Mathematics Question Paper With Detailed Solutions
Class 10 Mathematics
I. CHOOSE THE CORRECT ANSWER
1. Prime factorization of 91 is,
(A) 2×13×7
(B) 13×7 ✓
(C) 91×1
(D) 13×7
Solution:
$$
91 = 7 \times 13
$$
Since 7 and 13 are prime numbers, the prime factorization is:
$$
13 \times 7
$$
Correct Answer: (B)
2. The L.C.M of a and 18 is 36 and H.C.F of a and 18 is 2. The value of a is,
(A) 2
(B) 3
(C) 4 ✓
(D) 1
Using formula:
$$
LCM \times HCF = Product\ of\ numbers
$$
$$
36 \times 2 = a \times 18
$$
$$
72 = 18a
$$
$$
a = \frac{72}{18}
$$
$$
a = 4
$$
Correct Answer: (C)
3. The H.C.F and L.C.M of two numbers a and b are 5 and 200, then the product of a and b is,
(A) 205
(B) 1000 ✓
(C) 200
(D) 195
$$
Product = HCF \times LCM
$$
$$
= 5 \times 200
$$
$$
= 1000
$$
Correct Answer: (B)
4. The quadratic polynomial whose sum and product of the zeroes are -2 and 8 is,
(A) x² + 2x + 8 ✓
(B) x² – 2x – 8
(C) x² + 2x – 8
(D) x² – 2x + 8
Formula:
$$
x^2-(\alpha+\beta)x+\alpha\beta
$$
Given:
$$
\alpha+\beta=-2
$$
$$
\alpha\beta=8
$$
Substituting:
$$
x^2-(-2)x+8
$$
$$
x^2+2x+8
$$
Correct Answer: (A)
5. If one of the zeroes of the polynomial p(x)=x²−x+k is 2, then the value of k is,
(A) 2
(B) -2 ✓
(C) -6
(D) 6
Substitute x = 2:
$$
2^2-2+k=0
$$
$$
4-2+k=0
$$
$$
2+k=0
$$
$$
k=-2
$$
Correct Answer: (B)
6. If the graph of a polynomial P(x) passes through the coordinate points (-3,0), (-1,-5), (0,-6) and (2,0) then the zeroes of the quadratic polynomial are,
(A) -3 and -6
(B) 0 and -3
(C) -1 and -5
(D) -3 and 2 ✓
Zeroes occur where y = 0.
Points are:
$$
(-3,0)\ and\ (2,0)
$$
Therefore zeroes are:
$$
-3\ and\ 2
$$
Correct Answer: (D)
7. The lines representing the equations 2x+3y−9=0 and 4x+6y−18=0 are
(A) Intersecting lines
(B) Perpendicular lines
(C) Parallel lines
(D) Coincident lines ✓
Second equation is exactly double the first equation.
Hence both equations represent same line.
Therefore lines are coincident.
Correct Answer: (D)
8. If a pair of linear equations is inconsistent then the lines represented by them are
(A) Intersecting lines
(B) Perpendicular lines
(C) Parallel lines ✓
(D) Coincident lines
Inconsistent equations have no solution.
Parallel lines never meet.
Hence inconsistent equations represent parallel lines.
Correct Answer: (C)
9. If the lines represented by the pair of linear equations x+2y=3 and 2x+4y=k are coincident then the value of k is,
(A) 3
(B) 6 ✓
(C) -3
(D) -6
Multiply first equation by 2:
$$
2x+4y=6
$$
Comparing with second equation:
$$
2x+4y=k
$$
Therefore:
$$
k=6
$$
Correct Answer: (B)
10. The H.C.F of (12,15) is 3. The L.C.M of (12,15) is,
(A) 60 ✓
(B) 45
(C) 36
(D) 90
Prime factors:
$$
12=2^2 \times 3
$$
$$
15=3 \times 5
$$
LCM:
$$
2^2 \times 3 \times 5
$$
$$
=60
$$
Correct Answer: (A)
II. SOLVE THE FOLLOWING
11. State the Fundamental Theorem of Arithmetic.
Every composite number can be expressed as a product of prime numbers and this factorization is unique apart from the order of the prime factors.
12. If 200 = 2แต × 5โฟ , then find the value of m and n.
Prime factorization of 200:
$$
200=2^3 \times 5^2
$$
Comparing:
$$
m=3,\ n=2
$$
13. What is the maximum number of zeroes of a cubic polynomial?
A cubic polynomial has degree 3.
Hence maximum number of zeroes is:
$$
3
$$
14. Write the degree of the polynomial p(x)=x(x²+2x)+3x−5
Simplifying:
$$
p(x)=x^3+2x^2+3x-5
$$
Highest power is 3.
Therefore degree = 3.
15. How many solutions does the pair of linear equation in two variables have if the lines represented by them are intersecting?
Intersecting lines meet at exactly one point.
Therefore the equations have one unique solution.
III. SOLVE THE FOLLOWING
16. Prove that √3 + 5 is an irrational number.
We know that:
$$
\sqrt3
$$
is irrational.
Also:
$$
5
$$
is rational.
Sum of a rational number and irrational number is always irrational.
Hence:
$$
\sqrt3+5
$$
is irrational.
17. If ฮฑ and ฮฒ are the two zeroes of the polynomial P(x)=x²−4x+5 then find ฮฑ²+ฮฒ².
For polynomial:
$$
x^2-4x+5
$$
$$
\alpha+\beta=4
$$
$$
\alpha\beta=5
$$
Formula:
$$
\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta
$$
Substituting:
$$
=4^2-2(5)
$$
$$
=16-10
$$
$$
=6
$$
18. Find the polynomial whose zeroes are √3 and −√3
Polynomial:
$$
(x-\sqrt3)(x+\sqrt3)
$$
Using identity:
$$
(a-b)(a+b)=a^2-b^2
$$
$$
=x^2-3
$$
Required polynomial:
$$
x^2-3
$$
19. Solve the given pair of linear equations by elimination method.
3x−y=15 and 2x−y=5
Given:
$$
3x-y=15
$$
$$
2x-y=5
$$
Subtract second equation from first:
$$
x=10
$$
Substitute in:
$$
2x-y=5
$$
$$
2(10)-y=5
$$
$$
20-y=5
$$
$$
y=15
$$
Therefore:
$$
x=10,\ y=15
$$
20. Solve the given pair of linear equations by elimination method.
2x+5y=7 and 5x+2y=7
Multiply first equation by 5:
$$
10x+25y=35
$$
Multiply second equation by 2:
$$
10x+4y=14
$$
Subtract:
$$
21y=21
$$
$$
y=1
$$
Substitute in first equation:
$$
2x+5(1)=7
$$
$$
2x+5=7
$$
$$
2x=2
$$
$$
x=1
$$
Therefore:
$$
x=1,\ y=1
$$
IV. SOLVE THE FOLLOWING
21. Prove that √5 is an irrational number.
Assume √5 is rational.
Then:
$$
\sqrt5=\frac{a}{b}
$$
where a and b are co-prime integers.
Squaring both sides:
$$
5=\frac{a^2}{b^2}
$$
$$
a^2=5b^2
$$
Thus a² is divisible by 5.
Therefore a is divisible by 5.
Let:
$$
a=5k
$$
Substituting:
$$
25k^2=5b^2
$$
$$
5k^2=b^2
$$
Thus b² is divisible by 5.
Hence b is divisible by 5.
So both a and b are divisible by 5.
This contradicts that a and b are co-prime.
Hence:
$$
\sqrt5
$$
is irrational.
22. Find the H.C.F and L.C.M of the numbers 510 and 92 by prime factorization and verify that
LCM × HCF = Product of numbers.
Prime factorization:
$$
510=2 \times 3 \times 5 \times 17
$$
$$
92=2^2 \times 23
$$
HCF:
$$
HCF=2
$$
LCM:
$$
LCM=2^2 \times 3 \times 5 \times 17 \times 23
$$
$$
=23460
$$
Verification:
$$
LCM \times HCF
$$
$$
=23460 \times 2
$$
$$
=46920
$$
Product of numbers:
$$
510 \times 92
$$
$$
=46920
$$
Hence verified.
V. SOLVE THE FOLLOWING
23. Solve the pair of linear equations by graphical method.
x+3y=6 and 2x−3y=12
Adding equations:
$$
x+3y=6
$$
$$
2x-3y=12
$$
$$
3x=18
$$
$$
x=6
$$
Substitute in first equation:
$$
6+3y=6
$$
$$
3y=0
$$
$$
y=0
$$
Therefore intersection point is:
$$
(6,0)
$$
24. Solve the pair of linear equations by graphical method.
2x+y=8 and x−y=1
Given:
$$
2x+y=8
$$
$$
x-y=1
$$
From second equation:
$$
y=x-1
$$
Substitute in first equation:
$$
2x+(x-1)=8
$$
$$
3x-1=8
$$
$$
3x=9
$$
$$
x=3
$$
$$
y=2
$$
Therefore intersection point is:
$$
(3,2)
$$
IV. SOLVE THE FOLLOWING: 3 × 2 = 6
21. Prove that √5 is an irrational number.
Solution:
Assume that:
$$
\sqrt5
$$
is a rational number.
Then it can be written in the form:
$$
\sqrt5=\frac{a}{b}
$$
where a and b are integers having no common factor other than 1 and:
$$
b \ne 0
$$
Squaring both sides:
$$
5=\frac{a^2}{b^2}
$$
Multiplying both sides by:
$$
b^2
$$
we get:
$$
a^2=5b^2
$$
This means:
$$
a^2
$$
is divisible by 5.
Therefore:
$$
a
$$
is also divisible by 5.
So let:
$$
a=5k
$$
Substituting in:
$$
a^2=5b^2
$$
we get:
$$
(5k)^2=5b^2
$$
$$
25k^2=5b^2
$$
$$
5k^2=b^2
$$
Thus:
$$
b^2
$$
is divisible by 5.
Hence:
$$
b
$$
is also divisible by 5.
Therefore both:
$$
a \text{ and } b
$$
are divisible by 5.
But this contradicts the fact that:
$$
a \text{ and } b
$$
have no common factor.
Hence our assumption is wrong.
Therefore:
$$
\sqrt5
$$
is an irrational number.
22. Find the H.C.F and L.C.M of the numbers 510 and 92 by the method of prime factorization and verify that:
L.C.M × H.C.F = Product of the numbers
L.C.M × H.C.F = Product of the numbers
Solution:
Prime factorization of:
$$
510
$$
$$
510 = 2 \times 3 \times 5 \times 17
$$
Prime factorization of:
$$
92
$$
$$
92 = 2^2 \times 23
$$
H.C.F
Common prime factor is:
$$
2
$$
Therefore:
$$
H.C.F = 2
$$
L.C.M
Taking highest powers of all prime factors:
$$
LCM = 2^2 \times 3 \times 5 \times 17 \times 23
$$
$$
LCM = 23460
$$
Verification
$$
LCM \times HCF
$$
$$
= 23460 \times 2
$$
$$
= 46920
$$
Product of numbers:
$$
510 \times 92
$$
$$
= 46920
$$
Therefore:
$$
LCM \times HCF = Product\ of\ the\ numbers
$$
Hence verified.
V. SOLVE THE FOLLOWING: 4 × 2 = 8
23. Solve the given pair of linear equations by graphical method.
x + 3y = 6 and 2x - 3y = 12
x + 3y = 6 and 2x - 3y = 12
Solution:
Given equations:
$$
x + 3y = 6
$$
$$
2x - 3y = 12
$$
Adding both equations:
$$
x + 3y + 2x - 3y = 6 + 12
$$
$$
3x = 18
$$
$$
x = 6
$$
Substituting:
$$
x = 6
$$
in the first equation:
$$
6 + 3y = 6
$$
$$
3y = 0
$$
$$
y = 0
$$
Therefore the solution is:
$$
(x,y) = (6,0)
$$
Thus the two lines intersect at:
$$
(6,0)
$$
24. Solve the given pair of linear equations by graphical method.
2x + y = 8 and x - y = 1
2x + y = 8 and x - y = 1
Solution:
Given equations:
$$
2x + y = 8
$$
$$
x - y = 1
$$
From second equation:
$$
x - y = 1
$$
$$
y = x - 1
$$
Substituting in first equation:
$$
2x + (x - 1) = 8
$$
$$
3x - 1 = 8
$$
$$
3x = 9
$$
$$
x = 3
$$
Substituting:
$$
x = 3
$$
in:
$$
y = x - 1
$$
$$
y = 3 - 1
$$
$$
y = 2
$$
Therefore the solution is:
$$
(x,y) = (3,2)
$$
Thus the two lines intersect at:
$$
(3,2)
$$