Problem 1 :
Find the value of k, if x - 1 is a factor of
4x3 + 3x2 - 4x + k
(A) -3 (B) -5 (C) 0
Solution :
Let P(x) = 4x3 + 3x2 - 4x + k -----(1)
Given, x - 1 is a factor of P(x).
Then, x = 1
By applying x = 1 in equation (1), we get
4(1)3 + 3(1)2 - 4(1) + k = 0
4 + 3 - 4 + k = 0
3 + k = 0
k = -3
So, the value of k is -3.
Problem 2 :
What is the value of the expression given below ?
∛4 × ∛16
(A) 3 (B) 4 (C) 2
Solution :
Finding the value of ∛4 and ∛16.
By using the calculator, we get
∛4 = 1.59
∛16 = 2.52
Then,
∛4 × ∛16 = (1.59 × 2.52)
= 4
So, the answer is 4.
Problem 3 :
Find the area of the quadrilateral whose diagonal measures the length of 50m and the perpendicular distance measures 10m and 20m respectively.
(A) 7.50 sq.m (B) 75 sq.m (C) 750 sq.m
Solution :
Area of quadrilateral = 1/2 × d × (h1 + h2)
Here d = 50m, h1 = 10m and h2 = 20m
Then,
Area = 1/2 × 50 × (10 + 20)
= 1/2 × 50 × 30
= 750 sq.m
So, the area of the quadrilateral is 750 sq.m
Problem 4 :
Cost of leveling a land is $12 per square meter. A land is in the form of a trapezium whose parallel sides are of lengths 18m and 12m. If its other two sides are each of length 5m, find the total cost incurred in leveling the land.
(A) 230 (B) 540 (C) 900
Solution :
Area of trapezium = 1/2 × h × (a + b)
Here h = 5m, a = 18m and b = 12m
Area = 1/2 × 5 × (18 + 12)
= 1/2 × 5 × 30
Area = 75 sq.m
To find the total cost of leveling land,
= cost of leveling land × Area of land
= 12 × 75
= 900
So, the answer is 900.
Problem 5 :
The perimeter of a rhombus is 20cm. One of the diagonals is of length 8cm. Find the length of the other diagonal and the area of the rhombus.
(A) 50 sq.m (B) 32 sq.m (C) 24 sq.m
Solution :
Given, perimeter of a rhombus is 20cm.
Length of the one diagonal d1 is 8cm.
Let the other diagonal be d2.
Perimeter of a rhombus = 4 sides
So, side = 20/4 = 5cm
We know that the diagonals of a rhombus bisect each other at right angles.
Since the diagonals bisect each other, so half the length of the diagonal d1.
That is, d1/2 = 8/2 = 4cm
To find the other diagonal,
Using the Pythagorean theorem :
Here d1/2 = 4cm, side = 5cm and d2/2 = ?
(d1/2)2 + (d2/2)2 = side2
42 + (d2/2)2 = 52
16 + (d2/2)2 = 25
(d2/2)2 = 9
d2/2 = 3
d2 = 6cm
So, length of the other diagonal d2 is 6cm.
Now,
Area of the rhombus using diagonals = 1/2 × d1 × d2
= 1/2 × 8 × 6
Area = 24cm2.
So, area of the rhombus is 24cm2.
Problem 6 :
Evaluate the following log2 64 + log2 4
(A) 4 (B) 8 (C) 6
Solution :
By using the rule loga (mn) = loga m + loga n, we get
log2 64 + log2 4 = log2 (64 . 4)
= log2 (26 . 22)
= log2 2(6 + 2)
= log2 28
= 8log2 2
= 8(1)
= 8
So, the answer is 8.
Problem 7 :
Two sets are called ---------- if they have same elements.
(A) Equivalent sets (B) Null sets (C) Equal sets
Solution :
So, the answer is equal sets.
Problem 8 :
Find the perimeter of the sector whose area is 924 sq.cm and the central angle is 240.
(A) 100cm (B) 115cm (C) 130cm
Solution :
Area of the sector = πr2 × θ/360
Here A = 924 and θ = 240
924 = πr2 × 240/360
924 = 22/7 × r2 × 2/3
(924 × 21)/44 = r2
r2 = 441
r = 21cm
Perimeter of the sector = 2r + θ/360 × 2πr
= 2(21) + (240/360) × 2 × 22/7 × 21
= 42 + (2/3) × 132
= 42 + 88
= 130cm
So, the perimeter of the sector is 130cm.
Problem 9 :
Factorise 3x2 - 18xy
(A) 3x(x - 6y) (B) 3x(x - 18y) (C) 3(x2 - 18y)
Solution :
3x2 - 18xy = 3x(x - 6y)
So, the answer is 3x(x - 6y)
Problem 10 :
Solve 3(x + 4) + 2(y - 3) = 17 and 7(x - 3) - 3(y + 4) = 8
(A) (8, 1) (B) (5, -2) (C) (0, 6)
Solution :
3(x + 4) + 2(y - 3) = 17 -----(1)
7(x - 3) - 3(y + 4) = 8 -----(2)
From (1), we get
3x + 12 + 2y - 6 = 17
3x + 2y = 17 - 6
2y = 11 - 3x
y = (11 - 3x)/2 -----(3)
From (2), we get
7x - 21 - 3y - 12 = 8
7x - 3y = 41
-3y = 41 - 7x
y = (7x - 41)/3 -----(4)
Equating (3) and (4), we get
(11 - 3x)/2 = (7x - 41)/3
3(11 - 3x) = 2(7x - 41)
33 - 9x = 14x - 82
33 + 82 = 14x + 9x
115 = 23x
x = 5
By applying x = 5 in equation (3), we get
y = (11 - 3(5))/2
y = (-4)/2
y = -2
So, the solution is (5, -2)
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
Aug 31, 24 08:51 PM
Aug 31, 24 08:47 PM
Aug 31, 24 06:59 PM