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SAT (Stage-1) Mock Test - 76

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SAT (Stage-1) Mock Test - 76
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Organisms like jellyfish, viperfish, tube worms and sea cucumbers are found in which of the following zones?

    Solution

    Hadopelagic zone extends from 6,000 m (19,686 feet) to the bottom of the deepest parts of the ocean. Invertebrates such as starfish, jellyfish, viperfish, tube worms and sea cucumbers thrive at these depths.

     

  • Question 2
    1 / -0

    Which of the following statements is/are false?

    a. The fusion of a sperm with an egg to form a zygote during sexual reproduction is called fertilisation.
    b. The ovum has a head, a middle piece and a tail.
    c. All multicellular animals start their life from a single cell called zygote.a

    Solution

    Statement b is false . Ovum does not have a head, middle piece and tail. This is the structure of the sperm.

     

  • Question 3
    1 / -0

    Five stimuli give five types of tropism. Which of the following is the stimulus in the process of thigmotropism?

    Solution

    The stimulus in the process of thigmotropism is touch.

     

  • Question 4
    1 / -0

    The undigested and unabsorbed food from small intestine enters into organ B. The organ B absorbs most of the water from the undigested waste food material making it semi-solid. This semisolid waste is stored in part C of organ B for some time. It is passed out from the body through part D in the form of feces.

    What are B, C and D respectively in the above passage?

    Solution

    The organ B, C and D in the above passage are large intestine, rectum and anus, respectively.

     

  • Question 5
    1 / -0

    Which of the following parts occur(s) in the reproductive system of flowering plants as well as that of human females?

    Solution

    Ovary occurs in the reproductive system of flowering plants as well as that of females.

     

  • Question 6
    1 / -0

    If p(x) = x– k and x = 7 is a zero of p(x), then find the value of k.

    Solution

    Zero of a polynomial p(x) is the value c such that p(c) = 0

    Here, p(x) = x2 – k and x = 7 is the zero of p(x).

    ⇒ p (7) = 0

    ⇒ 72 – k = 0

    ⇒ k = 49

     

  • Question 7
    1 / -0

    The graph of y = p(x), for some polynomial p(x), is given below:

    What is the number of zeros of p(x)?

    Solution

    Number of zeros = Number of times the graph intersects or touches the x-axis

    ∴ Number of zeros = 0

     

  • Question 8
    1 / -0

    If A(0, 2), B(-2, -2) and C(2, -2) are plotted on a grid, which type of triangle will be formed?

    Solution

    A(0, 2), B(-2, -2) and C(2, -2)

    Using distance formula, we get

     

  • Question 9
    1 / -0

    Mid-point of the line segment joining (2, 4) and (6, -4) is ________.

    Solution

     

  • Question 10
    1 / -0

    If point (x, 0) is equidistant from points (7, 6) and (9, 4), the value of x is __________.

    Solution

    Solving by distance formula



    Squaring Both Sides,

    ⇒ x2 – 14x + 85 = x2 – 18x + 97

    ⇒ 4x = 12

    ⇒ x = 3

     

  • Question 11
    1 / -0

    The sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 square metres. The total cost of fencing the park at 50 paise per metre is

    Solution

    Assume that the sides of the rectangle are 3a and 2a.
    Area = 3a × 2a = 6a2 = 3750 square metres
    a2 = 625
    ⇒ a = 25 m
    Perimeter = 2 × (3a + 2a) = 2 × 5a = 2 × 5 × 25 = 250 m
    Total cost of fencing = 250 × 1/2 = Rs. 125

     

  • Question 12
    1 / -0

    A√2 metre wide rectangular plank is placed symmetrically on the diagonal of a square of side 8 metres as shown in the figure.

    The area of the plank is

    Solution

    AL = AP = x (say), PL = √2

     x2 + x2 = 2
    2x2 = 2
    ∴ x = 1
    LB = AB - AL = 8 - 1 = 7
    Similarly, MB = 7

    The area of the plank = 7√2 × √2 = 14 m2

     

  • Question 13
    1 / -0

    The radius of a right circular cylinder is increased by 25%. By what percent should its height be changed so that its volume remains the same?

    Solution

    Let the radius be r and the height be h.
    Volume of cylinder = πr2 h
    As radius is increased by 25%
    Hence, the new radius = 1.25 r
    and let the new height be H.
    As volume remains the same,
    So, πr2 h = π(1.25r)2 H

    So, the height should be decreased by 36%.

     

  • Question 14
    1 / -0

    Length and width of a rectangle are in the ratio 5 : 2. If the sum of length and width is 63 units, what will be the width and length of the rectangle respectively?

    Solution

    Let the length and width of a rectangle be 5x and 2x.
    2x + 5x = 63
    7x = 63
    x = 9
    So, width = 2 x 9 = 18 units
    Length = 5 x 9 = 45 units

     

  • Question 15
    1 / -0

    Length of a rectangular field is 100 m and Breadth is 50 m. Find the perimeter and area of the field. Also, find the cost to fence the rectangular field if the cost of fencing per meter is Rs. 80

    Solution

    Length,  = 100 m

    Breadth, b = 50 m

    Perimeter = 2 ( + b) = 2(100 + 50) = 300 m

    Area = 100 × 50 = 5000 m2

    Cost of fencing = Rs. 80 × 300 = Rs. 24000

     

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