REAL NUMBERS.
1. The decimal expansion of 22/7 is
(a) Terminating
(b) Non-terminating and repeating
(c) Non-terminating and Non-repeating
(d) None of the above
Answer: (b) Non-terminating and repeating
Explanation: 22/7 = 3.14285714286..
2. For some integer n, the odd integer is represented in the form of:
(a) n
(b) n + 1
(c) 2n + 1
(d) 2n
Answer: (c) 2n + 1
Explanation: Since 2n represents the even numbers, hence 2n + 1 will always represent the odd numbers. Suppose if n = 2, then 2n = 4 and 2n + 1 = 5.
3. HCF of 26 and 91 is:
(a) 15
(b) 13
(c) 19
(d) 11
Answer: (b) 13
Explanation: The prime factorisation of 26 and 91 is;
26 = 2 x 13
91 = 7 x 13
Hence, HCF (26, 91) = 13
4. Which of the following is not irrational?
(a) (3 + √7)
(b) (3 – √7)
(c) (3 + √7) (3 – √7)
(d) 3√7
Answer: (c) (3 + √7) (3 – √7)
Explanation: If we solve, (3 + √7) (3 – √7), we get;
(3 + √7) (3 – √7) = 32 – (√7)2 = 9 – 7 = 2 [By a2 – b2 = (a – b) (a + b)]
5. The addition of a rational number and an irrational number is equal to:
(a) rational number
(b) Irrational number
(c) Both
(d) None of the above
Answer: (b) Irrational number
The addition of a rational number and an irrational number is equal to irrational number.
6. The multiplication of two irrational numbers is:
(a) irrational number
(b) rational number
(c) Maybe rational or irrational
(d) None
Answer: (c) Maybe rational or irrational
The multiplication of two irrational numbers is maybe rational or irrational.
7. If set A = {1, 2, 3, 4, 5,…} is given, then it represents:
(a) Whole numbers
(b) Rational Numbers
(c) Natural numbers
(d) Complex numbers
Answer: (c) Natural numbers
If set A = {1, 2, 3, 4, 5,…} is given, then it represents natural numbers.
8. If p and q are integers and is represented in the form of p/q, then it is a:
(a) Whole number
(b) Rational number
(c) Natural number
(d) Even number
Answer: (b) Rational number
If p and q are integers and is represented in the form of p/q, then it is a rational number.
9. The largest number that divides 70 and 125, which leaves the remainders 5 and 8, is:
(a) 65
(b) 15
(c) 13
(d) 25
Answer: (c) 13
Explanation: 70 – 5 = 65 and 125 – 8 = 117
HCF (65, 117) is the largest number that divides 70 and 125 and leaves remainder 5 and 8.
HCF (65, 117) = 13
10. The least number that is divisible by all the numbers from 1 to 5 is:
(a) 70
(b) 60
(c) 80
(d) 90
Answer: (b) 60
Explanation: The least number will be LCM of 1, 2, 3, 4, 5.
Hence, LCM (1, 2, 3, 4, 5) = 2 x 2 x 3 x 5 = 60
11. The sum or difference of of two irrational numbers is always
(a) rational
(b) irrational
(c) rational or irrational
(d) not determined
Answer: (b) irrational
12. The decimal expansion of the rational number 23/(22 . 5) will terminate after
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) more than 3 decimal places
Answer: (b) two decimal places
Explanation:
23/(22 . 5) = (23 × 5)/(22 . 52) = 115/(10)2 = 115/100 = 1.15
Hence, 23/(22 . 5) will terminate after two decimal places.
13. Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy
(a) 1 < r < b
(b) 0 < r ≤ b
(c) 0 ≤ r < b
(d) 0 < r < b
Correct option: (c) 0 ≤ r < b
Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy 0 ≤ r < b.
14. For some integer m, every even integer is of the form
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1
Answer: (c) 2m
For some integer m, every even integer is of the form 2m.
15. Using Euclid’s division algorithm, the HCF of 231 and 396 is
(a) 32
(b) 21
(c) 13
(d) 33
Answer: (d) 33
Explanation:
396 > 231
Using Euclid’s division algorithm,
396 = 231 × 1 + 165
231 = 165 × 1 + 66
165 = 66 × 2 + 33
66 = 33 × 2 + 0
16. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is
(a) 4
(b) 2
(c) 1
(d) 3
Answer: (b) 2
Explanation:
117 > 65
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
Therefore, HCF(65, 117) = 13
According to the given,
65m – 117 = 13
65m = 117 + 13
65m = 130
m = 130/65 = 2
17. The prime factorisation of 96 is
(a) 25 × 3
(b) 26
(c) 24 × 3
(d) 24 × 32
Answer: (a) 25 × 3
Explanation:
The prime factorisation of 96 is:
96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3
18. n² – 1 is divisible by 8, if n is
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Answer: (c) an odd integer
Explanation:
We know that an odd number in the form (2Q + 1) where Q is a natural number ,
so, n² -1 = (2Q + 1)² -1
= 4Q² + 4Q + 1 -1
= 4Q² + 4Q
Substituting Q = 1, 2,…
When Q = 1,
4Q² + 4Q = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8.
When Q = 2,
4Q² + 4Q = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 .
When Q = 3,
4Q² + 4Q = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8
It is concluded that 4Q² + 4Q is divisible by 8 for all natural numbers.
Hence, n² -1 is divisible by 8 for all odd values of n.
19. For any two positive integers a and b, HCF (a, b) × LCM (a, b) =
(a) 1
(b) (a × b)/2
(c) a/b
(d) a × b
Answer: (d) a × b
For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.
20. The values of the remainder r, when a positive integer a is divided by 3 are
(a) 0, 1, 2
(b) Only 1
(c) Only 0 or 1
(d) 1, 2
Answer: (a) 0, 1, 2
Explanation:
According to Euclid’s division lemma,
a = 3q + r, where 0 ≤ r < 3 and r is an integer.
Therefore, the values of r can be 0, 1 or 2.
Polynomials
1. The zeroes of x2–2x –8 are:
(a) (2,-4)
(b) (4,-2)
(c) (-2,-2)
(d) (-4,-4)
Answer: (b) (4,-2)
Explanation: x2–2x –8 = x2–4x + 2x –8
= x(x–4)+2(x–4)
= (x-4)(x+2)
Therefore, x = 4, -2.
2. What is the quadratic polynomial whose sum and the product of zeroes is √2, ⅓ respectively?
(a) 3x2-3√2x+1
(b) 3x2+3√2x+1
(c) 3x2+3√2x-1
(d) None of the above
Answer: (a) 3x2-3√2x+1
Explanation: Sum of zeroes = α + β =√2
Product of zeroes = α β = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the polynomial is;
x2–(α+β)x +αβ
= x2 –(√2)x + (1/3)
= 3x2-3√2x+1
3. If the zeroes of the quadratic polynomial ax2+bx+c, c≠0 are equal, then
(a) c and b have opposite signs
(b) c and a have opposite signs
(c) c and b have same signs
(d) c and a have same signs
Answer: (d) c and a have same signs
Explanation:
For equal roots, discriminant will be equal to zero.
b2 -4ac = 0
b2 = 4ac
ac = b2/4
ac>0 (as square of any number cannot be negative)
4. The degree of the polynomial, x4 – x2 +2 is
(a) 2
(b) 4
(c) 1
(d) 0
Answer: (b) 4
Explanation: Degree is the highest power of the variable in any polynomial.
5. If one of the zeroes of cubic polynomial is x3+ax2+bx+c is -1, then product of other two zeroes is:
(a) b-a-1
(b) b-a+1
(c) a-b+1
(d) a-b-1
Answer: (b) b-a+1
Explanation: Since one zero is -1, hence;
P(x) = x3+ax2+bx+c
P(-1) = (-1)3+a(-1)2+b(-1)+c
0 = -1+a-b+c
c=1-a+b
Product of zeroes, αβγ = -constant term/coefficient of x3
(-1)βγ = -c/1
c=βγ
βγ = b-a+1
6. If p(x) is a polynomial of degree one and p(a) = 0, then a is said to be:
(a) Zero of p(x)
(b) Value of p(x)
(c) Constant of p(x)
(d) None of the above
Answer: (a) Zero of p(x)
Explanation: Let p(x) = mx+n
Put x = a
p(a)=ma+n=0
So, a is zero of p(x).
7. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial is:
(a) Intersects x-axis
(b) Intersects y-axis
(c) Intersects y-axis or x-axis
(d) None of the above
Answer: (a) Intersects x-axis
8. A polynomial of degree n has:
(a) Only one zero
(b) At least n zeroes
(c) More than n zeroes
(d) At most n zeroes
Answer: (d) At most n zeroes
Explanation: Maximum number of zeroes of a polynomial = Degree of the polynomial
9. The number of polynomials having zeroes as -2 and 5 is:
(a) 1
(b) 2
(c) 3
(d) More than 3
Answer: (d) More than 3
Explanation: The polynomials x2-3x-10, 2x2-6x-20, (1/2)x2-(3/2)x-5, 3x2-9x-30, have zeroes as -2 and 5.
10. Zeroes of p(x) = x2-27 are:
(a) ±9√3
(b) ±3√3
(c) ±7√3
(d) None of the above
Answer: (b) ±3√3
Explanation: x2-27 = 0
x2=27
x=√27
x=±3√3
11. Given that two of the zeroes of the cubic polynomial ax3 + bx2 + cx + d are 0, the third zero is
(a) -b/a
(b) b/a
(c) c/a
(d) -d/a
Answer: (a) -b/a
Explanation:
Let α be the third zero.
Given that two zeroes of the cubic polynomial are 0.
Sum of the zeroes = α + 0 + 0 = -b/a
α = -b/a
12. If one zero of the quadratic polynomial x2 + 3x + k is 2, then the value of k is
(a) 10
(b) –10
(c) 5
(d) –5
Answer: (b) -10
Explanation:
Given that 2 is the zero of the quadratic polynomial x2 + 3x + k.
⇒ (2)2 + 3(2) + k = 0
⇒ 4 + 6 + k = 0
⇒ k = -10
13. A quadratic polynomial, whose zeroes are –3 and 4, is
(a) x² – x + 12
(b) x² + x + 12
(c) (x²/2) – (x/2) – 6
(d) 2x² + 2x – 24
Answer: (c) (x²/2) – (x/2) – 6
Explanation:
Let the given zeroes be α = -3 and β = 4.
Sum of zeroes, α + β= -3 + 4 = 1
Product of Zeroes, αβ = -3 × 4 = -12
Therefore, the quadratic polynomial = x² – (sum of zeroes)x + (product of zeroes)
= x² – (α + β)x + (αβ)
= x² – (1)x + (-12)
= x² – x – 12
Dividing by 2,
= (x²/2) – (x/2) – 6
14. The zeroes of the quadratic polynomial x2 + 99x + 127 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
Answer: (b) both negative
Explanation:
Given quadratic polynomial is x2 + 99x + 127.
By comparing with the standard form, we get;
a = 1, b = 99 and c = 127
a > 0, b > 0 and c > 0
We know that in any quadratic polynomial, if all the coefficients have the same sign, then the zeroes of that polynomial will be negative.
Therefore, the zeroes of the given quadratic polynomial are negative.
15. The zeroes of the quadratic polynomial x2 + 7x + 10 are
(a) -4, -3
(b) 2, 5
(c) -2, -5
(d) -2, 5
Answer: (c) -2, -5
Explanation:
x2 + 7x + 10 = x2 + 2x + 5x + 10
= x(x + 2) + 5(x + 2)
= (x + 2)(x + 5)
Therefore, -2 and -5 are the zeroes of the given polynomial.
16. If the discriminant of a quadratic polynomial, D > 0, then the polynomial has
(a) two real and equal roots
(b) two real and unequal roots
(c) imaginary roots
(d) no roots
Answer: (b) two real and unequal roots
If the discriminant of a quadratic polynomial, D > 0, then the polynomial has two real and unequal roots.
17. If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, then the relation between the degrees of p(x) and g(x) is
(a) degree of p(x) < degree of g(x)
(b) degree of p(x) = degree of g(x)
(c) degree of p(x) > degree of g(x)
(d) nothing can be said about degrees of p(x) and g(x)
Answer: (a) degree of p(x) < degree of g(x)
Explanation:
We know that, p(x)= g(x) × q(x) + r(x)
Given that, q(x) = 0
When q(x) = 0, then r(x) = 0
So, now when we divide p(x) by g(x),
Then p(x) should be equal to zero.
If r(x) = 0, then the degree of p(x) < degree of g(x).
18. By division algorithm of polynomials, p(x) =
(a) g(x) × q(x) + r(x)
(b) g(x) × q(x) – r(x)
(c) g(x) × q(x) × r(x)
(d) g(x) + q(x) + r(x)
Answer: (a) g(x) × q(x) + r(x)
By division algorithm of polynomials, p(x) = g(x) × q(x) + r(x).
19. The product of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is
(a) -b/a
(b) c/a
(c) -d/a
(d) -c/a
Answer: (c) -d/a
The product of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is -d/a.
20. If the graph of a polynomial intersects the x-axis at three points, then it contains ____ zeroes.
(a) Three
(b) Two
(c) Four
(d) More than three
Answer: (a) Three
If the graph of a polynomial intersects the x-axis at three points, then it contains three zeroes.
Quadratic Equations
1. Equation of (x+1)2-x2=0 has number of real roots equal to:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Explanation: (x+1)2-x2=0
X2+2x+1-x2 = 0
2x+1=0
x=-½
Hence, there is one real root.
2. The roots of 100x2 – 20x + 1 = 0 is:
(a) 1/20 and 1/20
(b) 1/10 and 1/20
(c) 1/10 and 1/10
(d) None of the above
Answer: (c) 1/10 and 1/10
Explanation: Given, 100x2 – 20x + 1=0
100x2 – 10x – 10x + 1 = 0
10x(10x – 1) -1(10x – 1) = 0
(10x – 1)2 = 0
∴ (10x – 1) = 0 or (10x – 1) = 0
⇒x = 1/10 or x = 1/10
3. The sum of two numbers is 27 and product is 182. The numbers are:
(a) 12 and 13
(b) 13 and 14
(c) 12 and 15
(d) 13 and 24
Answer: (b) 13 and 14
Explanation: Let x is one number
Another number = 27 – x
Product of two numbers = 182
x(27 – x) = 182
⇒ x2 – 27x – 182 = 0
⇒ x2 – 13x – 14x + 182 = 0
⇒ x(x – 13) -14(x – 13) = 0
⇒ (x – 13)(x -14) = 0
⇒ x = 13 or x = 14
4. If ½ is a root of the quadratic equation x2-mx-5/4=0, then value of m is:
(a) 2
(b) -2
(c) -3
(d) 3
Answer: (b) -2
Explanation: Given x=½ as root of equation x2-mx-5/4=0.
(½)2 – m(½) – 5/4 = 0
¼-m/2-5/4=0
m=-2
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:
(a) Base=10cm and Altitude=5cm
(b) Base=12cm and Altitude=5cm
(c) Base=14cm and Altitude=10cm
(d) Base=12cm and Altitude=10cm
Answer: (b) Base=12cm and Altitude=5cm
Explanation: Let the base be x cm.
Altitude = (x – 7) cm
In a right triangle,
Base2 + Altitude2 = Hypotenuse2 (From Pythagoras theorem)
∴ x2 + (x – 7)2 = 132
By solving the above equation, we get;
⇒ x = 12 or x = – 5
Since the side of the triangle cannot be negative.
Therefore, base = 12cm and altitude = 12-7 = 5cm
6. The roots of quadratic equation 2x2 + x + 4 = 0 are:
(a) Positive and negative
(b) Both Positive
(c) Both Negative
(d) No real roots
Answer: (d) No real roots
Explanation: 2x2 + x + 4 = 0
⇒ 2x2 + x = -4
Dividing the equation by 2, we get
⇒ x2 + 1/2x = -2
⇒ x2 + 2 × x × 1/4 = -2
By adding (1/4)2 to both sides of the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2
⇒ (x + 1/4)2 = 1/16 – 2
⇒ (x + 1/4)2 = -31/16
The square root of negative number is imaginary, therefore, there is no real root for the given equation.
7. The value of 6+6+6–√………−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√ is:
(a) 4
(b) 3
(c) 3.5
(d) -3
Answer: (b) 3
Explanation: Let, 6+6+6–√………−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√=x
Hence, we can write, √(6+x) = x
6+x=x2
x2-x-6=0
x2-3x+2x-6=0
x(x-3)+2(x-3)=0
(x+2) (x-3) = 0
x=-2,3
Since, x cannot be negative, therefore, x=3
8. The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3. The present age of Rehman is:
(a) 7
(b) 10
(c) 5
(d) 6
Answer: (a) 7
Explanation: Let, x is the present age of Rehman
Three years ago his age = x – 3
Five years later his age = x + 5
Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.
∴ 1/x-3 + 1/x-5 = 1/3
(x+5+x-3)/(x-3)(x+5) = 1/3
(2x+2)/(x-3)(x+5) = 1/3
⇒ 3(2x + 2) = (x-3)(x+5)
⇒ 6x + 6 = x2 + 2x – 15
⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
⇒ x = 7, -3
We know age cannot be negative, hence the answer is 7.
9. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
(a) 30 km/hr
(b) 40 km/hr
(c) 50 km/hr
(d) 60 km/hr
Answer: (b) 40 km/hr
Explanation: Let x km/hr be the speed of train.
Time required to cover 360 km = 360/x hr.
As per the question given,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 – x + 1800-5/x = 360
⇒ x2 + 5x + 10x – 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ x = 40, -45
Negative value is not considered for speed hence the answer is 40km/hr.
10. If one root of equation 4x2-2x+k-4=0 is reciprocal of the other. The value of k is:
(a) -8
(b) 8
(c) -4
(d) 4
Answer: (b) 8
Explanation: If one root is reciprocal of others, then the product of roots will be:
α x 1/α = (k-4)/4
k-4=4
k=8
11. Which one of the following is not a quadratic equation?
(a) (x + 2)2 = 2(x + 3)
(b) x2 + 3x = (–1) (1 – 3x)2
(c) (x + 2) (x – 1) = x2 – 2x – 3
(d) x3 – x2 + 2x + 1 = (x + 1)3
Answer: (c) (x + 2) (x – 1) = x2 – 2x – 3
Explanation:
We know that the degree of a quadratic equation is 2.
By verifying the options,
(a) (x + 2)2 = 2(x + 3)
x2 + 4x + 4 = 2x + 6
x2 + 2x – 2 = 0
This is a quadratic equation.
(b) x2 + 3x = (–1) (1 – 3x)2
x2 + 3x = -1(1 + 9x2 – 6x)
x2 + 3x + 1 + 9x2 – 6x = 0
10x2 – 3x + 1 = 0
This is a quadratic equation.
(c) (x + 2) (x – 1) = x2 – 2x – 3
x2 + x – 2 = x2 – 2x – 3
x2 + x – 2 – x2 + 2x + 3 = 0
3x + 1 = 0
This is not a quadratic equation.
12. Which of the following equations has 2 as a root?
(a) x2 – 4x + 5 = 0
(b) x2 + 3x – 12 = 0
(c) 2x2 – 7x + 6 = 0
(d) 3x2 – 6x – 2 = 0
Answer: (c) 2x2 – 7x + 6 = 0
Explanation:
If 2 is a root then substituting the value 2 in place of x should satisfy the equation.
Let us verify the given options.
(a) x2 – 4x + 5 = 0
(2)2 – 4(2) + 5 = 1 ≠ 0
So, x = 2 is not a root of x2 – 4x + 5 = 0
(b) x2 + 3x – 12 = 0
(2)2 + 3(2) – 12 = -2 ≠ 0
So, x = 2 is not a root of x2 + 3x – 12 = 0
(c) 2x2 – 7x + 6 = 0
2(2)2 – 7(2) + 6 = 0
Here, x = 2 is a root of 2x2 – 7x + 6 = 0
13. A quadratic equation ax2 + bx + c = 0 has no real roots, if
(a) b2 – 4ac < 0
(b) b2 – 4ac = 0
(c) b2 – 4ac < 0
(d) b2 – ac < 0
Answer: (c) b2 – 4ac < 0
A quadratic equation ax2 + bx + c = 0 has no real roots, if b2 – 4ac < 0. That means, the quadratic equation contains imaginary roots.
14. The product of two consecutive positive integers is 306. To find the integers, this can be represented in the form of quadratic equation as
(a) x2 + x + 360 = 0
(b) x2 + x – 360 = 0
(c) 2x2 + x – 360
(d) x2 – 2x – 360 = 0
Answer: (b) x2 + x – 360 = 0
Explanation:
Let x and (x + 1) be the two consecutive integers.
According to the given,
x(x + 1) = 360
x2 + x = 360
x2 + x – 360
15. The equation which has the sum of its roots as 3 is
(a) 2x2 – 3x + 6 = 0
(b) –x2 + 3x – 3 = 0
(c) √2x2 – 3/√2x + 1 = 0
(d) 3x2 – 3x + 3 = 0
Answer: (b) –x2 + 3x – 3 = 0
Explanation:
The sum of the roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0 is given by,
Coefficient of x / coefficient of x2 = –(b/a)
Let us verify the options.
(a) 2x2 – 3x + 6 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(b) -x2 + 3x – 3 = 0
Sum of the roots = – b/a = -(3/-1) = 3
(c) √2x2 – 3/√2x + 1=0
2x2 – 3x + √2 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(d) 3x2 – 3x + 3 = 0
Sum of the roots = – b/a = -(-3/3) = 1
16. The quadratic equation 2x2 – √5x + 1 = 0 has
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than 2 real roots
Answer: (c) no real roots
Explanation:
Given,
2x2 – √5x + 1 = 0
Comparing with the standard form of a quadratic equation,
a = 2, b = -√5, c = 1
Now,
b2 – 4ac = (-√5)2 – 4(2)(1)
= 5 – 8
= -3 < 0
Therefore, the given equation has no real roots.
17. The equation (x + 1)2 – 2(x + 1) = 0 has
(a) two real roots
(b) no real roots
(c) one real root
(d) two equal roots
Answer: (a) two real roots
Explanation:
(x + 1)2 – 2(x + 1) = 0
x2 + 1 + 2x – 2x – 2 = 0
x2 – 1 = 0
x2 = 1
x = ± 1
18. The quadratic formula to find the roots of a quadratic equation ax2 + bx + c = 0 is given by
(a) [-b ± √(b2-ac)]/2a
(b) [-b ± √(b2-2ac)]/a
(c) [-b ± √(b2-4ac)]/4a
(d) [-b ± √(b2-4ac)]/2a
Answer: (d) [-b ± √(b2-4ac)]/2a
The quadratic formula to find the roots of a quadratic equation ax2 + bx + c = 0 is given by [-b ± √(b2-4ac)]/2a.
19. The quadratic equation x2 + 7x – 60 has
(a) two equal roots
(b) two real and unequal roots
(b) no real roots
(c) two equal complex roots
Answer: (b) two real and unequal roots
Explanation:
Given,
x2 + 7x – 60 = 0
Comparing with the standard form,
a = 1, b = 7, c = -60
b2 – 4ac = (7)2 – 4(1)(-60) = 49 + 240 = 289 > 0
Therefore, the given quadratic equation has two real and unequal roots.
20. The maximum number of roots for a quadratic equation is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
The maximum number of roots for a quadratic equation is equal to 2 since the degree of a quadratic equation is 2.
Arithmetic Progression
1. In an Arithmetic Progression, if a = 28, d = -4, n = 7, then an is:
(a) 4
(b) 5
(c) 3
(d) 7
Answer: (a) 4
Explanation: For an AP,
an = a+(n-1)d
= 28+(7-1)(-4)
= 28+6(-4)
= 28-24
an=4
2. If a = 10 and d = 10, then first four terms will be:
(a) 10, 30, 50, 60
(b) 10, 20, 30, 40
(c) 10, 15, 20, 25
(d) 10, 18, 20, 30
Answer: (b) 10, 20, 30, 40
Explanation: a = 10, d = 10
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
3. The first term and common difference for the A.P. 3, 1, -1, -3 is:
(a) 1 and 3
(b) -1 and 3
(c) 3 and -2
(d) 2 and 3
Answer: (c) 3 and -2
Explanation: First term, a = 3
Common difference, d = Second term – First term
⇒ 1 – 3 = -2
⇒ d = -2
4. 30th term of the A.P: 10, 7, 4, …, is
(a) 97
(b) 77
(c) -77
(d) -87
Answer: (c) -77
Explanation: Given,
A.P. = 10, 7, 4, …
First term, a = 10
Common difference, d = a2 − a1 = 7−10 = −3
As we know, for an A.P.,
an = a +(n−1)d
Putting the values;
a30 = 10+(30−1)(−3)
a30 = 10+(29)(−3)
a30 = 10−87 = −77
5. 11th term of the A.P. -3, -1/2, 2 …. Is
(a) 28
(b) 22
(c) -38
(d) -48
Answer: (b) 22
Explanation: A.P. = -3, -1/2, 2 …
First term a = – 3
Common difference, d = a2 − a1 = (-1/2) -(-3)
⇒(-1/2) + 3 = 5/2
Nth term;
an = a+(n−1)d
a11 = 3+(11-1)(5/2)
a11 = 3+(10)(5/2)
a11 = -3+25
a11 = 22
6. The missing terms in AP: __, 13, __, 3 are:
(a) 11 and 9
(b) 17 and 9
(c) 18 and 8
(d) 18 and 9
Answer: (c)
Explanation: a2 = 13 and
a4 = 3
The nth term of an AP;
an = a+(n−1) d
a2 = a +(2-1)d
13 = a+d ………………. (i)
a4 = a+(4-1)d
3 = a+3d ………….. (ii)
Subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
Now put value of d in equation 1
13 = a+(-5)
a = 18 (first term)
a3 = 18+(3-1)(-5)
= 18+2(-5) = 18-10 = 8 (third term).
7. Which term of the A.P. 3, 8, 13, 18, … is 78?
(a) 12th
(b) 13th
(c) 15th
(d) 16th
Answer: (d) (d) 16th
Explanation: Given, 3, 8, 13, 18, … is the AP.
First term, a = 3
Common difference, d = a2 − a1 = 8 − 3 = 5
Let the nth term of given A.P. be 78. Now as we know,
an = a+(n−1)d
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
8. The 21st term of AP whose first two terms are -3 and 4 is:
(a) 17
(b) 137
(c) 143
(d) -143
Answer: (b) 137
Explanation: First term = -3 and second term = 4
a = -3
d = 4-a = 4-(-3) = 7
a21=a+(21-1)d
=-3+(20)7
=-3+140
=137
9. If 17th term of an A.P. exceeds its 10th term by 7. The common difference is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Explanation: Nth term in AP is:
an = a+(n-1)d
a17 = a+(17−1)d
a17 = a +16d
In the same way,
a10 = a+9d
Given,
a17 − a10 = 7
Therefore,
(a +16d)−(a+9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.
10. The number of multiples of 4 between 10 and 250 is:
(a) 50
(b) 40
(c) 60
(d) 30
Answer: (c) 60
Explanation: The multiples of 4 after 10 are:
12, 16, 20, 24, …
So here, a = 12 and d = 4
Now, 250/4 gives remainder 2. Hence, 250 – 2 = 248 is divisible by 2.
12, 16, 20, 24, …, 248
So, nth term, an = 248
As we know,
an = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60
11. 20th term from the last term of the A.P. 3, 8, 13, …, 253 is:
(a) 147
(b) 151
(c) 154
(d) 158
Answer: (d) 158
Explanation: Given, A.P. is 3, 8, 13, …, 253
Common difference, d= 5.
In reverse order,
253, 248, 243, …, 13, 8, 5
So,
a = 253
d = 248 − 253 = −5
n = 20
By nth term formula,
a20 = a+(20−1)d
a20 = 253+(19)(−5)
a20 = 253−95
a20 = 158
12. The sum of the first five multiples of 3 is:
(a) 45
(b) 55
(c) 65
(d) 75
Answer: (a) 45
Explanation: The first five multiples of 3 is 3, 6, 9, 12 and 15
a=3 and d=3
n=5
Sum, Sn = n/2[2a+(n-1)d]
S5 = 5/2[2(3)+(5-1)3]
=5/2[6+12]
=5/2[18]
=5 x 9
= 45
13. The 10th term of the AP: 5, 8, 11, 14, … is
(a) 32
(b) 35
(c) 38
(d) 185
Answer: (a) 32
Explanation:
Given AP: 5, 8, 11, 14,….
First term = a = 5
Common difference = d = 8 – 5 = 3
nth term of an AP = an = a + (n – 1)d
Now, 10th term = a10 = a + (10 – 1)d
= 5 + 9(3)
= 5 + 27
= 32
14. In an AP, if d = -4, n = 7, an = 4, then a is
(a) 6
(b) 7
(c) 20
(d) 28
Answer: (d) 28
Solution;
Given,
d = -4, n = 7, an = 4
We know that,
an = a + (n – 1)d
4 = a + (7 – 1)(-4)
4 = a + 6(-4)
4 = a – 24
⇒ a = 4 + 24 = 28
15. The list of numbers –10, –6, –2, 2,… is
(a) an AP with d = –16
(b) an AP with d = 4
(c) an AP with d = –4
(d) not an AP
Answer: (b) an AP with d = 4
Explanation:
–10, –6, –2, 2,…
Let a1 = -10, a2 = -6, a3 = -3, a4 = 2
a2 – a1 = -6 – (-10) = 4
a3 – a2 = -2 – (-6) = 4
a4 – a3 = 2 – (-2) = 4
The given list of numbers is an AP with d = 4.
16. If the 2nd term of an AP is 13 and the 5th term is 25, then its 7th term is
(a) 30
(b) 33
(c) 37
(d) 38
Answer: (b) 33
Explanation:
Given,
a2 = 13
a + d = 13
a = 13 – d….(i)
a5 = 25
a + 4d = 25….(ii)
Substituting (i) in (ii),
13 – d + 4d = 25
3d = 12
d = 4
So, a = 13 – 4 = 9
a7 = a + 6d = 9 + 6(4) = 9 + 24 = 33
17. Which term of the AP: 21, 42, 63, 84,… is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Answer: (b) 10th
Explanation:
Given AP:
21, 42, 63, 84,…
a = 21
d = 42 – 21 = 21
an = 210
a + (n – 1)d = 210
21 + (n – 1)(21) = 210
21 + 21n – 21 = 210
21n = 210
n = 10
18. What is the common difference of an AP in which a18 – a14 = 32?
(a) 8
(b) -8
(c) -4
(d) 4
Answer: (a) 8
Explanation:
Given,
a18 – a14 = 32
We know that, an = a + (n – 1)d
So,
a + 17d – (a + 13d) = 32
17d – 13d = 32
4d = 32
d = 8
19. The famous mathematician associated with finding the sum of the first 100 natural numbers is
(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid
Answer: (c) Gauss
Explanation:
The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.
20. The sum of first 16 terms of the AP: 10, 6, 2,… is
(a) –320
(b) 320
(c) –352
(d) –400
Answer: (a) -320
Explanation:
Given AP: 10, 6, 2,…
Here, a = 10, d = -4
Sum of first n terms = Sn = (n/2)[2a + (n – 1)d]
The sum of first 16 terms = S16 = (16/2)[2(10) + (16 – 1)(-4)]
= 8[20 + 15(-4)]
= 8(20 – 60)
= 8(-40)
= -320
Triangles
1. Which of the following triangles have the same side lengths?
(a) Scalene
(b) Isosceles
(c) Equilateral
(d) None of these
Answer: (c) Equilateral
Explanation: Equilateral triangles have all its sides and all angles equal.
2. Area of an equilateral triangle with side length a is equal to:
(a) √3/2a
(b) √3/2a2
(c) √3/4 a2
(d) √3/4 a
Answer: (c) √3/4 a2
Area of an equilateral triangle with side length a = √3/4 a2
3. D and E are the midpoints of side AB and AC of a triangle ABC, respectively and BC = 6 cm. If DE || BC, then the length (in cm) of DE is:
(a) 2.5
(b) 3
(c) 5
(d) 6
Answer: (b) 3
Explanation: By midpoint theorem,
DE=½ BC
DE = ½ of 6
DE=3 cm
4. The diagonals of a rhombus are 16 cm and 12 cm, in length. The side of rhombus in length is:
(a) 20 cm
(b) 8 cm
(c) 10 cm
(d) 9 cm
Answer: (c) 10 cm
Explanation: Here, half of the diagonals of a rhombus are the sides of the triangle and side of the rhombus is the hypotenuse.
By Pythagoras theorem,
(16/2)2+(12/2)2=side2
82+62=side2
64+36=side2
side=10 cm
5. Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of small triangle is 48 sq.cm, then the area of large triangle is:
(a) 230 sq.cm.
(b) 106 sq.cm
(c) 107 sq.cm.
(d) 108 sq.cm
Answer: (d) 108 sq.cm
Solution: Let A1 and A2 are areas of the small and large triangle.
Then,
A2/A1=(side of large triangle/side of small triangle)
A2/48=(3/2)2
A2=108 sq.cm.
6. If perimeter of a triangle is 100 cm and the length of two sides are 30 cm and 40 cm, the length of third side will be:
(a) 30 cm
(b) 40 cm
(c) 50 cm
(d) 60 cm
Answer: (a) 30 cm
Solution: Perimeter of triangle = sum of all its sides
P = 30+40+x
100=70+x
x=30 cm
7. If triangles ABC and DEF are similar and AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm, the perimeter of triangle is:
(a) 22 cm
(b) 20 cm
(c) 21 cm
(d) 18 cm
Answer: (d) 18 cm
Explanation: ABC ~ DEF
AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm
AB/DE = BC/EF = AC/DF
4/6 = BC/9 = AC/12
BC = (4.9)/6 = 6 cm
AC = (12.4)/6 = 8 cm
Perimeter = AB+BC+AC
= 4+6+8
=18 cm
8. The height of an equilateral triangle of side 5 cm is:
(a) 4.33 cm
(b) 3.9 cm
(c) 5 cm
(d) 4 cm
Answer: (a) 4.33 cm
Explanation: The height of the equilateral triangle ABC divides the base into two equal parts at point D.
Therefore,
BD=DC= 2.5 cm
In triangle ABD, using Pythagoras theorem,
AB2=AD2+BD2
52=AD2+2.52
AD2 = 25-6.25
AD2=18.75
AD=4.33 cm
9. If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if
(a) ∠A=∠F
(b) ∠B=∠D
(c) ∠A=∠D
(d) ∠B=∠E
Answer: (b) ∠B=∠D
If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if ∠B=∠D.
10. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio
(a) 2: 3
(b) 4: 9
(c) 81: 16
(d) 16: 81
Answer: (d) 16: 81
Explanation: Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16: 81
11. Which of the following are not similar figures?
(a) Circles
(b) Squares
(c) Equilateral triangles
(d) Isosceles triangles
Answer: (d) Isosceles triangles
Explanation:
All circles, squares, and equilateral triangles are similar figures.
12. In triangle ABC, ∠BAC = 90° and AD ⊥ BC. Then
(A) BD . CD = BC2
(B) AB . AC = BC2
(C) BD . CD = AD2
(D) AB . AC = AD2
Answer: (c) BD . CD = AD2
Explanation:
In ΔADB and ΔADC,
∠D = ∠D = 90°
∠DBA = ∠DAC
By AAA similarity criterion,
ΔADB ~ ΔADC
BD/AD = AD/CD
BD.CD = AD2
13. If in two triangles ABC and PQR, AB/QR = BC/PR = CA/PQ, then
(a) ΔPQR ~ ΔCAB
(b) ΔPQR ~ ΔABC
(c) ΔCBA ~ ΔPQR
(d) ΔBCA ~ ΔPQR
Answer: (a) ΔPQR ~ ΔCAB
Explanation:
Given that, in triangles ABC and PQR, AB/QR = BC/PR = CA/PQ
If sides of one triangle are proportional to the side of the other triangle, and their corresponding angles are also equal, then both the triangles are similar by SSS similarity. Therefore, ΔPQR ~ ΔCAB
14. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Answer: (b) similar but not congruent
Explanation:
In ΔABC and ΔDEF,
∠B = ∠E, ∠F = ∠C and AB = 3 DE
By AA similarity criterion,
ΔABC ~ ΔDEF
AB = 3DE
⇒ AB/DE = 3
⇒ AB/DE = BC/EF = AC/DF = 3
For triangles to be congruent, the ratio of sides must be 1
Therefore, triangles are similar but not congruent.
15. It is given that ΔABC ~ ΔPQR, with BC/QR = 1/4 then, ar(ΔPRQ)/ar(ABC) is equal to
(a) 16
(b) 4
(c) 1/4
(d) 1/16
Answer: (a) 16
Explanation:
Given,
ΔABC ~ ΔPQR
and BC/QR = 1/4
Ratio of area of similar triangles is equal to the square of its corresponding sides.
So, ar(ΔPRQ)/ar(ABC) = (QR/BC)2 = (4/1)2 = 16
16. It is given that ΔABC ~ ΔDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true:
(a) DE = 12 cm, ∠F = 50°
(b) DE = 12 cm, ∠F = 100°
(c) EF = 12 cm, ∠D = 100°
(d) EF = 12 cm, ∠D = 30°
Answer: (b) DE = 12 cm, ∠F = 100°
Explanation:
Given,
ΔABC ~ ΔDFE, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm
In triangle ABC,
∠A + ∠B + ∠C = 180°
∠B = 180° – 30° – 50° = 100°
Since ΔABC ~ ΔDFE, the corresponding angles are equal.
Thus, ∠D = ∠A = 30°
∠F = ∠B = 100°
∠E = ∠C = 50°
And
AB/DF = AC/DE
5/7.5 = 8/DE
DE = (8 × 7.5)/5 = 12 cm
17. If triangle ABC is similar to triangle DEF, then,
(a) AB/FD = BC/EF = CA/DE
(b) AB/DE = BC/DF = CA/EF
(c) AB/DE = BC/EF = CA/FD
(d) AB/BC = CA/DE = EF/FD
Answer: (c) AB/DE = BC/EF = CA/FD
Explanation:
If two triangles are similar, i.e. when ΔABC ~ ΔDEF, then
(i) their corresponding angles are equal and
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and
(ii) their corresponding sides are in the same ratio (or proportion).
AB/DE = BC/EF = CA/FD
18. Which of the following is not a similarity criterion for two triangles?
(a) AAA
(b) SAS
(c) SSS
(d) ASA
Answer: (d) ASA
Explanation:
The main criteria for similarity of two triangles are AAA, AA, SAS and SSS.
19. The ratio of the areas of two similar triangles is equal to
(a) square of the ratio of their corresponding sides
(b) cube of the ratio of their corresponding sides
(c) square root of the ratio of their corresponding sides
(d) twice the ratio of their corresponding sides
Answer: (a) square of the ratio of their corresponding sides
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
20. In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is
(a) 120°
(b) 60°
(c) 90°
(d) 45°
Answer: (c) 90°
Explanation:
Given,
In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
Here, AC is the longest side.
If the square of the hypotenuse is equal to the square of the other two sides, then it is a right angled triangle.
So, AC2 = AB2 + BC2
(12)2 = (6√3)2 + (6)2
144 = 108 + 36
144 = 144
∴ ∆ABC is a right angled triangle and angle opposite to hypotenuse, i.e. opposite to AC is ∠B and is equal to 90°.
