Subject: Mathematics
Grade: X
Marks: 30
Time: 1 Hour
General Instructions:
i) All Questions are compulsory
ii) The question paper consists of 14 questions divided into four sections A, B, C and D. Section A comprises of 5 questions of 1 mark each, Section B comprises of 4 questions of 2 marks each, Section C comprises of 3 questions of 3 marks each and Section D comprises of 2 questions of 4 marks each.
iii) There is no overall choice. However, an internal choice has been provided. You have to attempt only one of the alternatives in all such questions.
iv) Use of calculator is not permitted.
Section A
Q1-Q5 are multiple choice questions. Select the most appropriate answer from the given options.
1. The quadratic polynomial whose zeroes are 5 and – 8 is
a) x2 + 13x – 40
b) x2 – 13x + 40
c) x2 + 3x – 40
d) x2 – 3x – 40
View AnswerAns. (c)
2. If two positive integers p and q are written as p = a2b3 and q = a4b; a, b are prime numbers, then HCF (p, q) is
a) ab
b) a2b
c) a4b3
d) 1
View AnswerAns. (b)
3. The pair of equations 3x + y + 4 = 0 and -3x -6y + 1 = 0 have
a) A unique solution
b) exactly two solutions
c) infinitely many solutions
d) no solution
View AnswerAns. (a)
4. The graph of y = f(x) is given, for some polynomial f(x). Find the number of zeroes of f(x)
a) 0
b) 1
c) 2
d) 3
View AnswerAns. (b)
5. The pair of equations x = a and y = b graphically represents line which are
a) parallel
b) coincident
c) intersecting at (b, a)
d) intersecting at (a, b)
View AnswerAns. (d)
Section B
6. For what value of k, does the pair of equations 2x + 3y = 4 and (k + 2) x + 6y = 3k + 2 will have infinite number of solutions?
Ans.
7. Given that HCF (240, 228) = 12. Find LCM (240, 228)
View AnswerAns. Prime factorization of 240 = 2 × 2 × 2 × 2 × 3 × 5
Prime factorization of 228 = 2 × 2 × 3 × 19
LCM = 2 × 2 × 2 × 2 × 3 × 5 × 19 = 4560
8. If one zero of 2x2 – 3x + k is the reciprocal of the other, find the value of k.
View AnswerAns. k = 2
9. On a morning walk, three persons step off together and their steps measure 40cm, 42cm and 45cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
View AnswerAns. The LCM of 40 cm, 42 cm and 45 cm has to be found to get the required minimum distance.
Now,
40 = 2 × 2 × 2 × 5
42 = 2 × 3 × 7
45 = 3 × 3 × 5
LCM (40, 42, 45) = 2 × 3 × 5 × 2 × 2 × 3 × 7
= 2520.
Therefore, the minimum distance each should walk is 2520 cm so that each can cover the same distance in complete steps.
Section C
10. If α and β are the zeroes of the polynomials 2x2 – 7x + 3, find α2 + β2
View AnswerAns. α, β are the zeroes of the polynomial p(x)=2x2−7x+3
Sum of zeroes (α+β) = −b/a =−(−7)/2=7/2
Product of zeroes αβ = c/a = 3/2
∴α2 + β2 = (α+β)2 − 2αβ
= (7/2)2 – 2(3/2)
= 49/4 – 12/4
= 37/4
11. Solve for x and y: 99x + 101y = 499; 101x + 99y = 501
View Answer99x + 101y = 499 —– (1)
101x + 99y = 501 —– (2)
Adding (1) and (2)
200x+200y=1000
∴ x + y = 5—–(3)
Subtracting (1) from (2)
2x−2y=2
∴ x – y = 1−−−(4)
Adding (3) and (4)
2x=6
∴x=3
Putting x=3 in equation (4)
3−y=1∴ y=2
∴ y=2, x=3
OR
Solve for x and y: ax + by = a – b; bx – ay = a + b
View Answerax + by = a – b —- (1) x b
bx – ay = a + b —– (2) x a
abx + b2y = ab – b2
abx + a2y = a2 + ab
Subtracting
b2y + a2y = ab – b2 – a2 – ab
b2y +a2y = -(a2 + b2)
y(b2 + a2)= – (a2 + b2)
y = -1
Putting the value of y = -1
ax + by = a – b
ax + b(-1) = a – b
ax – b = a – b
x = 1
12. Find the zeroes of the quadratic polynomial 4x2 + 4x – 3 and verify the relationship between the zeroes and the coefficients.
View AnswerAns. Let f(x) = 4 x2 – 4x – 3
By splitting the middle term, we get
f(x) = 4x2 – 6x + 2x – 3
= 2x(2x – 3) + 1(2x – 3)
= (2x + 1) (2x – 3)
On putting f(x) = 0, we get
(2x + 1) (2x – 3) = 0
⇒ 2x + 1 = 0 or 2x – 3 = 0
x = -1/2 or x = 3/2
Section D
13. Prove that √3 is an irrational number.
14. A fraction becomes 4/5, if 1 is added to bot numerator and denominator. If 5 is subtracted from both numerator and denominator it becomes ½. Find the fraction.
View AnswerAns. 7/9
OR
The sum of a two-digit number and the number formed by interchanging the digits is 132. If 12 is added to the first number, the new number becomes 5 times the sum of its digits. Find the number.
View AnswerAns. Let the digits at units and tens place in the given number be x and y respectively.
Then, number = 10y + x . . . . (i)
Number formed by interchanging the digits = 10x + y
According to the given condition, we have (10y + x) + (10x + y) = 132
and (10y + x) + 12 = 5(x + y)
⇒ 11x + 11y = 132 and 4x – 5y = 12
⇒ x + y – 12 = 0 . . . (ii) and 4x – 5y – 12 = 0 . . . (iii)
Solving these two equations:
Multiplying eq. (i) × 4, we get
4x + 4y = 48 . . . (iv)
On subtracting (iii) from (iv), we get
9y = 36
⇒ y = 4
On substituting y = 4 in eq. (ii), we get
x + 4 – 12 = 0
⇒ x = 8
⇒ x = 8 and y = 4
On substituting the values of x and y in equation (1), we have
10y + x
= 10 × 4 + 8 = 48
∴ The number is 48.