(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
Let the length of model ship be x Length of actual ship (in m) 12 28 Length of model ship (in cm) 9 x Here length of mast and actual length of ship are in direct proportion. ∴ 12/9 = 28/x ⇒ xX12=28x9 ⇒ x=28x9/12=21 cm Hence, the length of the model ship is 21 cm. Class 8 Maths Chapter 11 Exercise 13Read more
Let the length of model ship be x
Length of actual ship (in m) 12 28
Length of model ship (in cm) 9 x
Here length of mast and actual length of ship are in direct proportion.
∴ 12/9 = 28/x ⇒ xX12=28×9 ⇒ x=28×9/12=21 cm
Hence, the length of the model ship is 21 cm.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Let enlarged length of bacteria be x . Actual length of bacteria = 5/5000 = 1/10000 cm=10⁻⁴ Length 5 x Enlarged length 50,000 20,000 Here length and enlarged length of bacteria are in direct proportion. ∴ 5/50000 = x/20000 ⇒ xX50000 = 5x20000 ⇒ x= 5x20000/50000 = 2cm Hence, the enlarged length of baRead more
Let enlarged length of bacteria be x .
Actual length of bacteria = 5/5000 = 1/10000 cm=10⁻⁴
Length 5 x
Enlarged length 50,000 20,000
Here length and enlarged length of bacteria are in direct proportion.
∴ 5/50000 = x/20000 ⇒ xX50000 = 5×20000 ⇒ x= 5×20000/50000 = 2cm
Hence, the enlarged length of bacteria is 2 cm.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Let the number of bottles filled in five hours be x. Hours 1 x Bottles 75 1800 Here ratio of hours and bottles are in direct proportion. ∴ 6/840 = 5/x ⇒ 6Xx = 5x840 ⇒ x=5x840/6 = 700 bottles Hence, machine will fill 700 bottles in five hours. Class 8 Maths Chapter 11 Exercise 13.1 Solution in VideoRead more
Let the number of bottles filled in five hours be x.
Hours 1 x
Bottles 75 1800
Here ratio of hours and bottles are in direct proportion.
∴ 6/840 = 5/x ⇒ 6Xx = 5×840 ⇒ x=5×840/6 = 700 bottles
Hence, machine will fill 700 bottles in five hours.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Let the parts of red pigment mix with 1800 mL base be x. Parts of red pigment 1 x Parts of base 75 1800 Since it is in direct proportion. ∴ 1/75 = x/1800 ⇒ 75Xx=1x1800 ⇒ x= 1x1800/75=24 Hence, with base 1800 mL, 24 parts red pigment should be mixed. Class 8 Maths Chapter 11 Exercise 13.1 Solution inRead more
Let the parts of red pigment mix with 1800 mL base be x.
Parts of red pigment 1 x
Parts of base 75 1800
Since it is in direct proportion.
∴ 1/75 = x/1800 ⇒ 75Xx=1×1800 ⇒ x= 1×1800/75=24
Hence, with base 1800 mL, 24 parts red pigment should be mixed.
Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video
Let the ratio of parts of red pigment and parts of base be a/b Here a₁=1,b₁=8 ⇒ a₁/b₁=1/8=k (say) When a₂=4,b₂=? k= a₂/b₂ ⇒ b₂= a₂/k =4/1/8 = 4/8 =32 When a₃=7,b₃=? k=a₃/b₃ ⇒ a₃=a₃/k = 7/1/8 = 7x8=56 When a₄=12,b₄=? k=a₄/b₄ ⇒ b₄=a₄/k= 12/1/8=12x8 =96 When a₅=20, b₅=? k=a₅/b₅ ⇒ b₅ = a₅/k =20/1/8 = 20Read more
Let the ratio of parts of red pigment and parts of base be a/b
Here a₁=1,b₁=8 ⇒ a₁/b₁=1/8=k (say)
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/
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length if the ship is 28 m, how long is the model ship?
Let the length of model ship be x Length of actual ship (in m) 12 28 Length of model ship (in cm) 9 x Here length of mast and actual length of ship are in direct proportion. ∴ 12/9 = 28/x ⇒ xX12=28x9 ⇒ x=28x9/12=21 cm Hence, the length of the model ship is 21 cm. Class 8 Maths Chapter 11 Exercise 13Read more
Let the length of model ship be x
Length of actual ship (in m) 12 28
Length of model ship (in cm) 9 x
Here length of mast and actual length of ship are in direct proportion.
∴ 12/9 = 28/x ⇒ xX12=28×9 ⇒ x=28×9/12=21 cm
Hence, the length of the model ship is 21 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/
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Let enlarged length of bacteria be x . Actual length of bacteria = 5/5000 = 1/10000 cm=10⁻⁴ Length 5 x Enlarged length 50,000 20,000 Here length and enlarged length of bacteria are in direct proportion. ∴ 5/50000 = x/20000 ⇒ xX50000 = 5x20000 ⇒ x= 5x20000/50000 = 2cm Hence, the enlarged length of baRead more
Let enlarged length of bacteria be x .
Actual length of bacteria = 5/5000 = 1/10000 cm=10⁻⁴
Length 5 x
Enlarged length 50,000 20,000
Here length and enlarged length of bacteria are in direct proportion.
∴ 5/50000 = x/20000 ⇒ xX50000 = 5×20000 ⇒ x= 5×20000/50000 = 2cm
Hence, the enlarged length of bacteria is 2 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/
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Let the number of bottles filled in five hours be x. Hours 1 x Bottles 75 1800 Here ratio of hours and bottles are in direct proportion. ∴ 6/840 = 5/x ⇒ 6Xx = 5x840 ⇒ x=5x840/6 = 700 bottles Hence, machine will fill 700 bottles in five hours. Class 8 Maths Chapter 11 Exercise 13.1 Solution in VideoRead more
Let the number of bottles filled in five hours be x.
Hours 1 x
Bottles 75 1800
Here ratio of hours and bottles are in direct proportion.
∴ 6/840 = 5/x ⇒ 6Xx = 5×840 ⇒ x=5×840/6 = 700 bottles
Hence, machine will fill 700 bottles in five hours.
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/
In Question 2, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Let the parts of red pigment mix with 1800 mL base be x. Parts of red pigment 1 x Parts of base 75 1800 Since it is in direct proportion. ∴ 1/75 = x/1800 ⇒ 75Xx=1x1800 ⇒ x= 1x1800/75=24 Hence, with base 1800 mL, 24 parts red pigment should be mixed. Class 8 Maths Chapter 11 Exercise 13.1 Solution inRead more
Let the parts of red pigment mix with 1800 mL base be x.
Parts of red pigment 1 x
Parts of base 75 1800
Since it is in direct proportion.
∴ 1/75 = x/1800 ⇒ 75Xx=1×1800 ⇒ x= 1×1800/75=24
Hence, with base 1800 mL, 24 parts red pigment should be mixed.
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 mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Let the ratio of parts of red pigment and parts of base be a/b Here a₁=1,b₁=8 ⇒ a₁/b₁=1/8=k (say) When a₂=4,b₂=? k= a₂/b₂ ⇒ b₂= a₂/k =4/1/8 = 4/8 =32 When a₃=7,b₃=? k=a₃/b₃ ⇒ a₃=a₃/k = 7/1/8 = 7x8=56 When a₄=12,b₄=? k=a₄/b₄ ⇒ b₄=a₄/k= 12/1/8=12x8 =96 When a₅=20, b₅=? k=a₅/b₅ ⇒ b₅ = a₅/k =20/1/8 = 20Read more
Let the ratio of parts of red pigment and parts of base be a/b
Here a₁=1,b₁=8 ⇒ a₁/b₁=1/8=k (say)
When a₂=4,b₂=?
k= a₂/b₂ ⇒ b₂= a₂/k =4/1/8 = 4/8 =32
When a₃=7,b₃=?
k=a₃/b₃ ⇒ a₃=a₃/k = 7/1/8 = 7×8=56
When a₄=12,b₄=?
k=a₄/b₄ ⇒ b₄=a₄/k= 12/1/8=12×8 =96
When a₅=20, b₅=?
k=a₅/b₅ ⇒ b₅ = a₅/k =20/1/8 = 20×8=160
Parts of red pigment 1 4 7 12 20
Parts of base 8 32 56 96 160
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/