Problem 1 :
Find greatest common factor of the following terms
20x3, 36x6
Solution :
20 = 22 ⋅ 5
36 = 22 ⋅ 32
Greatest common factor = 22 ⋅ 32 ⋅ 5
= 180
So, the greatest common factor is 180x6.
Problem 2 :
Simplify the following
Solution :
p2 - 1 = (p+1) (p-1)
= [(p+1) (p-1)/p]⋅ [p3/(p-1)] ⋅ [1/(p+1)]
= p2
Problem 3 :
Find the square root of 9801 by factor method.
Solution :
So, the square root of 9801 is 99.
Problem 4 :
Solve 6x2 + x - 1 = 0
Solution :
6x2 + x - 1 = 0
(3x - 1) (2x + 1) = 0
3x - 1 = 0 3x = 1 x = 1/3 |
2x + 1 = 0 2x = -1 x = -1/2 |
So the solution is {-1/2, 1/3}.
Problem 5 :
The product of two consecutive odd number is 323. Find them.
Solution :
Let x and x + 2 are two consecutive odd numbers.
Product of two consecutive odd numbers = 323
x (x + 2) = 323
x2 + 2x = 323
x2 + 2x - 323 = 0
(x - 17)(x + 19)
x = 17 and x = -19
So, the two odd numbers are 17 and 19.
Problem 6 :
Find two consecutive even integers whose product is 224.
Solution :
Let x and x + 2 are two consecutive even integers.
Product of even integers = 224
x(x + 2) = 224
x2 + 2x = 224
x2 + 2x - 224 = 0
(x + 16) (x - 14) = 0
x = -16 and x = 14
So, two consecutive even numbers are 14 and 16.
Problem 7 :
The length of the hall is 3 m more than its width. The numerical value of its area is equal to the numerical value of its perimeter. Find the length and width of the hall.
Solution :
Let x be the width of the hall
length = x + 3
Area of the hall = Perimeter of the hall
x(x+3) = 2(x+x+3)
x2 + 3x = 2(2x+3)
x2 + 3x - 4x - 6 = 0
x2 - 1x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 and x = -2
x + 3 = 6
So, the width and length of the rectangle are 3 and 6 m.
Problem 8 :
Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector approximately (use π = 3.14).
Solution :
Area of major sector = πr2 - (θ/360)πr2
= π42 - (30/360)π42
= π42(1 - 1/12)
= π42(11/12)
= (176/12)(3.14)
= 46.05 cm2
Problem 9 :
OACB is a quadrant of a circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.
Solution :
Area of shaded region = Area of quadrant - Area of triangle ODB
= πr2 - (1/2) ⋅ base ⋅ height
= (22/7)(3.5)2 - (1/2) ⋅ 3.5 ⋅ 2
= 38.5 - 3.5
= 35 cm2
Problem 10 :
The wheels of a car are of diameter 80 cm each. Find how many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour.
Solution :
Speed of the car = 66 km/hr
1000 m = 1 km
100 cm = 1 m
66 km = 6600000 cm
66 km/hr = 6600000/60
= 110000 cm/min
Distance covered = Time (Speed)
= 10(110000)
= 1100000
Radius of the wheel = 40 cm
Number of revolutions
= Distance covered / Distance covered by 4 wheels
Distance covered by 4 wheels = 2πr
= 1100000 / [2 (3.14) ⋅ 40]
= 1100000/251.2
= 4379
So, each wheel has to revolve 4379 times.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 20, 24 12:02 AM
Apr 19, 24 11:58 PM
Apr 19, 24 11:45 PM