# 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.    Get next Exercise 22A solutions click here

Updated: March 29, 2019 — 3:24 pm

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