Algebra


  1. If the number of vertices, edges and faces of a rectangualr parallelopiped are denoted by v, e and f respectively, the value of (v– e+f )is









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    Vertices of parallel to piped = v = 8
    Edges = e = 12
    Surfaces = f = 6
    ∴ v – e + f = 8 – 12 + 6 = 2

    Correct Option: B

    Vertices of parallel to piped = v = 8
    Edges = e = 12
    Surfaces = f = 6
    ∴ v – e + f = 8 – 12 + 6 = 2


  1. The area of the triangle formed by the graphs of the equations x = 0, 2x+3y = 6 and x+y = 3 is :









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    x = 0 ⇒ Equation of y – axis Putting x = 0 in 2x + 3y = 6
    0 + 3y = 6 ⇒ y = 2
    ∴ Co-ordinates of point of intersection on y – axis
    = (0, 2)
    Again, putting y = 0, x = 3
    ∴ Point of intersection onx – axis = (3, 0)
    In x + y = 3
    Putting x = 0, y = 3
    and on putting y = 0, x = 3

    ∴ Required area = ∆OAC – ∆OAB = OCD =
    1
    × 3 × 3 -
    1
    × 3 × 2
    22

    =
    9
    -
    6
    =
    3
    222

    = 1
    1
    sq. units
    2

    Correct Option: C


    x = 0 ⇒ Equation of y – axis Putting x = 0 in 2x + 3y = 6
    0 + 3y = 6 ⇒ y = 2
    ∴ Co-ordinates of point of intersection on y – axis
    = (0, 2)
    Again, putting y = 0, x = 3
    ∴ Point of intersection onx – axis = (3, 0)
    In x + y = 3
    Putting x = 0, y = 3
    and on putting y = 0, x = 3

    ∴ Required area = ∆OAC – ∆OAB = OCD =
    1
    × 3 × 3 -
    1
    × 3 × 2
    22

    =
    9
    -
    6
    =
    3
    222

    = 1
    1
    sq. units
    2



  1. If 5x + 9y = 5 and 125x³ + 729y3 = 120 then the vlaue of the product of x and y is









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    5x + 9y = 5
    On cubing both sides, (5x)³ + (9y)³ + 3 × 5x × 9y (5x + 9y) = (5)³
    [∵ (a + b)³ = a³ + b³ + 3ab (a + b)]
    ⇒ 125x³ + 729y³ + 135xy × 5 = 125
    ⇒ 120 + 135 × 5xy = 125
    ⇒ 135 × 5xy = 125 – 120 = 5

    ⇒ xy =
    5
    =
    1
    135 × 5135

    Correct Option: B

    5x + 9y = 5
    On cubing both sides, (5x)³ + (9y)³ + 3 × 5x × 9y (5x + 9y) = (5)³
    [∵ (a + b)³ = a³ + b³ + 3ab (a + b)]
    ⇒ 125x³ + 729y³ + 135xy × 5 = 125
    ⇒ 120 + 135 × 5xy = 125
    ⇒ 135 × 5xy = 125 – 120 = 5

    ⇒ xy =
    5
    =
    1
    135 × 5135


  1. A point in the 4th quadrant is 6 unit away from x–axis and 7 unit away from y–axis. The point is at









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    Correct Option: A



  1. The straight line y = 3x must pass through the point :









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    y = 3x, passes through the origin (0, 0).

    Correct Option: A

    y = 3x, passes through the origin (0, 0).