S.chand publication New Learning Composite mathematics solution of class 7 Chapter 15 Data Handling Exercise 15C

 Exercise 15C

Question 1

(a) 5, 3, 5, 2, 3, 5, 17

Mode – 5

(b) 1, 7, 13, 19, 25

= No Mode

(c) 4, 6, 8, 4, 10, 4, 6, 12, 6, 10

= Modes – 4, 6

(d) 51, 68, 70, 76, 55, 70, 73, 68, 70, 81

= Mode – 70

(e) 8, 4, 8, 4, 8, 4, 8, 4, 8, 4, 8, 4

= Modes – 8, 4

(f) 29, 29, 10, 10, 3, 3, 14, 10, 2, 2

= Mode – 10

(g) 26, 32, 26, 21, 83, 26, 83, 67, 53, 83, 85

= Modes – 26, 83

Question 2

(a)

Size	2	3	4	5	6	7
Frequency	4	6	2	5	3	1

Mode – 3

(b)

Size of shoes	6	7	8	9	10
No. of persons	12	20	40	15	7

Mode – 8

Question 3

(a) 1, 6, 8, 11, 11, 11, 14, 16, 21

(i) The mode = 11

(ii) The median = 11

(iii) The mean $=\frac{(1+6+8+11+11+11+14+16+21)}{9}$

$=\frac{99}{9}$

= 11

(b) 3,10,10,12,14,16,16,18,18,25

(i) The mode=10,16,18

(ii) The median $=\frac{1}{2} \left(14+16\right)$

$=\frac{1}{2} \times 30$

=15

(iii) The mean $=\frac{3+10+10+12+14+16+16+18+18+25}{10}$

$=\frac{142}{10}$

=14.2

(c) 2,6,9,11,15,23,29,36,42

(i) No mode

(ii) The median =15

(iii) The mean$=\frac{2+6+9+11+15+23+29+36+42}{9}$

$=\frac{173}{9}=19\frac{2}{9}$

(d) 8,8,8,8,18,18,18,18

(i) The mode=8, 18

(ii) The median $=\frac{1}{1} (8+18)$

$=\frac{1}{2} \times 26$

=13

(iii) The mean $=\frac{8+8+8+8+18+18+18+18}{8}$

$=\frac{104}{8}$

=13

Question 4

Number of fiction books read by 28 grade VII pupils:

4,6,1,7,3,9,5,7,8,5,4,6,10,6,9,5,6,6,8,6,8,5,10,7,2,5,3,7

Mode = 6

Mean $=\frac{ (4+6+1+7+3+9+5+7+8+5+4+6+10+6+9+5+6+6+8+6+8+5+10+7+2+5+3+7)}{25}$

$=\frac{168}{28}$

= 6

Question 5

20,21,19,22,18,23,25,22,23,20,23,20,22,21,24,21,22,23,19,21,22,22,24,26,22

Mode = 22

Mean $=\frac{(20+21+19+22+18+23+25+22+23+20+23+20+22+21+24+21+22+23+19+21+22+22+24,26+22)}{25}$

$=\frac{550}{25}$

= 22

Question 6

(a) T

(b) F

(c) T