Selina Concise Class 7 Math Chapter 21 Data Handling Exercise 21B Solutions
EXERCISE 21B
(1) Sum of all observation = 53 + 61 + 60 + 67 + 64 = 305
Number of observation = 5
∴ Required mean = 305/5 = 61
(2) First 6 natural number = 1, 2, 3, 4, 5, 6
Sum of all observation = 1 + 2 + 3 + 4 + 5 + 6 = 21
Number of observation = 6
∴ Required mean = 21/6 = 3.5
(3) First ten odd natural number = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Sum of all observation = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
Number of observation = 10
∴ Required mean = 100/10 = 10
(4) Factors of 10 = 2, 5
Sum of all observation = 2 + 5 = 7
Number of observation = 2
∴ Required mean = 7/2 = 3.5
(5) Sum of all observation = x + 3 + x + 5 + x + 7 + x + 9 + x + 11 = 5x + 35 = 5 (x + 7)
Number of observation = 5
∴ Required mean = [5 (x+7)]/5 = x + 7
(6) Sum of all observation = 19.8 + 15.4 + 13.7 + 11.71 + 11.8 + 12.6 + 12.8 + 18.6 + 20.5 + 21.1 = 158.01
Number of observation = 10
∴ Required mean = 158.01/10 = 15.801
(7)(i) Increased by = 3
Observed mean = 32
∴ Resulting mean = 32 + 3 = 35
(ii) Decreased by 7 = 7
Observed mean = 32
∴ Resulting mean = 32 – 7 = 25
(iii) Multiplied by = 2
Observed mean = 32
∴ Resulting mean = (32 × 2) = 64
(iv) Divided by = 0.5
Observed mean = 32
∴ Resulting mean = 32/0.5 = 64
(v) Increased by = 60%
Observed mean = 32
⇒ x = 45 – 33 = 12
(10) First five prime numbers = 3, 6, 9, 12, 15, 18
Sum of observation = 3 + 6 + 9 + 12 + 15 + 18 = 63
Number of observation = 6
∴ Required mean = 63/6 = 10.5
Sum of first four numbers = (13 – 5) + (13 – 1) + 13 + (13 + 2)
= 13 – 5 + 13 – 1 + 13 + 13 + 2 = 48
∴ Mean = 48/4 = 12
(13) Squares of first five whole numbers = (0)2, (1)2, (2)2, (3)2, (4)2
= 0, 1, 4, 9, 16
∴ Mean = (0+1+4+9+16)/5 = 30/5 = 6
(15) First six multiples of 5 = 5, 10, 15, 20, 25, 30
∴ Mean = (5+10+15+20+25+30)/6 = 105/6 = 17.5
(16) Sum of observation = 0.5 + 2.7 + 2.6 + 0.5 + 2 + 5.8 + 1.5 = 15.6
∴ Mean = 15.6/7 = 2.2 mm
(17) Mean of 40 observations = 100
Total sum of 40 observations = 100 × 40 = 4000
Incorrect total of 40 observation is = 4000
Correct total of 40 observations = 4000 – 83 + 53 = 3970
∴ Correct mean = 3970/100 = 39.70
(18) Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On excluding an observation, the mean of remaining 6 observations = 25
⇒ Total of remaining 4 observation = 25 × 4 = 100
⇒ Excluded observation = 135 – 100 = 35
(19) Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On including an observation, the mean of remaining 6 observations = 25 × 6 = 150
⇒ Included observation = 150 – 135 = 15
(20) Mean of 5 numbers = 20
Mean of other 5 numbers = 30
∴ Mean = (20+30)/2 = 50/2 = 25
(22)
Term (xi) | Frequency (fi) | (fi xi) |
18 | 3 | 54 |
22 | 5 | 110 |
26 | 10 | 260 |
30 | 2 | 60 |
34 | 8 | 276 |
38 | 2 | 76 |
Total | 30 | 832 |
(23) (i) Since, in the given data, the number 6 appears maximum times.
∴ Mode = 6
(ii) Since, in the given data, the number 10 appears maximum times.
∴ Mode = 10
(24) (i) Since, the frequency of number 18 is maximum times.
∴ Mode = 18.
(ii) Since, the frequency of number 41 is maximum times.
∴ Mode = 41.
T H A N K Y O U!!!