Mathematics Class X

Question 1 of 30

1. Question
1 point(s)
The area of three adjacent faces of a cuboid are $x,y,z$ if the volume is $V $, then $ V^2$ will be equal to
- $\dfrac{xy}{z} $
- $\dfrac{yz}{x^2} $
- $\dfrac{x^2y^2}{z^2} $
- $xyz $
Correct

Incorrect

Hint

$l.\ b=x\ …(1)\\ \\ b.\ h = y\ …(2)\\ \\ h. \ l= z\ …(3)\\ \\ l^2b^2h^2=xyz\\ \\ v^2=xyz$
Question 2 of 30

2. Question
1 point(s)
The volume of a sphere of diameter 2p cm is given by
- $pp^2\ cm^3 $
- $pp^3\ cm^3 $
- $4pp^3\ cm^3 $
- $\dfrac{4}{3}\pi\ p^3\ cm^3 $
Correct

Incorrect

Hint

Diameter $= 2p\ cm$
Radius $= p\ cm$
Volume $=\dfrac{4}{3}\pi p^3$
Question 3 of 30

3. Question
1 point(s)
The maximum length of pencil which can be put inside the box whose dimensions are 8cm, 6cm, 2cm
- $2\sqrt{13}\ cm $
- $2\sqrt{14}\ cm $
- $2\sqrt{26}\ cm $
- $10\sqrt2\ cm $
Correct

Incorrect

Hint

$L=8\ cm, b=6\ cm, h= 2\ cm$
Max. length $=\sqrt{l^2+b^2+h^2}=\sqrt{8^2+6^2+2^2}$
$=\sqrt{104}=2\sqrt{26}\ cm$
Question 4 of 30

4. Question
1 point(s)
One room’s breadth is double its height and half its length. If volume of room is $512\ m^3 $ then length of the room will be
- 12 m
- 16 m
- 20 m
- 32 m
Correct

Incorrect

Hint

Let the breadth be $x\ m$
$h=\dfrac{x}{2}\ m\\ \\ l=2x\ m$
Volume $= lbh=512\ m^3$
$\Rightarrow x\times\dfrac{x}{2}\times 2x=512\Rightarrow x=8m\\ \\ l=2\times8=16\ m$
Question 5 of 30

5. Question
1 point(s)
A rectangular wooden block whose dimensions are 15cm long, 12cm wide and 6 cm high. How many minimum cubes to be made by its block.
- 6
- 11
- 33
- 40
Correct

Incorrect

Hint

Volume $=15\times12\times6=1080\ cm^3$
$1080=x\times$ volume of one cube
$1080=2\times2\times10\times27\\ \\ =40\times3^3$
So, 40 minimum cube can be made
Question 6 of 30

6. Question
1 point(s)
Diameter of a cylinder vessel is 60cm. it is filled with water such that a sphere of diameter 30cm is mersed fully into it. Then what is the increase in height of the water surface after putting the sphere in the vessels.
- 2 cm
- 3 cm
- 4 cm
- 5 cm
Correct

Incorrect

Hint

Volume of the sphere $=\dfrac{4}{3}\pi(15)^3$
Volume of cylinder $=\pi\times(30)^2\times h$
$\pi\times(30)^2\times h=\dfrac{4}{3}\pi(15)^3\\ \\ h=\dfrac{\frac{4}{3}\pi\times15\times15\times15}{\pi\times30\times30}\\ \\ h=5\ cm$
Question 7 of 30

7. Question
1 point(s)
A village having a population of 4000 requires 150 liter per head per day. It has a tank measuring 20 m by 15m by 6m. for how many days the water of this tank will last.
- 6 days
- 5 days
- 4 days
- 3 days
Correct

Incorrect

Hint

Total requirement of water in a day
$=4000\times150\\ \\ =600000\ \text{litre}\\ \\ =600\ m^3$
No. of days $=\dfrac{20\times15\times6}{600}=3\ days$
Question 8 of 30

8. Question
1 point(s)
A drum of water is $\dfrac{3}{5} $ full. When 57 liters are drawn from it, it is just $\dfrac{1}{8} $ full. Find the total capacity of drum.
- 120 ml
- 120 litre
- 100 litre
- 240 litre
Correct

Incorrect

Hint

Let the total capacity of drum be $x$
$\left(\dfrac{3}{8}-\dfrac{1}{8}\right)x=57\\ \\ \dfrac{19}{40}x=57\\ \\ x=120\ \text{litres}$
Question 9 of 30

9. Question
1 point(s)
If the length, breadth and height of a cube are increased, decreased and increased by 1%, 3%, 2% respectively then the volume of solid.
- increased
- decreased
- increases or decreases depending upon the original dimension
- None of these
Correct

Incorrect

Hint

Volume $= x y z$
New volume $=\dfrac{101x}{100}\times\dfrac{97y}{100}\times\dfrac{102z}{100}$
$= 0.999294\ xyz$
So, the volume will decrease
Question 10 of 30

10. Question
1 point(s)
Solids of various shapes are cost from equal volumes of metal. The least surface will be possessed by
- cylinder
- sphere
- cube
- hemisphere
Correct

Incorrect

Hint

The least surface will be possessed by the cylinder
Question 11 of 30

11. Question
1 point(s)
A wire in the shape of an equilateral triangle has an area S sq cm. If the same wire is bent to form a circle, the area of the circle will be
- $PS^2/\pi $
- $3S/\pi $
- $3S^2/\pi $
- $3\sqrt3s/\pi $
Correct

Incorrect

Hint

$\dfrac{\sqrt3}{4}a^2=S\$ sq cm
$a^2=\dfrac{4S}{\sqrt3}…(1)$
Perimeter of triangle = perimeter of circle.
$3a=2\pi r\\ \\ r=\dfrac{3a}{2\pi}$
Area of circle $\pi r^2=\pi\left(\dfrac{3a}{2\pi}\right)^2\ .....(2)$
Solving we get
$\pi r^2=\dfrac{3\sqrt3S}{\pi}$
Question 12 of 30

12. Question
1 point(s)
A wire of length 2p is bent to form a circle, a triangle, a rectangle and a polygon. State which figure he greatest area.
- circle
- rectangle
- triangle
- polygon
Correct

Incorrect

Hint

Rectangle has greatest area
Question 13 of 30

13. Question
1 point(s)
A right circular cylinder, a right circular cone and a hemisphere are all having equal base area and height. These volumes are in the ratio.
- 1:2:3
- 3:1:2
- 3:2:1
- 2:1:3
Correct

Incorrect

Hint

$\pi r^2h:\dfrac{1}{3}\pi r^2h:\dfrac{2}{3}\pi r^3\\ \\ 1:\dfrac{1}{3}:\dfrac{2}{3}\\ \\ 3:1:2$
Question 14 of 30

14. Question
1 point(s)
A rectangular sheet of cardboard is 9cm by 6cm. If greatest possible circle is cut off from cardboard, the remaining area is
- $36\pi $
- $9(6-\pii) $
- $6-\pi $
- $6(9-\pi) $
Correct

Incorrect

Hint

$l=9\ cm\\ \\ b=6\ cm$
Area of circle $=\pi r^2=\pi\times3^2=9\pi\ cm^2$
Area of cardboard $=9\times6=54\ cm^2$
Remaining area $=(54-9\pi)\ cm^2$
Question 15 of 30

15. Question
1 point(s)
A cone is cut half way through its axis and parallel to the base the volume of two portions are in the ratio.
- 1:1
- 1:3
- 1:7
- 1:4
Correct

Incorrect

Hint

$V_1=\dfrac{1}{3}\pi(r^2_1)\times h$
Volume of frustum, $V_2=\dfrac{1}{3}\pi h[r^2_1+r^2_2+r_1r_2]$
$\dfrac{r_1}{r_2}=\dfrac{2h}{h}\\ \\ r_1=2r_2\\ \\ V_2=\dfrac{1}{3}\pi h[7r^2_2]\\ \\ V_2=\dfrac{1}{3}\pi[r_2]^2h\\ \\ \dfrac{V_1}{V_2}=\dfrac{\dfrac{1}{3}\pi(r_2)^2h}{\dfrac{1}{3}\pi(7r_2)^2h}=\dfrac{1}{7}$
Question 16 of 30

16. Question
1 point(s)
Three cubes of metal of edges 6cm, 8cm and 10cm are melted to form a new cube. Find the diagonal of this cube
- 8 cm
- 12 cm
- 20.8 cm
- 21.8 cm
Correct

Incorrect

Hint

Let the edge of new cube be $x$
$x^3=6^3+8^3+10^3=1728\\ \\ x=12\ cm$
Diagonal of new cube $=\sqrt3x$
$=\sqrt3\times12=20.8\ cm$
Question 17 of 30

17. Question
1 point(s)
The dimension of a brick are 24 cm × 12 cm × 8 cm. How many bricks will be required to build a wall 24m × 8m × 6m if 20% of the wall is filled with mortar?
- 400000
- 450000
- 500000
- 550000
Correct

Incorrect

Hint

Volume of brick $=(24\times12\times8)\ cm^3$
Volume of wall $=(24\times8\times6)\ m^3$
$=24\times8\times6\times10^6\ cm^3$
Volume left $=\dfrac{80}{100}\times(24\times8\times6)\times10^6$
Number of bricks $=\dfrac{8\times10^5\times24\times8\times6}{24\times8\times12}=400000$
Question 18 of 30

18. Question
1 point(s)
The area of 4 walls of the hall whose breadth is 15m and height is 8m is 1068 sqm, the length of the hall is
- 18m
- 17m
- 16m
- None of these
Correct

Incorrect

Hint

Area of $4$ walls $=2h(l+b)=1068$
$\Rightarrow 2h(l+b)=1068\\ \\ \Rightarrow l=\dfrac{1068}{2h}-b\\ \\ \Rightarrow l=\dfrac{531-120}{8}=\dfrac{411}{8}m$
Question 19 of 30

19. Question
1 point(s)
The perimeter of the following shaded portion is
- 40 m
- 40.07 m
- 40.28 m
- 35 m
Correct

Incorrect

Hint

$r=1\ m\\ \\ \theta=90^o$
length of arc LE = arc FG = arc HI = arc JK
$=2\pi r\times\dfrac{\theta}{360}\\ \\ =2\pi r\times\dfrac{90}{360}=\dfrac{\pi}{2}m\\ \\ KL=GH=4-(1+1)=2m\\ \\ JI=EF=17-(1+1)=15\ m$
Perimeter of the shaded portion
$=\dfrac{\pi}{2}+15+\dfrac{\pi}{2}+2+\dfrac{\pi}{2}+15+\dfrac{\pi}{2}+2 \\ \\ =2\pi+34\\ \\ =34+6.28=40.28\ m$
Question 20 of 30

20. Question
1 point(s)
The area of the shaded region is
- $\dfrac{\pi}{3} $ sq unit
- $\dfrac{\pi}{2} $ sq unit
- $\dfrac{\pi}{4} $ sq unit
- $\pi^2 $ sq unit
Correct

Incorrect

Hint

$r=1$ unit
$\theta=180-60=120^o$
Area of the shaded region $=\pi r^2\dfrac{\theta}{360^o}$
$=\pi(1)^2\times\dfrac{120}{360}$
$\dfrac{\pi}{3}$ sq unit
Question 21 of 30

21. Question
1 point(s)
The internal and external diameters of a hollow hemispherical vessel are 42 cm and 45.5 cm respectively. Find the capacity
- 19.404 liters
- 19.508 liters
- 20.486 liters
- 2004 liters
Correct

Incorrect

Hint

$r_1=21\ cm\\ \\ r_2=22.75\ cm$
capacity $=\dfrac{2}{3}\pi r^2_1$
capacity $=\dfrac{2}{3}\times\dfrac{22}{7}\times21\times21\times21$
$=19404\ cm^3\\ \\ =19.404\ \text{litres}$
Question 22 of 30

22. Question
1 point(s)
The volume of a versal in the form of a right circular cylinder is $448\pi\ cm^3 $ and height is $ 7\ cm$. Find the radius
- 6 cm
- 8 cm
- 9 cm
- 10 cm
Correct

Incorrect

Hint

$\pi r^2h=448\pi\\ \\ \Rightarrow r^2h=448\\ \\ \Rightarrow r^2=64\\ \\ r=8\ cm$
Question 23 of 30

23. Question
1 point(s)
A cubical block of side 7 cm is sumounted by a hemisphere. What is the greatest diameter the hemisphere can have?
- 7 cm
- 8 cm
- 7.5 cm
- 3.5 cm
Correct

Incorrect

Hint

Diameter is equal to the cube’s edge ie 7 cm
Question 24 of 30

24. Question
1 point(s)
$2 $ cubes each of volume $64\ cm^3 $ are joined end to end. Find the surface area of the resulting cuboids.
- $160\ cm^3 $
- $140\ cm^2 $
- $150\ cm^2 $
- $130\ cm^2 $
Correct

Incorrect

Hint

Edge $=4\ cm$
Surface area of cuboid $= 2(lb+bh+hl)$
$=2\times(4\times4+4\times8+4\times8)\\ \\ =2\times80=160\ cm^2$
Question 25 of 30

25. Question
1 point(s)
A cuboidal metal of dimensions $44\ cm \times 30\ cm \times 15\ cm$ was melted and cast into a cylinder of height $28\ cm$. find its radius
- 20 cm
- 15 cm
- 10 cm
- 18 cm
Correct

Incorrect

Hint

Volume of cuboid = volume of cylinder
$lbh=\pi r^2h\\ \\ r^2=\dfrac{44\times30\times15\times7}{22\times28}=15\ cm$
Question 26 of 30

26. Question
1 point(s)
The L.S.A of a cylinder is $176\ cm^2 $ and base area is $38.5\ cm^2 $. What is its volume?
- $830\ cm^3 $
- $380\ cm^3 $
- $308\ cm^3 $
- $803\ cm^3 $
Correct

Incorrect

Hint

$\pi r^2=38.5\Rightarrow r=\dfrac{7}{2}\ cm\\ \\ \text{L.S.A}=2\pi rh=176\\ \\ \Rightarrow h=8\ cm\\ \\ \text{volume}=\pi r^2h=\dfrac{22}{7}\times\left(\dfrac{7}{2}\right)^2\times8\\ \\ =308\ cm^3$
Question 27 of 30

27. Question
1 point(s)
The side of a solid metallic cube is 50 cm. It is melted and recast into 8000 similar solid cubical dice. Find the side of each die
- 2.5 cm
- 6 cm
- 1.9 cm
- 5.2 cm
Correct

Incorrect

Hint

Volume $= (50)^3 = 125000\ cm^3$
Let $x$ cm be the side of a cubical dice volume of $8000$ dice $=8000\times x^3$
A/Q $8000\times x^3=125000$
$x=\dfrac{5}{2}2.5\ cm$
Question 28 of 30

28. Question
1 point(s)
A cylindrical vessel contains 49.896 liters of liquid. The cost of painting its C.S.A at 2 paise/sq cm is ‘ 95.04. Find its total surface area.
- $5724\ cm^2 $
- $ 7524\ cm^2 $
- $5742\ cm^2 $
- $7254\ cm^2 $
Correct

Incorrect

Hint

$V=49.896\times1000\ cm^3\\ \\ \pi r^2h=49896\\ \\ \text{CSA}=\dfrac{9504}{2}\\ \\ \therefore2\pi rh=4752\\ \\ \dfrac{\pi r^2h}{2\pi rh}=\dfrac{49896}{4752}\\ \\ r=21,h=36\\ \\ TSA=2\pi r(h+r)=7524\ cm^2$
Question 29 of 30

29. Question
1 point(s)
A right angled triangle $ABC $, where $\angle B=90^o $ is rotated about $BC $. If $BC=16\ cm $ and $AC=20\ cm $. Find the volume of the right circular cone.
- $2413\ cm^3 $
- $2311\ cm^3 $
- $1254\ cm^3 $
- $1725\ cm^3 $
Correct

Incorrect

Hint

$r=\sqrt{l^2-h^2}=\sqrt{20^2-16^2}=12\ cm\\ \\ \text{volume}=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times\dfrac{22}{7}\times12^2\times16\\ \\ =\dfrac{16896}{7}=2413\ cm^3$
Question 30 of 30

30. Question
1 point(s)
A cylindrical can whose base is horizontal and of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed in the can, water just covers the sphere. Find the depth of water in the can before the sphere was put into it.
- 2.3 cm
- 5.1 cm
- 1.5 cm
- 3.2 cm
Correct

Incorrect

Hint

Let the initial depth of water be $x$ cm.
Volume $=\pi r^2h=\pi\times(3.5)^2\times x\ cm^3$
volume of the sphere $=\dfrac{4}{3}\pi r^3$
$=\dfrac{4}{3}\pi(3.5)^3$
A/Q
$\pi\times(3.5)^2\times x\dfrac{4}{3}\pi\times(3.5)^3=\pi\times\left(\dfrac{7}{2}\right)^2\times7\\ \\ \Rightarrow x+\dfrac{4}{3}\times3.5=7\\ \\ \Rightarrow x+\dfrac{14}{3}=7\\ \\ \Rightarrow x=\dfrac{7}{3}=2.3\ cm$