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Aptitude miscellaneous
  1. If angles of measure (5y + 62°) and (22° + y) are supplementary, then value of y is :








  1. View Hint View Answer Discuss in Forum

    As we know that Sum of two supplementary angles = 180°
    ∴ ( 5y + 62° ) + ( 22° + y ) = 180°
    ⇒ 6y + 84° = 180°
    ⇒ 6y = 180° – 84° = 96°

    Correct Option: A

    As we know that Sum of two supplementary angles = 180°
    ∴ ( 5y + 62° ) + ( 22° + y ) = 180°
    ⇒ 6y + 84° = 180°
    ⇒ 6y = 180° – 84° = 96°

    ∴ y =
    96
    		 = 16°
    6

  1. In, ΔABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ΔADE and ΔABC.












  1. View Hint View Answer Discuss in Forum

    Clearly DE || BC (by converse of BPT)
    ∴ ΔADE ∼ ABC (∠A = ∠A and ∠ADE = ∠B)

    ∴Area (∆ADE)=AD2(Area Theorem)
    Area (∆ABC)AB2

    Correct Option: B

    Clearly DE || BC (by converse of BPT) ∴ ΔADE ∼ ABC (∠A = ∠A and ∠ADE = ∠B)

    ∴area (∆ADE)=AD2(Area Theorem)
    area (∆ABC)AB2

    =AD2=1(∴AB = 2AD)
    (2AD)24

  1. A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Find the height of the tower.










  1. View Hint View Answer Discuss in Forum

    Draw a figure as per given question,
    In ΔACB and PCQ
    ∠C = ∠C (common)
    ∠ABC = ∠PQC (each 90°)
    ∴ ΔACB ∼ ΔPC (AA Similarity)


    Correct Option: C

    In ΔACB and PCQ
    ∠C = ∠C (common)
    ∠ABC = ∠PQC (each 90°)
    ∴ ΔACB ∼ ΔPC (AA Similarity)

    ∴AB=BC
    PQQC

    h=4000
    128

    h = 60m


  1. In the given figure, find the length of BD.












  1. View Hint View Answer Discuss in Forum

    In Δs ADE and ΔABC
    ∠A = ∠A [common]
    ∠ADE = ∠ACB = x° (Given)
    ∴ ΔADE ∼ ΔACB ( AA Similarly)

    Correct Option: A

    In ΔsADE and ΔABC
    ∠A = ∠A [common]
    ∠ADE = ∠ACB = x°(Given)
    ∴ ΔADE ∼ ΔACB (AA Similarly)

    AD = AE(corresponding side of ⁓ ∆s are proportional)
    ACAB

    6 = 9
    13AB

    AB = 39 = 19.5 cm
    2

    Hence BD = AB - AD = 19.5 - 6 = 13.5 cm.

  1. ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120° and ∠BAC = 30°, then ∠BCD is :








  1. View Hint View Answer Discuss in Forum

    As per the given in question , we draw a figure of a quadrilateral ABCD inscribed in a circle with centre O

    Given , ∠COD = 120°
    ∠BAC = 30°

    ∠CAD =
    1
    		 × ∠COD
    2

    ∠CAD =
    1
    		 × 120° = 60°
    2

    Correct Option: B

    As per the given in question , we draw a figure of a quadrilateral ABCD inscribed in a circle with centre O

    Given , ∠COD = 120°
    ∠BAC = 30°

    ∠CAD =
    1
    		 × ∠COD
    2

    ∠CAD =
    1
    		 × 120° = 60°
    2

    ∴ ∠BAD = 90°
    ∴ ∠BCD = 180° - ∠BAD
    ∴ ∠BCD = 180° – 90° = 90°