Mensuration Actual Question in CMAT 2024 Slot – 2

  1. QUESTION

Ice-cream, completely filled in a cylinder of diameter 35 cm and height 32 cm, is to be served by completely filling identical disposable cones of diameter 4 cm and height 7 cm. The maximum number of cones that can be used in this way is


A)950
B)1000
C)1050
D)1100

Explanation

(C)

Given:

Diameter of cylinder = 35cm

Height of cylinder = 32cm

Diameter of cone = 4 cm

Height of cone = 7 cm

Formula used:

Volume of cylinder = pi * r ^ 2 * h

Volume of cone = 1/3 * pi * r ^ 2 * h

Calculation:

Radius of cylinder = 35/2 cm

Radius of cone = 4/2 = 2 cm

Now,

Let n persons can be served

Volume of the cylinder = n * (Volume of a disposable cone

= [πα (35/2) × (35/2) x 32]

= n x (π/3 × 4 × 7)

=n = 350 * 3 = 1050



QUESTION-2

    Which two figures will have same area?

    (A) Circle with diameter of 14 cm

    (B) Rectangle with length 12 cm and breadth 10 cm.

    (C) Rectangle with length 14 cm and breadth 11 cm.

    (D) Square with side 12 cm

    Choose the correct answer from the options given below :


    A)A and B only
    B)B and D only
    C)A and C only
    D)A and D only

    Explanation

    (C)

    =A and C only

    (A) Area of the circle with diameter \( 14 \) cm:

    \[ \text{Area} = \pi \times \left( \frac{d}{2} \right)^2 \]

    \[ \text{Area} = \pi \times \left( \frac{14}{2} \right)^2 = 49\pi \, \text{cm}^2 \]

    (C) Area of the rectangle with length \( 14 \) cm and breadth \( 11 \) cm:

    \[ \text{Area} = \text{length} \times \text{breadth} = 14 \times 11 = 154 \, \text{cm}^2 \]

    You’re right; I made a mistake in my previous response. The correct comparison shows that the areas of options (A) and (C) are equal:

    \[ 49\pi \, \text{cm}^2 = 154 \, \text{cm}^2 \]

    So, the correct answer is option (C) – A and C only.



    QUESTION-3

      Given below are two statements

      Statement I -The perimeter of a triangle is greater than the sum of its three medians

      Statement II -In any triangle ABC, if D is any point on BC, then AB+BC+CA > 2 AD

      In the light of the above statements, choose the correct answer from the options given below:


      A)Both Statement I and Statement II are true
      B)Both Statement I and Statement II are false
      C)Statement I is true but Statement II is false
      D)Statement I is false but Statement II is true

      Explanation

      (A)

      =

      Let A, BC be the triangle & D, E & F are midpoints

      Now sum of two sides of a triangle is greater than twice the radius

      So, AB + AC > 2AD

      Same for

      BC + AC > 2CF

      BC + AB > 2BE

      Odd above 3 equations

      2(A.B+BC+AC)> 2(AD + BE + CF)

      AB + BC + AC > AD + BE + CF

      So, perimeters of triangle is greater than sum of its medians

      Statement 1 is correct

      D is a midpoint on BC means AD is a median

      We know that,

      Sum of two sides a triangle is greater the twice of the median

      AB + AC > 2AD

      Since BC is positive number

      AB + AC + BC > 2AD

      So, statement 2 is correct

      = Both 1 & 2 are correct

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