Let, the lenght of hall = l m, breadth = b m and height = h m Permeter of floor = 2(l + b) According of question, 2(l + b) = 250 m ...(1) Area of four walls of hall = 2(l + b) h So, the cost of painting the four walls at the rate of Rs 10 per m² = 2(l + b) h × Rs 10 = 250h × Rs 10 [∵ From equation (Read more

Let, the lenght of hall = l m, breadth = b m and height = h m
Permeter of floor = 2(l + b)
According of question, 2(l + b) = 250 m …(1)
Area of four walls of hall = 2(l + b) h
So, the cost of painting the four walls at the rate of Rs 10 per m² = 2(l + b) h × Rs 10
= 250h × Rs 10 [∵ From equation (1)]
= Rs 2500h
According to question, Rs 2500h = 15000
⇒ h = 15000/2500 = 6 m
Hence, the height of the hall is 6 m.

Lenght of one brick l = 22.5 cm, breadth b = 10 cm and height h = 7.5 cm Total surface area of brick = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm² = 2(225 + 75 + 168.75) cm² = 937.5 cm² = 0.09375 m² Number of bricks to be painted = Total paint available/paint for one brick = 9.375 m²/Read more

Lenght of one brick l = 22.5 cm, breadth b = 10 cm and height h = 7.5 cm
Total surface area of brick = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2(225 + 75 + 168.75) cm² = 937.5 cm² = 0.09375 m²
Number of bricks to be painted
= Total paint available/paint for one brick = 9.375 m²/0.09375 m² = 100
Hence, the paint of this container can paint 100 bricks.

Five examples of data that we can collect from our day-to-day life. 1. Height of our classmates or weight of our classmate. 2. Height of first 100 plants near by our locality. 3. Maximum or minimum temperature of a particular month. 4. Time spend for watching TV in a particular week. 5. Rainfall inRead more

Five examples of data that we can collect from our day-to-day life.
1. Height of our classmates or weight of our classmate.
2. Height of first 100 plants near by our locality.
3. Maximum or minimum temperature of a particular month.
4. Time spend for watching TV in a particular week.
5. Rainfall in our city in last 10 years.

Total number of balls = 30 Number of balls having boundary = 6 Therefore, the number of balls not having boundary = 30 - 6 = 24 P (she did not hit a boundary) = 24/30 = 4/5 = 0.8 Hence, the probability of not hitting a boundary by her is 0.8.

Total number of balls = 30
Number of balls having boundary = 6
Therefore, the number of balls not having boundary = 30 – 6 = 24
P (she did not hit a boundary) = 24/30 = 4/5 = 0.8
Hence, the probability of not hitting a boundary by her is 0.8.

Example 5, section 14.4, chapter 14 is given below: In a particular section of class ix, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained: Total number of student = 40 Number of students born in August = 6 Therefore, p(Student bornRead more

Example 5, section 14.4, chapter 14 is given below:
In a particular section of class ix, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:
Total number of student = 40
Number of students born in August = 6
Therefore,
p(Student born in August) = 6/40 = 3/20
Hence, the probability that the student born in August is 3/20.

The frequency table of the two-wheelers, three-wheelers and four-wheelers going past during 2:30 pm to 3:30 pm in front of our school is given below: Vehicle Two-wheelers Three-wheelers Four-wheelers Total Frequency 12 7 9 28 Total number of vehicles = 28 and number of two- wheelers = 12 Therefore,Read more

The frequency table of the two-wheelers, three-wheelers and four-wheelers going past during 2:30 pm to 3:30 pm in front of our school is given below:
Vehicle Two-wheelers Three-wheelers Four-wheelers Total
Frequency 12 7 9 28
Total number of vehicles = 28 and number of two- wheelers = 12
Therefore,
P(two – wheelers ) = 12/28 = 3/7
Hence, the probability of two- wheelers is 3/7.

Given 3- digit numbers = 100, 101, 102 ... 999 and total number of 3-digit numbers = 999 - 99 = 900 Numbers divisible by 3 = 102, 105, 108 ... 999. So, the number of numbers divisible by 3 = 300 Therefore, P(number divisible by 3) = 300/900 = 1/3 Hence, the probability that the number written by herRead more

Given 3- digit numbers = 100, 101, 102 … 999 and total number of 3-digit numbers = 999 – 99 = 900
Numbers divisible by 3 = 102, 105, 108 … 999. So, the number of numbers divisible by 3 = 300
Therefore,
P(number divisible by 3) = 300/900 = 1/3
Hence, the probability that the number written by her/him is divisible by 3 is 1/3.

Arranging the data in ascending order: 4.97, 4.98, 5.00. 5.03, 5.04, 5.05, 5.06, 5.07, 5.08, Total number of flour bang = 11 and number of bags containing more than 5 kg of flour = 7 Therefore, P(A bag contains more than 5 kg of flour) = 7/11 Hence, the probality that any of these bags chosen at ranRead more

Arranging the data in ascending order: 4.97, 4.98, 5.00. 5.03, 5.04, 5.05, 5.06, 5.07, 5.08,
Total number of flour bang = 11 and number of bags containing more than 5 kg of flour = 7
Therefore,
P(A bag contains more than 5 kg of flour) = 7/11
Hence, the probality that any of these bags chosen at random contains more than 5 kg of flour is 7/11.

## The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m² is Rs 15000, find the height of the hall.

Let, the lenght of hall = l m, breadth = b m and height = h m Permeter of floor = 2(l + b) According of question, 2(l + b) = 250 m ...(1) Area of four walls of hall = 2(l + b) h So, the cost of painting the four walls at the rate of Rs 10 per m² = 2(l + b) h × Rs 10 = 250h × Rs 10 [∵ From equation (Read more

Let, the lenght of hall = l m, breadth = b m and height = h m

See lessPermeter of floor = 2(l + b)

According of question, 2(l + b) = 250 m …(1)

Area of four walls of hall = 2(l + b) h

So, the cost of painting the four walls at the rate of Rs 10 per m² = 2(l + b) h × Rs 10

= 250h × Rs 10 [∵ From equation (1)]

= Rs 2500h

According to question, Rs 2500h = 15000

⇒ h = 15000/2500 = 6 m

Hence, the height of the hall is 6 m.

## The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Lenght of one brick l = 22.5 cm, breadth b = 10 cm and height h = 7.5 cm Total surface area of brick = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm² = 2(225 + 75 + 168.75) cm² = 937.5 cm² = 0.09375 m² Number of bricks to be painted = Total paint available/paint for one brick = 9.375 m²/Read more

Lenght of one brick l = 22.5 cm, breadth b = 10 cm and height h = 7.5 cm

See lessTotal surface area of brick = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²

= 2(225 + 75 + 168.75) cm² = 937.5 cm² = 0.09375 m²

Number of bricks to be painted

= Total paint available/paint for one brick = 9.375 m²/0.09375 m² = 100

Hence, the paint of this container can paint 100 bricks.

## Give five examples of data that you can collect from your day-to-day life.

Five examples of data that we can collect from our day-to-day life. 1. Height of our classmates or weight of our classmate. 2. Height of first 100 plants near by our locality. 3. Maximum or minimum temperature of a particular month. 4. Time spend for watching TV in a particular week. 5. Rainfall inRead more

Five examples of data that we can collect from our day-to-day life.

See less1. Height of our classmates or weight of our classmate.

2. Height of first 100 plants near by our locality.

3. Maximum or minimum temperature of a particular month.

4. Time spend for watching TV in a particular week.

5. Rainfall in our city in last 10 years.

## In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Total number of balls = 30 Number of balls having boundary = 6 Therefore, the number of balls not having boundary = 30 - 6 = 24 P (she did not hit a boundary) = 24/30 = 4/5 = 0.8 Hence, the probability of not hitting a boundary by her is 0.8.

Total number of balls = 30

See lessNumber of balls having boundary = 6

Therefore, the number of balls not having boundary = 30 – 6 = 24

P (she did not hit a boundary) = 24/30 = 4/5 = 0.8

Hence, the probability of not hitting a boundary by her is 0.8.

## Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.

Example 5, section 14.4, chapter 14 is given below: In a particular section of class ix, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained: Total number of student = 40 Number of students born in August = 6 Therefore, p(Student bornRead more

Example 5, section 14.4, chapter 14 is given below:

See lessIn a particular section of class ix, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:

Total number of student = 40

Number of students born in August = 6

Therefore,

p(Student born in August) = 6/40 = 3/20

Hence, the probability that the student born in August is 3/20.

## Activity : Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

The frequency table of the two-wheelers, three-wheelers and four-wheelers going past during 2:30 pm to 3:30 pm in front of our school is given below: Vehicle Two-wheelers Three-wheelers Four-wheelers Total Frequency 12 7 9 28 Total number of vehicles = 28 and number of two- wheelers = 12 Therefore,Read more

The frequency table of the two-wheelers, three-wheelers and four-wheelers going past during 2:30 pm to 3:30 pm in front of our school is given below:

See lessVehicle Two-wheelers Three-wheelers Four-wheelers Total

Frequency 12 7 9 28

Total number of vehicles = 28 and number of two- wheelers = 12

Therefore,

P(two – wheelers ) = 12/28 = 3/7

Hence, the probability of two- wheelers is 3/7.

## Activity : Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Given 3- digit numbers = 100, 101, 102 ... 999 and total number of 3-digit numbers = 999 - 99 = 900 Numbers divisible by 3 = 102, 105, 108 ... 999. So, the number of numbers divisible by 3 = 300 Therefore, P(number divisible by 3) = 300/900 = 1/3 Hence, the probability that the number written by herRead more

Given 3- digit numbers = 100, 101, 102 … 999 and total number of 3-digit numbers = 999 – 99 = 900

See lessNumbers divisible by 3 = 102, 105, 108 … 999. So, the number of numbers divisible by 3 = 300

Therefore,

P(number divisible by 3) = 300/900 = 1/3

Hence, the probability that the number written by her/him is divisible by 3 is 1/3.

## Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00 Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Arranging the data in ascending order: 4.97, 4.98, 5.00. 5.03, 5.04, 5.05, 5.06, 5.07, 5.08, Total number of flour bang = 11 and number of bags containing more than 5 kg of flour = 7 Therefore, P(A bag contains more than 5 kg of flour) = 7/11 Hence, the probality that any of these bags chosen at ranRead more

Arranging the data in ascending order: 4.97, 4.98, 5.00. 5.03, 5.04, 5.05, 5.06, 5.07, 5.08,

See lessTotal number of flour bang = 11 and number of bags containing more than 5 kg of flour = 7

Therefore,

P(A bag contains more than 5 kg of flour) = 7/11

Hence, the probality that any of these bags chosen at random contains more than 5 kg of flour is 7/11.