516 + Interview Questions in MENSURATION IN GENERAL AWARENESS Page 8 InterviewSolution

351.	Find the total surface area of a cuboid with l= 4m, b = 3m and h = 1.5 m
Answer» T.S.A = 2 (lb + bh + lh) = 2(4 × 3 + 3 × 1.5 + 1.5 × 4) = 2(12 + 4.5 + 6) = 2(22.5) T.S.A = 45 m²

351.

Find the total surface area of a cuboid with l= 4m, b = 3m and h = 1.5 m

Answer»

T.S.A = 2 (lb + bh + lh)

= 2(4 × 3 + 3 × 1.5 + 1.5 × 4)

= 2(12 + 4.5 + 6)

= 2(22.5)

T.S.A = 45 m²

Discussion

352.	Find the area of Fig. as the sum of the areas of two trapezium and a rectangle.
Answer» Area of figure = Area of two trapeziums + area of rectangle Length of rectangle = 50 cm Breadth of rectangle = 10 cm Length of parallel sides of trapezium = 30 cm and 10 cm Distance between parallel sides of trapezium = \(\frac{70-50}{2}\) = \(\frac{20}{2}\) = 10 cm Area of figure = 2 x \(\frac{1}{2}\)(sum of sides) x distance between parallel sides + length x breadth Area of figure = 2 × \(\frac{1}{2}\)(30 + 10) x 10 + 50 x 10 Area of figure = 400 + 500 = 900 cm²

352.

Find the area of Fig. as the sum of the areas of two trapezium and a rectangle.

Answer»

Area of figure = Area of two trapeziums + area of rectangle

Length of rectangle = 50 cm

Breadth of rectangle = 10 cm

Length of parallel sides of trapezium = 30 cm and 10 cm

Distance between parallel sides of trapezium = \(\frac{70-50}{2}\) = \(\frac{20}{2}\) = 10 cm

Area of figure = 2 x \(\frac{1}{2}\)(sum of sides) x distance between parallel sides + length x breadth

Area of figure = 2 × \(\frac{1}{2}\)(30 + 10) x 10 + 50 x 10

Area of figure = 400 + 500 = 900 cm²

Discussion

353.	Find the area enclosed by each of the following figures [fig. (i) - (ii)] as the sum of the areas of a rectangle and a trapezium.
Answer» Figure (i) Area of given figure = Area of trapezium + area of rectangle Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth Area of given figure = \(\frac{1}{2}\)x (18 + 7) x 8 + 18 x 18 Area of given figure = 100 +324 = 424 Therefore area of given figure is 424 cm² Figure (ii) Area of given figure = Area of trapezium + area of rectangle Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth Area of given figure = \(\frac{1}{2}\)x (15 + 6) x 8 + 15 x 20 Area of given figure = 84 + 300 = 384 Therefore area of given figure is 384 cm² Figure (iii) Using pythagorous theorem in the right angled triangle 5² = 4² + x² x² = 25 – 16 x² = 9 x = 3 cm Area of given figure = Area of trapezium + area of rectangle Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth Area of given figure = \(\frac{1}{2}\)x (14 + 6) x 3 + 4 x 6 Area of given figure = 30 + 24 = 54 Therefore area of given figure is 54 cm²

353.

Find the area enclosed by each of the following figures [fig. (i) - (ii)] as the sum of the areas of a rectangle and a trapezium.

Answer»

Figure (i)

Area of given figure = Area of trapezium + area of rectangle

Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth

Area of given figure = \(\frac{1}{2}\)x (18 + 7) x 8 + 18 x 18

Area of given figure = 100 +324 = 424

Therefore area of given figure is 424 cm²

Figure (ii)

Area of given figure = Area of trapezium + area of rectangle

Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth

Area of given figure = \(\frac{1}{2}\)x (15 + 6) x 8 + 15 x 20

Area of given figure = 84 + 300 = 384

Therefore area of given figure is 384 cm²

Figure (iii)

Using pythagorous theorem in the right angled triangle

5² = 4² + x²

x² = 25 – 16

x² = 9

x = 3 cm

Area of given figure = Area of trapezium + area of rectangle

Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth

Area of given figure = \(\frac{1}{2}\)x (14 + 6) x 3 + 4 x 6

Area of given figure = 30 + 24 = 54

Therefore area of given figure is 54 cm²

Discussion

354.	If the angle of a sector is `30^(@)` and the radius of the sector is 21 cm, then length of the arc of the sector is ______.A. 9 cmB. 11 cmC. 10 cmD. 13 cm
Answer» Correct Answer - B

Discussion

355.	There is a pentagonal shaped park as shown in Fig. Jyoti and Kavita divided it in two different ways.Find the area of this park using both ways. Can you suggest some another way of finding its area?
Answer» Area of given figure = Area of trapezium + area of rectangle Area of Jyoti’s diagram = 2 x \(\frac{1}{2}\) x (sum of parallel sides) x altitude Area of given figure = 2 x \(\frac{1}{2}\) x (15 + 30) x 7.5 Area of given figure = 337.5 cm² Therefore area of given figure is 54 cm² Area of given pentagon = Area of triangle + area of rectangle Area of given pentagon = \(\frac{1}{2}\) x base x altitude + length x breadth Area of given pentagon = \(\frac{1}{2}\) x 15 x 15 + 15 x 15 Area of given pentagon = 112.5 + 225 = 337.5 Therefore area of given pentagon is 337.5 cm²

355.

There is a pentagonal shaped park as shown in Fig. Jyoti and Kavita divided it in two different ways.Find the area of this park using both ways. Can you suggest some another way of finding its area?

Answer»

Area of given figure = Area of trapezium + area of rectangle

Area of Jyoti’s diagram = 2 x \(\frac{1}{2}\) x (sum of parallel sides) x altitude

Area of given figure = 2 x \(\frac{1}{2}\) x (15 + 30) x 7.5

Area of given figure = 337.5 cm²

Therefore area of given figure is 54 cm²

Area of given pentagon = Area of triangle + area of rectangle

Area of given pentagon = \(\frac{1}{2}\) x base x altitude + length x breadth

Area of given pentagon = \(\frac{1}{2}\) x 15 x 15 + 15 x 15

Area of given pentagon = 112.5 + 225 = 337.5

Therefore area of given pentagon is 337.5 cm²

Discussion

356.	There is a regular hexagon MNOPQR of side 5 cm . aman and ridhima divided it into two different ways . Find the area of this hexagon both ways.
Answer» for ridhima total area = area 1 + area 2 + area 3 `= 2 xx area 1` + area 2 `= 2 xx 1/2 xx 8 xx 3 + 8 xx 5` `= 24+40 = 64 cm^2` for aman total area = area of 1 + area of 2 = area of `1` `= 2 xx 1/2 xx (5 +11) xx 4` `= 64 cm^2` answer

Discussion

357.	Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm. AO = 60 cm and AD = 90 cm.
Answer» AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm given LM = AM – AL = 20 – 10 = 10 cm MN = AN – AM = 50 – 20 = 30 cm OD = AD – AO = 90 – 60 = 30 cm ON = AO – AN = 60 – 50 = 10 cm DN = OD + ON = 30 + 10 = 40 cm OM = MN + ON = 30 + 10 = 40 cm LN = LM + MN = 10 + 30 = 40 cm Area of given figure = area of triangle AMF + area of trapezium FMNE + area of triangle END + area of triangle ALB + area of trapezium LBCN + area of triangle DNC Area of right angled triangle = \(\frac{1}{2}\)x base x altitude Area of trapezium = \(\frac{1}{2}\)x (sum of parallel sides) x altitude Area of given hexagon = \(\frac{1}{2}\)x AM x FM + \(\frac{1}{2}\) x (MF + OE) x OM + \(\frac{1}{2}\) x OD x OE + \(\frac{1}{2}\)x AL x BL + \(\frac{1}{2}\)x (BL + CN) x LN + \(\frac{1}{2}\)x DN x CN Area of given hexagon = \(\frac{1}{2}\) x 20 x 20 + \(\frac{1}{2}\) x (20 + 60) x 40 + \(\frac{1}{2}\) x 30 x 60 + \(\frac{1}{2}\)x 10 x 30 + \(\frac{1}{2}\)x (30 + 40) x 40 + \(\frac{1}{2}\)x 40 x 40 Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050 cm² Therefore area of given hexagon is 5050 cm²

357.

Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm. AO = 60 cm and AD = 90 cm.

Answer»

AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm given

LM = AM – AL = 20 – 10 = 10 cm

MN = AN – AM = 50 – 20 = 30 cm

OD = AD – AO = 90 – 60 = 30 cm

ON = AO – AN = 60 – 50 = 10 cm

DN = OD + ON = 30 + 10 = 40 cm

OM = MN + ON = 30 + 10 = 40 cm

LN = LM + MN = 10 + 30 = 40 cm

Area of given figure = area of triangle AMF + area of trapezium FMNE + area of triangle END + area of triangle ALB + area of trapezium LBCN + area of triangle DNC

Area of right angled triangle = \(\frac{1}{2}\)x base x altitude

Area of trapezium = \(\frac{1}{2}\)x (sum of parallel sides) x altitude

Area of given hexagon = \(\frac{1}{2}\)x AM x FM + \(\frac{1}{2}\) x (MF + OE) x OM + \(\frac{1}{2}\) x OD x OE + \(\frac{1}{2}\)x AL x BL + \(\frac{1}{2}\)x (BL + CN) x LN + \(\frac{1}{2}\)x DN x CN

Area of given hexagon = \(\frac{1}{2}\) x 20 x 20 + \(\frac{1}{2}\) x (20 + 60) x 40 + \(\frac{1}{2}\) x 30 x 60 + \(\frac{1}{2}\)x 10 x 30 + \(\frac{1}{2}\)x (30 + 40) x 40 + \(\frac{1}{2}\)x 40 x 40

Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050 cm²

Therefore area of given hexagon is 5050 cm²

Discussion

358.	Find the area of the following regular hexagon
Answer» NQ = 23 cm given ND + QC = 23 -13 = 10 ND = QC = 5 cm OM = 24 cm Area of given hexagon = 2 times area of triangle MNO + area of rectangle MOPR Area of right angled triangle = \(\frac{1}{2}\)x base x altitude Area of rectangle = length × breadth Area of regular hexagon = \(\frac{1}{2}\)x OM x ND + MO x OP + \(\frac{1}{2}\) x RP x QC Area of given hexagon = \(\frac{1}{2}\) x 24 x 5 + 24 x 13 + \(\frac{1}{2}\) x 24 x 5 = 432 cm² Therefore area of given hexagon is 432 cm²

358.

Find the area of the following regular hexagon

Answer»

NQ = 23 cm given

ND + QC = 23 -13 = 10

ND = QC = 5 cm

OM = 24 cm

Area of given hexagon = 2 times area of triangle MNO + area of rectangle MOPR

Area of right angled triangle = \(\frac{1}{2}\)x base x altitude

Area of rectangle = length × breadth

Area of regular hexagon = \(\frac{1}{2}\)x OM x ND + MO x OP + \(\frac{1}{2}\) x RP x QC

Area of given hexagon = \(\frac{1}{2}\) x 24 x 5 + 24 x 13 + \(\frac{1}{2}\) x 24 x 5 = 432 cm²

Therefore area of given hexagon is 432 cm²

Discussion

359.	Find the volume of a cuboid whose(i) length = 12 cm, breadth = 8 cm, height = 6 cm(ii) length = 1.2 m, breadth = 30 cm, height = 15 cm(iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm.
Answer» (i) Length of cuboid = 12 cm Breadth of cuboid = 8 cm Height of cuboid = 6 cm Hence, Volume of cuboid = length × breadth × height = 12 × 8 ×6 = 576 cm³ (ii) Length = 1.2 m = 120 cm Breadth = 30 cm Height = 15 cm Hence, Volume of cuboid = 120 × 30 × 15 = 54000 cm³ (iii) Length = 15 cm Breadth = 2.5 dm = 25 cm Height = 8 cm Hence, Volume of cuboid = 15 × 25 × 8 = 3000 cm³

359.

Find the volume of a cuboid whose(i) length = 12 cm, breadth = 8 cm, height = 6 cm(ii) length = 1.2 m, breadth = 30 cm, height = 15 cm(iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm.

Answer»

(i) Length of cuboid = 12 cm

Breadth of cuboid = 8 cm

Height of cuboid = 6 cm

Hence,

Volume of cuboid = length × breadth × height = 12 × 8 ×6 = 576 cm³

(ii) Length = 1.2 m = 120 cm

Breadth = 30 cm

Height = 15 cm

Hence,

Volume of cuboid = 120 × 30 × 15 = 54000 cm³

(iii) Length = 15 cm

Breadth = 2.5 dm = 25 cm

Height = 8 cm

Hence,

Volume of cuboid = 15 × 25 × 8 = 3000 cm³

Discussion

360.	Find the volume of a cube whose side is(i) 4 cm(ii) 8 cm(iii) 1.5 dm(iv) 1.2 m(v) 25 mm
Answer» (i) Given, side of cube = 4 cm volume of cube = (side)³ = 4³ = 64 cm³ (ii) side of cube = 8 cm volume of cube =(side)³ = 8³ = 512 cm³ (iii) side of cube = 1.5 dm volume of cube = (side)³= 1.5³ = 3.375 dm³ = 3375 cm³ (iv) side of cube = 1.2 m volume of cube = 1.2 3 = 1.728 m³ (v) side of cube = 25 mm volume of cube = 25³ = 15625 mm³ = 15.625 cm³

360.

Find the volume of a cube whose side is(i) 4 cm(ii) 8 cm(iii) 1.5 dm(iv) 1.2 m(v) 25 mm

Answer»

(i) Given,

side of cube = 4 cm

volume of cube = (side)³ = 4³ = 64 cm³

(ii) side of cube = 8 cm

volume of cube =(side)³ = 8³ = 512 cm³

(iii) side of cube = 1.5 dm

volume of cube = (side)³= 1.5³ = 3.375 dm³ = 3375 cm³

(iv) side of cube = 1.2 m

volume of cube = 1.2 3 = 1.728 m³

(v) side of cube = 25 mm

volume of cube = 25³ = 15625 mm³ = 15.625 cm³

Discussion

361.	Find the height of a cuboid of volume 100 cm3, whose length and breadth are 5 cm and 4 cm respectively.
Answer» Given, Volume of cuboid = 100 cm³ Length = 5 cm Breadth = 4 cm Let height of cuboid = h cm So, = l x b x h = 100 cm = h = \(\frac{100}{5\times 4}\) = 5 cm

361.

Find the height of a cuboid of volume 100 cm3, whose length and breadth are 5 cm and 4 cm respectively.

Answer»

Given, Volume of cuboid = 100 cm³

Length = 5 cm

Breadth = 4 cm

Let height of cuboid = h cm

So,

= l x b x h = 100 cm

= h = \(\frac{100}{5\times 4}\) = 5 cm

Discussion

362.	A cuboidal vessel is 10 cm long and 5 cm wide. How high it must be made to hold 300 cm3 of a liquid?
Answer» Given, Length of cuboidal vessel = 10 cm Width = 5 cm Volume of liquid in it = 300 cm³ Let height of vessel = h cm So, = l x b x h = 300 = h = \(\frac{300}{10\times 5}\) = 6 cm

362.

A cuboidal vessel is 10 cm long and 5 cm wide. How high it must be made to hold 300 cm3 of a liquid?

Answer»

Given,

Length of cuboidal vessel = 10 cm

Width = 5 cm

Volume of liquid in it = 300 cm³

Let height of vessel = h cm

So,

= l x b x h = 300

= h = \(\frac{300}{10\times 5}\) = 6 cm

Discussion

363.	The volume of cone of diameter’d’ and height ‘h’ is ……… units3.A) πd2/2B) πd2h/12C) πdh2/12D) πd2/12
Answer» Correct option is: B) \(\frac{\pi d^2 h}{12}\) Diameter of the cone is d. \(\therefore\) Radius of the cone is r = \(\frac d2\) Height of the cone is h. \(\therefore\) Volume of cone = \(\frac 13 \pi r^2h\) = \(\frac 13 \pi \) (\(\frac d2\))\(^2\) h = \(\frac {\pi \, d^2h}{12}\) cubic units. Correct option is: B) \(\frac{\pi\,d^2h}{12}\)

363.

The volume of cone of diameter’d’ and height ‘h’ is ……… units3.A) πd2/2B) πd2h/12C) πdh2/12D) πd2/12

Answer»

Correct option is: B) \(\frac{\pi d^2 h}{12}\)

Diameter of the cone is d.

\(\therefore\) Radius of the cone is r = \(\frac d2\)

Height of the cone is h.

\(\therefore\) Volume of cone = \(\frac 13 \pi r^2h\) = \(\frac 13 \pi \) (\(\frac d2\))\(^2\) h = \(\frac {\pi \, d^2h}{12}\) cubic units.

Correct option is: B) \(\frac{\pi\,d^2h}{12}\)

Discussion

364.	Geometry box is an example of A) sphere B) cone C) cube D) cuboid
Answer» Correct option is: D) cuboid Geometry box is an example of cuboid. Correct option is: D) cuboid

364.

Geometry box is an example of A) sphere B) cone C) cube D) cuboid

Answer»

Correct option is: D) cuboid

Geometry box is an example of cuboid.

Correct option is: D) cuboid

Discussion

365.	The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres.
Answer» l = 20.5 m b = 16 m h = 8 m ∴ Capacity = Volume = lb h = (20.5 x 16 x 8) m³ = 2624 m³ 1 m³ = 1000 litres ∴ 2624 m³ = 2624000 litres

365.

The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres.

Answer»

l = 20.5 m

b = 16 m

h = 8 m

∴ Capacity = Volume = lb h = (20.5 x 16 x 8) m³

= 2624 m³

1 m³ = 1000 litres

∴ 2624 m³ = 2624000 litres

Discussion

366.	A cubical milk tank holds 125000 litres of milk. Find the length of its side in metres.
Answer» V = a³ = 125000 litres = 125 m³ a = \(\sqrt[3]{125m^3}=\sqrt[3]{5\times\,5\times\,5}\) ∴ length of its side = 5 m.

366.

A cubical milk tank holds 125000 litres of milk. Find the length of its side in metres.

Answer»

V = a³ = 125000 litres = 125 m³

a = \(\sqrt[3]{125m^3}=\sqrt[3]{5\times\,5\times\,5}\)

∴ length of its side = 5 m.

Discussion

367.	Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volumes V1 and V2 of the cubes and compare them.
Answer» Given, Edge of one cube a₁= 2 cm Edge of second cube a₂ = 4 cm Hence , = v₁ = 2³= 8 cm³ = v₂ = 4³= 64 cm³ = v₂ = 8v₁

367.

Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volumes V1 and V2 of the cubes and compare them.

Answer»

Given,

Edge of one cube a₁= 2 cm

Edge of second cube a₂ = 4 cm

Hence ,

= v₁ = 2³= 8 cm³

= v₂ = 4³= 64 cm³

= v₂ = 8v₁

Discussion

368.	A tea - packet measures 10 cm × 6 cm × 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm × 30 cm × 0.2 m?
Answer» Given, Dimensions of tea packet = 10 cm × 6 cm × 4 cm Dimension of cardboard box = 50 cm × 30 cm × 0.2 m So, Number of tea packets can be put in cardboard box = \(\frac{volume\,of\,cardboard\,box}{volume\, of \,tea\,packet}\) = \(\frac{50\times 30\times 20}{10\times 6\times 4}\) = 125 packets

368.

A tea - packet measures 10 cm × 6 cm × 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm × 30 cm × 0.2 m?

Answer»

Given,

Dimensions of tea packet = 10 cm × 6 cm × 4 cm

Dimension of cardboard box = 50 cm × 30 cm × 0.2 m

So,

Number of tea packets can be put in cardboard box = \(\frac{volume\,of\,cardboard\,box}{volume\, of \,tea\,packet}\) = \(\frac{50\times 30\times 20}{10\times 6\times 4}\) = 125 packets

Discussion

369.	If the radius of a wheel is 42 cm and it makes 20 rotations in 1 h, then find the time taken by the wheel to cover a distance of 528 m. The following steps are involved in solving the above problem. Arrange them in sequential order. (A) Time taken to complete 200 rotations `=200xx(1)/(2)=10h` (B) The number of rotations required to cover a distance of 528m `=("Distance")/("Circumference")=(528m)/(264cm)=(52800)/(264)=200` (C) The distance covered by the wheel in `"1 rotation"="Circumference of the wheel"=` `2pir=2xx(22)/(7)xx42=264cm` (D) Time taken to make 20 rotations = 1 hA. CBDAB. CDBAC. BCDAD. CABD
Answer» Correct Answer - A (C), (B), (D), and (A) is the required sequential order.

Discussion

370.	The angle of a sector of a circle is `108^(@)` and radius of the circle is 15 m. The ratio of the circumference of the circle and the length of the arc of the sector is _______A. `3:10`B. `10:3`C. `1:3`D. `3:1`
Answer» Correct Answer - B `"The required ratio"=2pir:(2pirxx108)/(360)` `=360:108=10:3`

Discussion

371.	Find the cost of levelling a ground which is in the shape of a sector with angle `40^(@)` and radius 63 m at the rate of Rs. 15 per square metre. The following steps are involved in solving the above problem. Arrange them in sequential order. (A)The are of the sector `=(theta)/(360^(@))pir^(2)` (B) Cost of levelling the ground per square metre = Rs. 15 (C) `therefore "The area of the ground"=(40^(@))/(360^(@))xx(22)/(7)xx63xx63=1386m^(2)` (D) Cost levelling `1386m^(2)=1386xx15="Rs."20,790`A. ADBCB. ABCDC. CABDD. ACBD
Answer» Correct Answer - D (A), (C), (B), and (C) is the required sequential order.

Discussion

372.	A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Answer» Volume of the wood in the pen stand = Volume of cuboid – Total volume of three depressions. Length of the cuboid (l) = 15 cm Breadth of the cuboid (b) = 10 cm Height of the cuboid (h) = 3.5 cm Volume of the cuboid (V) = lbh = 15 × 10 × 3.5 = 525 cm³. Radius of each depression (r) = 0.5 cm Height / depth (h) = 1.4 cm Volume of each depressions V = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4 = \(\frac{7.7}{3\times7}\) = \(\frac{1.1}{3}\) cm³ Total volume of the three depressions = 3 × \(\frac{1.1}{3}\) = 1.1 cm³ ∴ Volume of the wood = 525 – 1.1 = 523.9 cm³

372.

A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Answer»

Volume of the wood in the pen stand = Volume of cuboid – Total volume of three depressions.

Length of the cuboid (l) = 15 cm

Breadth of the cuboid (b) = 10 cm

Height of the cuboid (h) = 3.5 cm

Volume of the cuboid (V) = lbh = 15 × 10 × 3.5 = 525 cm³.

Radius of each depression (r) = 0.5 cm Height / depth (h) = 1.4 cm

Volume of each depressions V = \(\frac{1}{3}\)πr²h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4

= \(\frac{7.7}{3\times7}\) = \(\frac{1.1}{3}\) cm³

Total volume of the three depressions = 3 × \(\frac{1.1}{3}\) = 1.1 cm³

∴ Volume of the wood = 525 – 1.1 = 523.9 cm³

Discussion

373.	If the sum of the length and the breadth of rectangle is 1009 cm, then the perimeter of the reactangle is
Answer» Correct Answer - 2018 cm `l+b=1009 cm` Perimeter of the rectangle `=2(l+b)=2xx1009=2018cm`

Discussion

374.	The volumes of two cylinders of radii R, r and heights H, h respectively are equal. If `R:r=2:3,` then `H:h=`_______.
Answer» Correct Answer - `9:4`

Discussion

375.	Bob wants to cover the floor of a room 3 m wide and 4 m longby squared tiles. If each square tile is of side 0.5 m, then find the number of tiles required to cover the floor of the room.
Answer» Area of the floor = sideside =34=12`m^2` Area of the tile = sideside =0.50.5=0.25`m^2` numbers of the tiles required = area of the floor/area of the tile =12/0.25= 48 tiles.

Discussion

376.	Find the area in square metre of a piece of cloth 1m 25 cm wideand 2 m long.
Answer» Here, length of piece of cloth, `l = 1` m `25` cm `= 1.25` m Breadth of piece of cloth, `b = 2` m So, Area of piece of cloth, `A = l*b = 21.25 = 2.5m^2` So, required area will be `2.5` square meters.

Discussion

377.	If the radius of a wheel is 21 cm, then find how many times it should rotate to cover a distance of 264 m. (A) The circumference of the wheel `=2pirxx2xx(22)/(7)xx21` (given r = 21 cm) (C) `therefore` The circumference of the wheel = 132 cm (C) `"Number of rotations"=(26400cm)/(132cm)=200` (D) Number of rotatins `=("Distance to be covered")/("Circumference of the wheel")` (It is given that distance = 264 m and circumference = 132 cm)A. ABDCB. ACBDC. ABCDD. BDAC
Answer» Correct Answer - B (A), (B), (D) and (C) is required sequential order.

Discussion

378.	`{:(,"Column A",,"Column B"),(14.,"Area of a trapezium is_______.",(a),4m^(2)),(15.,"If the volulme of a cube is" 512 cm^(3)", then the lateral surface are is ________.",(b),64 cm),(16., "If the diagonal of a square is" sqrt(8)m", then its area is ________.",(c ),((1)/(2))"(Sum of the parallel sides) (Distance between them)"),(17., "If the radius of the circle of angle" 45^(@)"is ______.",(d),((1)/(2))"(Product of the diagonals)"),(,,(e ),256cm^(2)),(,,(f),19.5 m^(2)):}`
Answer» Correct Answer - C 14.Option (c) : `"Area of trapezium"=((1)/(2))("Sum of the parallel sides")("Distance between them")` 15. Option (e): `"Volume of cube "a^(3)=512` `rArra^(3)=8^(3)rArra=8cm` `"Later surface area"=4a^(2)=4xx8^(2)=4xx64=256cm^(2)` 16. Option (a) : Diagonal of square `(d)=sqrt8cm` `therefore"Area"=(d^(2))/(2)=((sqrt8)^(2))/(4)=4m^(2)` 17. Option (f) : Perimeter of a sector = `l+2r` `=(theta)/(360^(@))2pir+2r=(45^(@))/(360^(@))xx2xx(22)/(7)xx7+2xx7` `=(1)/(2)xx2xx22+14` `=(22)/(4)+14=5.5+14=19.5cm`

Discussion

379.	A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?
Answer» Length of one piece = Perimeter of the square = 4 × 2 cm = 8 cm Let the number of pieces be n. We have given, total area of n squares = 28 cm² ⇒ n (side)² = 28 ⇒ n (2)² = 28 ⇒ n × 4 = 28 ⇒ n = 28/4 = 7 The original length of the wire = Number of pieces × length of one piece = 7 × 8 cm = 56 cm

379.

A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?

Answer»

Length of one piece

= Perimeter of the square

= 4 × 2 cm = 8 cm

Let the number of pieces be n.

We have given, total area of n squares = 28 cm²

⇒ n (side)² = 28

⇒ n (2)² = 28

⇒ n × 4 = 28

⇒ n = 28/4 = 7

The original length of the wire

= Number of pieces × length of one piece

= 7 × 8 cm = 56 cm

Discussion

380.	A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?
Answer» Original length of the wire = 56 cm.

Discussion

381.	Two spheres are of radii in the ratio 3 : 5, then ratio of their surface areas is A) 16 : 1 B) 25 : 9 C) 9 : 16 D) 9 : 25
Answer» Correct option is: D) 9 : 25 Given that \(\frac {r_1}{r_2} = \frac35\) \(\therefore\) \(\frac {S_1}{S_2} = \frac {4\pi r_1^2}{4 \pi r_2^2} = (\frac {r_1}{r_2} )^2 = (\frac 35)^2 = \frac {9}{25} = 9:25\) Hence, the ratio of their surface areas is 9 : 25. Correct option is: D) 9 : 25

381.

Two spheres are of radii in the ratio 3 : 5, then ratio of their surface areas is A) 16 : 1 B) 25 : 9 C) 9 : 16 D) 9 : 25

Answer»

Correct option is: D) 9 : 25

Given that \(\frac {r_1}{r_2} = \frac35\)

\(\therefore\) \(\frac {S_1}{S_2} = \frac {4\pi r_1^2}{4 \pi r_2^2} = (\frac {r_1}{r_2} )^2 = (\frac 35)^2 = \frac {9}{25} = 9:25\)

Hence, the ratio of their surface areas is 9 : 25.

Correct option is: D) 9 : 25

Discussion

382.	A floor is 5 m long and 4 m wide. A square carpet of sides 3m is laid on the floor. Find the area of the floor that is not carpeted.
Answer» Length (l) = 5 m, Breadth (b) = 4 m Area = l × b = 5 × 4 = 20 m² Area covered by the carpet = (side)² = (3)² = 9 m² Area not covered by the carpet = 20 – 9 = 11 m²

382.

A floor is 5 m long and 4 m wide. A square carpet of sides 3m is laid on the floor. Find the area of the floor that is not carpeted.

Answer»

Length (l) = 5 m,

Breadth (b) = 4 m

Area = l × b = 5 × 4 = 20 m²

Area covered by the carpet = (side)² = (3)² = 9 m²

Area not covered by the carpet = 20 – 9 = 11 m²

Discussion

383.	Five square flower beds each of sides 1 m are dug on a piece of land 5m long and 4m wide. What is the area of the remaining part of the land?
Answer» Area of the land = 5 × 4 = 20 m² Area occupied by 5 flower beds = 5 × (side)² = 5 × (1)² = 5m² Areas of the remaining part = 20 – 5 = 15 m²

383.

Five square flower beds each of sides 1 m are dug on a piece of land 5m long and 4m wide. What is the area of the remaining part of the land?

Answer»

Area of the land = 5 × 4 = 20 m²

Area occupied by 5 flower beds = 5 × (side)² = 5 × (1)² = 5m²

Areas of the remaining part = 20 – 5 = 15 m²

Discussion

384.	Four identical strips in the form of right isosceles triangles are removed from the four corners of a square sheet of side 8 cm. Find the area of the remaining sheet, if the length of each, equal sides of the triangles is 4 cm,
Answer» Correct Answer - 32 sq.cm Side of the square sheet `=8 cm` `Area of the square `=s^2=64sq.cm` Side of the isosceles triangel, `(a)=4cm` Area of the isosceles triangle `=(1)/(2)a^2` `=(1)/(2)xx4xx4` `=8sq.cm` Area of the remaining paper. = Area of the square `-4xx` Area of the isosceles triangle `=64-4(8)=64-32` `=32 sq.cm`

Discussion

385.	The side of a square metal sheet is 12 m. Four squares each of side 3 m are cut and removed from the four corners of the sheet and the remaining part is folded to form a cuboid (without top face). Find the volume of the cuboid so formed.A. `108m^(3)`B. `36m^(3)`C. `118m^(3)`D. `54m^(3)`
Answer» Correct Answer - A The required volume of the cuboid `=(12-2xx3)xx(12-2xx3)xx3=6xx6xx3=108m^(3)`

Discussion

386.	Which of the following vessel can be filled with more water (A, B are in cylindrical shape) ?(A) A (B) B (C) Both are equal(D) Cannot be determined
Answer» Correct option is: (B) B

Discussion

387.	The volume of a cube will be ………………… (in cm3), whose total surface area is 216 cm2.(A) 216 (B) 196 (C) 212 (D) 144
Answer» Correct option is: (A) 216 Let side of the cube be a. Given that total surface area of cube is 216 \(cm^2\) \(\therefore\) \(6\, a^2 = 216\) = \(a^2 = \frac {216}6 = 36 = 6^2\) \(\therefore\) a = 6 cm Now, the volume of cube = \(a^3=6^3=216\,cm^3.\) Correct option is: (A) 216

387.

The volume of a cube will be ………………… (in cm3), whose total surface area is 216 cm2.(A) 216 (B) 196 (C) 212 (D) 144

Answer»

Correct option is: (A) 216

Let side of the cube be a.

Given that total surface area of cube is 216 \(cm^2\)

\(\therefore\) \(6\, a^2 = 216\)

= \(a^2 = \frac {216}6 = 36 = 6^2\)

\(\therefore\) a = 6 cm

Now, the volume of cube = \(a^3=6^3=216\,cm^3.\)

Correct option is: (A) 216

Discussion

388.	The inner, length, breadth and height of an open box are 95cm, 75cm and 97.5 cm respectively. If the wood is 25 mm thick the cost of painting outside of the box(the bottom is not to be painted) at 50 paise per 100 sq. cm is :
Answer» inner length`= 95 cm` inner breadth `= 75cm` outer length`= PQ= AB+ 2 + (2.5)` `= 95+5 = 100cm` outer breadth = `AD+ 2(2.5) = 80cm` outer height =`97.5+2.5 = 100cm` Surface area of wall = `2(l+h+ b+h)` `= 2[100 xx 100 + 80 xx 100]` `= 3600 cm^2` total cost =`36000/100 xx 0.5` `= 180rs` answer

Discussion

389.	The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm3)
Answer» Given: For the cuboidal can, length (l) = 20 cm, breadth (b) = 20 cm, height (h) = 30 cm To find: Oil that can be contained in the can. Volume of cuboid = l × b × h = 20 × 20 × 30 = 12000 cm³ = 12000/1000 litres = 12 litres ∴ The oil can will contain 12 litres of oil.

389.

The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm3)

Answer»

Given: For the cuboidal can,

length (l) = 20 cm,

breadth (b) = 20 cm,

height (h) = 30 cm

To find: Oil that can be contained in the can.

Volume of cuboid = l × b × h

= 20 × 20 × 30

= 12000 cm³

= 12000/1000 litres

= 12 litres

∴ The oil can will contain 12 litres of oil.

Discussion

390.	The adjoining figure shows the measures of a Joker’s dap. How much cloth is needed to make such a cap?
Answer» Given: For the conical cap, radius (r) = 10 cm, slant height (l) = 21 cm To find: Cloth required to make the cap. Cloth required to make the cap = Curved surface area of the conical cap = πrl = 22/7 × 10 × 21 =22 × 10 × 3 = 660 cm² ∴ 660 cm² of cloth will be required to make the cap.

390.

The adjoining figure shows the measures of a Joker’s dap. How much cloth is needed to make such a cap?

Answer»

Given: For the conical cap,

radius (r) = 10 cm,

slant height (l) = 21 cm

To find: Cloth required to make the cap.

Cloth required to make the cap = Curved surface area of the conical cap

= πrl = 22/7 × 10 × 21

=22 × 10 × 3

= 660 cm²

∴ 660 cm² of cloth will be required to make the cap.

Discussion

391.	A goat is tied to one corner of a field of dimensions 16 m `xx` 10m with a rope 7 m long. Find the area of the field that the goat can graze. The following are the steps involved in solving the above problem. Arrange them in sequential order. (A) Required area = 38.5 `m^(2)` (B) Area of the field that the goat can graze = Area of the sector of radius 7 m and a sector angle of `90^(@)` (C ) `(90^(@))/(360)xx(22)/(7)xx(7)^(2)`A. ABCB. BCAC. BACD. CBA
Answer» Correct Answer - B BCA

Discussion

392.	The given figure shows a rectangle. Find the area of the shaded region within the rectangle.
Answer» We can provide the points to the given rectangle. Let `ABCD` is the given rectangle. Please refer to video for the complete figure. From the figure, Area of shaded region = Area of rectangle ABCD - (Area of triangle AEF + Area of triangle DEC) We are given `AE = ED` Also, `AD = 12` cm Now, `AE =ED = (AD)/2 = 12/2 = 6` cm So, Area of `Delta AEF(A_1) = 1/2bh` Here, `b = AE = 6` cm, `h = 7` cm `=> A_1 = 1/267 = 21cm^2` Similarly, Area of `Delta DEC(A_2) = 1/267 = 21cm^2` Now, Area of rectangle ` = lb = 127 = 84cm^2` So, Area of shaded region `= 84-(21+21) = 42cm^2`

Discussion

393.	There si a closed rectangular shed of shed of dimensions `10m xx 4m` inside a field. A cow is tied at one corner of outside of the shed with a 6 m long rope. What is the area that the cow can graze in the field?A. 66 `m^(2)`B. 88 `m^(2)`C. 0.8`pi m^(2)`D. 27`pi m^(2)`
Answer» Correct Answer - B Draw the diagram and proceed.

Discussion

394.	In the shown figure, two circles of radii of 7 cm each, are shown. ABCD is rectangle and AD and BC are the radii. Find the area of the shaded region (in `cm^(2)`). A. 20B. 21C. 19D. 18
Answer» Correct Answer - B Required area is the difference of areas of rectangle and sum of areas sum of areas of two sectors.

Discussion

395.	Area of a sector of a circle of radius 15 cm is 30 cm2 . Find the length of the arc of the sector.
Answer» Given: Radius (r) =15 cm, area of sector = 30 cm² To find: Length of the arc (l). Area of sector = ((lengh of the arc) x radius)/2 ∴ 30 = (l x 15)/2 ∴ l = (30 x 2)/5 = 4cm ∴ The lenght of the arc is 4cm.

395.

Area of a sector of a circle of radius 15 cm is 30 cm2 . Find the length of the arc of the sector.

Answer»

Given: Radius (r) =15 cm,

area of sector = 30 cm²

To find: Length of the arc (l).

Area of sector = ((lengh of the arc) x radius)/2

∴ 30 = (l x 15)/2

∴ l = (30 x 2)/5 = 4cm

∴ The lenght of the arc is 4cm.

Discussion

396.	A large sphere of radius 3.5 cm is carved from a cubical solid. Find the difference between their surface areas.A. 122 `cm^(2)`B. 80.5 `cm^(2)`C. 144.5 `cm^(2)`D. 140 `cm^(2)`
Answer» Correct Answer - D Diameter of the sphere = Length of the edge of a cube.

Discussion

397.	The base radius of a conical tent is 120 cm and itsslant height is 750 cm. Find the area of the canvasrequired to make 10 such tents (in m2). (Take pi=3.14)
Answer» Correct Answer - 282.6 `m^(2)`

Discussion

398.	Find the volume of a cube of side 0.01 cm. (A) 1 cm2 (B) 0.001 cm2 (C) 0.0001 cm2(D) 0.000001 cm2
Answer» Correct answer is (D) 0.000001 cm² Volume of cube = (side)³ = (0.01)³ = 0.000001 cm³

398.

Find the volume of a cube of side 0.01 cm. (A) 1 cm2 (B) 0.001 cm2 (C) 0.0001 cm2(D) 0.000001 cm2

Answer»

Correct answer is

(D) 0.000001 cm²

Volume of cube = (side)³

= (0.01)³ = 0.000001 cm³

Discussion

399.	A cylindrical tank of radius 7 m, has water to some level. If 110 cubes of side 7 dm are completely immersed in it, then find the rise in the water level in the tank. (in meters)
Answer» Correct Answer - 0.245

Discussion

400.	A drum is in the shape of a frustrum of a cone with radii 24ft and 15ft and height 5ft is full of water.The drum is emptied into a rectangular tank of base 99ft×43ft.Find the rise in the height of the water level in the tank.
Answer» Correct Answer - `1(3)/(7)`ft

Discussion

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