(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work. (ii) Time and distance covered in direct proportion. (iii) It is a direct proportion becauseRead more
(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work.
(ii) Time and distance covered in direct proportion.
(iii) It is a direct proportion because more are of cultivated land will yield morecrops.
(iv) Time and speed are inverse proportion because if time is less, speed is
more.
(v) It is a inverse proportion. If the population of a country increases, the area
of land per person decreases.
Class 8 Maths Chapter 11 Exercise 13.2 Solution in Video
Let distance covered in 5 hours be x km. ∴ 1 hour = 60 minutes ∴ 5 hours = 5 x 60 = 300 minutes Distance (in km) 14 x Time (in minutes) 25 300 Here distance covered and time in direct proportion. ∴ 14/25 = x/300 ⇒ xX25=14x300 ⇒ x= 14x300/25 = 168 km Hence, the distance covered in 5 hours is 168 km.Read more
Let distance covered in 5 hours be x km.
∴ 1 hour = 60 minutes ∴ 5 hours = 5 x 60 = 300 minutes
Distance (in km) 14 x
Time (in minutes) 25 300
Here distance covered and time in direct proportion.
∴ 14/25 = x/300 ⇒ xX25=14×300 ⇒ x= 14×300/25 = 168 km
Hence, the distance covered in 5 hours is 168 km.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Here height of the pole and length of the shadow are in direct proportion. And 1 m = 100 cm 5 m 60 cm = 5 x 100 + 60 = 560 cm 3 m 20 cm = 3 x 100 + 20 = 320 cm 10 m 50 cm = 10 x 100 + 50 = 1050 cm 5 m = 5 x 100 = 500 cm (i). Let the length of the shadow of another pole be x. Height of pole (in cm) 5Read more
Here height of the pole and length of the shadow are in direct proportion.
And 1 m = 100 cm
5 m 60 cm = 5 x 100 + 60 = 560 cm
3 m 20 cm = 3 x 100 + 20 = 320 cm
10 m 50 cm = 10 x 100 + 50 = 1050 cm
5 m = 5 x 100 = 500 cm
(i). Let the length of the shadow of another pole be x.
Height of pole (in cm) 560 1050
Length of shadow (in cm) 320 x
∴ 560/320 = 1050/x ⇒ xX560 = 1050 x 320 ⇒ x=1050×320/560 = 600cm = 6m
Hence, the length of the shadow of another pole is 6 m.
(ii). Let the height of the pole be x.
Height of pole (in cm) 560 x
Length of shadow
(in cm) 320 500
∴ 560/320 = x/500 ⇒ xX320 = 560×500 ⇒ x=560X500/320 = 875 cm = 8 m 75 cm
Hence, the height of the pole is 8 m 75 cm.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Let distance covered in the map be x. Actual distance (in km) 18 72 Distance covered in map (in cm) 1 x Here actual distance and distance covered in the map are in direct proportion. ∴ 18/1 = 72/x ⇒ xX18=72x1 ⇒x=72x1/18 = 4cm Hence, the distance covered in the map is 4 cm. Class 8 Maths Chapter 11 ERead more
Let distance covered in the map be x.
Actual distance (in km) 18 72
Distance covered in map (in cm) 1 x
Here actual distance and distance covered in the map are in direct proportion.
∴ 18/1 = 72/x ⇒ xX18=72×1 ⇒x=72×1/18 = 4cm
Hence, the distance covered in the map is 4 cm.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
(i) Let sugar crystals be x. Weight of sugar (in kg) 2 5 No. of crystals 9x10⁶ x Here weight of sugar and number of crystals are in direct proportion. ∴ 2x/9x10⁶ = 5/x ⇒ xX2=5x9x10⁶ ⇒ x=5x9x10⁶/2 = 22.5x10⁶ =2.25x10⁷ Hence, the number of sugar crystals is 2.25x10⁷. (ii) Let sugar crystals be x. WeigRead more
(i) Let sugar crystals be x.
Weight of sugar (in kg) 2 5
No. of crystals 9×10⁶ x
Here weight of sugar and number of crystals are in direct proportion.
∴ 2x/9×10⁶ = 5/x ⇒ xX2=5x9x10⁶ ⇒ x=5x9x10⁶/2 = 22.5×10⁶ =2.25×10⁷
Hence, the number of sugar crystals is 2.25×10⁷.
(ii) Let sugar crystals be x.
Weight of sugar (in kg) 2 1.2
No. of crystals 9×10⁶ x
Here weight of sugar and number of crystals are in direct proportion.
∴ 2/9×10⁶ = 1.2/x ⇒ xX2=1.2x9x10⁶ ⇒ x=1.2x9x10⁶/2 = 0.6x9x10⁶=5.4×10⁶
Hence, the number of sugar crystals is 5.4×10⁶.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Which of the following are in inverse proportion: (i) The number of workers on a job and the time to complete the job. (ii) The time taken for a journey and the distance travelled in a uniform speed. (iii) Area of cultivated land and the crop harvested. (iv) The time taken for a fixed journey and the speed of the vehicle. (v) The population of a country and the area of land per person.
(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work. (ii) Time and distance covered in direct proportion. (iii) It is a direct proportion becauseRead more
(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work.
(ii) Time and distance covered in direct proportion.
(iii) It is a direct proportion because more are of cultivated land will yield morecrops.
(iv) Time and speed are inverse proportion because if time is less, speed is
more.
(v) It is a inverse proportion. If the population of a country increases, the area
of land per person decreases.
Class 8 Maths Chapter 11 Exercise 13.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Let distance covered in 5 hours be x km. ∴ 1 hour = 60 minutes ∴ 5 hours = 5 x 60 = 300 minutes Distance (in km) 14 x Time (in minutes) 25 300 Here distance covered and time in direct proportion. ∴ 14/25 = x/300 ⇒ xX25=14x300 ⇒ x= 14x300/25 = 168 km Hence, the distance covered in 5 hours is 168 km.Read more
Let distance covered in 5 hours be x km.
∴ 1 hour = 60 minutes ∴ 5 hours = 5 x 60 = 300 minutes
Distance (in km) 14 x
Time (in minutes) 25 300
Here distance covered and time in direct proportion.
∴ 14/25 = x/300 ⇒ xX25=14×300 ⇒ x= 14×300/25 = 168 km
Hence, the distance covered in 5 hours is 168 km.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.
Here height of the pole and length of the shadow are in direct proportion. And 1 m = 100 cm 5 m 60 cm = 5 x 100 + 60 = 560 cm 3 m 20 cm = 3 x 100 + 20 = 320 cm 10 m 50 cm = 10 x 100 + 50 = 1050 cm 5 m = 5 x 100 = 500 cm (i). Let the length of the shadow of another pole be x. Height of pole (in cm) 5Read more
Here height of the pole and length of the shadow are in direct proportion.
And 1 m = 100 cm
5 m 60 cm = 5 x 100 + 60 = 560 cm
3 m 20 cm = 3 x 100 + 20 = 320 cm
10 m 50 cm = 10 x 100 + 50 = 1050 cm
5 m = 5 x 100 = 500 cm
(i). Let the length of the shadow of another pole be x.
Height of pole (in cm) 560 1050
Length of shadow (in cm) 320 x
∴ 560/320 = 1050/x ⇒ xX560 = 1050 x 320 ⇒ x=1050×320/560 = 600cm = 6m
Hence, the length of the shadow of another pole is 6 m.
(ii). Let the height of the pole be x.
Height of pole (in cm) 560 x
Length of shadow
(in cm) 320 500
∴ 560/320 = x/500 ⇒ xX320 = 560×500 ⇒ x=560X500/320 = 875 cm = 8 m 75 cm
Hence, the height of the pole is 8 m 75 cm.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Let distance covered in the map be x. Actual distance (in km) 18 72 Distance covered in map (in cm) 1 x Here actual distance and distance covered in the map are in direct proportion. ∴ 18/1 = 72/x ⇒ xX18=72x1 ⇒x=72x1/18 = 4cm Hence, the distance covered in the map is 4 cm. Class 8 Maths Chapter 11 ERead more
Let distance covered in the map be x.
Actual distance (in km) 18 72
Distance covered in map (in cm) 1 x
Here actual distance and distance covered in the map are in direct proportion.
∴ 18/1 = 72/x ⇒ xX18=72×1 ⇒x=72×1/18 = 4cm
Hence, the distance covered in the map is 4 cm.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/
Suppose 2 kg of sugar contains 9 x 10⁶ crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
(i) Let sugar crystals be x. Weight of sugar (in kg) 2 5 No. of crystals 9x10⁶ x Here weight of sugar and number of crystals are in direct proportion. ∴ 2x/9x10⁶ = 5/x ⇒ xX2=5x9x10⁶ ⇒ x=5x9x10⁶/2 = 22.5x10⁶ =2.25x10⁷ Hence, the number of sugar crystals is 2.25x10⁷. (ii) Let sugar crystals be x. WeigRead more
(i) Let sugar crystals be x.
Weight of sugar (in kg) 2 5
No. of crystals 9×10⁶ x
Here weight of sugar and number of crystals are in direct proportion.
∴ 2x/9×10⁶ = 5/x ⇒ xX2=5x9x10⁶ ⇒ x=5x9x10⁶/2 = 22.5×10⁶ =2.25×10⁷
Hence, the number of sugar crystals is 2.25×10⁷.
(ii) Let sugar crystals be x.
Weight of sugar (in kg) 2 1.2
No. of crystals 9×10⁶ x
Here weight of sugar and number of crystals are in direct proportion.
∴ 2/9×10⁶ = 1.2/x ⇒ xX2=1.2x9x10⁶ ⇒ x=1.2x9x10⁶/2 = 0.6x9x10⁶=5.4×10⁶
Hence, the number of sugar crystals is 5.4×10⁶.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/