NCTB Class 8 Math Chapter Three Exercise 3 Solution by Math Expert. Bangladesh Board Class 8 Math Solution Chapter 3 Measurements Exercise 3 Solution.
Board |
NCTB |
Class |
8 |
Subject |
Mathematics |
Chapter |
3 |
Chapter Name |
Measurements |
Exercise |
3 Solution |
(10) The length of a pond is 60 meters and the breadth is 40 meters. If the breadth of its bank is 3 meters, find the area of the bank.
Solution:
Length of pond = 60m
Breadth of pond = 40m
Area = (60X40) Sqm
= 2400 Sqm
Length with bank of 3m = (60+ 3X2) m
= 66m.
Therefore, Breadth with bank of 3m = (40+ 3X2) m
= 46m.
Therefore, New area = (66X46) Sqm= 3036 Sqm
Therefore, area of the bank = (3036-2400) Sqm
= 636 Sqm.
(11) The area of a rectangle is 10 acres and its length is 4 times the breadth. What is the length of the rectangle in meters?
Solution:
We know, 1 acre = 4046.86 Sqm.
Therefore, 10 acre = (4046.86X10) Sqm
= 40468.6 Sqm
Let breadth = x m.
Length = 4x m.
Therefore, Area = (x X 4x) Sqm
= 4x^{2} Sqm.
Therefore, 4x^{2} = 40468.6
= x^{2} = 40468.6/4
= 10117.15
= x= √10117.15
= 100.5840
Therefore, Length (4X100.5840)
= 402.3361
= 402.34m (Answer)
(12) The length of a rectangular house is one and a half time its breadth. If the area of the house is 216 sq. meters, what is its perimeter?
Solution:
Let, breadth of house = X m.
Therefore, length of house = (x X 1 ½) = 3x/2 m.
Therefore, Area = (x X 3x/2) Sqm
= 3x^{2}/2 Sqm.
Therefore, 3x^{2}/2 = 216.
= 3x^{2} = 216X2
= x^{2} = 216X2/3
= x= √144
= x= 12
Therefore, Breadth = 12m
Length = 12X 1 ½
= 12X 3/2
= 18m.
Therefore, Perimeter = 2 (l+b)
= 2 (12+18)
= 60m (Answer)
(13) The base of triangular region is 24 metres and the height is 15 metres 50 cm. Find its area.
Solution:
Triangular region, base = 24m.
Height = 15m 50cm.
= (15+ 50/100)
= 1500+50/100
= 31/2 m.
Therefore, Area = ½ X b X h
= ½ X 24 X 31/2
= 186 Sqm (Answer).
(14) The length of a rectangle is 48 metres and is breadth is 32 metres 80 cm. there is a 3 metres wide road around outside. What is the area of the road?
Solution:
Length of rectangle = 48m.
Breadth of rectangle = 32m 80cm.
= (32+ 80/100) m
= 32.8m
Therefore, Area (l X b)
= (48X32.8)
= 1574.4 Sqm.
There is a road outside of breadth 3m.
Length of rectangle with road
= (48+ 3X2)
= 54m
Therefore, Breadth of rectangle = (32.8+ 3X2)
= 38.8m
Therefore, New Area = (54X38.8) Sqm
= 2055.2 Sqm
Therefore, Area of road = (2095.2-1574.4) Sqm
= 520.8 Sqm. (Answer)
(15) The length of one side of a square is 300 metres and around its outside, there is a 4 metres road wide. Find the area of the road.
Solution:
Length of one side square = 300m
Area = (300X300) Sqm
= 90.000 Sqm
Therefore, Length of the square with 4m road
(300+ 4X2) m
= 308m.
Therefore, New Area = (308X308)
= 94864 Sqm
Therefore, area of road = (94864-90000) Sqm
= 4864 Sqm. (Answer)
(16) The area of a triangle land is 264 sq. metres. Find the height if the base is 22 metres.
Solution:
Let, height of the triangular land = x m.
Therefore, Base of the triangular land = 22m.
Therefore, Base of the area = ½ X b X h
= ½ X 22 X x
= 11x Sqm.
Therefore, 11x = 264
= x = 264/11
= 24m.
Therefore, Height of the triangular land = 24m (Answer).
(17) A reservoir contains 19200 litres of water. Its depth is 2.56 metres and its breadth is 2.5 metres. What is its length?
Solution:
Reserve contains 19200 lit.
Therefore, we know, 1 lit = 1000 Cu^{2}
Therefore, 19200 lit = (19200X1000) Cu^{2}
= 19200000 Cu^{2}
= 19.2 Cu^{2}
Therefore, depth = 2.56m
Therefore, breadth = 2.5m.
Let, Length = x m.
Therefore, Area = (2.56X2.5X x) Cu^{2}
= 6.4x Cu^{2}
Therefore, 6.4x = 19.2
= x= 19.2/6.4 = 3
Therefore, length = 3m (Answer)
(18) Gold is 19.3 times heavier than water. The length of a rectangular gold bar is 7.8 cm, the breadth is 6.4 cm and the height is 2.5 cm. What is the weight of the gold bar?
Solution:
Rectangular gold bar,
Length = 7.8 cm
Breadth = 6.4 cm
Height = 2.5 cm
Therefore, Area = (7.8X6.4X2.5)
= 124.8 Cm^{3}
We know, 1 Cm^{3} water weight = 1gm.
124.8 Cm^{3} water weight = (1X124.8)
= 124.8 gm
Therefore, Weight of gold bar = (124.8X19.3) gm
= 2408.64 gm.
(19) The length of a small box is cm 2.4 mm, the breadth is 7 cm 6.2 mm and the height is 5 cm 8 mm. What is the volume of the box in cubic centimetre.
Solution:
BOX, Length = 15cm 24cm
= (15+ 24/10) cm
= (15+0.24)
= 15.24 cm.
Breadth = (7cm + 6.2/10)
= (7+0.62)
= 7.62 cm.
Height = 5cm 8mm
= (5+0.8) = 5.8cm.
Area = (l X b X h) cm^{3}
= (15.24X4.62X5.8)
= 673.547 cm^{3} (Answer)
(20) The length of a rectangular reservoir is 5.5 metres, the breadth is 4 metres and the height is 2 metres. If the reservoir is full of water, what is the volume of water in litres and its weight in kg?
Solution:
Rectangular reservoir, length = 5.5 m.
= (5.5X100) = 550 cm.
Breadth = 4 m
= (4X100)
= 400 cm.
Height = 2m.
= 200cm
Therefore, Area = (550X400X200) cm^{3}
= 44000000
We know, 1000 cm^{3} = 1lit
1 cm^{3} = 1/1000 lit
Therefore, 44000000 cm^{3} = 44000000X1/1000
= 44000 lit
Again, 1 lit water weight 1 kg.
44000 lit water weight (44000X1)
= 44000 kg. (Answer)
(21) The length of a rectangular field is 1.5 times its breadth. An amount of Tk. 10260 is spent to plant grass at Tk. 1.90 per sq. metres. How much money will be spent at Tk. 2.50 per metre to erect a fence around that field?
Solution:
1.90 Tk spend in 1 Sqm.
1 Tk spend in 1/1.90 Sqm
Therefore, 10260 Tk spend in 10260X1/1.90 = 5400 Sqm
Therefore, Let breadth = x m.
Therefore, Length = (x X 1.5) m= 1.5 cm.
Therefore, from question, x X 1.5x = 5400
= 1.5x^{2} = 5400/1.5
Therefore, x^{2} = 5400/1.5
= 3600
Therefore, x = 60m
Therefore, length = (60X1.5) = 90m.
Breadth = 60 cm.
Therefore, perimeter = 2(90+60) m
= 300 m.
Therefore, To erect a fence around the field
= (300X2.50) Tk
= 750 Tk.
(22) An amount of Tk. 7200 is spent to cover the floor of a room by carpet. An amount of Tk. 576 would be saved if the breadth were 3 metres less. What is the breadth of the room?
Solution:
3m spent 576 Tk.
Therefore, 1m spent 576/3 Tk = 192 Tk.
Now, 192 Tk spent when breadth 1m.
Therefore, 1 Tk spend when breadth 1/192 m.
Therefore, 7200 Tk spent when breadth 1X7200/192 m
= 37.5m.
Therefore, Breadth 37.5m
(23) Around inside a rectangular garden of length 80 metres and breadth 60 metres, there is a road of breadth 4 metres. How much money will be spent to construct that road at Tk. 7.25 per square metre?
Solution:
Length of the garden with road = 80m.
Breadth of the garden with road = 60m.
Therefore, Area = (80X60) Sqm
= 4800 Sqm.
Without the road, length of garden = {80-(4+4)} m
= 72 m
Without the road Breadth of garden = 60-(2X4) m
= 52m
Therefore, Area = (72X52) Sqm
= 3744 Sqm
Therefore, Area of road = (4800-3744) Sqm
= 1056 Sqm.
Total cost to construct the road
= (1056X7.25) Tk.
= 7656 Tk (Answer)
(24) A square open reservoir of depth 2.5 metres contains 28,900 litres of water inside. How much money will be spent to put a lead sheet in the innerside at Tk. 12 50 per sq. metres?
Solution:
28900 lit = 28900/1000 m^{3}
= 28.9 m^{3}
Let, length of the reservoir = x m.
Therefore, Area = x^{2} Sqm.
And, depth = 2.5m
Therefore, Area = (x^{2} X 2.5) = 2.5x^{2} .m^{3}
Therefore, 2.5x^{2} = 28.9
= x^{2}= 28.9/2.5
= 11.56
Therefore, x= √11.56 = 3.4m
Therefore, Area of (1 side surface) innerside = (3.4X2.5) Sqm
Therefore, Area of 4 side surface innerside = (4X3.4X2.5) Sqm
= 34 Sqm.
Area of lower surface = (3.4X3.4) Sqm
= 11.56 Sqm
Therefore, Total money spent = (34+11.56) Sqm X12.50 Rs.
= (45.56X12.50)
= 569.50 Tk (Answer)
(25) The length of the floor of a house is 26 metres and breadth is 20 metres. How many mats of length 4 metres and breadth 2.5 metres will be required to cover the floor completely? How money will be spent if the price of each mat is Tk. 27.50?
Solution:
Area of the floor = (26X20) Sqm
= 520 Sqm
Therefore, Area of 1 mat = (4X2.5) m^{2}
= 10 m^{2}
Therefore, Mat required = 520/10 = 52 pc.
Therefore, 1 pc mat cost = 27.50 Tk.
52 pc mat cost = (52X27.50) Tk
= 1430 Tk.
(26) The length of a book is 25 cm and the breadth is 18 cm. The number of pages of the book is 200 and the thickness of each page is 0.1 mm. Find the volume of the book.
Solution:
200 page = 100 sheet [Therefore, 2pg= 1 sheet]
Thickness of 1 page is 0.1 mm.
Therefore, Thickness of 100 page (100X0.1)
= 10 mm
= 1 Cm.
Therefore, Volume = (2.5X18X1 cm)
= .450 cm^{3} (Answer)
(27) The length of a pond is 32 metres, breadth is 20 metres and the depth of water of the pond is 3 metres. The pond is being made empty by a machine which can remove 0.1 cubic metres of water per second. How much time will be required to make the pond empty?
Solution:
Volume of the pond= (32X20X3) cm^{3}
= 1920 m^{3}
Therefore, 0.1m^{3} water removes in 1 sec.
Therefore, 1 m^{3} water removes in 1/0.1 sec
Therefore 1920 m^{3} water remove in 1920/0.1 sec
= 19200 sec.
[Therefore, 1hr = 60X60 = 3600 sec]
= 19200/60X60
= 16/3 hr.
= 5hr 20m (Answer)
(28) A solid cube of sides 50cm is kept in an empty reservoir of length 3 metres, breadth 2 metres and height 1 metre. The cube is taken out after filling the reservoir with water. What is the depth of water now ?
solution :
volume of reservoir = (3x2x1)m^{3 }= 6m^{3}
Another cube of 50 cm length ,volume is= (50)^{3 }cm^{3}
= 125000cm^{3}
^{ }= 125000/1000000
^{ }= 0.125 m^{3}
^{ }
(29) A breadth of a room is 2/3 times of its length. The length and height of the room Are 15m and 4m respectively. The floor of the room are set with stone of the size 50 sq.cm leaving 1m margin in all sides. Air is 0.00129 times heavier than water.
a) find out the perimeter of the room
b) how many pieces of stone will be needed ?
c) how much air is their in the room ?
Solution :
a) Given , length of room = 15m
Breadth of room = 15 x 2/3= 10m.
Therefore, perimeter = 2(15+10)= 50m.
b) Without blank space , length of the room = 15- (1+1)m
= 13 m
Breadth of the room = 10- (1+1)m
= 8m
Area = (13×8) sq.m = 10452m.
Length of the stone of the square size = 50 cm.
= 0.5m.
Therefore, Area= (0.5)^{2}
= 0.25 sqm.
Therefore, pieces of stone will be needed = 104/0.25
= 416 pcs.
c) volume of the room = (15x10x4) m^{3}
= 600m^{3}
= 600 x 1000000 cm^{3}
^{ }= 60000000cm^{3}
Weight of the 1cm^{3 }air = 0.00129gm.
Therefore quantity of the air in the room
= 60000000 x 0.00129
= 774000gm.
= 774 kg.
(31) The area of a rectangular school campus is 10 acres. Its length is four times the breadth. The size of the auditorium is 40m x 35m x 10m and the thickness of the wall is 15 cm.
Solution:
a) We know , 2.47 acre = 1 hector
Therefore 1 acre = ½.47 hector
Therefore, 10 acre = 10/2.47 hector
= 4.048582996 hector
= 4.05 hector
b) Let the breadth of school campus = X m.
Length ; ; ; = 4x m.
Therefore, area = (X . 4x ) sqm
= 4x^{2}
^{ }
Given the area of campus = 10 acre
= (10×4046.86) sqm
= 40468.6 sqm
Therefore 4x^{2 = }40468.6
X^{2 }= 40468.6/4
X^{2= }10117.15
X= √10117.15
= 100.58m
Therefore, length = (4 x 100.58) m = 402.32m.
Therefore length of the boundary wall = 2(L+b) m
= 2 (402.32 + 100.58) m
= 1005.8 m.
c) Thickness of the wall = 15 cm.
= 0.15 m.
therefore volume of two wall to the length = ( 40+0.15+10+2) m^{3}
^{ }= 120m^{3}
to, Breadth volume of two wall = { 35- (0.15 x2 )} x0.15 x10 x 2 m^{3}
^{ }= 34.7 x 0.5 x 10 x 2
^{ }= 104.1 m^{3}
^{ }
therefore volume of 4 walls = (120+104.1)m^{3}
^{ }= 224.1m^{3}