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 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.