Mensuration for CTET Exam

Solids
Important Questions

Plan figures

Figures	Perimeter	Area	Image
Triangle	= a + b + c	$\sqrt{s(s-a)(s-b)(s-c)}$ Where s = $\frac{a+b+c}{2}$ $\frac{1}{2}\: bh$
Right angled triangle	$a+b+\sqrt{a^{2}+b^{2}}$	$\frac{1}{2}\: ab$
Equilateral triangle	3a	$\frac{\sqrt{3}}{4}\: a^{2}$
Isosceles triangle	2a + b	$\frac{b}{4}\sqrt{4a^{2}-b^{2}}$
Circle	$2\pi r$	$\pi r^{2}$
Sector of a circle	$\frac{\theta}{360}\times2\pi r+2r$ ( $\theta$ is in degrees)	$\frac{\theta }{360}\times \pi r^{2}$
Square	4a	$a^{2}$
Rectangle	2(l + b)	lb
Trapezium	a + b + c + d	$\frac{1}{2}(a+b)h$
Parallelogram	2(a + b)	bh or ab $sin\theta$
Cyclic quadrilateral	a + b + c + d	$\sqrt{(s-a)(s-b)(s-c)(s-d)}$ S = $\frac{a+b+c+d}{2}$
Rhombus	4a	$\frac{1}{2}\times product\:of\:diagonal$
Ring	$2\pi (R+r)$	$\pi R^{2}-\pi r^{2}$

Solids

Figure	Lateral Surface Area	Total Surface Area	Volume	Images
Cube	$4a^{2}$	$6a^{2}$	$a^{3}$
Cuboid	2h(l + b)	2(lb + bh + lh)	lbh
Cylinder	$2\pi rh$	$2\pi r(r+h)$	$\pi r^{2}h$
Cone	$\pi rl$	$\pi r(l+r)$	$\frac{1}{3}\pi r^{2}h$
Sphere		$4\pi r^{2}$	$\frac{4}{3}\pi r^{3}$
Hemisphere	$2\pi r^{2}$	$3\pi r^{2}$	$\frac{2}{3}\pi r^{3}$

Important Questions for Mensuration

Mensuration

Question 1 of 50

1. Question
1 points
The number of revolutions made by a bicycle wheel 56 cm in diameter in covering a distance of 1.1 km is
- 31.25
- 312.5
- 62.5
- 625
Correct

Number of revolutions = $\frac{Distance}{Circumstance\:of\:wheel}$

Circumstance of wheel = $2\times \frac{22}{7}\times 28=176\: cm=1.76\: m$

No. of revolutions = $\frac{1.1\times 1000}{1.76}=625$

Incorrect

Number of revolutions = $\frac{Distance}{Circumstance\:of\:wheel}$

Circumstance of wheel = $2\times \frac{22}{7}\times 28=176\: cm=1.76\: m$

No. of revolutions = $\frac{1.1\times 1000}{1.76}=625$
Question 2 of 50

2. Question
1 points
The area of a circular field is 13.86 hectare. The cost of fencing it @ Rs. 0.60/m is
- Rs. 784
- Rs. 792
- Rs. 972
- None of the above
Correct

A = $13.86\times 10000\:m^{2}$

= $138600 \:m^{2}$

$\pi r^{2}=138600$

$r^{2}=\frac{138600}{22}\times 7=6300\times 7$

R = 210

Length of fence = $2\pi r=2\times \frac{22}{7}\times 210=1320$

Cost = $1320 \times 0.60= Rs.792$

Incorrect

A = $13.86\times 10000\:m^{2}$

= $138600 \:m^{2}$

$\pi r^{2}=138600$

$r^{2}=\frac{138600}{22}\times 7=6300\times 7$

R = 210

Length of fence = $2\pi r=2\times \frac{22}{7}\times 210=1320$

Cost = $1320 \times 0.60= Rs.792$
Question 3 of 50

3. Question
1 points
The base of an isosceles triangle is 8 cm and its perimeter is 18 cm. The area of triangle is
- $12\:cm^{2}$
- $24\:cm^{2}$
- $18\:cm^{2}$
- $14\:cm^{2}$
Correct

a + a + 8 = 18

$\Rightarrow$ a = 5

DC = $\frac{1}{2}\times 8=4$

$\Delta ADC$ is a right triangle. Hence

$h^{2}|4^{2}=5^{2}$

$\Rightarrow$ h = 3

Area of $\Delta ABC=\frac{1}{2}\times 8\times 3\:cm^{2}$

= $12\:cm^{2}$

Incorrect

a + a + 8 = 18

$\Rightarrow$ a = 5

DC = $\frac{1}{2}\times 8=4$

$\Delta ADC$ is a right triangle. Hence

$h^{2}|4^{2}=5^{2}$

$\Rightarrow$ h = 3

Area of $\Delta ABC=\frac{1}{2}\times 8\times 3\:cm^{2}$

= $12\:cm^{2}$
Question 4 of 50

4. Question
1 points
Each square of a chess board has an area of $9\:cm^{2}$ . Also the chess board has a margin of 3 cm around it. The side of the chess board is
- 24 cm
- 33 cm
- 27 cm
- 30 cm
Correct

Side of each square of chess board = $\sqrt{9}$

= 3 cm

Side of board = $8\times side\:of\:square+2 \times width\:of\:margin.$

Side of board = $8\times 3+2 \times 3$

= 24 + 6 = 30 cm

Incorrect

Side of each square of chess board = $\sqrt{9}$

= 3 cm

Side of board = $8\times side\:of\:square+2 \times width\:of\:margin.$

Side of board = $8\times 3+2 \times 3$

= 24 + 6 = 30 cm
Question 5 of 50

5. Question
1 points
The area of the quadratic of a circle whose circumference is 22 cm is
- $38.5\:cm^{2}$
- $19.25\:cm^{2}$
- $9.625\:cm^{2}$
- None of the above
Correct

$2\pi r=22\: cm$

$\Rightarrow r=22 \times \frac{7}{24}\: cm=\frac{7}{2}\: cm$

Area quadrant = $\frac{1}{4}\times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\:cm^{2}=9.625\:cm^{2}$

Incorrect

$2\pi r=22\: cm$

$\Rightarrow r=22 \times \frac{7}{24}\: cm=\frac{7}{2}\: cm$

Area quadrant = $\frac{1}{4}\times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\:cm^{2}=9.625\:cm^{2}$
Question 6 of 50

6. Question
1 points
If AB = 12 cm and BC = 5 cm the area of the shaded portion is (use $\pi$ = 3.14)
- $72.665\:cm^{2}$
- $145.33\:cm^{2}$
- $36.3325\:cm^{2}$
- None of the above
Correct

BD = $\sqrt{5^{2}+12^{2}}\;cm=13\:cm$

Also BD is diameter

$\Rightarrow$ r = 6.5 cm

Required = Area of circle – area of rectangle

= $(3.14\times 6.5^{2}-12 \times 5)\:cm^{2}=72.665\:cm^{2}$

Incorrect

BD = $\sqrt{5^{2}+12^{2}}\;cm=13\:cm$

Also BD is diameter

$\Rightarrow$ r = 6.5 cm

Required = Area of circle – area of rectangle

= $(3.14\times 6.5^{2}-12 \times 5)\:cm^{2}=72.665\:cm^{2}$
Question 7 of 50

7. Question
1 points
PQRS is diameter of a circle of radius 6 cm such that PQ = OR = RS. Semicircles are drawn on PQ and QS as diameter as shown in figure. Area of the shaded region is
- $24\pi cm^{2}$
- $8\pi cm^{2}$
- $6\pi cm^{2}$
- $12\pi cm^{2}$
Correct

Required area = area semicircle PS + area semicircle PQ – area semicircle QS

= $\frac{1}{2}\pi (6^{2}+2^{2}-4^{2})\: cm^{2}$

= $\frac{1}{2}\pi (36+4-16)\: cm^{2}$

= $12\pi \:cm^{2}$

Incorrect

Required area = area semicircle PS + area semicircle PQ – area semicircle QS

= $\frac{1}{2}\pi (6^{2}+2^{2}-4^{2})\: cm^{2}$

= $\frac{1}{2}\pi (36+4-16)\: cm^{2}$

= $12\pi \:cm^{2}$
Question 8 of 50

8. Question
1 points
A rectangular field has area equal to $150\:m^{2}$ and perimeter 50 m. Its length and breadth respectively must be
- 30 m, 5m
- 25 m, 6m
- 15 m, 10 m
- None of the above
Correct

$l \times b=150$

And 2(l + b) = 50 $\Rightarrow$ l + b = 25

Clearly l = 15, b = 10

Incorrect

$l \times b=150$

And 2(l + b) = 50 $\Rightarrow$ l + b = 25

Clearly l = 15, b = 10
Question 9 of 50

9. Question
1 points
A rectangular park is 46 m by 36 m. A path 2 m wide is built all around outside it. The cost of constructing these paths @ Rs. 50 per $m^{2}$ is
- Rs. 17,200
- Rs. 16,080
- Rs. 15,920
- Rs. 15, 940
Correct

Area of path = $(50 \times 40-46 \times 36)m^{2}$

= $344 \:m^{2}$

Cost = Rs. $344 \times 50$

= Rs. 17,200

Incorrect

Area of path = $(50 \times 40-46 \times 36)m^{2}$

= $344 \:m^{2}$

Cost = Rs. $344 \times 50$

= Rs. 17,200
Question 10 of 50

10. Question
1 points
The length of hypotenuse of a right angled triangle is 5 cm and its area is 6 $cm^{2}$ . The lengths of the remaining sides are
- 3 cm and 4 cm
- 4 cm and 2 cm
- 3 cm and 2 cm
- 1 cm and 3 cm
Correct

$\frac{1}{2}ab=6\Rightarrow ab=12$

$a^{2}+b^{2}=5^{2}=25$

Use a + b = $\sqrt{a^{2}+b^{2}+2ab}\:\:and\:a-b=\sqrt{a^{2}+b^{2}-2ab}$

Incorrect

$\frac{1}{2}ab=6\Rightarrow ab=12$

$a^{2}+b^{2}=5^{2}=25$

Use a + b = $\sqrt{a^{2}+b^{2}+2ab}\:\:and\:a-b=\sqrt{a^{2}+b^{2}-2ab}$
Question 11 of 50

11. Question
1 points
The perimeter of a triangle is 450 m and its sides are in the ratio 13 : 12 : 5. The area of the triangle is
- $675\:m^{2}$
- $6750\:m^{2}$
- $6250\:m^{2}$
- None of the above
Correct

13x + 12 x + 5x = 450

X = 15

A = 195

B = 180

C = 75

Area = $\sqrt{225 \times 30 \times 45 \times 150}\:\:=6750\:m^{2}$

Incorrect

13x + 12 x + 5x = 450

X = 15

A = 195

B = 180

C = 75

Area = $\sqrt{225 \times 30 \times 45 \times 150}\:\:=6750\:m^{2}$
Question 12 of 50

12. Question
1 points
A circle has area which is 100 times the area of another circle. The ratio of their circumferences is
- 1 : 10
- 100 : 1
- 10 : 1
- None of the above
Correct

$A_{1}=100A_{2}$

$\Rightarrow \pi r\: _{1}\: ^{2}=100\pi r\: _{2} \: ^{2}$

$\Rightarrow r\: _{1}\: ^{2}=100r\: _{2}\: ^{2}$

$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{10}{1}$

$\Rightarrow \frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{r_{1}}{r_{2}}=\frac{10}{1}$

$\Rightarrow C_{1}:C_{2}=10:1$

Incorrect

$A_{1}=100A_{2}$

$\Rightarrow \pi r\: _{1}\: ^{2}=100\pi r\: _{2} \: ^{2}$

$\Rightarrow r\: _{1}\: ^{2}=100r\: _{2}\: ^{2}$

$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{10}{1}$

$\Rightarrow \frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{r_{1}}{r_{2}}=\frac{10}{1}$

$\Rightarrow C_{1}:C_{2}=10:1$
Question 13 of 50

13. Question
1 points
Four equal circles of radius 7 cm touch each other as shown in figure. The area of the shaded part is
- $84cm^{2}$
- $21cm^{2}$
- $42cm^{2}$
- None of the above
Correct

Join centres to obtain a square of side 14 cm.

Incorrect

Join centres to obtain a square of side 14 cm.
Question 14 of 50

14. Question
1 points
The number of square tiles of side 20 cm are required to pave a footpath 1 m wide around a rectangular plot $28\:m \times 18\:m$ is
- 1200
- 2400
- 4800
- 9600
Correct

Area of foot path = $(30 \times 20-28 \times 18)m^{2}=96\:m^{2}$

Number of tiles = $\frac{960000}{20 \times 20}=2400$

Incorrect

Area of foot path = $(30 \times 20-28 \times 18)m^{2}=96\:m^{2}$

Number of tiles = $\frac{960000}{20 \times 20}=2400$
Question 15 of 50

15. Question
1 points
If each side of a rectangle is increased by 50%, its area will increase by
- 125%
- 75%
- 100%
- 225%
Correct

Original area = $l \times b$

New area = $\frac{3}{2}l \times \frac{3}{2}b=\frac{9}{4}lb$

Increase % = $\frac{\frac{9}{4}lb-lb}{lb}\times 100\%$

= 125%

Incorrect

Original area = $l \times b$

New area = $\frac{3}{2}l \times \frac{3}{2}b=\frac{9}{4}lb$

Increase % = $\frac{\frac{9}{4}lb-lb}{lb}\times 100\%$

= 125%
Question 16 of 50

16. Question
1 points
Circular disc of area $A_{1}$ is given. With its radius as diameter a circular disc of area $A_{2}$ is cut out of it. The area of the remaining disc is denoted by $A_{3}$ . Then
- $A_{1}A_{3}< 12\:(A_{2})^{2}$
- $A_{1}A_{3}> 16\:(A_{2})^{2}$
- $A_{1}A_{3}=16\:(A_{2})^{2}$
- $A_{1}A_{3}> 2\:(A_{2})^{2}$
Correct

$A_{1}=\pi r^{2},A_{2}=\pi \left ( \frac{r}{2} \right )^{2}=\frac{\pi r^{2}}{4}$

$A_{3}=A_{1}-A_{2}=\frac{3}{4}\pi r^{2}$

Incorrect

$A_{1}=\pi r^{2},A_{2}=\pi \left ( \frac{r}{2} \right )^{2}=\frac{\pi r^{2}}{4}$

$A_{3}=A_{1}-A_{2}=\frac{3}{4}\pi r^{2}$
Question 17 of 50

17. Question
1 points
A circular ground has a radius of 26 m. A 4 m wide road runs on the outside around it. The area of the road is (use $\pi$ = 3.14)
- $700\:m^{2}$
- $703.36\:m^{2}$
- $713.65\:m^{2}$
- $716\:m^{2}$
Correct

Area of road = 3.14 $\left ( 30^{2}-26^{2} \right )\: m^{2}$

= 3.14 (30 + 26) (30 – 26) $m^{2}$

= 703.36 $m^{2}$

Incorrect

Area of road = 3.14 $\left ( 30^{2}-26^{2} \right )\: m^{2}$

= 3.14 (30 + 26) (30 – 26) $m^{2}$

= 703.36 $m^{2}$
Question 18 of 50

18. Question
1 points
Diagonals of a rhombus are in the ratio 3 : 4 and its area is 2400 $cm^{2}$ . Side of the rhombus is
- 20 cm
- 40 cm
- 50 cm
- 100 cm
Correct

$d_{1}=3x,d_{2}=4x$

$\frac{1}{2} \times 3x \times 4x=2400$

$\Rightarrow x^{2}=400\Rightarrow x=20$

Use : side = $\frac{1}{2}\sqrt{d_{1}^{2}+d_{2}^{2}}$

Incorrect

$d_{1}=3x,d_{2}=4x$

$\frac{1}{2} \times 3x \times 4x=2400$

$\Rightarrow x^{2}=400\Rightarrow x=20$

Use : side = $\frac{1}{2}\sqrt{d_{1}^{2}+d_{2}^{2}}$
Question 19 of 50

19. Question
1 points
AC is a diameter of circle with centre O. If BD $\perp$ at O and OA = 7 cm. Area of shaded region is
- $66.5\:cm^{2}$
- $133\:cm^{2}$
- $53.2\:cm^{2}$
- None of the above
Correct

Required area = Area of smaller circle + $2 \times area\:of\:minor\:segment\:AB$

= $\pi \times \left ( \frac{7}{2} \right )^{2}+2\times \frac{7^{2}}{2}\left ( \frac{90 \times \pi }{180}-sin\: 90 \right )$

= $66.5\:cm^{2}$

Incorrect

Required area = Area of smaller circle + $2 \times area\:of\:minor\:segment\:AB$

= $\pi \times \left ( \frac{7}{2} \right )^{2}+2\times \frac{7^{2}}{2}\left ( \frac{90 \times \pi }{180}-sin\: 90 \right )$

= $66.5\:cm^{2}$
Question 20 of 50

20. Question
1 points
O is the centre of bigger circle and AC its diameter. Another circle with AB as diameter is drawn. If AC = 54 cm and BC = 10 cm, find area of shaded region.
- 770 $cm^{2}$
- 385 $cm^{2}$
- 1155 $cm^{2}$
- 777 $cm^{2}$
Correct

AC = 54 cm; BC = 10 cm

$\Rightarrow$ AB = (54 – 10) cm = 44 cm

Radius of larger circle = 27 cm

Radius of smaller circle = 22 cm

Required area = $\pi (27^{2}-22^{2})\: cm^{2}$

= $\frac{22}{7}(27+22))(27-22)\:cm^{2}=770\:cm^{2}$

Incorrect

AC = 54 cm; BC = 10 cm

$\Rightarrow$ AB = (54 – 10) cm = 44 cm

Radius of larger circle = 27 cm

Radius of smaller circle = 22 cm

Required area = $\pi (27^{2}-22^{2})\: cm^{2}$

= $\frac{22}{7}(27+22))(27-22)\:cm^{2}=770\:cm^{2}$
Question 21 of 50

21. Question
1 points
Two card boards pieces in form of equilateral triangles having a side of 3 cm each are symmetrically glued to form a regular star as shown in the figure. The area of the star is
- $3\sqrt{3}\: cm^{2}$
- $6\: cm^{2}$
- $4\: cm^{2}$
- $\frac{3\sqrt{3}}{2}\: cm^{2}$
Correct

Area of star = $Area\: of\: \Delta ABC+3 \times area\: \Delta DEF.\: \Delta DEF$

is an equilateral triangle with each side equal to 1 cm.

Incorrect

Area of star = $Area\: of\: \Delta ABC+3 \times area\: \Delta DEF.\: \Delta DEF$

is an equilateral triangle with each side equal to 1 cm.
Question 22 of 50

22. Question
1 points
ABC is an equilateral $\Delta \: with\: area\: 36\sqrt{3}\: cm^{2}$ . The area of the inscribed circle is
- $36\pi \: cm^{2}$
- $48\pi \: cm^{2}$
- $24\pi \: cm^{2}$
- $12\pi \: cm^{2}$
Correct

$Side^{2}\times \frac{\sqrt{3}}{4}=36\sqrt{3}$

$\Rightarrow$ side = 12 cm

Semi-Perimeter = $\frac{12\times 3}{2}=18\:cm$

Use r = $\frac{\Delta }{S}$

Incorrect

$Side^{2}\times \frac{\sqrt{3}}{4}=36\sqrt{3}$

$\Rightarrow$ side = 12 cm

Semi-Perimeter = $\frac{12\times 3}{2}=18\:cm$

Use r = $\frac{\Delta }{S}$
Question 23 of 50

23. Question
1 points
The length of one diagonal of a rhombus is 80% of the other diagonal. The area of the rhombus is how many times the square of the length of the longer diagonal?
- $\frac{4}{5}$
- $\frac{2}{5}$
- $\frac{3}{4}$
- $\frac{1}{4}$
Correct

Let the length of the longer diagonal be L.

$\therefore$ Length of shorter diagonal = 80% of L = $\frac{4}{5}L$

Area of rhombus = $\frac{1}{2} \times L \times \frac{4}{5}L$

$\Rightarrow Area\:of\:rhombus =\frac{2}{5}L^{2}$

Incorrect

Let the length of the longer diagonal be L.

$\therefore$ Length of shorter diagonal = 80% of L = $\frac{4}{5}L$

Area of rhombus = $\frac{1}{2} \times L \times \frac{4}{5}L$

$\Rightarrow Area\:of\:rhombus =\frac{2}{5}L^{2}$
Question 24 of 50

24. Question
1 points
OABC is a rhombus whose vertices A, B and C lie on a circle with centre at O. If radius of circle is 10 cm then area of the rhombus
- $100\sqrt{3}\: cm^{2}$
- $50\:cm^{2}$
- $50\sqrt{3}\: cm^{2}$
- $25\sqrt{3}\: cm^{2}$
Correct

OA = OB = OC = AB = BC = 10 cm

Clearly $\Delta OBA\:and\:\Delta OBC$ are both equilateral and also congruent.

Then area Rhombus = $2 \times area\:equilateral\:\Delta OAB=2 \times \left ( 10^{2}\times \frac{\sqrt{3}}{4} \right )\: cm^{2}$

= $50\sqrt{3}\: cm^{2}$

Incorrect

OA = OB = OC = AB = BC = 10 cm

Clearly $\Delta OBA\:and\:\Delta OBC$ are both equilateral and also congruent.

Then area Rhombus = $2 \times area\:equilateral\:\Delta OAB=2 \times \left ( 10^{2}\times \frac{\sqrt{3}}{4} \right )\: cm^{2}$

= $50\sqrt{3}\: cm^{2}$
Question 25 of 50

25. Question
1 points
$\Delta ABC$ is an equilateral triangle with area $17300\:cm^{2}$ . With each vertex as centre a circle is described with radius equal to half the length of the said of the triangle. [Use $\sqrt{3}=1.73\:and\:\pi =3.14$ ] The area of the shaded part is
- $800\:m^{2}$
- $1600\:m^{2}$
- $2400\:m^{2}$
- None of the above
Correct

$\Delta ABC\:is\:an\:equilateral\Delta$

$\Rightarrow each\:interior\:angle\:is\:60^{\circ}$

$Side^{2}\times \frac{\sqrt{3}}{4}=17300$

Side = 200 m

Radius of each circle = 100 m

Shaded area = area of $\Delta$ - 3(area of any sector).

Incorrect

$\Delta ABC\:is\:an\:equilateral\Delta$

$\Rightarrow each\:interior\:angle\:is\:60^{\circ}$

$Side^{2}\times \frac{\sqrt{3}}{4}=17300$

Side = 200 m

Radius of each circle = 100 m

Shaded area = area of $\Delta$ - 3(area of any sector).
Question 26 of 50

26. Question
1 points
If both radius and heights of a cone are increased by 50%, then the volume of the cone will increase by
- 225%
- 200%
- 100%
- 237.5%
Correct

$r_{1} =r,r_{2}=r+50\%\:of\:r=\frac{3}{2}r,$

$h_{1}=h;h_{2}=\frac{3h}{2}$

$\frac{V_{2}}{V_{1}}=\frac{\frac{1}{3}\pi \left ( \frac{3}{2}r \right )^{2}\left ( \frac{3}{2}h \right )}{\frac{1}{3}\pi r^{2}h}=\frac{27}{8}=3.375$

Volume increasing by $(3.375-1)\times 100\%=237.5\%$

Incorrect

$r_{1} =r,r_{2}=r+50\%\:of\:r=\frac{3}{2}r,$

$h_{1}=h;h_{2}=\frac{3h}{2}$

$\frac{V_{2}}{V_{1}}=\frac{\frac{1}{3}\pi \left ( \frac{3}{2}r \right )^{2}\left ( \frac{3}{2}h \right )}{\frac{1}{3}\pi r^{2}h}=\frac{27}{8}=3.375$

Volume increasing by $(3.375-1)\times 100\%=237.5\%$
Question 27 of 50

27. Question
1 points
If both the height and the radius of the cone are doubled, then the ratio of volume of the bigger cone to that of the smaller cone will be
- 10 : 1
- 6 : 1
- 8 : 1
- 4 : 1
Correct

$\frac{V_{2}}{V_{1}}=\frac{\frac{1}{3}\pi (2r)^{2}(2h)}{\frac{1}{3}\pi r^{2}h}$

$V_{2}:V_{1}=8:1$

Incorrect

$\frac{V_{2}}{V_{1}}=\frac{\frac{1}{3}\pi (2r)^{2}(2h)}{\frac{1}{3}\pi r^{2}h}$

$V_{2}:V_{1}=8:1$
Question 28 of 50

28. Question
1 points
A cylindrical tub of radius 12 cm contains water to the height of 20 cm. A spherical iron ball is dropped into the tub. The ball gets completely immersed and the level of water is raised by 6.75 cm. The radius of the ball is
- 9.1 cm
- 4.5 cm
- 9 cm
- 6.3 cm
Correct

Radius of ball : r = ?; Radius of tube : R = 12 cm

H = 6.75 cm

Volume of ball = Volume of cylinder ABCD

$\frac{4}{3}\pi r^{3}=\pi R^{2}h$

$\frac{4}{3}\times r^{3}=12 \times 12 \times \frac{675}{100}$

$\Rightarrow$ r= 9 cm

Incorrect

Radius of ball : r = ?; Radius of tube : R = 12 cm

H = 6.75 cm

Volume of ball = Volume of cylinder ABCD

$\frac{4}{3}\pi r^{3}=\pi R^{2}h$

$\frac{4}{3}\times r^{3}=12 \times 12 \times \frac{675}{100}$

$\Rightarrow$ r= 9 cm
Question 29 of 50

29. Question
1 points
Diameter of a roller 60 m long is 42 cm. If it takes 700 complete revolutions to level a pay ground, the cost of leveling the play ground @ 45 p per square m is nearly
- Rs. 300
- Rs. 250
- Rs. 125
- Rs. 400
Correct

h = 42 cm; r = 30 cm, n = 700

Area leveled = $700 \times 2 \times \frac{22}{7} \times 30 \times 42\:cm^{2}$

= $5544000\:cm^{2}$

= $554.4\:m^{2}$

Cost = Rs. $554.4 \times 0.45=RS.\: 249.48\approx Rs.\: 250$

Incorrect

h = 42 cm; r = 30 cm, n = 700

Area leveled = $700 \times 2 \times \frac{22}{7} \times 30 \times 42\:cm^{2}$

= $5544000\:cm^{2}$

= $554.4\:m^{2}$

Cost = Rs. $554.4 \times 0.45=RS.\: 249.48\approx Rs.\: 250$
Question 30 of 50

30. Question
1 points
2 cubes of each of side 12 cm are joined end to end. Volume of the resulting cuboid is
- $3546\: cm^{3}$
- $3556\: cm^{3}$
- $3457\: cm^{3}$
- $3456\: cm^{3}$
Correct

L = 24 cm, B = 12 cm, H = 12 cm

V = $24 \times 12 \times 12\: cm^{3}=3456\: cm^{3}$

Incorrect

L = 24 cm, B = 12 cm, H = 12 cm

V = $24 \times 12 \times 12\: cm^{3}=3456\: cm^{3}$
Question 31 of 50

31. Question
1 points
A circus tent is cylindrical up to a height of 5 m and conical above it. If its diameter is 105 m and slant height of the conical portion is 53 m. The area of the canvas used to build the tent is
- $10395\:m^{2}$
- $13095\:m^{2}$
- $10400\:m^{2}$
- $9775\:m^{2}$
Correct

r = $\frac{105}{2}$ m; l = 53 m

H = 3 m

Area of canvas

= C.S.A. cone + C.S.A. of cylindrical

= $\pi rl+2\pi rh$

= $\pi r(l+2h)$

= $\frac{22}{7}\times \frac{105}{2}(53+10)\: m^{2}$

= $10395\:m^{2}$

Incorrect

r = $\frac{105}{2}$ m; l = 53 m

H = 3 m

Area of canvas

= C.S.A. cone + C.S.A. of cylindrical

= $\pi rl+2\pi rh$

= $\pi r(l+2h)$

= $\frac{22}{7}\times \frac{105}{2}(53+10)\: m^{2}$

= $10395\:m^{2}$
Question 32 of 50

32. Question
1 points
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and diameter of the cylinder is 7 cm. The S.A. of the solid is
- $418 \: cm^{2}$
- $836 \: cm^{2}$
- $627 \: cm^{2}$
- None of the above
Correct

r = $\frac{7}{2}\: cm$

H + 2r = 19 cm

$\Rightarrow$ h = 12 cm

Surface area = $2\pi rh+2 \times 2\pi r^{2}$

= $2\pi r(h+2r)$

= $2 \times \frac{22}{7} \times \frac{7}{2} \times 19\:cm^{3}$

= $418\:cm^{2}$

Incorrect

r = $\frac{7}{2}\: cm$

H + 2r = 19 cm

$\Rightarrow$ h = 12 cm

Surface area = $2\pi rh+2 \times 2\pi r^{2}$

= $2\pi r(h+2r)$

= $2 \times \frac{22}{7} \times \frac{7}{2} \times 19\:cm^{3}$

= $418\:cm^{2}$
Question 33 of 50

33. Question
1 points
The radius of a sphere is increased by 100%, then the increase in surface area of the sphere will be
- 400%
- 200%
- 100%
- 300%
Correct

$r_{1}=r,r_{2}=2r$

$\frac{A_{1}}{A_{1}}=\frac{(2r)^{2}}{r^{2}}=4$

Increase in area = 300%

Incorrect

$r_{1}=r,r_{2}=2r$

$\frac{A_{1}}{A_{1}}=\frac{(2r)^{2}}{r^{2}}=4$

Increase in area = 300%
Question 34 of 50

34. Question
1 points
Diameter of a cooper sphere is 6 cm. The sphere is melted and drawn into a wire of uniform circular cross section which is 72 cm long. The diameter of the wire is nearly
- 2.8 cm
- 2.7 cm
- 1.4 cm
- 1.5 cm
Correct

Volume of wire (cylinder) = Volume of sphere

$\pi r^{2} \times 72=\frac{4}{3}\pi \times 3^{3}$

$\Rightarrow r=\frac{1}{\sqrt{2}}\: cm$

$\Rightarrow diameter=\frac{2}{\sqrt{2}}\: cm$

= $\sqrt{2}\: cm$

$\approx 1.4\: cm$

Incorrect

Volume of wire (cylinder) = Volume of sphere

$\pi r^{2} \times 72=\frac{4}{3}\pi \times 3^{3}$

$\Rightarrow r=\frac{1}{\sqrt{2}}\: cm$

$\Rightarrow diameter=\frac{2}{\sqrt{2}}\: cm$

= $\sqrt{2}\: cm$

$\approx 1.4\: cm$
Question 35 of 50

35. Question
1 points
Three cubes of edges 12 cm, 16 cm and 20 cm are melted and a new cube is made. The side of the new cube is
- 12 cm
- 24 cm
- 48 cm
- 36 cm
Correct

$x^{3}=12^{3}+16^{3}+20^{3}$

$x^{3}=13824=24^{3}$

Side of new cube =24 cm

Incorrect

$x^{3}=12^{3}+16^{3}+20^{3}$

$x^{3}=13824=24^{3}$

Side of new cube =24 cm
Question 36 of 50

36. Question
1 points
The cost of white washing the four walls of a room is Rs. 100. The cost of white washing a room twice in length and breadth but one fourth in height is
- Rs. 20
- Rs. 50
- Rs. 25
- Cannot be determined
Correct

$2(l+b)h \times rate=100$

$2(2l+2b)\frac{h}{4} \times rate$

$\Rightarrow$ required cost = Rs. 50

Incorrect

$2(l+b)h \times rate=100$

$2(2l+2b)\frac{h}{4} \times rate$

$\Rightarrow$ required cost = Rs. 50
Question 37 of 50

37. Question
1 points
The volume of a cylinder is $6358\:cm^{3}$ and its height is 28 cm. Its curved surface area is
- $2244 \: cm^{2}$
- $112 \: cm^{2}$
- $748 \: cm^{2}$
- $1496 \: cm^{2}$
Correct

$\frac{22}{7}\times r^{2}\times 28=6358$

$\Rightarrow r=\frac{17}{2}\: cm$

C.S.A. = $2\times \frac{22}{7}\times \frac{17}{2}\times 28=1496\:cm^{2}$

Incorrect

$\frac{22}{7}\times r^{2}\times 28=6358$

$\Rightarrow r=\frac{17}{2}\: cm$

C.S.A. = $2\times \frac{22}{7}\times \frac{17}{2}\times 28=1496\:cm^{2}$
Question 38 of 50

38. Question
1 points
A hemispherical bowl is made of steel 12.1 cm thick. The inside radius of the bowl is 5 cm. The volume of steel used in making the bowl is nearly
- $366 \: cm^{3}$
- $244 \: cm^{3}$
- $488 \: cm^{3}$
- $976 \: cm^{3}$
Correct

r = 5 cm

R = (5 + 2.1) cm = 7.1

Volume of material = $\frac{2}{3}\times \frac{22}{7}(7.1^{3}-5^{3})\:cm^{3}$

= $488.004\: cm^{3}$

$\approx 488\: cm^{3}$

Incorrect

r = 5 cm

R = (5 + 2.1) cm = 7.1

Volume of material = $\frac{2}{3}\times \frac{22}{7}(7.1^{3}-5^{3})\:cm^{3}$

= $488.004\: cm^{3}$

$\approx 488\: cm^{3}$
Question 39 of 50

39. Question
1 points
A spherical balloon was deflated till its radius was halved. If its volume becomes n times its original volume then value of n is
- $\frac{1}{4}$
- $\frac{1}{16}$
- $\frac{1}{8}$
- $\frac{1}{2}$
Correct

$\frac{V_{2}}{V_{1}}=\frac{\frac{4}{3}\pi \left ( \frac{r}{2} \right )^{3}}{\frac{4}{3}\pi \left ( r \right )^{3}}=\frac{1}{8}$

$V_{2}=\frac{1}{8}V_{1}$

Incorrect

$\frac{V_{2}}{V_{1}}=\frac{\frac{4}{3}\pi \left ( \frac{r}{2} \right )^{3}}{\frac{4}{3}\pi \left ( r \right )^{3}}=\frac{1}{8}$

$V_{2}=\frac{1}{8}V_{1}$
Question 40 of 50

40. Question
1 points
Three cubes of edges 12 cm, 16 cm and 20 cm are melted and a new cube is made. The diagonal of the new cube is
- $48\sqrt{3}\: cm$
- $12\sqrt{3}\: cm$
- 36 cm
- $24\sqrt{3}\: cm$
Correct

$x^{3}=12^{3}+16^{3}+20^{3}$

$x^{3}=13824=24^{3}$

$\therefore$ Side of new cube = 24 cm

$\therefore diagonal=24\sqrt{3}\:cm$

Incorrect

$x^{3}=12^{3}+16^{3}+20^{3}$

$x^{3}=13824=24^{3}$

$\therefore$ Side of new cube = 24 cm

$\therefore diagonal=24\sqrt{3}\:cm$
Question 41 of 50

41. Question
1 points
A rectangular sheet of area $88 \: cm^{2}$ and length 8 cm is rolled along its breadth to make a hollow cylinder. The volume of this cylinder is
- $4982 \: cm^{2}$
- $4950 \: cm^{2}$
- $4900 \: cm^{2}$
- $4928 \: cm^{2}$
Correct

Circumference of cylinder = 88 cm

$\Rightarrow 2\pi r=88$

$\Rightarrow r=14$

Also h = 8

$\therefore Volume\: of\: cylinder=\frac{22}{7}\times 14\times 14\times 8\:cm^{3}$

= $4928\: cm^{3}$

Incorrect

Circumference of cylinder = 88 cm

$\Rightarrow 2\pi r=88$

$\Rightarrow r=14$

Also h = 8

$\therefore Volume\: of\: cylinder=\frac{22}{7}\times 14\times 14\times 8\:cm^{3}$

= $4928\: cm^{3}$
Question 42 of 50

42. Question
1 points
Curved surface area of a cone whose volume is $12936 \: cm^{3}$ and the diameter of base is 42 cm, is
- $1100 \: cm^{2}$
- $115 \: cm^{2}$
- $4620 \: cm^{2}$
- $2310 \: cm^{2}$
Correct

r = $\frac{42}{2}\: cm=21\: cm$

$\frac{1}{3}\times \frac{22}{7}\times 21^{2}\times h=12936$

$\Rightarrow h=28\:cm$

L = $\sqrt{28^{2}+21^{2}}=35$

C.S.A. = $\frac{22}{7}\times 21\times 35\: cm^{2}$

= $2310\:cm^{2}$

Incorrect

r = $\frac{42}{2}\: cm=21\: cm$

$\frac{1}{3}\times \frac{22}{7}\times 21^{2}\times h=12936$

$\Rightarrow h=28\:cm$

L = $\sqrt{28^{2}+21^{2}}=35$

C.S.A. = $\frac{22}{7}\times 21\times 35\: cm^{2}$

= $2310\:cm^{2}$
Question 43 of 50

43. Question
1 points
If each side of a cube is tripped then its volume increases by
- 800%
- 700%
- 100%
- None of the above
Correct

$\frac{V_{2}}{V_{1}}=\frac{(3x)^{3}}{x^{3}}=\frac{27x^{3}}{x^{3}}=\frac{27}{1}$

Increse in vol = $V_{2}-V_{1}$ = 27 -1 = 26

$\therefore$ increase % = 2600%

Incorrect

$\frac{V_{2}}{V_{1}}=\frac{(3x)^{3}}{x^{3}}=\frac{27x^{3}}{x^{3}}=\frac{27}{1}$

Increse in vol = $V_{2}-V_{1}$ = 27 -1 = 26

$\therefore$ increase % = 2600%
Question 44 of 50

44. Question
1 points
A spherical ball was melted and made into smaller balls of half the radius of the original. The ratio of the sum total of the surface of all the smaller balls to that of the original ball is
- 8 : 1
- 4 : 1
- 1 : 2
- 2 : 1
Correct

Number of small balls = $\frac{Volume\:of\:original\:ball}{Volume\:of\:small\:ball}$

= $\frac{\frac{4}{3}\times \pi \times r^{3}}{\frac{4}{3}\times \pi \times \left ( \frac{r}{2} \right )^{3}}=8$

$\frac{CSA\: of\: original\: ball}{Sum\: of\: CSA\: of\: 8\: small\: balls}=\frac{4\pi r^{2}}{8\times \pi \times \left ( \frac{r}{2} \right )^{2}}=\frac{2}{1}$

Or 2 : 1

Incorrect

Number of small balls = $\frac{Volume\:of\:original\:ball}{Volume\:of\:small\:ball}$

= $\frac{\frac{4}{3}\times \pi \times r^{3}}{\frac{4}{3}\times \pi \times \left ( \frac{r}{2} \right )^{3}}=8$

$\frac{CSA\: of\: original\: ball}{Sum\: of\: CSA\: of\: 8\: small\: balls}=\frac{4\pi r^{2}}{8\times \pi \times \left ( \frac{r}{2} \right )^{2}}=\frac{2}{1}$

Or 2 : 1
Question 45 of 50

45. Question
1 points
A cooper sphere of diameter 6 cm is drawn into a wire of diameter 0.4 cm. The length of the wire is
- 6 m
- 9 m
- 8 m
- None of the above
Correct

Volume wire (cyl) =Vol. of sphere

$\Rightarrow \pi \times (0.2)^{2}h=\frac{4}{3}\times \pi \times 3^{3}$

$\Rightarrow h=900\:cm=9\:m$

Incorrect

Volume wire (cyl) =Vol. of sphere

$\Rightarrow \pi \times (0.2)^{2}h=\frac{4}{3}\times \pi \times 3^{3}$

$\Rightarrow h=900\:cm=9\:m$
Question 46 of 50

46. Question
1 points
Radii of two cylinder are in ratio 4 : 3 and their heights are in ratio 3 : 4. Ratio of their volumes is
- 16 : 9
- 4 : 3
- 9 : 16
- 3 : 4
Correct

$r_{1}=4x,r_{2}=3x$

$h_{1}=3y;h_{2}=4y$

$\frac{V_{2}}{V_{1}}=\frac{\pi (4x)^{2}(3y)}{\pi (3x)^{2}(4y)}=\frac{4}{3}$

Incorrect

$r_{1}=4x,r_{2}=3x$

$h_{1}=3y;h_{2}=4y$

$\frac{V_{2}}{V_{1}}=\frac{\pi (4x)^{2}(3y)}{\pi (3x)^{2}(4y)}=\frac{4}{3}$
Question 47 of 50

47. Question
1 points
The curved surface area of a cylinder is $264 \: cm^{2}$ and its volume is $924 \: cm^{3}$ . The height of the cylinder is
- 8 m
- 8.5 m
- 6 m
- None of the above
Correct

$\frac{\pi r^{2}h}{2\pi rh}=\frac{924}{264}$

$\Rightarrow \frac{r}{2}=\frac{7}{2}$

$\Rightarrow r=7$

$2\times \frac{22}{7}\times 7\times h=264$

$\Rightarrow h=6\:m$

Incorrect

$\frac{\pi r^{2}h}{2\pi rh}=\frac{924}{264}$

$\Rightarrow \frac{r}{2}=\frac{7}{2}$

$\Rightarrow r=7$

$2\times \frac{22}{7}\times 7\times h=264$

$\Rightarrow h=6\:m$
Question 48 of 50

48. Question
1 points
The circumference of edge of a hemispherical bowl is 132 cm. Its capacity is
- $18404 \: cm^{3}$
- $19400 \: cm^{3}$
- $19444 \: cm^{3}$
- $19404 \: cm^{3}$
Correct

$2\pi r=132$

$\Rightarrow r=21\:cm$

Volume = $\frac{2}{3}\times \frac{22}{7}\times 21^{3}\: cm^{3}$

= $19404\:cm^{3}$

Incorrect

$2\pi r=132$

$\Rightarrow r=21\:cm$

Volume = $\frac{2}{3}\times \frac{22}{7}\times 21^{3}\: cm^{3}$

= $19404\:cm^{3}$
Question 49 of 50

49. Question
1 points
The radii of two right circular cylinder are in ratio 2 : 3 and their heights are in ratio 5 : 4. The ratio of their curved surface areas is
- 9 : 5
- 5 : 9
- 4 : 3
- 3 : 4
Correct

$\frac{CSA_{1}}{CSA_{2}}=\frac{2\pi (2x)^{2}(5y)}{2\pi (3x)^{2}(4y)}=\frac{5}{9}$

Incorrect

$\frac{CSA_{1}}{CSA_{2}}=\frac{2\pi (2x)^{2}(5y)}{2\pi (3x)^{2}(4y)}=\frac{5}{9}$
Question 50 of 50

50. Question
1 points
A vessel is in form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the base of the hemispherical bowl is 6 cm and the total height of the vessel is 11 cm. The capacity of the vessel is (Use $\pi$ = 3.14)
- 282.8 cc
- 282.9 cc
- 282.6 cc
- None of the above
Correct

r = 3 cm

H = (11 – 3) cm = 8 cm

Capacity of vessel = $\frac{2}{3}\pi r^{3}+\pi r^{2}h$

= $\pi r^{2}\left ( \frac{2}{3}r+h \right )$

= $3.14\times 3^{2}\left ( \frac{2}{3}\times 3+8 \right )cm^{3}$

= $3.14\times 9 \times 10\: cm^{3}$

= $282.6\: cm^{3}$

Incorrect

r = 3 cm

H = (11 – 3) cm = 8 cm

Capacity of vessel = $\frac{2}{3}\pi r^{3}+\pi r^{2}h$

= $\pi r^{2}\left ( \frac{2}{3}r+h \right )$

= $3.14\times 3^{2}\left ( \frac{2}{3}\times 3+8 \right )cm^{3}$

= $3.14\times 9 \times 10\: cm^{3}$

= $282.6\: cm^{3}$