b) 28 (c) 32 (d) None of these
Answer: (b) 28
Sol: Total numbers = 9, Mean = 12
⇒ Total sum = 9 × 12 = 108
Sum of known terms = 5 + 8 + 10 + 15 + 11 + 13 + 18 = 80
So, p + q = 108 − 80 = 28
Que-4: If the mean of x, x+3, x+5, x+7 and x+10 is 9, then the mean of the last three observations is:
(a) 10 1/3 (b) 10 2/3 (c) 11 1/3 (d) 11 2/3
Answer: (c) 11 1/3
Sol: Mean = 9
⇒ Total sum = 5 × 9 = 45
Sum = x + (x+3) + (x+5) + (x+7) + (x+10) = 5x + 25
5x + 25 = 45 ⇒ x = 4
Last three terms = 9, 11, 14
Mean = (9+11+14)/3
= 34/3 = 11 1/3
Que-5: The mean of 10 observations is 16.3. By error one value is taken as 32 instead of 23. Find corrected mean.
(a) 15.4 (b) 15.6 (c) 15.7 (d) 15.8
Answer: (a) 15.4
Sol: Wrong total = 16.3 × 10 = 163
Correct total = 163 − 32 + 23 = 154
Correct mean = 154 / 10 = 15.4
Que-6: The A.M. of 9 terms is 15. If one more term is added, mean becomes 16. Find added term.
(a) 23 (b) 25 (c) 27 (d) 29
Answer: (b) 25
Sol: Sum of 9 terms = 9 × 15 = 135
New sum = 10 × 16 = 160
Added term = 160 − 135 = 25
Que-7: If mean of numbers 27+x, 31+x, 89+x, 107+x, 156+x is 82, then find mean of 130+x, 126+x, 68+x, 50+x, 1+x.
(a) 75 (b) 79 (c) 80 (d) 82
Answer: (a) 75
Sol: Mean = 82
⇒ Sum = 5 × 82 = 410
Sum = (27+31+89+107+156) + 5x = 410
410 + 5x = 410 ⇒ x = 0
New mean = (130+126+68+50+1)/5
= 375/5 = 75
Que-8: For the given frequency distribution, mean is 7.5. Find missing frequency p.
| x | 2–4 | 4–6 | 6–8 | 8–10 | 10–12 | 12–14 |
|---|---|---|---|---|---|---|
| f | 6 | 8 | 15 | p | 8 | 4 |
(a) 9 (b) 6 (c) 3 (d) 5
Answer: (c) 3
Sol: First, we find the class marks (midpoints) of each class interval:
2–4 → 3, 4–6 → 5, 6–8 → 7, 8–10 → 9, 10–12 → 11, 12–14 → 13
Σf = 6 + 8 + 15 + p + 8 + 4 = 41 + p
Σfx = 18 + 40 + 105 + 9p + 88 + 52 = 303 + 9p
Given mean = 7.5
Using formula:
Mean = Σfx / Σf
7.5 = (303 + 9p) / (41 + p)
Now solving:
7.5(41 + p) = 303 + 9p
307.5 + 7.5p = 303 + 9p
307.5 − 303 = 9p − 7.5p
4.5 = 1.5p
p = 3
Que-9: On dividing each entry by a non-zero number ‘a’, the mean:
(a) does not change (b) is multiplied by a (c) is diminished by a (d) is divided by a
Answer: (d) is divided by a
Sol: Let the original observations be x₁, x₂, x₃, …, xₙ
Original mean = (x₁ + x₂ + … + xₙ) / n
After dividing each value by a:
New values = x₁/a, x₂/a, …, xₙ/a
New mean = (x₁/a + x₂/a + … + xₙ/a) / n
= (1/a)(x₁ + x₂ + … + xₙ) / n
= (1/a) × (Original Mean)
∴ New mean = Original mean / a
Que-10: A student gets 75%, 80%, 85%. After adding one more subject, average cannot be less than:
(a) 60% (b) 65% (c) 70% (d) 75%
Answer: (a) 60%
Sol: Initial sum = 75+80+85 = 240
For minimum average, take lowest marks = 0
New mean = 240/4 = 60%
Que-11: If mean of first n odd numbers is n²/81, find n.
(a) 9 (b) 18 (c) 27 (d) 81
Answer: (d) 81
Sol: First n odd numbers = 1, 3, 5, …
Sum of first n odd numbers = n²
Mean = (Sum / n) = n² / n = n
Given mean = n² / 81
So,
n = n² / 81
Multiply both sides by 81:
81n = n²
n² − 81n = 0
n(n − 81) = 0
Since n ≠ 0 (natural number),
n = 81
Que-12: If for some x∈R, the frequency distribution of the marks obtained by 20 students in a test is
| Marks (x) | 2 | 3 | 5 | 7 |
| Frequency (f) | (x + 1)2 | 2x − 5 | x2 − 3x | x |
then the mean of the marks is
(a) 3.0 (b) 2.5 (c) 2.8 (d) 3.2
Answer: (c) 2.8
Sol: Total frequency = 20
(x + 1)2 + (2x − 5) + (x2 − 3x) + x = 20
x2 + 2x + 1 + 2x − 5 + x2 − 3x + x = 20
2x2 + 2x − 4 = 20
⇒ 2x2 + 2x − 24 = 0
⇒ x2 + x − 12 = 0
⇒ (x + 4)(x − 3) = 0
x = 3 (Reject x = −4 as frequency cannot be negative)
Total frequency = 20
Mean = Σfx / Σf = 56 / 20
= 2.8
–: End Measures of Central Tendency Class 11 OP Malhotra Exe-20C ISC Maths Ch-20 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
Thanks
Please share with your friends
