3 Marks Question

Question 13 Marks

Aditi likes solving puzzles. She recently started attempting the ‘Easy’ level Sudoku puzzles. The time she took (in seconds) to solve these puzzles is — 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, 280, 270, 230, 220, 240. The first nine values correspond to Week 1 and the rest to Week 2.

(a) Construct a dot plot below showing the data for both weeks.
(b) Describe the mean, median, and any observations you may have about the data.

Answer

Arranging in ascending order:
220, 230, 240, 270, 280, 290, 310, 320, 320, 330, 340, 360, 370, 380, 400, 400, 410
With 17 values, the median is the 9th value.
So median = 320 sec.

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Question 23 Marks

Standing tall in the storm.

(a) Write estimated values for the number of skyscrapers in New York, Tokyo, and London.
(b) Are the following statements valid?
(i) Only 12 cities have more skyscrapers than Mumbai.
(ii) Only 7 cities have fewer skyscrapers than Mumbai.
(iii) The tallest building in the world is in Hong Kong.

Answer

(а) Estimated values for the number of skyscrapers in
New York – 38
Tokyo – 160
London – 305
(b) (i) Valid
(ii) Valid
(iii) Invalid

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Question 33 Marks

Consider a group of 17 students with the following heights (in cm): 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101. The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with heights less than a particular height, and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the ‘Telling Tall Tales’ section?

Answer

17 is an odd number that cannot be divided into two perfectly equal groups.
Arranging the heights of students in increasing order.
101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125
Mean $=\left(\frac{17+1}{2}\right)^{ th }$ term
= 9th term
= 115 cm
According to the conditions given in question, the height used to divide the students into two groups is 112 < height < 115
If we take 114 cm as the required height, then
Group 1 (Less than 114 cm) – 101, 102, 106, 109, 110, 112
Total = 7 Students
Group 2 (More than 114 cm) – 115, 115, 115, 115, 115, 117, 120, 120, 123, 125
Total = 10 students.

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Question 43 Marks

The following table shows the points scored by each player in four games:

Player	Game 1	Game 2	Game 3	Game 4
A	14	16	10	10
B	0	8	6	4
C	8	11	Did not play	13

Now answer the following questions:
(a) Find the average number of points scored per game by A.
(b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B?
(c) Who is the best performer?

Answer

(a) Average number scored per game by $A=\frac{\text { Sum of points }}{\text { Number of games played }}$
$=\frac{14+16+10+10}{4}$
$=\frac{50}{4}$
= 12.5
(b) To find the mean number of points scored per game by C, we divide the total points by 3 because player ‘C’ did not play game 3 and played only 3 games.
For player (B), we divide the total points by 4 because the player played all four games.
(c) Player ‘A’ is the best performer.

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Question 53 Marks

The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls.

Based on the dot plots, which of the following statements are true?
(a) The data varies more for the boys than for the girls.
(b) The median number of pockets for the boys is more than that for the girls.
(c) The mean number of pockets for the girls is more than that for the boys.
(d) The maximum number of pockets for boys is greater than that for the girls.

Answer

(a) False
(b) True
(c) False
(d) True

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Question 63 Marks

The temperature variation over two days in different months in Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use the scale 1 unit = 4°C. Can you guess which two months these days might belong to?

	12 am	3 am	6 am	9 am	12 pm	3 pm	6 pm	9 pm
Day 1	20°C	18°C	16°C	20°C	26°C	34°C	30°C	24°C
Day 2	37°C	34°C	30°C	33°C	37°C	43°C	42°C	39°C

Answer

These days might belong to December and May.

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Question 73 Marks

Preyashi asked her students ‘If you were to get a super power to become aquatic (water-borne), aerial (air-borne), or spaceborne, which one would you choose?’. The responses are shown below. Some chose none. Draw a double-bar graph comparing how both grades chose each option. Choose an appropriate scale.

Grade 5	w, a, a, a, w, n, s, a, n, w, a, a, a, a, a, w, w, s, a, a, n, w, a, a, n
Grade 9	n, w, s, a, s, w, s, s, a, a, w, s, s, a, s, a, n, w, s, s, a, w, a, w, a

Answer

	W	a	S	n
Grade 5	6	13	2	4
Grade 9	6	8	9	2

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Question 83 Marks

The dot plots of the heights of another section of Grade 5 students of the same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures?

Answer

We infer from the dot plots and the central tendency measures.
The girls’ heights are more spread and are between 126 and 158.
The boys’ height lies between 130 and 148.
Both the tallest and shortest in the class are girls.
Yet, the girls’ average height is less than the whole class average and also less than the boys’ average height.
We can say that boys are taller than girls in this class.
For boys’ heights mean < median (142.05 < 143) indicates a small influence of values on the lower side.
For girls’ heights mean > median (140.14 > 140), indicating a small influence of values on the higher side.
On comparing the data of heights given here and in the previous example of the section of grade 5 section, we find that the boys and girls of this section are shorter in height.

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Question 93 Marks

The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data.

Boys	3.5	4.1	2.6	3.2	3.4	3.8
Girls	4.0	3.1	3.4	3.7	2.5	3.4

Answer

The weight of boys is between 2.6 kg and 4.1 kg.
The weight of girls lies between 2.5 kg and 4 kg.
The heaviest baby is a boy, and the lightest is a girl.

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Question 103 Marks

Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, —, 2, 4. Find the mean and median. How would you describe this data?

Answer

Ignoring the missing values, the total data values are 20.
Arranging daUj values in ascending order:
0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25
Since there are 20 values (an even number), the median is the average of the 10th and 11th values.
10th value = 2
11th value = 2
Median $=\frac{2+2}{2}=\frac{4}{2}=2$
Mean $=\frac{\text { Sum of the values }}{\text { Total number of values }}$
$=\frac{72}{20}$
= 3.6
Mean = 3.6
The data shows the number of domestic animals/pets students have at home.
Most students have 0 – 4 pets.
There is a wide range, with an outlier of 25 at the higher end of the data.

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Question 113 Marks

Find the median of onion prices in Yahapur and Wahapur.

Month	Yahapur
January	25
February	24
March	26
April	28
May	30
June	35
July	39
August	43
September	49
October	56
November	59
December	44

Month	Wahapur
January	19
February	17
March	23
April	30
May	38
June	35
July	42
August	39
September	53
October	60
November	52
December	42

Answer

Monthly onion prices in Yahapur in ascending order:
24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59
Since there are 12 numbers (an even count), the median is the average of the 6th and 7th numbers.
6th number = 35
7th number = 39
Median $=\frac{35+39}{2}=\frac{74}{2}=37$
Median of onion prices in Yahapur = ₹ 37/kg
Monthly onion prices in Wahapur in ascending order:
17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60
Since there are 12 numbers (an even count), the median is the average of the 6th and 7th numbers.
6th number = 38
7th number = 39
Median $=\frac{38+39}{2}=\frac{77}{2}=38.5$
Median of onion prices in Wahapur = ₹ 38.5/kg.

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Question 123 Marks

Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil: 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who, on average, ran quicker?

Answer

Nikhil’s running times over the past week in seconds: 17, 18, 17, 16, 19, 17, 18
Number of days in a week = 7
Average running time of Nikhil $=\frac{\text { Total number of seconds }}{\text { Number of days }}$
$=\frac{17+18+17+16+19+17+18}{7}$
$=\frac{122}{7}$
= 17.43 seconds.
Sunil’s running time over the past week in seconds: 20, 18, 18, 17, 16, 16, 17.
Average running time of Sunil $=\frac{20+18+18+17+16+16+17}{7}$
$=\frac{122}{7}$
= 17.43 seconds
Both Nikhil and Sunil have the same average running time.

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Question 133 Marks

Shreyas is playing with a bat and a ball—but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas can make with his bat.

Answer

Number of bounces of the ball in 8 attempts = 6, 2, 9, 5, 4, 6, 3, 5
Number of attempts = 8
Average number of bounces of the ball $=\frac{6+2+9+5+4+6+3+5}{8}$
$=\frac{40}{8}$
= 5
So, the average number of bounces of the ball is 5.

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MATHS 3 Marks Question

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