Solution

CLASS 10 MATH TEST PAPER 14

Class 10 - Mathematics
Section A
1. (a) 3
Explanation:
2 x
a= 2 × 3
2
b = 2 × 3 × 5

2
c = 2 × 3 × 7

LCM (a, b, c) = 3780

3780 = 2 × 3 × 5 × 7 2 3 1 1

In LCM, we consider highest power

m
So, x = 3
2.

.co
(b) 2 × 3 × 7 2 2

Explanation:
2 882

3 441

7
147

49

1
ath
nm
882 = 2 × 3 2
× 7
2

3.
lca

(c) 1 : 2
Explanation:
Least composite number is 4 and the least prime number is 2.
LCM (4, 2) = 4
al

HCF (4, 2) = 2
The ratio of HCF to LCM = 2 : 4 or 1 : 2.
w.

4. (a) qx2 + px + 1
ww

Explanation:
Let α and β be the zeros of the polynomial f (x) = x 2
+ px + q .Then,
− Coefficient of x p
α + β = = − = −p
Coefficient of x2 1

Constant term q
And αβ = = =q
Coefficient of x2 1

Let S and R denote respectively the sum and product of the zeros of a polynomial whose zeros are 1

α
and 1

β
, then
1 1 α+β −p
S = + = =
α β αβ q

1 1 1 1
R= × = =
α β αβ q

Hence, the required polynomial g(x) whose sum and product of zeros are S and R is given by
2
x − Sx + R = 0

2 P 1
x + x + = 0
q q
2
qx +P x+1
= 0
q

2
⇒ qx + px + 1

So g(x) = qx 2
+ px + 1

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 5. (a) 4 (x 2
− 5)

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

6.
(c) 10x2 - x -3
Explanation:
3 1 1 3 −1 −3
α + β = ( − ) = , αβ = × ( ) =
5 2 10 5 2 10

, i.e., 10x2 - x - 3
1 3
Required polynomial is x

m
2
− x −
10 10

7. (a) a = 1
,c=5

.co
2

Explanation:
P(x) = x2 - (sum of roots)x + (product of roots)
x2 - (10)x + 10 = 0
1

2
x2 - 5x + 5 = 0 ...(1)
Given, ax2 - 5x + c = 0 ...(2)
comparing (1) and (2), we get,
ath
nm
a= ,c=5 1

8. (a) 2
Explanation:
For infinite Solution
lca

a1 b1 c1
= =
a2 b2 c2

3 −1 8
= =
6 −k 16

3 −1
al

=
6 −k
−6
k = = 2
−3

k=2
w.

9.
(d) Parallel
ww

Explanation:
a1 b1

a2
=
3

6
and b2
=
4

8
c1 5
=
c2 7

a1 b1 c1
= ≠
a2 b2 c2

The above lines are parallel.

10.
(c) Has no solution.
Explanation:
We have, ax + by - c and lx + my = n
Now, = ≠ (given)
a

l
b

m
c

∴ The given system of equations has no solution.

11.
(b) 3

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 Explanation:
Given:
3x2 - 10x + 3 = 0
1
One root of the equation is 3
.
Let the other root be α .
We know that:
Product of the roots = c

a
1 3
⇒ × α =
3 3

⇒ α =3

−9
12. (a) k ≥ 2

Explanation:
For real roots, we must have, b 2
− 4ac ≥ 0 .
2
(−6) − 4 × k × (−2) ≥ 0 ⇒ 36 + 8k ≥ 0

m
−9
⇒ 8k ≥ −36 ⇒ k ≥
2
.
13.

.co
(d) -1
Explanation:
2
p(x) = 3x − 2x + c

D = 16
b
2

(−2)
− 4ac = 16

4 − 12c = 16
2
− 4(3)(c) = 16
ath
nm
4 − 16 = 12c

4 − 16 = 12c
−12
= c
12

c = -1
lca

14.
(b) 5
al

Explanation:
2 (3p + 5) = 3p - 1 + 5p + 1
(If a, b, c are in A.P., b - a = c - b ⇒ 2b = a + c)
w.

⇒ 6p + 10 = 8p

⇒ 10 = 2p
ww

⇒ p=5

15. (a) -32

Explanation:
The sequence given is 10, 7, 4, ....-62.
a = 10, d = -3, l = -62
tn = a + (n - 1)d
⇒ - 62 = 10 + (n - 1)(-3)
⇒ - 62 = 13 - 3n
⇒ 3n = 75 or n = 25.

The number of terms in the sequence is 25 .

The 11th term from the end is same as the 15th term from the first.
So, t15 = 10 + (15 - 1)(-3)
⇒ t15 = 10 + (14)(-3)

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 ⇒ t15 = -32

The 11th term from the end is -32.

−−
16. (a) √72
Explanation:
−− −− −−
Given: √18 , √32 , √50
– – –
⇒ 3√2, 4√2, 5√2
– – –
∴ d = 4√2 - 3√2 = √2
– –
Therefore, next term is 5√2 + √2
– −−
= 6√2 = √72
17.
(c) 6

Explanation:
Given: △ABC ∼ △P QR
Therefore,

m
Perimeter of △ABC AB
=
Perimeter of △P OR PQ

co
56 AB
= =
48 PQ

PQ 6
⇒ =
AB 7

.
ath
18.
(b) 7.5 cm
Explanation:
∵ △ABC ∼ △DEF
nm
Perimeter (△ABC) AB
∴ =
Perimeter (△DEF) DE

32 10
⇒ =
24 DE

⇒ DE =
10×24

32
=7.5 cm
ca

19. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
all

Both are correct. Reason is the correct explanation.

Assertion,
an = 7 - 4n
w.

d = an - 1 - an
= 7 - 4(n + 1) - (7 - 4n)
ww

= 7 - 4n - 4 - 7 + 4n = -4
20. (a) Both A and R are true and R is the correct explanation of A.
Explanation:

In △BDC
GF||DC
BG

GD
=
BF

FC
...(1) (By BPT)
In △DAB
EF || AB
GD

GB
=
DE

AE
(By BPT)

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 GB

GD
=
AE

DE
...(2)
from (1) and (2)
AE BF
=
DE FC

Section B
21. Let the two zeroes of x 2
− 8x + k be α, α + 2
∴ 2α + 2 = 8

⇒ α = 3, other zero is 5
∴ k = 15

22. In given figure, ABCD is a rectangle. Here we have to find out the values of x and y.
From given fig., x + y = 22 .. .(i)
and x - y = 16 ....(ii)
Adding (i) and (ii), we get
(x + y) + (x - y) = 22 + 16
x + y + x - y = 38

m
2x = 38
∴ x = 19

Substituting the value of x in equation (i), we get

.co
19 + y = 22
y = 22 -19
∴ y = 3

ath
Hence, x = 19 and y = 3
23. −
1

x
=3
1

x−2
x−2−x
=3
x(x−2)

-2 = 3x(x - 2)
nm
3x2 - 6x + 2 = 0
6± √36−24
x= 6

6±2√3
x=
lca

3± √3
x= 3

24. Here it is given that Sn = 3n2 + 5n

Sn - 1 = 3(n - 1)2 + 5(n - 1)

al

= 3(n2 - 2n + 1) + 5n - 5
w.

= 3n2 - 6n + 3 + 5n - 5
= 3n2 - n - 2
Sn - Sn - 1 = 3n2 + 5n - (3n2 - n - 2)
ww

⇒ (a1 + a2 + a3 + - - - + an - 1 + an) - (a1 + a2 + a3 + - - - + an - 1) = 6n + 2

⇒ ​an = 6n + 2
Then
ak = 6k + 2 = 164
6k = 164 - 2 = 162
k = 27
25. △OBE ∼ △ODA
⇒ =
OB

OD
BE

AD

⇒
1

2
= BE
(AD = BC)
BC

BE = 1.5 cm ⇒ BC = 3 cm

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