mensuration area volumes Model Questions & Answers, Practice Test for ssc mts paper 1 2023

Answer: (a)

Surface Area of rectangular blocks

= 2(3 × 2 + 2 × 1.75 + 3 × 1.75) = 29.5 $m^2$

Paint required for 0.1 mm thickness

= 29.5 × $1/{10,000} = 0.00295 m^3$

Volume of cubical boxes

= ${10}/{100} × {10}/{100} × {10}/{100} = 1/{1000} cm^3$

So boxes required = ${0.00295}/{0.001}$ = 2.95 ≈ 3

Answer: (a)

Volume of block = (6 × 9 × 12) $cm^3 = 648 cm^3$.

Side of largest cube = H.C.F. of 6 cm, 9 cm, 12 cm = 3 cm.

Volume of the cube = (3 × 3 × 3) = 27 $cm^3$.

∴ Number of cubes = $({648}/{27})$ = 24.

Answer: (a)

Let arc BED of a circle with center A and arc BFD is of a circle of centre C.

Side length of the square CD = a

BD is a diagonal of square ABCD of side length = a.

Then, Area of ΔABD = $1/2 × a × a = a^2/2$

Now, Area of sector ABED = $π/4 × (AB)^2 = π/4 (a)^2$

Area of BDEB = $π/4 a^2 - a^2/2 = {(π - 2) a^2}/4$

∴ Area of BEDFB = 2 × ${(π - 2) a^2}/4 = a^2 (π/2 - 1)$

Answer: (c)

We can divide the regular hexagon into 6 equilateral triangles. Since the hexagon is in a circle the radius r is the side of the equilateral triangle.

∴ Area of the hexagon

= 6 × ${√3}/{4} r^2 = {3 √3}/{2} r^2$ sq. units.

Answer: (a)

Let the kerosene level of cylindrical jar be h.

Now, volume of conical vessel = $1/3 π r^2 h$

Since, radius (r) = 2 cm and height (h) = 3cm of conical vessel.

∴ Volume = $1/3$ π × 4 × 3 = 4π

Now, volume of cylinderical jar = π $r^2h = π (b)^2$ h = 4 π h

Now, volume of conical vessel = Volume of cylindrical Jar

⇒4 π = 4 π h

h = 1cm

Hence, kerosene level in jar is 1 cm.