Solution

REAL NO, POLYNOMIALS, LINEAR EQUATIONS & QUADRATIC EQUATIONS

Class 10 - Mathematics
Section A
1.
(d) composite number
Explanation:
We have 7 × 11 × 13 + 13 = 13 (77 + 1) = 13 × 78. Since the given number has 2 more factors other than 1 and itself,
therefore it is a composite number.

2.
(c) 1680
Explanation:
​LCM = Product of greatest power of each prime factor involved in the numbers
= 24 × 3 × 5 × 7
= 16 × 3 × 5 × 7

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= 1680

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3.
(b) p q3 2

Explanation:
tor
We know that LCM = product of the highest powers of all the prime factors of the numbers pq2, p3q2
LCM = p3q2
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4. (a) an irrational number

Explanation:
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Let a be rational and √b is irrational.

If possible let a + √b be rational.
Then a + √b is rational and a is rational.
⇒ [(a + √b) − a] is rational [Difference of two rationals is rational]
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⇒ √b is rational.
This contradicts the fact that √b is irrational.
The contradiction arises by assuming that a + √b is rational.
Therefore, a + √b is irrational.
5.
(b) a = 2, b = 1
Explanation:
– √3−1
a − b√3 =
√3+1

√3−1 √3−1
= ×
√3+1 √3−1
2
( √3−1)
= 2 2
( √3) −(1)

2 2
( √3) +(1) −2(1))( √3)
= 3−1

3+1−2√3
= 2

4−2√3
= 2
2(2− √3)
= 2
–
= 2- √3
– –
⇒ a − b√3 = 2 - √3
⇒ a = 2, b = 1

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 6.
(b) 1
Explanation:
Co prime number are number which have 1 as HCF.

7.
(d) 17 × 500
Explanation:
850 = 2 × 5 × 5 × 17
500 = 2 × 2 × 5 × 5 × 5
LCM (850, 500) = 2 × 2 × 5 × 5 × 5 × 17 = 17 × 500

8.
(b) H.C.F. = 1, L.C.M. = p(p + 1)
Explanation:
Since, p is prime
∴ p and p + 1 has no common factor other than 1.

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∴ H.C.F of p and p + 1 = 1

& L.C.M of p and p + 1 = p × ( p + 1 ) = p(p + 1)

9. (a) a rational number

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Explanation:
tor
– – – –
(1 + √2) + (1 - √2) = 1 + √2 + 1 - √2 = 1 + 1 = 2 And 2 is a rational number.
Therefore the given number is rational number.
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10.
(b) a rational number
Explanation:
p
It can be expressed in form
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2.35 = 235

100

so, 2.35 is a rational number

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11.
(c) 1
Explanation:
Since α and β are the zeros of quadratic polynomial
2
f (x) = x − p(x + 1) − c

or f (x) = x 2
− px − p − c

− Coefficient of x −p
α + β = = −( ) =p
Coefficient of x2 1

Constant term −p−c

αβ = = = -p-c
Coefficient of x2 1

We have
0 = (α + 1)(β + 1)

0 = αβ + (α + β) + 1

0 = -c+1
c=1
Therefore, value of c is1.

12.
(b) equal to 0
Explanation:

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 Given that two of the zeros of the cubic polynomial ax3 + bx2+ cx + d are 0,
i.e. α = 0, β = 0
b
α + β + γ = −
a
b
γ = −
a
c
αβ + βγ + γα = −
a
c
0 = −
a

c=0

13.
(d) 2
Explanation:
If α = 2β
α + β = -
b

3β = 3k (substituting both sides.)

β = K ...(1)

(α )(β ) = c

(2β )(β ) = 4k.

2β 2 = 4k

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β
2= 4k

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β = root of (2k) ...(2)
From 1 and 2, tor
k = root of (2k)
on squaring both sides,
K2 - 2K
2
K
=2
Tu
K

Therefore, K = 2.

2
b −2ac
14. (a) ac
pta

Explanation:
Since
2 2
α +β
= αβ
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2
(α+β ) −2αβ
= αβ
2
−b c
( ) −2×
a a

= c

a
2
b 2c
−
a
a2
= c

a
2
b −2ac
= 2
×
a

c
a
2
b −2ac
= ac

15. (a) 4 (x − 5) 2

Explanation:
– –
α = √5 β = − √5
– –
α + β = √5 − √5 = 0
– –
αβ = (√5)(− √5) = −5

req. poly is
2
x − (α + β)x + αβ = 0

2
x − 0x − 5 = 0

2
x − 5 = 0

16. (a) 0
Explanation:

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 Sum of zeroes of the quadratic equation
ax2 + bx + c = 0 is
−b

Sum of zeroes of x2 - 1 = x2 + 0x - 1 = 0 is
−0
∴
1
=0
∴ α + β =0
17.
(d) x2 + 5x - 24
Explanation:
α = -8, β = 3
α + β = −8 + 3 = −5

αβ = (−8)(3) = −24

required poly is
2
x − (α + β)x + αβ = 0

2
x − (−5)x + (−24) = 0

2
x + 5x − 24 = 0

18.
(c) 11

s
4

Explanation:

ial
α+β
Here a = 3,b = 11,c = - 4 Since 1

α
+
1

β
=
αβ

−11 −4
α + β =
3
, αβ = 3
−11 tor
So,
3 11
=
−4 4

19.
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(b) a - b
Explanation:
ax + by = a2 - b2 ...(1)
bx + ay = 0 ....(2)
pta

add (1) and (2)

x(a + b) + y(a + b) = (a + b) (a - b)
divide by (a + b)
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x+y=a-b

20.
(b) parallel
Explanation:
We have,
6x - 2y + 9 = 0
And, 3x - y + 12 = 0
Here, a1 = 6, b1 = - 2 and c1 = 9
a2 = 3, b2 = - 1 and c2 = 12
−2 c1
a1

a2
=
6

3
=
2

1
,
b1

b2
=
−1
=
2

1
and c2
=
9

12
=
3

4
a1 b1 c1
Clearly, a2
=
b2
≠
c2

Hence, the given system has no solution and the lines are parallel.

21.
(c) k = 3

Explanation:

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 kx + 2y - 5 = 0
3x + 4y - 1 = 0
For No Solution
−5
k

3
= 2

4
≠
−1
6 3
k= 4
= 2

k= 3

22.
(b) parallel
Explanation:
We know that, If a pair of linear equations is inconsistent then their graph lines do not intersect each other and there will be no
solution exists. Hence, the lines are parallel.

23. (a) 87o

Explanation:
Let x and y be the measures of ∠ A and ∠ B respectively.
Now,∠ A +∠ B +∠ C = 18o[By angle sum property]

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⇒ x + y + 50o = 180o[Given,∠ C = 50o]
⇒ x + y = 130o ...(i)

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Also,∠ A -∠ B = 44o⇒ x - y = 44o ...(ii)
Adding (i) and (ii), we get
2x = 174o⇒ x = 87o⇒∠ A = 87o
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24.
(c) 10
km\hr
Tu
3

Explanation:
Let the speed of X and Y be x km/hr and y km/hr respectively. Then,
30
Time taken by X to cover 30 km = hr x

and time taken by Y to cover 30 km = hr 30

pta

By the given conditions, we have

30

x
− = 3⇒ − = 1 ...(i)
30

y
10

x
10

If X doubles his pace, then speed of X is 2x km/hr

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∴ Time taken by X to cover 30 km = hr 30

2x

We have 30

y
−
30

2x
=1 1

2
⇒
30

y
−
30

2x
= 3

2
−10
⇒
10

y
−
5

x
= 1

2
⇒
x
+
20

y
= 1 ...(ii)
Putting = u and =v in equation (i) and (ii), we get
1

x
1

10u - 10v = 1 ⇒ 10u -10v - 1 = 0 ...(iii)

and -10u + 20v = 1⇒ -10u + 20v - 1 = 0 ...(iv)
Adding equations (iii) and (iv), we get
10v - 2 = 0⇒ v = 1

Putting v = in equation (iii), we get

1

10u - 3 = 0⇒ u = 3

10

Now, u = 10
3
⇒
1

x
= 3

10
⇒ x= 10

Hence, X's speed = 10

3
km/hr

25.
(b) 36
Explanation:
Let the digit at units place be x and the digit at tens place e be y, then the number = 10y + x
10y+x
Now, according to the question, y+x
= 4

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 ⇒ 10y + x = 4y + 4x
⇒ 6y = 3x⇒ x = 2y ...(i)
Also, x = 3 + y⇒ 2y = 3 + y [From (i)]
⇒ y = 3 and x = 6
∴ Required number = 36

26.
(b) intersecting
Explanation:
intersecting
The lines representing linear equations x = 6 and y = 6
x + 0y = 6 ...(1)
0x + y = 6 ...(2)
a1 1
=
a2 0

b1 0
=
b2 1

1 0
≠
0 1
a1 b1
≠
a2 b2

s
⇒ intersecting lines

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27. (a) x(x + 1) + 8 = (x + 2) (x - 2)
Explanation: tor
x(x + 1) + 8 = (x + 2)(x - 2)
x2 + x + 8 = x2 - 4
x + 8 =-4
The degree of above equation became 1. so it is not a quadratic equation.
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28.
3
(c) − 2

Explanation:
pta

If y = 1 is solution
p(1) = 0
p(1)2 + p(1) + 3 = 0
p+p+3
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2p = -3
−3
p =
2

–
29. (a) ±2√2
Explanation:
–
±2√2

30.
(b) > 0
Explanation:
The roots of ax2 + bx + c = 0,a ≠ 0 are real and unequal only when (b2 - 4 ac) > 0.

31.
(c) 999
Explanation:
Let the number of bangles in a side of square = x
According to the question,
x2 + 38 = Total no. of bangles ...(i)
Also, (x + 1)2 - 25 = Total no. of bangles ...(ii)

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 From (i) and (ii), we have
x2 + 38 = (x + 1)2 - 25
⇒ 38 + 24 = 2x ⇒ x = 31

∴ Total no. of bangles = (31)2 + 38 = 999

32. (a) x + x − 5 = 0
2

Explanation:
In equation x + x − 5 = 0
2

a = 1, b = 1, c = −5
2
∴b
2
− 4ac = (1) − 4 × 1 × (−5) = 1 + 20 = 21
Since b 2
− 4ac > 0 therefore, x
2
+ x − 5 = 0 has two distinct roots.

33. (a) a = ± 1
Explanation:
In the equation ax2 + 2x + a = 0
D = b2- 4ac = (2)2 - 4 × a × a = 4 - 4a2
Roots are real and equal
D=0

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⇒ 4 - 4a2 = 0
4 = 4a2

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⇒

⇒ 1 = a2
a2 = 1
⇒
tor
⇒ a2 = (± 1)2
⇒ a = ± 1

–
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34. (a) 12√2
Explanation:
Equation is 2x2 + ax + 32 = 0
Let one root beα , then other would be 2α
pta

–
Nowα× 2α = 16⇒α =± 2√2
−a
andα + 2α = ⇒ 3α =
2
​​⇒ 6α = -a
a

2
– –
⇒± 12√2= -a or a =± 12√2
35.
Gu

(c) -2 < b < 2

Explanation:
In the equation
x2 - bx + 1 = 0
D = b2 - 4ac = (-b)2 - 4 × 1 × 1
= b2 - 4
∵ it is given that the roots are not real, D < 0
⇒ b2 - 4 < 0
⇒ b2 < 4 ⇒ b2 < (± 2)2
∴ b < 2 and b > -2 or -2 < b

∴ -2 < b < 2

36. (a) -2 < k < 2

Explanation:
For no real roots, we must have: b2 - 4ac < 0.
k − 4 < 0 ⇒ k < 4 ⇒ −2 < k < 2 .
2 2

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37.
(d) A is false but R is true.
Explanation:
We know that for any two numbers, Product of the two numbers = HCF × LCM
HCF × LCM = 18 × 169 = 3042 ≠ 3072
So, A is false but R is true.

38.
(c) A is true but R is false.
Explanation:
–
Here reason is not true. √4 = ±2 , which is not an irrational number.

39. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Explanation:
– –
As irrational roots/zeros always occurs in pairs, therefore, when one zero is (2 − √3) then other will be (2 + √3). So, both
Assertion and Reason are correct and Reason explains Assertion.
40. (a) Both A and R are true and R is the correct explanation of A.
Explanation:

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Both A and R are true and R is the correct explanation of A.

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41. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
On solving given equations, we get x = 0 and y = 0.
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42.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
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a1 2 b1 3 c1 8 a1 b1 c1
Clearly, a2
=
1
,
b2
=
5
and c2
=
6
, we know that for infinitely many solutions a2
=
b2
=
c2
.
a1 2 c1 8 a1 c1
But here, a2
=
1
and c2
=
6
, clearly a2
=
c2
, so what may be the value of ‘k’ systems will never has infinitely many
solutions.
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43. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
8x2 + 3kx + 2 = 0
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Discriminant, D = b2 - 4ac
D = (3k)2 - 4 × 8 × 2 = 9k2 - 64
For equal roots, D = 0
9k2 - 64 = 0
9k2 = 64
k2 = 64

k=± 8

So, both A and R are true and R is the correct explanation of A.

44. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Assertion 2x2 - 3x + 5 = 0
−b
α + β =
a

−(−3)
= 2
=
3

and αβ = c

a
=
5

45.
(b) 1
Explanation:

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 The number of zeroes is 1 as the graph intersects the x-axis at one point only.

46.
(c) -5, 0, 7
Explanation:
The graph intersect the x-axis at three distinct Points -5, 0, 7. So, there are three zeroes of P(x) which are -5, 0, 7.

47.
(c) no solution
Explanation:
Given, equations are
x + 2y + 5 = 0, and
- 3x - 6y + 1 = 0.
Comparing the equations with general form:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Here, a1 = 1, b1 = 2, c1 = 5

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And a2 = - 3, b2 = - 6, c2 = 1
Taking the ratio of coefficients to compare

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a1 −1 b1 −1 c1 5
= , = , =
a2 3 b2 3 c2 1

a b c
So a
1

2
=
1

b2
≠
c
1

2
tor
This represents a pair of parallel lines.
Hence, the pair of equations has no solution.
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48. (a) (4, 4)

Explanation:
(4, 4)
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Section B
49. i. HCF (96, 240, 336) = 48
ii. Number of stacks = =7 336

48

iii. Total number of stacks = 96

+
240
+
336
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48 48 48

= 14
OR
Height of each stack of History = 48 × 1.8 = 86.4 cm
Height of each stack of Science = 48 × 2.2 = 105.6 cm
Height of each stack of Mathematics = 48 × 2.5 = 120 cm
50. i. Zeroes are -2 and 8
α + β = -2 + 8 = 6

αβ =-2 × 8 = -16

expression of polynomial
x2 - (α + β)x + αβ
x2 - 6x - 16
ii. P(x) = x2 - 6x - 16
P(4) = 42 - 6(4) - 16
= 16 - 24 - 16
= -24
iii. P(x) = -x2 + 3x - 2
−3
α + β =
−1

α + β = 3 ...(i)

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 −2
αβ = −1

αβ = 2 ...(ii)
(α − β )
2
= (α + β )
2
- 4αβ
(α − β)
2
= (3)2 - 4(2)
(α − β)
2
=9-8
–
α − β = ± √1

α − β = ±1

Taking
α − β = 1

α + β = 3

2α = 4
α = 2

Put α = 2 in, α − β = 1
2-β=1
β = 1

OR
−3
α + β = −1
=3
−2
αβ = = 2

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−1

51. i. Let the fixed charge be ₹ x and per kilometer charge be ₹ y

x + 10y = 105 ...(i)

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∴

x + 15y = 155 ...(ii)

From (i) and (ii)
5y = 50
tor
y= = 10
50

From equation (i)

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x + 100 = 105
x = 105 - 100 = 5
Fixed charges = ₹ 5
ii. Let the fixed charge be ₹ x and per kilometer charge be ₹ y
pta

∴ x + 10y = 105 ...(1)

x + 15y = 155 ...(2)

From (1) and (2)
5y = 50
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y= = 10
50

From equation (1)

x + 100 = 105
x = 105 - 100 = 5
Per km charges = ₹ 10
iii. Let the fixed charge be ₹ a and per kilometer charge be ₹ b
a + 10b
20 + 10 × 10 = ₹ 120
OR
Total amount = x + 10y + x + 25y
= 2x + 35y
= 2 × 5 + 35 × 10
= 10 + 350
= ₹ 360

52. i. 200x2 = 128(x + 1)2

ii. 25x2 = 16x2 + 32x + 16
⇒ 9x2 - 32x - 16 = 0

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 iii. a. 9x2 - 32x - 16 = 0
⇒ (9x + 4)(x - 4) = 0
−4
x ≠
9
so, x = 4
OR
32± √1024+576
b. x = 18
=
32±40

18
−4
x ≠
9
so, x = 4
Section C
53. The given equations are
x - y + 1 = 0 ...(1)
3x + 2y - 12 = 0 ...(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of these equations. These two solutions of these
equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1) x - y + 1 = 0
⇒ y=x+1
Table 1 of solutions
x 0 -1

y 1 0

ls
12−3x
For equation (2) 3x + 2y - 12 = 0 ⇒ y = 2

Table 2 of solutions

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x tor 4 0

y 0 6
We plot the points A(0, 1) and B(-1, 0) on a graph paper and join these points to form the line AB representing the equation (1) as
shown in the figure. Also, we plot the points C(4, 0) and D(0, 6) on the same graph paper and join these points to form the line CD
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representing the equation (2) and shown in the same figure.
In the figure, we observe that the coordinates of the vertices of the triangle formed by these given lines and the x-axis are E(2, 3 ),
B(-1, 0) and C(4, 0)
pta
Gu

The triangular region EBC has been shaded and the area of triangular region EBC = 1

2
(5)(3) =
15

54. Let S1 and S2 be two squares. Let the side of the square S2 be x cm in length. Then,
the side of square S1 is (x + 4) cm.

Therefore, area of square S1 = (x + 4)2 [Because, Area = (side)2]

and, Area of square S2 = x2

It is given that
Area of square S1 + Area of square S2 =400 cm2

⇒ (x + 4)2 + x2 = 400
⇒ (x2 + 8x + 16) + x = 400
2

⇒ 2x2 + 8x - 384 = 0
⇒ x2 + 4x -192 = 0

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⇒ x2 + 16x - 12x -192 = 0
⇒ x(x + 16) - 12(x + 16) = 0

⇒ (x + 16) (x - 12) = 0

⇒ x = 12 or, x = -16
As the length of the side of a square cannot be negative. Therefore, x = 12.
Therefore, side of square S1 = x + 4 = 12 + 4 = 16 cm and, Side of square S2 = 12 cm.Hence the side of square S1 and S2 are 16
cm and 12 cm respectively.

ia ls
tor
Tu
pta
Gu

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