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)
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/