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IBPS RRB Mensuration Quiz 1

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IBPS RRB Mensuration Quiz 1

shape Introduction

Mensuration  is a topic in Geometry which is a branch of mathematics. Mensuration deals with length, area, and volume of different kinds of shape- both 2D and 3D. The article IBPS RRB Mensuration Quiz 1 provides information about Mensuration, an important topic of Mathematics Consists of different types of Mensuration questions with solutions useful for candidates preparing for different competitive examinations perticularly IBPS, SBI PO, SBI Clerks, SSC MTS, SSC CGL, SSC CHSL, RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, CAT and etc.

shape Quiz


1. A right triangular pyramid XYZB is cut from cube as shown in figure. The side of cube is 16 cm. X, Y and Z are mid points of the edges of the cube. What is the total surface area (in \({cm}^{2}\)) of the pyramid?

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    A. 48[√3 + 1]
    B. 24[4 + √3]
    C. 28[6 + √3]
    D. 28[6 + √3]


Answer:Option D

Explanation:

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In the figure;

\({(XY)}^{2}\) = 64 + 64

XY = 8√2 cm

∴ YZ = ZX = 8√2 cm

∴ Base area of the pyramid = \(\frac{√3}{4}\) × 128 = 32√3 \({cm}^{2}\)

Height of rest three surfaces of the pyramid (as in above figure) = 4√2 cm

Area of 3 surfaces of pyramid = 3 × \(\frac{1}{2}\) × 4√2 × 8√2 = 96 \({cm}^{2}\)

Total surface area = 96 + 32√3 = 32 (3 + √3) \({cm}^{2}\)

2. In the given figure, two squares of sides 8 cm and 20 cm are given. What is the area (in cm2) of the shaded part?

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    A. \(\frac{120}{7}\)
    B. \(\frac{160}{7}\)
    C. \(\frac{180}{7}\)
    D. \(\frac{240}{13}\)


Answer:Option B

Explanation:

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Here ∆AXB and ∆AFG will be similar

⇒\(\frac{8}{28}\) =\(\frac{BX}{20}\)

⇒ BX =\(\frac{40}{7}\)

⇒ Area of ∆AXB = \(\frac{1}{2}\) × (AB × BX)

⇒ Area of ∆AXB = ½ × (8 × \(\frac{40}{7}\))

∴ Area of ∆AXB = \(\frac{160}{7}\) \({cm}^{2}\)

3. A hemisphere is kept on top of a cube. Its front view is shown in the given figure. The total height of the figure is 21 cm. The ratio of curved surface area of the hemisphere and total surface area of cube is 11 ∶ 42. What is the total volume (in cm 3) of figure?

    A. 3318.33
    B. 3462.67
    C. 3154.67
    D. 3154.67


Answer:Option B

Explanation:
Curved surface area of hemisphere/Total surface area of cube =\(\frac{2π{r}^{2}}{6{a}^{2}}\) = \(\frac{11}{42}\)

⇒ 14 × \(\frac{22}{7}\)× \({r}^{2}\) = 11\({a}^{2}\)

⇒ a = 2r

And given total height of the figure is 21 cm,

⇒ a + r = 21

⇒ r = 7 cm and a = 14 cm

∴ Volume of the figure = \({a}^{3}\) + \(\frac{2}{3}\)π\({r}^{3}\)

⇒ 143 + \(\frac{2}{3}\) × \(\frac{22}{7}\) ×\({7}^{3}\)

⇒ 2744 + 718.67 = 3462.67 \({cm}^{3}\)

4. The perimeter of a square is 40 cm, find its area?

    A. 100 sq cm
    B. 25 sq cm
    C. 50 sq cm
    D. 160 sq cm


Answer:Option A

Explanation:

From the given data,

Perimeter of a square = 40 cm

⇒ 4 × side = 40 cm

⇒ side = 10 cm

Area of the square = side × side = 10 cm × 10 cm = 100 sq.cm

5. The base of a prism is a trapezium. The length of the parallel sides of the trapezium are 5 cm and 15 cm respectively and the distance between the parallel sides is 12 cm. If the volume of trapezium prism is 1200 cm3. What the total surface area of the trapezoidal prism?

    A. 400 \({cm}^{2}\)
    B. 500 \({cm}^{2}\)
    C. 700 \({cm}^{2}\)
    D. 600 \({cm}^{2}\)


Answer:Option C

Explanation:

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As we know, volume of the trapezium prism = base area(area of trapezium) × height

⇒ 1200 = \(\frac{1}{2}\) × (5 + 15) × 12 × h

⇒ h = 10 cm

∵ Lateral surface area = perimeter of trapezium × height

Total surface area of the trapezoidal prism

⇒ 2 × base area(area of trapezium) + lateral surface area

⇒ 2 × \(\frac{1}{2}\) × (5 + 15) × 12 + perimeter of trapezium × height

⇒ 240 + (15 + 13 + 5 + 13) × 10

⇒ 240 + 460 = 700 \({cm}^{2}\)

∴ Total surface area of the trapezoidal prism = 700 \({cm}^{2}\)

1. Two right circular cylinders of equal volume have their heights in the ratio 1 : 2. The ratio of
their radii is?

    A. √2:1
    B. 2 : 1
    C. 1 : 2
    D. 1 : 4


Answer:Option A

Explanation:
V1 ∶ V2 = 1 ∶ 1
π\({r1}^{2}\)h1 ∶ π\({r2}^{2}\)h2 = 1 ∶ 1
\(\frac{1}{2}\) × \(\frac{{r1}^{2}}{{r2}^{2}}\) × \(\frac{1}{3}\)
=\(\frac{1}{1}\)

r1 ∶ r2 = √2 ∶ 1

2. A hollow iron pipe is 21 cm long and its exterior diameter is 8 cm. If the thickness of the pipe
is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is ?


    A. 3.696 kg
    B. 3.6 kg
    C. 36 kg
    D. 36.9 kg


Answer:Option A

Explanation:
External radius =\(\frac{8}{2}\) = 4 cm
Thickness = 1 cm
Internal Radius = 4 – 1 = 3 cm
Volume of Material = πh (\({R}^{2}\) − \({r}^{2}\)) = \(\frac{22}{7}\) × 21 × (\({4}^{2}\) − \({3}^{2}\)) = 462 \({cm}^{3}\)
Iron → 462 × \(\frac{8}{100}\) kg
= 3.696 kg


3. Two iron sheets spherical in shape each of diameters 6 cm are immersed in the water contained in a cylindrical vessel of radius 6 cm. the level of the water in the vessel will be raised

    A. 1 cm
    B. 2 cm
    C. 3 cm
    D. 6 cm


Answer:Option B

Explanation:
Let water raised be x.
2 × \(\frac{4}{3}\) π × \({3}^{3}\) = π × \({6}^{2}\) × x
72π = π × 36x
x = 2 cm


4. The radii of the base of two cylinders A and B are in the ratio 3 : 2 and their height in the ratio n: 1. If the volume of cylinder A is 3 times that of cylinder B, the value of n is?

    A.\(\frac{4}{3}\)
    B. \(\frac{2}{3}\)
    C. \(\frac{3}{4}\)
    D.\(\frac{3}{2}\)


Answer:Option A

Explanation:

\(π{(3)}^{2}\) × \(\frac{n}{π{(2)}^{2}}\) × 1 = \(\frac{3}{1}\)
\(\frac{9}{4}\) × \(\frac{n}{1}\) = 3
n = \(\frac{4}{3}\)


5. Water is being pumped out through a circular pipe whose internal diameter is 7 cm. If the flow of water is 12 cm per second. How many litres of water is being pumped out in one hour?

    A. 1663.2
    B. 1500
    C. 1747.6
    D. 2000


Answer:Option A

Explanation:
Volume of water flow per second = π\({r}^{2}\)h = \(\frac{22}{7}\) × 72 × 72× 12
= 462 cm³
Volume of water pumped out in 1 hour = 462 × 60 × 60 cm³
= 1663200 cm³ = 1663.2 liters

1. The lateral surface area of a cylinder is 1056 cm² and its height is 16 cm. Find its volume?

    A. 4545 cm³
    B. 4455 cm³
    C. 5445 cm³
    D. 5544 cm³


Answer:Option D

Explanation:
Let Radius = r cm, h = 16 cm
2πrh = 1056
=> 2 × \(\frac{22}{7}\) × r × 16 = 1056
r =\(\frac{21}{2}\) cm
Volume = π\({r}^{2}\)h = \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) × 16 = 5544 \({cm}^{3}\)


2. From a solid cylinder whose height is 12 cm and diameter 10 cm, a conical cavity of same height and same diameter of the base is hollowed out. The volume of the remaining solid is
approximately ?


    A. 942.86
    B. 314.29
    C. 628.57
    D.450.76


Answer:Option C

Explanation:
Volume of solid cylinder = πr2h
Volume of cone =\(\frac{1}{3}\) π\({r}^{2}\)h
Difference = π\({r}^{2}\)h − \(\frac{1}{3}\)π\({r}^{2}\)h = \(\frac{2}{3}\)π\({r}^{2}\)h = 628.57 cubic cm

3. The curved surface area and the total surface area of a cylinder are in the ratio 1 : 2.If the total
surface area of the right cylinder is 616 cm², then its volume is?


    A.1232 cm³
    B. 1848 cm³
    C. 1632 cm³
    D. 1078 cm³


Answer:Option D

Explanation:
\(\frac{2πrh}{{2πrh+ 2π{r}^{2}}}\)
=\(\frac{1}{2}\) × \(\frac{2πrh}{616}\)

2πrh = 308 … (i)
2πrh+ 2π\({r}^{2}\) = 616 … (ii)
From (i) and (ii)
r = 7 … (iii)
From (i) and (iii)
h = 7
Volume = π\({r}^{2}\)h =\(\frac{22}{7}\) × 7 × 7 × 7 = 1078 \({cm}^{3}\)

4. If diagonal of a cube is √12 cm, then its volume in cubic cm is?

    A. 8
    B. 12
    C. 24
    D. \({\sqrt[3]{2}}\)


Answer:Option A

Explanation:
Diagonal = \(\frac{√3a}{2}\)
√3a = 2√3 ,a = 2
Volume = \({a}^{3}\) = 8


5. If the volume of two cubes are in the ratio 27 : 1, the ratio of their edge is ?

    A. 3 : 1
    B. 27 : 1
    C. 1:3
    D. 1:27


Answer:Option A

Explanation:
Ratio of volume = 27 : 1
Ratio of edges =\({\sqrt[3]{27}}\) : \({\sqrt[3]{1}}\)
= 3 ∶ 1



Mensuration – Related Information
Mensuration Practice Set 1
Mensuration Practice Set 2
Mensuration Practice Set 3
Book for Quantitative Aptitude