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1. Width of the plate = 236.251/17.5 = 13.5 cm
Perimeter = 2 (length + width) = 2 (17.5 + 13.5) = 62 cm
2. Let the length of the plot be X m
So, Breadth = (75/100)x X = (3X/4)m
But, here we know that the Perimeter = 1050
=> 1050 = 2 (X + (3X/4)
=> X = 300 m
So, the Area of the plot = 300 x ((3 x 300)/4) = 67500 m2
3. Let the length be 1 and breadth also be 10.
New length (10% excess) = 11,
New breadth (5% excess) = 9.5
New area = 11 x 9.5 = 104.5
So, Change in area = 104.5 - 100 = 4.5
Percentage change = ((4.5/100) x 100) = 4.5%
4. Area of the floor of the room = 800 x 500 = 400000 cm2
Area of the Tile = 20 x 10 cm2
So, number of Tiles required = (400000 / (20 x 10)) = 2000
5. Let the radius of larger circle be 'R' and small circle be 'r'.
So, 2πR = 132
=> 2 x (22/7) x R = 132
=> R = 21
2πr = 88
=> 2 x (22/7) x R= 88
=> r = 14
Difference in the areas = πR2- πr2
= π (R2- r2)
=> (22/7)(212 - 142) => (22/7) (21+14)(21-14)
=> (22/7)(35)(7) = 770 sq metres
6. Volume of sphere = (4/3)πr3
Volume of cone = (1/3)πr2h
Let the number of cones be 'X'
=> (4/3)πr3 = ((1/3)πr2h )(X)
=> ((4/3) π 73 = (1/3) π (72)(2)(X)
=> X = 14
7. Length of the carpet = 10 - (0.5 + 0.5) = 9 m
Width of the carpet = 8 - (0.5+0.5) = 7 m
Area of carpet = 9 x 7 = 63
Required length of carpet = 63/0.25 = 252 m
So, Cost of carpet = 252 x Rs. 20 = Rs. 5,040.
8. Side of the square field = √ 56025 = 225 m
Length of the wire required = 4 x 225 = 900 m
So, Total cost = Rs. 15 x 900 = Rs. 13,500
9. Circumference of the circular plot
= (2640/12) = 220 m
2πr = 220 => 2 . (22/7) . r = 220
=> r = 35 m
Area of the plot
= πr2 = (22/7) . 35 . 35 = 3850 sq.m
So, cost of carpeting = 3850 x Rs. 85 = Rs. 3,27,250
10. Perimeter of rectangle = 2 (18+26) = 88 cm
Circumference of circle => 2πr = 88
=> 2 x (22/7) x r = 88
=> r = 14 cm
Area of circle = πr2
= (22/7) . 14 . 14 = 616 cm2
11. Radius of outer circle = 21 + 7 = 28 m
Road area = Area of outside circle - Area of inside circle
Road area = π.(28)2- π.(21) 2 = 1078 sq.m
So, the cost = 1078 x Rs. 5 = 5,390
12. Lateral surface area of cylinder = 2πrh
Base area of cylinder = πr2
So, 2πrh = 3πr2
=> h:r = 3:2
13. Let the sides of two cubes be a1 and a2
a13-a23 = 8:27 => a1:a2 = 2:3
So, surface areas ratio = 6.a12-6. a22 = (2)2-(3)2 = 4:9
14. Length of the rope will be the radius of the circle = 14 m
Area of the circle = πr2 = (22/7) . 14 . 14 = 616 sq.m
So, Area of the ground that can be grazed = 616 sq.m
15. Length of the stick = length of the diagonal of the room
√(122+92+82) = √289 = 17 m
16. Quantity of the earth taken from the tank = 30 x 20 x 12 cu.m
Area of the field for the earth to be spreaded
= (375 x 40) - (30 x 20) = 14400
So, The rise in the level of the field = ((30 x 20 x 12) / 14400) x 100 = 50 cm
17. Area of the road = Area of outside rectangle - Area of inside rectangle
= 70 x 50 - 60 x 40
= 3500 - 2400 = 1100 m2
So, the cost of making 1100 m2 = 500 x 1100 = Rs. 5,50,000
18. Area of four walls = 2 x 5 (7.5 + 5.5) = 130 m2
Area of required paper = 130 m2
Breadth of the paper = 40 cm = 0.4 m
So, length of the paper = 130/0.4 = 325 m
So, cost of paper at 75 paise per meter = 325 x 0.75 = Rs. 243.75
19. Let the length and breadth be 10 and 10
Area = 10 x 10 = 100
New length = 12 (20% increase)
New breadth = 11 (10% increase)
New area = 12 x 11 = 132
So, Increment in area = 132 - 100 = 32 %
20. Let the number of spherical balls be "X"
Volume of cylinder = (X). (Volume of sphere)
π. 72. 28 = X . (4/3). π. 72
X = 3
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than you
ReplyDeletethnk mam
ReplyDeletevolume of sphere is 4/3 pi r^3
ReplyDeleteCorrect Q20... (4/3)pi7^3
ReplyDeletethanks mam....
ReplyDeletethank u... in problem 3 there is a small mistake.. let length be 10 is typed as 1, zero is missing..
ReplyDeletethanx..
ReplyDelete