Important Questions for CBSE Class 11 Maths Chapter 15

CBSE Class 11 Maths Chapter-15 Important Questions - Free PDF Download

1 Marks Questions

1. In a test with a maximum marks 25, eleven students scored 3,9,5,3,12,10,17,4,7,19,21 marks respectively. Calculate the range.

Ans. The marks can be arranged in ascending order as 3,3,4,5,7,9,10,12,17,19,21.

Range = maximum value – minimum value

=21-3

= 18

2. Coefficient of variation of two distributions is 70 and 75, and their standard deviations are 28 and 27 respectively what are their arithmetic mean?

Ans. Given C.V (first distribution) = 70

Standard deviation = = 28

C.V

Similarly for second distribution

C.V

3. Write the formula for mean deviation.

Ans.MD

4. Write the formula for variance

Ans. Variance

5. Find the median for the following data.

579101215

862226

Ans.

5	7	9	10	12	15
8	6	2	2	2	6
8	14	16	18	20	26

Median is the average of 13th and 14th item, both of which lie in the c.f 14

6. Write the formula of mean deviation about the median

Ans.

7. Find the rang of the following series 6,7,10,12,13,4,8,12

Ans. Range = maximum value – minimum value

= 113-4

8. Find the mean of the following data 3,6,11,12,18

Ans. Mean =

9. Express in the form of a + ib (3i-7) + (7-4i) – (6+3i) + i23

Ans. Let

Z =

10. Find the conjugate of

Ans.

11. Solve for x and y, 3x + (2x-y) i= 6 – 3i

Ans.3x = 6

x = 2

2x – y = - 3

2 × 2 – y = - 3

- y = - 3 – 4

y = 7

12. Find the value of 1+i2 + i4 + i6 + i8 + ---- + i20

Ans.

13. Multiply 3-2i by its conjugate.

Ans. Let z = 3 – 2i

14. Find the multiplicative inverse 4 – 3i.

Ans. Let z = 4 – 3i

15. Express in term of a + ib

Ans.

16. Evaluate

Ans.

17. If 1, w, w2 are three cube root of unity, show that (1 – w + w2) (1 + w – w2) = 4

Ans. (1 – w + w2) (1 + w – w2)

(1 + w2 - w) (1 + w – w2)

18. Find that sum product of the complex number

Ans.

19. Write the real and imaginary part 1 – 2i2

Ans. Let z = 1 – 2i2

=1 – 2 (-1)

= 1 + 2

= 3

= 3 + 0.i

Re (z) = 3, Im (z) = 0

20. If two complex number z1, z2 are such that |z1| = |z2|, is it then necessary that z1 = z2

Ans. Let z1 = a + ib

21. Find the conjugate and modulus of

Ans. Let

22. Find the number of non zero integral solution of the equation |1-i|x = 2x

Ans.

Which is false no value of x satisfies.

23. If (a + ib) (c + id) (e + if) (g + ih) = A + iB then show that

Ans.

4 Marks Questions

1.The mean of 2,7,4,6,8 and p is 7. Find the mean deviation about the median of these observations.

Ans.Observations are 2, 7, 4, 6, 8 and p which are 6 in numbers

The near of these observations is 7

Arrange the observations in ascending order 2,4,6,7,8,15

Medias (M) =

Calculation of mean deviation about Median.

xi	xi-M	\|xi-M\|
2	-4.5	4.5
4	-2.5	2.5
6	-0.5	0.5
7	0.5	0.5
8	1.5	1.5
15	8.5	8.5
Total		18

Media’s deviation about median

2.Find the mean deviation about the mean for the following data!

1030507090

42428168

Ans. To calculate mean, we requirevalues then for mean deviation, we require || values and values.


10	4	4	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320
	80	4000		1280

Mean deviation about the mean

3.Find the mean, standard deviation and variance of the first natural numbers.

Ans. The given numbers are 1, 2, 3, ……, n

Mean

Variance

Standard deviation

4.Find the mean variance and standard deviation for following data

Ans.

	4	8	11	17	20	24	32
	3	5	9	5	4	3	1

Note: - 4th, 5th and 6th columns are filled in after calculating the mean.


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
Total	30	402			1374

Here

Mean

Variance

Standard deviation

= 6.77

5.The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Ans. Let be the six given observations

Then

Also

As each observation is multiplied by 3, new observations are

New near =

Let be the new standard deviation, then

6.Prove that the standard deviation is independent of any change of origin, but is dependent on the change of scale.

Ans. Let us use the transformation to change the scale and origin

Now

Also

Both,are positive which shows that standard deviation is independent of choice of origin, but depends on the scale.

7.Calculate the mean deviation about the mean for the following data

Expenditure0-100100-200200-300300-400400-500500-600600-700700-800

persons 489107543

Ans.

Expenditure	No. of persons	Mid point
0-100	4	50	200	308	1232
100-200	8	150	1200	208	1664
200-300	9	250	2250	108	972
300-400	10	350	3500	8	80
400-500	7	450	3150	92	644
500-600	5	550	2750	192	960
600-700	4	650	2600	292	1168
700-800	3	750	2250	392	1176
	50		17900		7896

mean =

8.Find the mean deviation about the median for the following data

Marks 0-1010-2020-3030-4040-5050-60

No. of boys 810101642

Ans.

Marks	No. of boys	Cumulative Frequency	Mid points
0-10	8	8	5	22	176
10-20	10	18	15	12	120
20-30	10	28	25	2	20
30-40	16	44	35	8	128
40-50	4	48	45	18	72
50-60	2	50	55	28	56
total	50				572

which is the median class.

Median

= 27

9.An analysis of monthly wages point to workers in two firms A and B, belonging to the same industry, given the following result. Find mean deviation about median.

Firm AFirm B

No of wages earns586648

Average monthly wagesRs 5253Rs 5253

Ans.For firm A, number of workers = 586

Average monthly wage is Rs 5253

Total wages = Rs 5253586

= Rs 3078258

For firm B, total wages = Rs 253648

=Rs 3403944

Hence firm B pays out amount of monthly wages.

10.Find the mean deviation about the median of the following frequency distribution

Class 0-66-1212-1818-2424-30

Frequency8101295

Ans.

Class	Mid value	Frequency
0-6	3	8	8	11	88
6-12	9	10	18	5	50
12-18	15	12	30	1	12
18-24	21	9	39	7	63
21-30	27	5	44	13	65

12-18 is the medias class

Medias =

Medias

Mean deviation about median =

11.Calculate the mean deviation from the median from the following data

Salary per week(in Rs) 10-2020-3030-4040-5050-6060-70

no. of workers 461020106

Ans.

Salary per Week (in Rs)	Mid value	Frequency
10-20	15	4	4	30	120
20-30	25	6	10	20	120
30-40	35	10	20	10	100
40-50	45	20	40	0	0
50-60	55	10	50	10	100
60-70	65	6	56	20	120
70-80	75	4	60	30	120

40-50 is the median class

Medias =

Mean deviation = =

12.Let values of a variable Y and let ‘a’ be a non zero real number. Then prove that the variance of the observations is also, find their standard deviation.

Ans.Let value of variables such that then

13.If

Ans.

Taking conjugate both side

14.If

Ans.

15.Solve

Ans.

16.Find the modulus

Ans.i25 + (1+3i)3

17.If

Ans. (i) (Given)

(ii) [taking conjugate both side

(i) × (ii)

18.Evaluate

Ans.

19.Find that modulus and argument

Ans.

20.For what real value of x and y are numbers equal (1+i) y2 + (6+i) and (2+i) x

Ans.(1+i) y2 + (6 + i) = (2 + i) x

y2 + iy2 + 6 + i = 2x + xi

(y2 + 6) + (y2 + 1) i = 2x + xi

y2 + 6 = 2x

y2 + 1 = x

y 2 = x – 1

x – 1 + 6 = 2x

5 = x

21.If x + iy =

Ans.

taking conjugate both side

x2 + y2 = 1

Proved.

22.Convert in the polar form

Ans.

23.Find the real values of x and y if (x - iy) (3 + 5i) is the conjugate of – 6 – 24i

Ans.

(x – iy) (3 + 5i) = - 6 + 24i

3x + 5xi – 3yi – 5yi2 = - 6 + 24i

24.If

Ans. If

6 Marks Questions

1.Calculate the mean, variance and standard deviation of the following data:

Classes	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	7	12	15	8	3	2

Ans.

Classes	Frequency	Mid Point
30-40	3	35	105	729	2187
40-50	7	45	315	289	2023
50-60	12	55	660	49	588
60-70	15	65	975	9	135
70-80	8	75	600	169	1352
80-90	3	85	255	529	1587
90-100	2	95	190	1089	2178
Total	50		3100		10050

Here

Mean

Variance

Standard deviation

2.The mean and the standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who mistook one observation as 50 instead of 40. What are the correct mean and standard deviation?

Ans. Given that

Incorrect mean

Incorrect S.D

= incorrect sum of observation =4000

= correct sum of observations = 4000-50+40

= 3990

So correct mean =

Also

Using incorrect values,

= 162601

= incorrect

= correct

= 162601-2500+1600=161701

Correct

Hence, correct mean is 39.9 and correct standard deviation is 5.

3.200 candidates the mean and standard deviation was found to be 10 and 15 respectively. After that if was found that the scale 43 was misread as 34. Find the correct mean and correct S.D

Ans.