Area & volume
1. The two sides of a triangle are 32 and 68. the area is 960 sq.cm. find the third side of triangle
Solution:
we see 68^2-32^2
=(68+32)*(68-32)
=100*36=60^2
now (1/2)*60*32
=960(match with given options)
(i.e area of a right angled triangle whose sides are 32,60,68)
third side=60
2. Smita was making a cube with dimensions 5*5*5 using 1*1*1 cubes. What is the number of cubes needed to make a hollow cube looking of the same shape?
Solution:
first are so (5*5*5)=125
then to make hollow cube v muz take out the outr most cubes
so 5-2 and 5-2 and 5-2
so 125-[(5-2)*(5-2)*(5-2)]
=98
3. Arun makes a popular brand of icecream in a rectangular shaped bar 6cm long, 5cm wide and 4cm thick. To cut costs,the comapny had decided to reduce the volume of bar by19%. The thickness will remain the same, but the length and width will be decreased by the same percentage. the new width will be?
Solution:
volume=l*b*h
=6*5*2
=60cm^3
volume is reduced by 19%
new volume
=(100-19)100*60
=486000=48.6
thickness is same but length and breadth reduces by x%
So,new volume:
(x/100*6)(x/100*5)2=48.6
On Solving x=90
thus length and breadth reduced by 10%
new width=5-(10%of 5)
=4.5
4. There is a rectangular terrace whose length and width are 60m X 20m. There is a walkway of uniform width around terrace. Area of walkway is 516m^2. Find width of walkway:
Solution:
let the width be x
then area of walkway is
(60+2x)(20+2x)-(60x20)=510
-->4x2+160x-516=0
roots will be 3 and -43
since it cannot be negative so answer will be 3
5. A hall is 15m long and 12m broad.If the sum of the areas of the floor and the ceiling is equal to the sum of areas of the four walls, the volume of hall is?
Solution:
area of 4 wall=sum of area of floor and celling
2(l+b)*h=2(l*b)
(15+12)*h=2(12*15)
on solving h=20/3m
volume=l*b*h
=12*15*20/3
=1200m^3
6. The rainfall recorded on a certain day was 5 cm.find the volume of water that fell on 2 hectare field.
Solution:
20,000 m sq * 0.05m
=1000m cube
1m cube=1000 lit
1000m cube=10^6 lit
7. if a ladder is 100m long and distance between bottom of a ladder and wall is 60m. what is the maximum size of the cube that can be placed between the ladder and wall
Solution:
Hypoteneus = 100 and base is 60 so perpen will be 80..
Now putting a cube inside gives two similar triangles....
Comparing perp to the base ratio we get
80/60 = (80-x)/x where x is side of the cube.
Solving we get x= 34.28
8. 2 cm of rain has fallen on a sq.km of land. Assuming that 50 % of the raindrops could have been collected and contained in a pool having a 100 m x 10 m base, by what level could the water level in the pool have increased ?
Solution:
area of pool base = 100 m * 10 m
= 1000 m^2
volume of rain
= (1 km)^2 * 2 cm
= 1000*1000*2/100
= 20000 m^3
50% of rain collected
= 20000*50/100
= 10000 m^3
so, level of pool increased
= volume of rain collected / area of pool base
= 10000/1000
= 10 m
9. Find the length of the longest pole that can be placed in an indoor stadium 24 m long, 18 m wide and 16 m high
Solution:
answer is 34.
sqrt(1156)=34
becoz max length is cuboid diagonal d
=sqrt(L^2+B^2+H^2)
where L lenght
B breadth & H height
10. The diagonal of a square is twice the side of equilateral triangle the ratio of Area of the Triangle to the Area of Square is?
Solution:
√3/4*(a)^2:1/2*(2a)^2
Thus √3:8
11. A square garden has fourteen posts along each side at equal interval. Find how many posts are there in all four sides:
Solution:
each side 14 means total will be calculated as
14+13+13+12
=52
12. Arun makes a popular brand of ice-cream in a rectangular shaped bar 6 cm long, 5 cm wide and 2 cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain the same, but the length and width will be decreased by the same percentage. The new width will be
Solution:
volume v=l*b*t
new volume is 81% of old v.
therefore new vol = 81%(v)=l'*b'*t
=> 81% (l*b*t) = l'*b'*t
=> 81% (l*b) = l' * b'
since the change in new l and b is the same. hence consider change to be x.
=> 81% (l*b) = xl*xb
81%=x^2
x=9%.
9% change in 5cm width is 0.45
therefore new width is 5-0.45
=4.55cm
13. A cow is tied to the rope of 19mts length. what is the increase in the area that the cow can move when the length of the rope is increase from 19 to 30
Solution:
reqd area
=pi*(R^2-r^2)
=pi*(30^2-19^2)
=pi*49*11
= 22/7 *49*11
= 1694 m^2
14. A circular swimming pool is surrounded by a 10 m concrete walk feet wide. If the area of the walk is 11/25 of the area of the pool. Then the radius of the pool in feet is
Solution:
let radius of pool be =r m,
area of wall
= pi*(r+10)^2-pi*r^2
= pi*[(r+10)^2-r^2]
given, pi*[(r+10)^2-r^2]
= (11/25)*pi*r^2
=> (100+20r)*25 = 11r^2
=> 11r^2-500r-2500=0
=> 11r^2-550r+50r-2500= 0
=> 11r(r-50)+50(r-500)=0
=> (11r+50)(r-50)=0
=> r= 50 m
15. ONE RECTANGULAR PLATE WITH LENGTH 8INCHES,BREADTH 11 INCHES AND 2 INCHES THICKNESS IS THERE. WHAT IS THE LENGTH OF THE CIRCULAR ROD WITH DIAMETER 18 INCHES AND EQUAL TO VOLUME OF RECTANGULAR PLATE?
Solution:
volume of plate = 8*11*2
=176 cubic inches
volume of plate (l*b*h*) will be
= volume of rod (pi*r*r*h)
Length of rod (h)
= 176/(pi*r*r)
= 176/(pi*9*9)
= .69 inch
16. A triangle was cut out of a square which was originally cut out of a circle.if the diameter of circle was 4cms, the area of triangle is at the most?
Solution:
(1/2)*2*4 = 4
height will be radius of circle
base will be diam
17. Raj drives slowly along the perimeter of a rectangular park at 24 kmph and completes one full round in 4 minutes. If the ratio of the length to the breadth of the park is 3:2, what are its dimensions?
Solution:
let the length be=3x,and breadth be=2x
so the perimeter=2*(3x+2x)
=10x
we know that,24km/hr=24*5/18 m/s
=20/3 m/s
ram is needed 240 seconds to drive 10x m
so he is needed 1 sec to drive
=10x/240 m
= x/24m
so,
x/24=20/3
x=160
so the length of the park
=160*3=480m
and the breadth of the park
=160*2=320m
18. If a rectangle is divided into four parts of different size of rectangles. Area of the 1st part is 76, area of 2nd part is 4, area of 3rd part is 8. Then find out the area of fourth part of the rectangle.
Solution:
let area of 4th part is X then
if part 1,2,3,4 are in clockwise then
76/x = 4/8
=> x = 152
19. If the area of a square region having sides of length 6 cms is equal to the area of a rectangular region having width 2.5 cms, then the length of the rectangle, in cms, is
Solution:
Area of sq=6*6=36
Area of rect=l*2.5
Then 36=l*2.5
L=36/2.5
=14.4
20. Two full tanks, one shaped like a cylinder and the other like a cone. contain liquid fuel. The cylindrical held 500 litres more than the conical tank After 200 litres of fuel is pumped out from each tank the cylindrical tank now contains twice the amount of fuel in the conical tank Howmany litres of fuel did the cylindrical tank have when it was full.
Solution:
Let conical tank has x ltr of fuel then cylindrical tank hold x+500 ltr
A/Q
x-200=x+500-200
2(x-200)=x-300
2x-400=x-300
x=700
so
cylindrical tank has x+500=1200
21. He volume of water inside a swimming pool doubles every hour. If the pool is filled to its full capacity with in 8 hrs ,in how many hours was it filled to one quarter of its capacity?
Solution:
The volume of water inside a swimming pool doubles every hour,so
lets move in reverse direction
in 8th hour its full
in 7th hour it is 1/2
in 6th hour it will be 1/4
so ans is 6 hour
22. A swimming pool whos surface forms a rectangle measures 25 m long by 15 m wide The pool is 2 m deep at the shallow end and the depth increases at a constant rate to 4 m deep at the other end How many cubic meters of water will pool hold
Solution:
ool Dimensions:
Length 25 metres
Width 15 metres
Depth 2 metres to 4 metres
(average 3 metres) Volume
= 25 × 15 x 3
= 1125 cubic metres
23. Rectangular tile each of size 70cm by 30cm must be laid horizontally on a rectangular floor of size 110cm by 130cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is:
Solution:
Area of tile
= 70*30 = 2100
Area of floor
= 130*110 = 14300
No of tiles
= 14300/2100
= 6.8
So, the no of tile = 6
24. arun makes a popular a brand of icecream in a rectangular shapeds bar 6cm long,5cm wide and 2cm thick to cut costs the company has decided to reduce the volume of the bar by 19% the thickness will remain the same but the length and width will be reduced by same percentage the new width will be
Solution:
Volume =l×b×h = 6×5×2 = 60 cm3
Now volume is reduced by 19%.
Therefore, new volume = (100−19)100×60=48.6
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume:
(x100×6)(x100×5)2=48.6
Solving we get x =90
thus length and width is reduced by 10%
New width
= 5-(10% of 5)
=4.5
25. Aman walking at the speed of 4 km/h crosses a square field diagonally in 3 minutes. The area of the field (in m2) is:
Solution:
4KMPH=10/9 M/SEC
DIST=S*T
=10/9 * 3*60
=200 METERS
AREA OF SQUARE
= 1/2 DIAGONAL SQUARE
=20000 M2
26. Two sides of a plot measure 48m 64m and the angle between them is a right angle.the other two sides measure 50m each and the other three angle are not right angle.if the plot is not convex what is the area of the plot
Solution:
area of 1st triangle
=1/2*64*48=1536
hyp of 1st triangle
=root under 48^2+64^2
=80
2nd triangle is isosceles so height
=sqrt 2500-1600
=30
so are of 2nd triangle
=1/2*80*30=1200
so total area=1536+1200
=2736
27. Leena cut small cubes of 3 cubic cm each. She joined it to make a cuboid of length 10 cm, width 3 cm and depth 3 cm. How many more cubes does she need to make a perfect cube?
Solution:
10*3*3=90 cuboid,
to make a perfect cuboid it size should be of
10*10*10=1000
so extra cubes
=1000-90=910
28. A boy want to make a cuboid of dimension 5m, 6m, 7m from small cubes of 0.04 m3. Later he realized he can make same cuboid by making it hollow. Then it takes some cubes less. What is the no. of these cubes?
Solution:
volume of cuboid=5*6*7=210m3
volume of inner cuboid
=(5-1)*(6-1)*(7-1)
=120m3
therefore,volume of hollow cuboid
=210-120
=90 m3
no of cubes required
=90/.04
=2250 cubes
29. A lady builds 9cm length, 12cm width, and 3cm height box using 3cubic cm cubes. What is the minimum number of cubes required to build the box?
Solution:
volume of cube =3
volume of new cube
=9*12*3=324
no.of cubes=324/3=108
30. Smita was making a cube with dimensions 5*5*5 using 1*1*1 cubes. What is the number of cubes needed to make a hollow cube looking of the same shape?
Solution:
because 5*5*5=125 cubes needed and to make it hollow we do not count inner once
therefore,
125-3*3*3
=125-27=98
31. Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square
Solution:
Let the radius of circle be r
let the side of square be a
then diagonal of square= a*sqrt(2)
This diagonal length = 12*r + 2r * sqrt(2)
(because the extreme circle's radius is perpendicular to side of square.)
Thus we get
12*r+2r*sqrt(2)=a*sqrt(2)
r(6*sqrt(2)+2)=a
r/a=1/(6*sqrt(2)+2)
Thus ratio: r:a = 1:(6*sqrt(2)+2)
Ans: 1:(6*sqrt(2)+2)
32. An equilateral triangle is inscribed in a square. The length of each side of the triangle is equal to half the length of the diagonals of the square. Find the ratio of the areas of the triangle to that of the square.
Solution:
length of side of triangle =1/2 diagonal of square
so diagonal =2(side)of triangle
let side be a
so diagonal=2a
now area of eqiulateral triangle
=sqroot(3)/4 x side*2
so ratio becomes root(3)/4 x side*2 / .5x diagonal*2
solving this we get sqroot(3)/8
33. In a building there are 5 rooms.Each having a equal area .The length of the room is 4m and breadht is 5m.The height of the rooms are 2m.If 17 bricks are needed to make a square meter then how many bricks are needed to make the floor of a particular room?
Solution:
as area of floor =area of wall
=4*5=20 sq m
1 sq m= 17 bricks
then 20 sq m
=20*17=340 bricks
34. a hollow cube of size 5 cm is taken,thickness=1 cm. it is made into smaller cubes of size 1 cm, if 1 face of the outer surface of the cube is painted, how many faces of the smaller cubes remain unpainted?
Solution:
total no of smaller cubes
= 25+25+15+15+9+9=98
painted face of bigger cube=1=25 small faces
total faces of smaller cube
=98*6=588
remaining unpainted faces
=588-25=563
35. Alvin, Ben, and Clinton run a race, with Alvin finishing 48 meters ahead of Ben,and 72 meters ahead of Clinton while runner Ben finishes 32 meters ahead of runner Clinton. Each runner travels the entire distance at constant speed. What is the length of the race?
Solution:
let whole distance is 'd' and speed of Alvin, Ben, and Clinton is A,B,C,,,
then
A*t=d .... 1
B*t=(d-48).....2
C*t=(d-72).....3
so
from eqt 1,2 & 3
A:B:C::d:(d-48):(d-72).....4
now second condition is..
B:C::d:(d-32)....5
fro eqtn 4 and 5..
(d-48)/(d-72)=d/(d-32)...
on solving(or by option) d= 192
36. A rectangular box has six surfaces. The area of its front is 54 square inches; the area of its side is 72 square inches; and the area of its top is 108 square inches. What is the volume of the box?
Solution:
l*b=108
b*h=72
l*h=54
Solving
l=9
h=6
b=12
Volume=l*b*h
=9*6*12=648
37. A hollow cube of size 5cm is taken having thickness of 1 cm. It is made up of small cubes of size 1 cm . if 1 face of outer surface of cube are painted then find total how many faces of cube remain unpainted.
Solution:
6(5*5*5-3*3*3)=6(125-27)
=6(98)
=588
outer surface 1 face contain 25 smaller face
Unpainted face is=588-25=563
Ans:563
38. A rectangle ice cream was prepared with length 6cm, thicknes 2cm, height 5cm. if the volume of the ice cream was reduced by 19% and thickness remained the same , and height and length was reduced by same %. Then what is the new height?
Solution:
ans:4.5cm
volume=6*2*5=60
new volume=81% of 60=48.6
let height and length are now x% of previous
new volume=2*(x/100)*5*(x/100)*6=48.6
x=90
new height =90% of 5
=4.5
39. A semi circle is drawn with ab as diameter. from c a point on ab a line perpendicular toa ab is drawn meeting the circumference at D. given that ac=2cm and cd =6cm the ared of the semicircle is sq cm
Solution:
from two similar tri(ADC) & tri(BDC),
ac/cd=cd/bc
bc= cd^2/ac
= 6*6/2
=18
ab= ac+bc= 2+18= 20cm
radius, r= 20/2=10cm
area of the semi circle
= pi* r^2/2= 50pi
40. Perimeter of a equilateral triangle is equal to the perimeter of Hexagon. What is the ratio of their areas?
Solution:
let side of eq. triangle be x & side of hexagon be y then
Perimeter of a eq. triangle = perimeter of Hexagon => 3x=6y => x= 2y
area of eq. triangle (√3/4) * x^2
---------------------- = ------------------- = 2/3 = 2:3 [put x=2y]
area of hexagon 6* (√3/4)* y^2
ans 2:3
41. A rectangular plate with length 8 inches, breadth 11 inches and thickness 2 inches is available. What is the length of the circular rod with diameter 10 inches and equal to the volume of the rectangular plate?
Solution:
Volume of rectangular plate
=8*11*2
=176 sq.inch
Volume of the rod(cylinder)with r=5 & length=h is given by
Pi*(r^2)*h=176
(22/7)*(5^2)*h=176,
h=176*7/(22*25)
=1232/550
= 2.24
42. Susan made a block with small cubes of 5 cubic cm volume to make a block 7 small cubes long, 7 small cubes wide and 6 small cubes deep. She realise that she has used more small cubes than she really needed. She realised that she could have glued a fewer number of cubes together ot look like a block with same dimensions, if it were made
Solution:
Volume of cube of dimensions 7 7 6 is
7*7*6=294
If it were to look hollow the imaginary inside cube volume is to be deducted
which is
(7-2)*(7-2)*(6-2)=100
(Here 2 is deducted because the inside cube which has to be hollow is 2 cubes more on every dimension)
so minimum number of cubes required to make hollow cube is
=294-100
=194 cubes
43. Two equilateral triangles of side 12cm are placed one on top of another, such that a 6 pointed star is formed .if the six verticals lie on a circle what is the area of the circle not enclosed by the star ?(Give the examples to the nearest sq cm)
Solution:
12 equilateral triangle
6root2+2root2=diameter of circle
area of circle
=pie(4root2)^2
=150.2
area of 12 triangles
=12(root 3/4)*4*4
=80.2
area not enclosed by stars
=150.2-80.2
=68
44. A parallelogram is circumscribed about a circle. The lengths of its diagonals are 30 cm and 16 cm. The perimeter of the parallelogram is?
Solution:
the parallelogrm is here a rhombos..so two diagonals intersect each other at 90 degrees
so we get 4 right angled triangles whose two sides are
(30/2)=15 &
(16/2)=8
so the other side is
=root(15^2+8^2)=17
so perimeter is
=4*17=68
45. Find the radius of the circle inscribed in the triangle ABC, having sides 10 cm, 10 cm and 16 cm.
Solution:
Radius of an incircle is given by 2a/p
where a is the area and
p is the perineter of the triangle.
Here p=10+10+16=36cm and
area=(s(s-10)(s-10)(s-16))^(1/2)
where s=p/2=18.
Hence area=48sq.cm.
So radius of incircle
=2*48/36=8/3
=2.66cm
46. From a point within an equilateral triangle, perpendiculars are drawn to the three sides and are 3, 7 and 11 cm in length. Then the perimeter of the triangle (in cm) is
Solution:
Taking 3 perpendiculars within the triangle from the same point and joining that point to the corners of the triangle, makes three triangles within the equilateral triangle.
Given the sides of the perpendicular as 3cm, 7cm and 11cm.
Let the side of the equilateral triangle be a
Then,
Area of the equilateral triangle = (3^(1/2)*(a^2))/4
So √3a^2/4 = 1/2*3*a + 1/2*7*a + 1/2*11*a
= 1/2(21a)
= 21a/2
=> a = 42/√3
=> 3a = 42*√3
Thus, the perimeter of the triangle is
42√3 cm.
47. Two equilateral triangles of sides 12cms are placed one on top of another,such that a six pointed star is formed. If the six vertices lie on the circle,what is the area of the circle not enclosed by the star?
Solution:
Eq. tri Area= side2*√3/4
Given Side=12 cm
One eq.triangle placed over other they cut the line in ratio 1:3
Side of smaller triangle
=12/3=4 cm
Total area covered by two triangles
=((12*12)+3(4*4))*√3/4
=48√3
Radius of circle
=side/√3=12/√3
Area of Circle=A
=πr^2=48π
Uncovered area
=48(π-√3)
48. Leena cut cubes of 10cm dimension each.She joined it to make a cuboid of length 100cm width 50cm and depth 50cm.How many more cubes does she need to make a perfect cube?
Solution:
volume of large cube= 100*100*100
volume of small cube = 10*10*10
total cubes needed for large cube= 1000
volume of cuboid= 100*50*50
number of small cubes used for building the cuboid = 250
remaining small cubes needed for the perfect cube
= 1000- 250
so the answer is 750.
Solution:
we see 68^2-32^2
=(68+32)*(68-32)
=100*36=60^2
now (1/2)*60*32
=960(match with given options)
(i.e area of a right angled triangle whose sides are 32,60,68)
third side=60
2. Smita was making a cube with dimensions 5*5*5 using 1*1*1 cubes. What is the number of cubes needed to make a hollow cube looking of the same shape?
Solution:
first are so (5*5*5)=125
then to make hollow cube v muz take out the outr most cubes
so 5-2 and 5-2 and 5-2
so 125-[(5-2)*(5-2)*(5-2)]
=98
3. Arun makes a popular brand of icecream in a rectangular shaped bar 6cm long, 5cm wide and 4cm thick. To cut costs,the comapny had decided to reduce the volume of bar by19%. The thickness will remain the same, but the length and width will be decreased by the same percentage. the new width will be?
Solution:
volume=l*b*h
=6*5*2
=60cm^3
volume is reduced by 19%
new volume
=(100-19)100*60
=486000=48.6
thickness is same but length and breadth reduces by x%
So,new volume:
(x/100*6)(x/100*5)2=48.6
On Solving x=90
thus length and breadth reduced by 10%
new width=5-(10%of 5)
=4.5
4. There is a rectangular terrace whose length and width are 60m X 20m. There is a walkway of uniform width around terrace. Area of walkway is 516m^2. Find width of walkway:
Solution:
let the width be x
then area of walkway is
(60+2x)(20+2x)-(60x20)=510
-->4x2+160x-516=0
roots will be 3 and -43
since it cannot be negative so answer will be 3
5. A hall is 15m long and 12m broad.If the sum of the areas of the floor and the ceiling is equal to the sum of areas of the four walls, the volume of hall is?
Solution:
area of 4 wall=sum of area of floor and celling
2(l+b)*h=2(l*b)
(15+12)*h=2(12*15)
on solving h=20/3m
volume=l*b*h
=12*15*20/3
=1200m^3
6. The rainfall recorded on a certain day was 5 cm.find the volume of water that fell on 2 hectare field.
Solution:
20,000 m sq * 0.05m
=1000m cube
1m cube=1000 lit
1000m cube=10^6 lit
7. if a ladder is 100m long and distance between bottom of a ladder and wall is 60m. what is the maximum size of the cube that can be placed between the ladder and wall
Solution:
Hypoteneus = 100 and base is 60 so perpen will be 80..
Now putting a cube inside gives two similar triangles....
Comparing perp to the base ratio we get
80/60 = (80-x)/x where x is side of the cube.
Solving we get x= 34.28
8. 2 cm of rain has fallen on a sq.km of land. Assuming that 50 % of the raindrops could have been collected and contained in a pool having a 100 m x 10 m base, by what level could the water level in the pool have increased ?
Solution:
area of pool base = 100 m * 10 m
= 1000 m^2
volume of rain
= (1 km)^2 * 2 cm
= 1000*1000*2/100
= 20000 m^3
50% of rain collected
= 20000*50/100
= 10000 m^3
so, level of pool increased
= volume of rain collected / area of pool base
= 10000/1000
= 10 m
9. Find the length of the longest pole that can be placed in an indoor stadium 24 m long, 18 m wide and 16 m high
Solution:
answer is 34.
sqrt(1156)=34
becoz max length is cuboid diagonal d
=sqrt(L^2+B^2+H^2)
where L lenght
B breadth & H height
10. The diagonal of a square is twice the side of equilateral triangle the ratio of Area of the Triangle to the Area of Square is?
Solution:
√3/4*(a)^2:1/2*(2a)^2
Thus √3:8
11. A square garden has fourteen posts along each side at equal interval. Find how many posts are there in all four sides:
Solution:
each side 14 means total will be calculated as
14+13+13+12
=52
12. Arun makes a popular brand of ice-cream in a rectangular shaped bar 6 cm long, 5 cm wide and 2 cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain the same, but the length and width will be decreased by the same percentage. The new width will be
Solution:
volume v=l*b*t
new volume is 81% of old v.
therefore new vol = 81%(v)=l'*b'*t
=> 81% (l*b*t) = l'*b'*t
=> 81% (l*b) = l' * b'
since the change in new l and b is the same. hence consider change to be x.
=> 81% (l*b) = xl*xb
81%=x^2
x=9%.
9% change in 5cm width is 0.45
therefore new width is 5-0.45
=4.55cm
13. A cow is tied to the rope of 19mts length. what is the increase in the area that the cow can move when the length of the rope is increase from 19 to 30
Solution:
reqd area
=pi*(R^2-r^2)
=pi*(30^2-19^2)
=pi*49*11
= 22/7 *49*11
= 1694 m^2
14. A circular swimming pool is surrounded by a 10 m concrete walk feet wide. If the area of the walk is 11/25 of the area of the pool. Then the radius of the pool in feet is
Solution:
let radius of pool be =r m,
area of wall
= pi*(r+10)^2-pi*r^2
= pi*[(r+10)^2-r^2]
given, pi*[(r+10)^2-r^2]
= (11/25)*pi*r^2
=> (100+20r)*25 = 11r^2
=> 11r^2-500r-2500=0
=> 11r^2-550r+50r-2500= 0
=> 11r(r-50)+50(r-500)=0
=> (11r+50)(r-50)=0
=> r= 50 m
15. ONE RECTANGULAR PLATE WITH LENGTH 8INCHES,BREADTH 11 INCHES AND 2 INCHES THICKNESS IS THERE. WHAT IS THE LENGTH OF THE CIRCULAR ROD WITH DIAMETER 18 INCHES AND EQUAL TO VOLUME OF RECTANGULAR PLATE?
Solution:
volume of plate = 8*11*2
=176 cubic inches
volume of plate (l*b*h*) will be
= volume of rod (pi*r*r*h)
Length of rod (h)
= 176/(pi*r*r)
= 176/(pi*9*9)
= .69 inch
16. A triangle was cut out of a square which was originally cut out of a circle.if the diameter of circle was 4cms, the area of triangle is at the most?
Solution:
(1/2)*2*4 = 4
height will be radius of circle
base will be diam
17. Raj drives slowly along the perimeter of a rectangular park at 24 kmph and completes one full round in 4 minutes. If the ratio of the length to the breadth of the park is 3:2, what are its dimensions?
Solution:
let the length be=3x,and breadth be=2x
so the perimeter=2*(3x+2x)
=10x
we know that,24km/hr=24*5/18 m/s
=20/3 m/s
ram is needed 240 seconds to drive 10x m
so he is needed 1 sec to drive
=10x/240 m
= x/24m
so,
x/24=20/3
x=160
so the length of the park
=160*3=480m
and the breadth of the park
=160*2=320m
18. If a rectangle is divided into four parts of different size of rectangles. Area of the 1st part is 76, area of 2nd part is 4, area of 3rd part is 8. Then find out the area of fourth part of the rectangle.
Solution:
let area of 4th part is X then
if part 1,2,3,4 are in clockwise then
76/x = 4/8
=> x = 152
19. If the area of a square region having sides of length 6 cms is equal to the area of a rectangular region having width 2.5 cms, then the length of the rectangle, in cms, is
Solution:
Area of sq=6*6=36
Area of rect=l*2.5
Then 36=l*2.5
L=36/2.5
=14.4
20. Two full tanks, one shaped like a cylinder and the other like a cone. contain liquid fuel. The cylindrical held 500 litres more than the conical tank After 200 litres of fuel is pumped out from each tank the cylindrical tank now contains twice the amount of fuel in the conical tank Howmany litres of fuel did the cylindrical tank have when it was full.
Solution:
Let conical tank has x ltr of fuel then cylindrical tank hold x+500 ltr
A/Q
x-200=x+500-200
2(x-200)=x-300
2x-400=x-300
x=700
so
cylindrical tank has x+500=1200
21. He volume of water inside a swimming pool doubles every hour. If the pool is filled to its full capacity with in 8 hrs ,in how many hours was it filled to one quarter of its capacity?
Solution:
The volume of water inside a swimming pool doubles every hour,so
lets move in reverse direction
in 8th hour its full
in 7th hour it is 1/2
in 6th hour it will be 1/4
so ans is 6 hour
22. A swimming pool whos surface forms a rectangle measures 25 m long by 15 m wide The pool is 2 m deep at the shallow end and the depth increases at a constant rate to 4 m deep at the other end How many cubic meters of water will pool hold
Solution:
ool Dimensions:
Length 25 metres
Width 15 metres
Depth 2 metres to 4 metres
(average 3 metres) Volume
= 25 × 15 x 3
= 1125 cubic metres
23. Rectangular tile each of size 70cm by 30cm must be laid horizontally on a rectangular floor of size 110cm by 130cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is:
Solution:
Area of tile
= 70*30 = 2100
Area of floor
= 130*110 = 14300
No of tiles
= 14300/2100
= 6.8
So, the no of tile = 6
24. arun makes a popular a brand of icecream in a rectangular shapeds bar 6cm long,5cm wide and 2cm thick to cut costs the company has decided to reduce the volume of the bar by 19% the thickness will remain the same but the length and width will be reduced by same percentage the new width will be
Solution:
Volume =l×b×h = 6×5×2 = 60 cm3
Now volume is reduced by 19%.
Therefore, new volume = (100−19)100×60=48.6
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume:
(x100×6)(x100×5)2=48.6
Solving we get x =90
thus length and width is reduced by 10%
New width
= 5-(10% of 5)
=4.5
25. Aman walking at the speed of 4 km/h crosses a square field diagonally in 3 minutes. The area of the field (in m2) is:
Solution:
4KMPH=10/9 M/SEC
DIST=S*T
=10/9 * 3*60
=200 METERS
AREA OF SQUARE
= 1/2 DIAGONAL SQUARE
=20000 M2
26. Two sides of a plot measure 48m 64m and the angle between them is a right angle.the other two sides measure 50m each and the other three angle are not right angle.if the plot is not convex what is the area of the plot
Solution:
area of 1st triangle
=1/2*64*48=1536
hyp of 1st triangle
=root under 48^2+64^2
=80
2nd triangle is isosceles so height
=sqrt 2500-1600
=30
so are of 2nd triangle
=1/2*80*30=1200
so total area=1536+1200
=2736
27. Leena cut small cubes of 3 cubic cm each. She joined it to make a cuboid of length 10 cm, width 3 cm and depth 3 cm. How many more cubes does she need to make a perfect cube?
Solution:
10*3*3=90 cuboid,
to make a perfect cuboid it size should be of
10*10*10=1000
so extra cubes
=1000-90=910
28. A boy want to make a cuboid of dimension 5m, 6m, 7m from small cubes of 0.04 m3. Later he realized he can make same cuboid by making it hollow. Then it takes some cubes less. What is the no. of these cubes?
Solution:
volume of cuboid=5*6*7=210m3
volume of inner cuboid
=(5-1)*(6-1)*(7-1)
=120m3
therefore,volume of hollow cuboid
=210-120
=90 m3
no of cubes required
=90/.04
=2250 cubes
29. A lady builds 9cm length, 12cm width, and 3cm height box using 3cubic cm cubes. What is the minimum number of cubes required to build the box?
Solution:
volume of cube =3
volume of new cube
=9*12*3=324
no.of cubes=324/3=108
30. Smita was making a cube with dimensions 5*5*5 using 1*1*1 cubes. What is the number of cubes needed to make a hollow cube looking of the same shape?
Solution:
because 5*5*5=125 cubes needed and to make it hollow we do not count inner once
therefore,
125-3*3*3
=125-27=98
31. Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square
Solution:
Let the radius of circle be r
let the side of square be a
then diagonal of square= a*sqrt(2)
This diagonal length = 12*r + 2r * sqrt(2)
(because the extreme circle's radius is perpendicular to side of square.)
Thus we get
12*r+2r*sqrt(2)=a*sqrt(2)
r(6*sqrt(2)+2)=a
r/a=1/(6*sqrt(2)+2)
Thus ratio: r:a = 1:(6*sqrt(2)+2)
Ans: 1:(6*sqrt(2)+2)
32. An equilateral triangle is inscribed in a square. The length of each side of the triangle is equal to half the length of the diagonals of the square. Find the ratio of the areas of the triangle to that of the square.
Solution:
length of side of triangle =1/2 diagonal of square
so diagonal =2(side)of triangle
let side be a
so diagonal=2a
now area of eqiulateral triangle
=sqroot(3)/4 x side*2
so ratio becomes root(3)/4 x side*2 / .5x diagonal*2
solving this we get sqroot(3)/8
33. In a building there are 5 rooms.Each having a equal area .The length of the room is 4m and breadht is 5m.The height of the rooms are 2m.If 17 bricks are needed to make a square meter then how many bricks are needed to make the floor of a particular room?
Solution:
as area of floor =area of wall
=4*5=20 sq m
1 sq m= 17 bricks
then 20 sq m
=20*17=340 bricks
34. a hollow cube of size 5 cm is taken,thickness=1 cm. it is made into smaller cubes of size 1 cm, if 1 face of the outer surface of the cube is painted, how many faces of the smaller cubes remain unpainted?
Solution:
total no of smaller cubes
= 25+25+15+15+9+9=98
painted face of bigger cube=1=25 small faces
total faces of smaller cube
=98*6=588
remaining unpainted faces
=588-25=563
35. Alvin, Ben, and Clinton run a race, with Alvin finishing 48 meters ahead of Ben,and 72 meters ahead of Clinton while runner Ben finishes 32 meters ahead of runner Clinton. Each runner travels the entire distance at constant speed. What is the length of the race?
Solution:
let whole distance is 'd' and speed of Alvin, Ben, and Clinton is A,B,C,,,
then
A*t=d .... 1
B*t=(d-48).....2
C*t=(d-72).....3
so
from eqt 1,2 & 3
A:B:C::d:(d-48):(d-72).....4
now second condition is..
B:C::d:(d-32)....5
fro eqtn 4 and 5..
(d-48)/(d-72)=d/(d-32)...
on solving(or by option) d= 192
36. A rectangular box has six surfaces. The area of its front is 54 square inches; the area of its side is 72 square inches; and the area of its top is 108 square inches. What is the volume of the box?
Solution:
l*b=108
b*h=72
l*h=54
Solving
l=9
h=6
b=12
Volume=l*b*h
=9*6*12=648
37. A hollow cube of size 5cm is taken having thickness of 1 cm. It is made up of small cubes of size 1 cm . if 1 face of outer surface of cube are painted then find total how many faces of cube remain unpainted.
Solution:
6(5*5*5-3*3*3)=6(125-27)
=6(98)
=588
outer surface 1 face contain 25 smaller face
Unpainted face is=588-25=563
Ans:563
38. A rectangle ice cream was prepared with length 6cm, thicknes 2cm, height 5cm. if the volume of the ice cream was reduced by 19% and thickness remained the same , and height and length was reduced by same %. Then what is the new height?
Solution:
ans:4.5cm
volume=6*2*5=60
new volume=81% of 60=48.6
let height and length are now x% of previous
new volume=2*(x/100)*5*(x/100)*6=48.6
x=90
new height =90% of 5
=4.5
39. A semi circle is drawn with ab as diameter. from c a point on ab a line perpendicular toa ab is drawn meeting the circumference at D. given that ac=2cm and cd =6cm the ared of the semicircle is sq cm
Solution:
from two similar tri(ADC) & tri(BDC),
ac/cd=cd/bc
bc= cd^2/ac
= 6*6/2
=18
ab= ac+bc= 2+18= 20cm
radius, r= 20/2=10cm
area of the semi circle
= pi* r^2/2= 50pi
40. Perimeter of a equilateral triangle is equal to the perimeter of Hexagon. What is the ratio of their areas?
Solution:
let side of eq. triangle be x & side of hexagon be y then
Perimeter of a eq. triangle = perimeter of Hexagon => 3x=6y => x= 2y
area of eq. triangle (√3/4) * x^2
---------------------- = ------------------- = 2/3 = 2:3 [put x=2y]
area of hexagon 6* (√3/4)* y^2
ans 2:3
41. A rectangular plate with length 8 inches, breadth 11 inches and thickness 2 inches is available. What is the length of the circular rod with diameter 10 inches and equal to the volume of the rectangular plate?
Solution:
Volume of rectangular plate
=8*11*2
=176 sq.inch
Volume of the rod(cylinder)with r=5 & length=h is given by
Pi*(r^2)*h=176
(22/7)*(5^2)*h=176,
h=176*7/(22*25)
=1232/550
= 2.24
42. Susan made a block with small cubes of 5 cubic cm volume to make a block 7 small cubes long, 7 small cubes wide and 6 small cubes deep. She realise that she has used more small cubes than she really needed. She realised that she could have glued a fewer number of cubes together ot look like a block with same dimensions, if it were made
Solution:
Volume of cube of dimensions 7 7 6 is
7*7*6=294
If it were to look hollow the imaginary inside cube volume is to be deducted
which is
(7-2)*(7-2)*(6-2)=100
(Here 2 is deducted because the inside cube which has to be hollow is 2 cubes more on every dimension)
so minimum number of cubes required to make hollow cube is
=294-100
=194 cubes
43. Two equilateral triangles of side 12cm are placed one on top of another, such that a 6 pointed star is formed .if the six verticals lie on a circle what is the area of the circle not enclosed by the star ?(Give the examples to the nearest sq cm)
Solution:
12 equilateral triangle
6root2+2root2=diameter of circle
area of circle
=pie(4root2)^2
=150.2
area of 12 triangles
=12(root 3/4)*4*4
=80.2
area not enclosed by stars
=150.2-80.2
=68
44. A parallelogram is circumscribed about a circle. The lengths of its diagonals are 30 cm and 16 cm. The perimeter of the parallelogram is?
Solution:
the parallelogrm is here a rhombos..so two diagonals intersect each other at 90 degrees
so we get 4 right angled triangles whose two sides are
(30/2)=15 &
(16/2)=8
so the other side is
=root(15^2+8^2)=17
so perimeter is
=4*17=68
45. Find the radius of the circle inscribed in the triangle ABC, having sides 10 cm, 10 cm and 16 cm.
Solution:
Radius of an incircle is given by 2a/p
where a is the area and
p is the perineter of the triangle.
Here p=10+10+16=36cm and
area=(s(s-10)(s-10)(s-16))^(1/2)
where s=p/2=18.
Hence area=48sq.cm.
So radius of incircle
=2*48/36=8/3
=2.66cm
46. From a point within an equilateral triangle, perpendiculars are drawn to the three sides and are 3, 7 and 11 cm in length. Then the perimeter of the triangle (in cm) is
Solution:
Taking 3 perpendiculars within the triangle from the same point and joining that point to the corners of the triangle, makes three triangles within the equilateral triangle.
Given the sides of the perpendicular as 3cm, 7cm and 11cm.
Let the side of the equilateral triangle be a
Then,
Area of the equilateral triangle = (3^(1/2)*(a^2))/4
So √3a^2/4 = 1/2*3*a + 1/2*7*a + 1/2*11*a
= 1/2(21a)
= 21a/2
=> a = 42/√3
=> 3a = 42*√3
Thus, the perimeter of the triangle is
42√3 cm.
47. Two equilateral triangles of sides 12cms are placed one on top of another,such that a six pointed star is formed. If the six vertices lie on the circle,what is the area of the circle not enclosed by the star?
Solution:
Eq. tri Area= side2*√3/4
Given Side=12 cm
One eq.triangle placed over other they cut the line in ratio 1:3
Side of smaller triangle
=12/3=4 cm
Total area covered by two triangles
=((12*12)+3(4*4))*√3/4
=48√3
Radius of circle
=side/√3=12/√3
Area of Circle=A
=πr^2=48π
Uncovered area
=48(π-√3)
48. Leena cut cubes of 10cm dimension each.She joined it to make a cuboid of length 100cm width 50cm and depth 50cm.How many more cubes does she need to make a perfect cube?
Solution:
volume of large cube= 100*100*100
volume of small cube = 10*10*10
total cubes needed for large cube= 1000
volume of cuboid= 100*50*50
number of small cubes used for building the cuboid = 250
remaining small cubes needed for the perfect cube
= 1000- 250
so the answer is 750.
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