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  1. Asked: In:

    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?

    Mahesa

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/

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  2. Asked: In:

    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?

    Mahesa

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/

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  3. Asked: In:

    A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

    Mahesa

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/

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  4. Asked: In:

    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?

    Mahesa

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/

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  6. Asked: In:

    The car parking charges near a railway station up to: 4 hours ₹60, 8 hours ₹100, 12 hours ₹140, 24 hours ₹180 Check if the parking charges are in direct proportion to the parking time.

    Mahesa

    C₁ = 60/40 = ₹15 C₂ = 100/80 = ₹12.50 C₃ = 140/12 = ₹11.67 C₄ = 180/24 = ₹7.50 Here, the charges per hour are not same, i.e., C₁≠C₂≠C₃≠C₄ Therefore, the parking charges are not in direct proportion to the parking time. Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video for more answers vist toRead more

    C₁ = 60/40 = ₹15
    C₂ = 100/80 = ₹12.50
    C₃ = 140/12 = ₹11.67
    C₄ = 180/24 = ₹7.50

    Here, the charges per hour are not same, i.e., C₁≠C₂≠C₃≠C₄
    Therefore, the parking charges are not in direct proportion to the parking time.

    Class 8 Maths Chapter 11 Exercise 13.1 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-13/

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