Transcript Term-1 - Rachna Sagar

Solutions to EAD Mock Paper (Mathematics–X)
Term-1
Section A
1. 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
\ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
2. For mode of the data to be 25, frequency of 25 should be maximum
\ When x = 25, frequency of 25 is maximum.
Hence, value of x = 25
3. Q
DABC ~ DDEF
\ Area DABC = BC2
Area DDEF
EF2
⇒
2
54
= (3) 2
Area D DEF
(4)
⇒
2
Area D DEF = 54 ×2 4 = 96 cm2
3
4. Let greater number be x and smaller number be y
A.T.Q.
x + y=35
...(i)
and
x – y=13
...(ii)
Adding (i) and (ii), we get 2x = 48 ⇒ x = 24
When
x= 24
Equation (i) becomes, 24 + y= 35
⇒
y= 11
\ Numbers are 24 and 11.
Section B
5. We have
Also
⇒
Now
sin (A + B) =
3
2
sin 60° =
3
2
A + B = 60°
cos(A – B) =
...(i)
3
2
Solutions to EAD Mock Paper (Mathematics–X) 1
Also
cos 30° =
⇒
3
2
A – B = 30°
...(ii)
Adding (i) and (ii), we get
A = 45°
When A = 45°, equation (i) becomes
45° + B = 60° ⇒ B = 15°
\ A = 45° and B = 15° A
6. Draw AP ^ BC and DQ ^ BC.
1 × BC × AP
Area (DABC) = 2
Now,
= AP ...(i)
1 × BC × DQ
DQ
Area (DDBC)
1 O
B
Q
2
P
C
2
In ∆APO and ∆DQO
∠1=∠2, ∠APO=∠DQO \ ∆ AOP~∆ DOQ
[Vertically opposite angles]
[Each 90°]
AO = AP DO
DQ
⇒
D
... (ii)
From (i) and (ii)
Area (D ABC) = AO
DO
Area (D DBC)
7.
Marks obtained
Number of students
0 – 10
14
10 – 20
8
20 – 30
15
30 – 40
21
40 – 50
9
50 – 60
Total
8
75
2
2
2
2
8. tan 60º + 4 sin 45º + 3 sec 30º 2+ 5 cos 90º
cosec 30° + sec 60°– cot 30°
2
2
( 3 ) 2 + 4c 1 m + 3c 2 m + 5 × 0
2
3
=
= 3+2+4 = 9
1
2 + 2 − ( 3 )2
9. Given system of linear equations are cx + 3y – 3 = 0
12x + cy – 6=0
Here,
a1= c, b1=3, c1=– 3
a2= 12,b2=c, c2=– 6
2 Solutions to EAD Mock Paper (Mathematics–X)
...(i)
...(ii)
For no solution
a1
b
c
= 1 ! 1
a2
b2 c2
⇒
c = 3 ! − 3
c
12
−6
⇒
⇒
c = 3 and 3 ! 3
c
c 6
12
2
c = 36 and c ≠ 6
⇒
c = ±6 and c ≠ 6
\
c = – 6
10. Let
p(x)=2x2 + x + k
If 3 is a zero of p(x) then p(3) = 0
⇒
⇒
2(3)2 + 3 + k=0 ⇒ 18 + 3 + k = 0
k=– 21
Section C
11. According to the question
627 – 2 , 3129 – 4, 15628 – 3, i.e. 625, 3125 and 15625 are exactly divisible by the required
number.
\ The greatest number which divides 625, 3125 and 15625 is HCF of these numbers.
Now
\
625=5 × 5 × 5 × 5
3125=5 × 5 × 5 × 5 × 5
15625=5 × 5 × 5 × 5 × 5 × 5
HCF=625
Hence, the greatest number is 625
12. Let 2 3 – 7 is a rational number equals to
⇒
2 3 –7=
⇒
2 3 =
⇒
Now
\
p
q
p
, where p and q are coprime and q ≠ 0
q
p
+7
q
p
3 = 1 e + 7o 2 q
...(i)
p
+ 7 =rational number + rational number = rational number
q
1 × e p + 7o =rational number × rational number
2
q
=rational number
⇒ RHS of equation (i) is a rational number.
LHS of equation (i) is
3 which is an irrational number.
But irrational number is not equal to rational number.
Solutions to EAD Mock Paper (Mathematics–X) 3
p
3 ≠ 1 e + 7o
2 q
This is contradiction for equation (i).
\
Hence, our assumption is wrong. Therefore 2 3 –7 is an irrational number.
13. Let digit at unit's place be x and at ten’s place be y
\ Number = 10y + x
According to the question x + y = 9 and ...(i)
7(10y + x) = 4(10x + y)
⇒ 70y + 7x = 40x + 4y ⇒ 66y – 33x = 0 ⇒ 2y – x = 0
...(ii)
Solving (i) and (ii) we get
x = 6, y = 3
Hence, required number = 36.
14.
Here,
p(x) = 2x2 – 5x + 7
a = 2, b = –5, c = 7
Since,
a, b are zeroes of p(x)
a + b = − b = − (− 5) = 5 2
2
a
\
c =7 a 2
Now, S = Sum of zeroes of the required polynomial
ab =
and = (2a + 3b) + (3a + 2b)
= 5a + 5b = 5(a + b)
= 5 × 5 = 25
2
2
and P = Product of zeroes of the required polynomial
= (2a + 3b) (3a + 2b)
= 6a2 + 4ab + 9ab + 6b2
= 6(a2 + b2) + 13ab
= 6[(a + b)2 – 2ab] + 13ab
= 6(a + b)2 + ab
2
= 6 b 5 l + 7 = 75 + 7 = 41
2
2
2 2
Required quadratic polynomial
f(x) = k(x2 – Sx + P)
= k b x2 − 25 x + 41l
2
Where k is any non-zero real number.
15. p = cosec q + cot q ⇒ p =
1 + cos q ⇒ p = 1 + cos q
sin q sin q
sin q
4 Solutions to EAD Mock Paper (Mathematics–X)
...(i)
...(ii)
b 1 + cos q l − 1
p −1
sin q
LHS = 2
=
1
cos q l2 + 1
+
p +1
b
sin q
2
2
2
2
2
2
= (1 + cos q ) 2 − sin2 q = 1 + cos2 q + 2 cos q − sin2 q
(1 + cos q ) + sin q 1 + cos q + 2 cos q + sin q
2
= 2 cos q + 2 cos q = 2 cos q (1 + cos q )
2 + 2 cos q
2 (1 + cos q )
= cos q = RHS
16. cos θ – sin θ =
⇒ cos θ =
2 sin θ
2 sin θ + sin θ ⇒ cos θ = ( 2 + 1) sin q
⇒ cos q = sin θ ⇒ cos q ( 2 − 1) = sin θ
2 +1
( 2 + 1 ) ( 2 − 1)
⇒
2 cos θ – cos θ = sin θ ⇒ 2 cos θ = sin θ + cos θ
17. In D ABD, PQ || AB
A
QD
\ DP =
AD
BD
C
and ∠ADB is common angle
\ DADB ~ DPDQ
DQ
PQ
BQ
BD – BQ
⇒
=
⇒
= z ⇒1−
= z
x
x
BD
AB
BD
BD
BQ
⇒
1 − z =
x
BD
Similarly, DBDC ~ DBQP
BQ
PQ
BQ z
\
=
⇒
= BD
CD
BD y
From (i) and (ii)
⇒
y
z
Q
B
D
...(i)
...(ii)
1 – z = z ⇒ 1 = z + z x
y
x y
⇒
18.
P
x
1 = z c 1 + 1 m ⇒ 1 + 1 = 1
x y
z
x y
PS = 4
6
= 4 , PT =
= 6 = 4
4+1
PQ
5 PR 6 + 1.5 7.5 5
P
\ PS = PT
PQ
PR
4 cm
6 cm
In DPST and DPQR,
∠P = ∠P
[Common]
\ DPST ~ DPQR
\
1 cm
S
Q
T
1.5 cm
R
ar (DPQR)
(PQ) 2
=
ar (DPST)
(PS) 2
Solutions to EAD Mock Paper (Mathematics–X) 5
2
ar (DPST) + ar (trap QRTS)
= (5) 2
ar (DPST)
(4)
ar
(
trap
QRTS
)
ar (trap QRTS)
1 +
= 25 ⇒
= 9
16
16
ar (DPST)
ar (D PST)
⇒
⇒
ar (DPST)
= 16
9
ar (trap QRTS)
19.
Marks
Number of students
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
70 – 80
3
5
16
12
13
20
6
5
Modal class = 50 – 60
Here, l = 50, f0 = 13, f1 = 20, f2 = 6, h = 10
f1 – f0 × h , we get
2f1 – f0 – f2
20 – 13
Mode = 50 +
×10 = 50 + 7 × 10 = 53.33
21
2 × 20 – 13 – 6
From formula mode = l +
20.
Number of letters
Number of surnames (f )
c.f.
0–5
5 – 10
10 – 15
15 – 20
20 – 25
20
60
80
32
8
20
80
160
192
200
n = 100
2
\ Median class = 10 – 15
Here l = 10, c = 80, n = 100, f = 80, h = 5
2
n –c
From formula median = l + 2
× h , we get
f
Median = 10 + 100 – 80 × 5 = 10 + 1.25 = 11.25
80
Section D
21. Q
3 and −
5
3 are zeroes of f(x)
5
\b x −
3 lb x + 3 l is a factor of f(x)
5
5
3
2
i.e. c x – m is a factor of f(x).
5
6 Solutions to EAD Mock Paper (Mathematics–X)
For other factors
3
x2 –
5
5x2 – 5x – 30
5x – 5x3 – 33x2 + 3x + 18
4
2
– 3x
5x
–
+
– 5x3 – 30x2 + 3x + 18
– 5x3
+ 3x
–
+
– 30x2 + 18
– 30x2 + 18
–
+
×
4
For other zeroes 5x2 – 5x – 30 = 0
⇒ 5(x2 – x – 6) = 0 ⇒ (x – 3)(x + 2) = 0
⇒ x = 3, x = – 2
\ Other zeroes are 3 and –2
22. In right DACB, AB2 =AC2 + BC2
⇒
A
(26)2 =(2x)2 + {2(x + 7)}2
⇒
676 =4x2 + 4(x2 + 14x + 49)
⇒
676 =4x2 + 4x2 + 56x + 196
⇒
8x2 + 56x – 480 =0
⇒
8(x + 7x – 60) =0
⇒
x2 + 7x – 60 =0
26 km
2x km
C
2(x+7) km
B
2
⇒x2 + 12x – 5x – 60= 0
⇒
⇒
(x + 12)(x – 5) =0
x = 5, x=– 12 (rejected) [Q Length can not be negative]
We have x = 5
\ AC = 2 × 5 = 10 km
BC = 2(x + 7) = 2(5 + 7) = 24 km
Total distance from A to B via C = AC + BC = 10 km + 24 km = 34 km
Distance saved=34 km – 26 km = 8 km
23.
LHS=
tan q + cot q
1 − cot q 1 − tan q
=
tan2 q −
1
tan q − 1 tan q (tan q − 1)
=
2
tan3 q − 1
= (tan q − 1) (tan q + tan q + 1)
tan q (tan q − 1)
tan q (tan q − 1)
Solutions to EAD Mock Paper (Mathematics–X) 7
2
= tan q + tan q + 1 = tanq + 1 + cotq
tan q
2
2
1
= sin q + cos q + 1 = sin q + cos q + 1 =
+1
cos q sin q
sin q $ cos q
sin q cos q
= secq.cosecq + 1 = RHS
2
º − q ) − cot2 q + 2 sin2 30º $ tan2 32º $ tan2 58º
24. sec (90
2 (sin2 25º + sin2 65º)
3 (sec2 33º − cot2 57º)
2 × b 1 l $ tan2 32° $ cot2 32°
cosec
q
q
cot
−
2
=
+
2 (sin2 25° + cos2 25°)
3 (sec2 33° − tan2 33°)
2
2
= 1 +
2
2
2 × 1 ×1
4
= 1+1 = 4 = 2
3 ×1
2 6 6
3
25. Table for equation x + 2y = 8.
x
0
8
6
y
4
0
1
Table for equation y – x = 1.
x
0
–1
1
y
1
0
2
Y
5
(0, 4)
4
y
3
X′
=
1
(1, 2)
2
(–1, 0)
–
x
(6, 1)
(0, 1)
1
0
–4 –3 –2 –1
–1
1
(8, 0)
2
3
4
5
6
7
8
x+
2y
X
=8
–2
–3
–4
Y′
Lines meet the y-axis at (0, 4) and (0, 1)
26.
Height (in metres)
0–8
8 – 16
16 – 24
24 – 32
32 – 40
40 – 48
Number of trees
3
7
13
9
6
2
Less than
c.f.
8
16
24
32
40
48
3
10
23
32
38
40
8 Solutions to EAD Mock Paper (Mathematics–X)
n = 40 ⇒ n = 20
2
From graph median = 22.1
27. Let
1 = A and 1 = B
x+y
x−y
\ Equations become 57A + 9B = 6
and
...(i)
38A + 21B = 9
...(ii)
Solving (i) and (ii) we get A = 1 , B = 1
19
3
\ x + y = 19 and x – y = 3 ⇒ x = 11, y = 8
28. Area ∆ABC = 49 cm2, Area ∆PQR = 64 cm2
P
PM – AD = 10 cm
Q ∆ABC ~ ∆PQR
A
2
∴Area DABC = AB2 Area DPQR
PQ
In ∆ADB and ∆PMQ,
∠B= ∠Q,
∠ADB= ∠PMQ
∴ ∆ADB~ ∆PMQ
⇒ AB = AD ⇒ AB2 = AD2 PQ
PM
PQ2
PM2
...(i)
B
D
C Q
M
R
[Each 90º]
...(ii)
Solutions to EAD Mock Paper (Mathematics–X) 9
From (i) and (ii)
Area DABC = AD2
Area DPQR
PM2
2
⇒ 49 = AD 2 ⇒ AD = 7 ⇒ AD = 7 PM
64
PM
8
8
PM
Now, PM – AD = 10 [Given]
∴PM – 7 PM = 10
8
⇒PM = 80 cm and AD = 70 cm
29.
Number of accidents(xi)
Frequency (fi)
fixi
0
1
2
3
4
5
46
x
y
25
10
5
0
x
2y
75
40
25
Total
200 = 86 + x + y
140 + x + 2y
Now ⇒
Also
86 + x + y= 200
x + y= 200 – 86 = 114 ⇒ x = 114 – y
Mean, x=
...(i)
Σfi x i
Σfi
140 + x + 2y
200
⇒
140 + x + 2y= 292 ⇒ x + 2y = 152
⇒
114 – y + 2y= 152
⇒
y= 152 – 114
⇒
y= 38
When y = 38 equation (i) becomes
x= 114 – 38
⇒
x= 76
⇒
1.46=
\
[Using (i)]
x= 76, y = 38
Now distribution becomes
Number of accidents
0
1
2
3
4
5
\
Frequency
46
76
38
25
10
5
n = 200 = 100
2
2
Median= 1
10 Solutions to EAD Mock Paper (Mathematics–X)
c.f.
46
122
160
185
195
200
30. LHS = 2x2 + y2 – 2xy
= 2(2cosq – sinq)2 + (cosq – 3sinq)2 – 2(2cosq – sinq) (cosq – 3sinq)
= 2(4cos2q + sin2q – 4 cosq sinq) + cos2q + 9sin2q – 6cosq.sinq
– 2(2cos2q – 6cosqsinq – sinq.cosq + 3sin2q)
= 8cos2q + 2sin2q – 8cos q.sin q + cos2q + 9sin2q – 6cosq.sin q – 4cos2q
= 5cos2q + 5sin2 q
= 5(cos2q + sin2q)
= 5 × 1 = 5
+ 12cosq. sinq + 2sinq.cosq – 6sin2q
31. (a) Number of students in each bus equals to HCF of 156, 208 and 260
156 = 2 × 2 × 3 × 13
208 = 2 × 2 × 2 × 2 × 13
260 = 2 × 2 × 5 × 13
\ HCF of 156, 208 and 260 = 2 × 2 × 13 = 52
Number of buses = 156 + 208 + 260 = 3 + 4 + 5 = 12
52
52
52
(b) Cooperation, helpfulness.
Solutions to EAD Mock Paper (Mathematics–X) 11