Set A MAHESH TUTORIALS...
Transcript of Set A MAHESH TUTORIALS...
MAHESH TUTORIALS I.C.S.E.
Model Answer Paper
Marks : 80
Time : 2½ hrs.
SUBJECT : MATHEMATICSICSE X
Exam No. : MT/ICSE/Semi Prelim I- Set-A-003
T18 I SP A003 Turn over
SECTION – I (40 Marks)
Attempt all questions from this Section. A.1
(a) 1 –12 =
12
32 – 1 =
12
The difference between the consecutive terms is constant The given sequence is an arithmetic progression.
where first term, a =12
and common difference, d =12
tn = a + (n – 1) d
t30 =12 + (30 – 1)
12
t30 =12 +
292
t30 =302
t30 = 15
30th term is 15 [3]
(b) f(x) = px3 + 4x2 – 3x + q f(x) is divisible by x2 – 1 f(x) is divisble by (x + 1) (x – 1) x + 1 = 0 x = –1Put x = –1 in f(x)
f(–1) = 0 p(–1)3 + 4(–1)2 – 3(–1) + q = 0 –p + 4 + 3 + q = 0 –p + q = –7 ... (i)Again x – 1 = 0 x = 1Put x = 1 in f(x)
f(1) = 0 p(1)3 + 4(1)2 – 3(1) + q = 0
Set A
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Set A
p + 4 – 3 + q = 0 p + q + 1 = 0 p + q = –1 ... (ii)
Adding (i) and (ii);–p + q = – 7
+ p + q = –12q = –8
q = –4Substituting in (i);
–p + (–4) = –7 –p = –7 + 4 –p = –3 p = 3Hence, p = 3 and q = –4 [4]
(c) Let ‘l’ be the slant height, ‘r’ be the base radius and ‘h’ be the height ofthe tent (cone)Diameter of cone = 48 m it’s radius (r) = 24 m
it’s height (h) = 7 m l2 = r2 + h2
l2 = (24)2 + (7)2
l2 = 576 + 49 l2 = 625 l = 25 m ...... [Taking square roots]
Curved surface area of the tent (cone)= rl
=227
× 24 × 25
=13200
7m2
Let the total area of the canvas be ‘A’Now, 10% of the canvas is used in folds and stitchings
Area of the canvas used in tent = A – 10% of A
= A –10100
A
=9A10
9A10
is the area of canvas required to make a tent
9A10
=13200
7Now, A = l × 1.5 (canvas is rectangular)
... 2 ...
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Set A... 3 ...
9 × × 1.5
10l
=13200
7
l =13200 109 1.5 7
l =13200 2 10
9 7 3
length of canvas =1396.825 cmLength of canvas required = 1396.83 cmCost of canvas at the rate of ` 24 per meter = 1396.83 × 24 = `33523.92
Cost of canvas at the rate of ` 24 per meter is ` 33523.92 [3]
A.2(a) AB = 4 cm
BX = 6 cmDX = 5 cm
Let,CD = x cm
AX = AB + BX [ A – B – X] AX = 4 + 6 AX = 10 cmand
CX = CD + DX [ C – D – X]CX = (x + 5) cm
Now,AX × BX = CX × DX [When two chords of a circle intersect
internally/externally, then the products ofthe lengths of segment are equal.]
10 × 6 = (x + 5) 5 60 = 5x + 25 60 – 25 = 5x 5x = 35
x =355
x = 7
CD = 7 cm [3]
(b) For the given G.P.a = 1
r =2
1
tt =
31
= 3tn = arn – 1
t7 = 1( 3 )7 – 1
= ( 3 )6
B
DA
X
C
4 cm
x
6 cm
5 cm
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Set A... 4 ...
= (123 )6
=1 623
[(am)n = amn]
= 33
t7 = 27 [3]
(c) Let the number to be added be x The nos. will be (16 + x), (7 + x), (79 + x) and (43 + x)
167
xx
=7943
xx
(16 + x)(43 + x) = (79 + x) (7 + x) 688 + 16x + 43x + x2 = 553 + 79x + 7x + x2
59x + 688 = 553 + 86x 86x – 59x = 688 – 553 27x = 135
x =13527
x = 5 [4]
A.3(a) Height (cm) No. of plants fx
x f
50 2 10055 4 22058 10 58060 f 60 f65 5 32570 4 28071 3 213
f = 28 + f fx = 1718 + 60f
Mean =fxf
60.95 =1718 + 60
28 +f
f 28 × 60.95 + 60.95 f = 1718 + 60f 1706.60 + 60.95 f = 1718 + 60f 60.95f – 60 f = 1718 – 1706.60 0.95 f = 11.40
f =1140
95 f = 12 [3]
Turn overT18 I SP A003
Set A... 5 ...
(b) 3x² + 12x + (m + 7) = 0a = 3, b = 12, c = m + 7D = b² – 4ac = (12)² – 4(3)(m+7)
= 144 – 12m – 84= 60 – 12m
Given equation has equal roots D = 0 60 – 12m = 0 12m = 60 m = 5Putting the value of m in the equation
3x² + 12x + (m + 7) = 0 3x² +12x + (5 + 7) = 0
3x² + 12x + 12 = 0 3(x² + 4x + 4) = 0 x² + 4x + 4 = 0 (x)² + 2(x) (2) + (2)² = 0 (x + 2)² = 0 x + 2 = 0 x = –2
Hence m = 5 and solution x = –2 [3]
(c) Let the three numbers in G.P. bea, ar, ar2
Sum =3910
a + ar + ar2 =3910
... (i)
Product = 1 a × ar × ar2 = 1 a3r3 = 1 ar = 1
a =1r
... (ii)
Substituting the value of ‘a’ in equation (i), we get,
1r
+1r
× r +1r
× r2 =3910
1r
+ 1 + r =3910
21+ r + r
r=
3910
10(r2 + r + 1) = 39 × r 10r2 + 10r + 10 = 39r 10r2 + 10r – 39r + 10 = 0 10r2 – 29r + 10 = 0
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Set A... 6 ...
10r2 – 25r – 4r + 10 = 0 5r(2r – 5) –2(2r – 5) = 0 (2r – 5) (5r – 2) = 0 2r – 5 = 0 or 5r – 2 = 0
r =52
or r =25
Case (i) : When r =52
,
a =1r
=152
=25
The numbers are : a, ar, ar2
=25
,25
×52
,25
×52
×52
=25
, 1,52
Case (ii) : When r =25
,
a =1r
=125
=52
The numbers are : a, ar, ar2
=52
,52
×25
,52
×25
×25
=52
, 1,25
The numbers in G.P. are25 , 1,
52 or
52 , 1,
25 [4]
A.4
(a)4
2
12
xx
=178
By Componendo Dividendo,4 2
4 2
1 21– 2
x xx x
=17 817 – 8
4 2
4 2
2 1– 2 1
x xx x
=
259
2 2 2 2
2 2 2 2
( ) 2. .1 (1)( ) – 2. .1 (1)x xx x
=
259
2 2
2 2
( 1)( –1)xx
=259
22
2
1–1
xx
=259
Taking square root on both sides, we get,
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Set A... 7 ...
2
2
1–1
xx
=259
2
2
1–1
xx
= 53
By Componendo – Dividendo, By Componendo – Dividendo,2 2
2 2
( 1) ( –1)( 1) – ( –1)x xx x =
5 35 – 3 2 2
2 2
( 1) ( –1)( 1) – ( –1)x xx x =
–5 3–5 – 3
2 2
2 2
1 –11– 1
x xx x
=82
22
2x
=–2–8
22
2x
= 4 x2 =14
x2 = 4 x = 14
x = 4 x = 12
x = 2 [3]
(b)
PB = 3.6 cm [4]
(c) Let son’s present age = xMan’s present age = x2
One year back,Son’s age = x – 1Father’s age = (x2 – 1)According to condition, (x2 – 1) = 8(x – 1) x2 – 1 = 8x – 8
X
M
A
l
B C
P
3.5
cm
6 cm60º
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Set A... 8 ...
x2 – 8x + 8 – 1 = 0 x2 – 8x + 7 = 0 x2 – 7x – x + 7 = 0 x (x – 7) –1(x – 7) = 0 (x – 7) (x – 1) = 0 x = 7 or x = 1 x2 = 49 or x2 = 1
But x2 = 1, is not possible x = 7 is the son present age, and x2 = 49 is the fathers present age Son’s present age = 7 years and father’s present age = 49 years. [3]
SECTION – II (40 Marks)Attempt any four questions from this Section.
A.5
(a)3 – 2 3 – 8
=2 – 3 + 4
x xx x (3x – 2) (x + 4) = (3x – 8) (2x – 3) 3x2 + 12x – 2x – 8 = 6x2 – 9x – 16x + 24 3x2 – 6x2 + 10x + 25x – 8 – 24 = 0 –3x2 + 35x – 32 = 0 3x2 – 35x + 32 = 0 3x2 – 3x – 32x + 32 = 0 3x(x – 1) – 32 (x – 1) = 0 (x – 1) (3x – 32) = 0 (x – 1) = 0 or (3x – 32) = 0
x – 1 = 0 or 3x = 32
x = 1 or x =323
x = 1 or x = 10 32
[3]
(b) Let f(x) = 2x3 – 7x2 + ax – 6Put (x – 2) = 0 x = 2 Remainder = f(2) = 2(2)3 – 7(2)2 + a(2) – 6
f(2) = 16 – 28 + 2a – 6f(2) = – 12 + 2a – 6f(2) = 2a – 18 .......... (i)
Let g(x) = x3 – 8x2 + (2a + 1) x – 16Put (x – 2) = 0 x = 2 Remainder = g(2) = (2)3 – 8(2)2 + (2a + 1)(2) – 16
g(2) = 8 – 32 + 4a + 2 – 16 g(2) = 4a – 38 .... (ii)
f(2) = g(2) [ both polynomials leave same remainder]
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Set A... 9 ...
2a – 18 = 4a – 38 38 – 18 = 4a – 2a 20 = 2a a = 10 The value of a is 10 [3]
(c) First instalment = ` 3000Every other instalment is ` 100 less than the previous one.There are total 12 monthly instalments.Hence the instalments in ` are :
3000, 3000 – 100, 3000 – 2(100), 3000 – 3(100) ...... 3000 – 11(100)= 3000, 2900, 2800, 2700, ......., 1900
This is an A.P.Here a = 3000 and d = –100
(i) Amount of instalment paid in 9th month= t9
= a + (9 – 1)d= a + 8d= 3000 + 8(–100)= 3000 – 800= ` 2200
(ii) The total amount paid in the instalment scheme= Total of 12 instalments= Sum of 12 terms= S12
=122
[2a + (12 – 1)d]
= 6 [2(3000) + 11 × (–100)]= 6 [6000 – 1100]= 6 [4900]= ` 29,400 [4]
A.6(a) (i) SRT = 90º
(Angle in a semi circle)PRS + SRT + TRQ = 180º [Angles in a straight line]
PRS + 90º + 30º = 180º PRS + 120º = 180º PRS = 180º – 120º PRS = 60º
(ii) RST = TRQ (Angles in alternate segment) RST = 30º
ROT = 2 RST [Angle at the centre is twice theangle at the circumference]
= 2 × 300
ROT = 60º [3]
RP
SO
T30º
Q
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Set A... 10 ...
(b) (i) Let number of rows originally = x Number of seats in each row = x Total number of seats = x × x = x2
Number of rows in 2nd case = 2xNumber of seats in each row = x – 10
Total number of seats = 2x (x – 10) = 2x2 – 20xAccording to the given condition,
2x2 – 20x = x2 + 300 2x2 – 20x – x2 – 300 = 0 x2 – 20x – 300 = 0 x2 – 30x + 10x – 300 = 0 x(x – 30) + 10(x – 3) = 0 (x – 30) (x + 10) = 0Either x – 30 = 0 or x + 10 = 0 x = 30 or x = –10But x = –10 is not possible for rows. x = 30Number of rows in original arrangement = 30(ii) Number of seats after re-arrangement in a row
= x – 10 = 30 – 10 = 20Hence (i) 30 (ii) 20. [3]
(c) x =6ab
a b ....(i)
From (i),3xa =
2b
a bBy componendo dividendo,
3–3x a
x a =2 ( )2 – ( )b a bb a b
3
–3x a
x a =3
–b a
b a ....(ii)
From (i),3xb =
2a
a bBy componendo dividendo,
3– 3x b
x b =2 +( + )2 – ( )a a ba a b
3
–3x b
x b =3
–a + ba b ....(iii)
Adding equation (ii) and (iii), we get3
– 3x a
x a +3
– 3x bx b
=3
–b+ a
b a +3
–a + b
a b
=3
–b+ a
b a – 3
–a + b
b a
= 3 – 3
–b+ a a+ b
b a
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Set A... 11 ...
=3 – 3
–b + a a – b
b a
=2 – 2
–b ab a
= 2 –
–b a
b a
+3– 3
x ax a +
+ 3– 3
x bx b = 2 [4]
A.7(a) x2 – 8x + 5 = 0
a =1, b = – 8, c = 5By formula,
2– ± –4=
2b b ac
xa
= 12– – 8 ± (– 8) – 4 ( ) (5)
2 (1)
=8± 64 – 20
2
=8 ± 44
2
=8 ±2 11
2
= 2 4± 11
2 = 4 11
= 4 3.317
x = 4 + 3.317 or x = 4 – 3.317
x = 7.317 or x = 0.683
x = 7.32 or x = 0.68 [Correct to 2 decimal places] [3]
(b) Given Arithmetic Progression is :1 + 4 + 7 + 10 + ....First term, a = 1Common difference, d = 4 – 1 = 3Let nth term be 52 tn = 52 a + (n – 1) d = 52 1 + (n – 1) 3 = 52 3(n – 1) = 51
n – 1 =513
n – 1 = 17
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Set A... 12 ...
n = 17 + 1 n = 18 18th term is 52 [3]
(c) Let f(x) = x3 + ax2 + bx – 12Put (x – 2) = 0 x = 2 Remainder = f(2) = 23 + a(2)2 + b(2) – 12 0= 8 + 4a + 2b – 12 [ x – 2 is a factor of f(x) f(2) =0] 0= 4a + 2b – 4
4a + 2b = 4Dividing throughout by 2
2a + b = 2 ..... (i)Put (x + 3) = 0 x = –3 Remainder = f(–3) = (–3)3 + a(–3)2 + b(– 3) – 12 0 = – 27 + 9a – 3b – 12 [ x + 3 is a factor of f(x)
f(-3) =0] 0 = 9a – 3b – 39 9a – 3b = 39
Dividing throughout by 3, 3a – b = 13 ..... (ii)
By adding (i) and (ii) we get,
+
2 2
3 13
5 = 15
a ba ba
a =155
a = 3Substituting a = 3 in equation (i) we get,
2a + b = 2 2(3) + b = 2
6 + b = 2 6 – 2 = – b
4 = – b
b = – 4 The value for a = 3 and b = –4 [4]
A.8(a) Let number of articles be x.
Cost price = ` 1200Profit = ` 60Selling price = ` (1200 + 60) = ` 1260Number of articles damaged = 10Remaining articles = x – 10
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Set A... 13 ...
Cost price per article = `1200
x and selling price per article = `
1260– 10x
As per condition,
1260– 10x
=1200
x+ 2
1260
– 10x–
1200x
= 2
1260 – 1200 + 12000
( – 10)x x
x x = 2
2
60 + 12000 – 10
xx x
= 2
2x2 – 20x = 60x + 12000 2x2 – 80x – 12000 = 0 x2 – 40x – 6000 = 0 [Dividing throughout by 2] x2 – 100x + 60x – 6000 = 0 x (x – 100) + 60 (x – 100) = 0 (x – 100) (x + 60) = 0
x = 100 or x = – 60But x = – 60 is not possible
No. of articles is 100 [4]
(b)
(i) Median
Median =N2
=802
= 40th term = 40
Median = 40
10
20
30
40
50
60
70
80
10X
Q1 = 18 Q3 = 66Md = 40
R
Y
20 30 40 50 60 70 80 90 100
(10,
5)
(20, 24)
(40, 40)
(50, 42) (60, 48)
(70, 70)
(80, 77)(90, 79)
(100, 80)
N2 = 40
N4 = 20
3N4 = 60
Marks (Less than)
c.f. (30, 37)
(0, 0)
Scale : On X axis : 2 cm = 10 units Y axis : 2 cm = 10 unitsOn
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Set A... 14 ...
(ii) Lower quartile, (Q1)
Q1 =N4
=804
= 20th term = 18
Q1 = 18
(iii) Upper quartile, (Q3)
Q3 =3N4
=3 × 80
4 = 60th term = 66
Q3 = 66Hence (i) 40 (ii) 18 (iii) 66 [6]
A.9(a) 5x – 3 < 5 + 3x < 4x + 2
5x – 3 < 5 + 3x 5 + 3x < 4x + 2 5x – 3x < 5 + 3 3x – 4x < 2 – 5 2x < 8 –x < –3
x <82 x > 3
x < 4 3 < x < 4Comparing with a < x < b
a = 3 and b = 4 [3]
(b) (i)4 34 3m nm n
=74
By componendo dividendo4 3 4 34 3 4 3
m n m nm n m n =
7 47 4
i.e.86mn =
113
mn =
11 63 8
mn =
114
(ii)mn =
114
2
2mn
=12116
(Squaring both sides)
2
2211
mn
=2 12111 16
(Multiplying both sides by211
)
2
2211
mn
=118
By componendo dividendo,2 2
2 22 112 –11m nm n
=11 811– 8
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Set A... 15 ...
2 2
2 22 112 –11m nm n
=193
2 2
2 22 -112 +11
m nm n =
319 [By invertendo] [3]
(c)
Mode = 23 [4]
A.10(a)
[3]
(b) Odd natural numbers less than 50 are :1, 3, 5, 7, ......................... 49This is an arithmetic progressiona = 1, d = 2, last term l = 49Let nth term be 49 tn = 49 a + (n – 1)d = 49 1 + (n – 1)2 = 49 2(n – 1) = 48 n – 1 = 24
P
C
A B
X
Y9 cm
6 cm
9 cm
4 cm
P
1
0
2
3
4
5
7
6
8
9
10
10
X
C
AY
20 30 40 50 60 70 80Class Interval
FRE
QU
EN
CY
MODE
D
B
L
K
Scale :X-axis : 2 cm = 10 unitsY-axis : 2 cm = 1 unit
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Set A
n = 25There are 25 terms
Sn = 2n
(a + l )
S25 =252 (1 + 49)
=252 × 50
= 25 × 25 S25 = 625 [3]
(c) (i) AOB = 2APB [ Angle at the centre is twice the angle at the circumference]
AOB = 2(75º)
AOB = 150º
(ii) OACB is a cyclic quadrilateral.
AOB + ACB = 180º [Opposite angles of a cyclic quadrilateral are supplementary]
150º + ACB = 180º
ACB = 30º(iii) ABCD is a cyclic quadrilateral.
ABD + ACD = 180º [Opposite angles of a cyclic quadrilateral are supplementary]
ABD + 30º + 40º = 180º [ ACB = 30º]
ABD = 180º – 70º
ABD = 110º(iv) ADB = ACB = 30º [Angles in the same segment]
ADB = 30º [4]
A.11(a) G.P. : 1, 4, 16, 64, ...............
In the given G.P., a = 1, |r| = 4 > 1Sn = 5461
( – 1)
– 1
na rr
= 5461 [ |r| > 1]
1(4 –1)
4 –1
n
= 5461
... 16 ...
B
P
CA
O
D
400
75o
Turn overT18 I SP A003
Set A
4 –14 –1
n
= 5461
4n – 1 = 5461 × 3 4n – 1 = 16383 4n = 16383 + 1 4n = 16384 4n = 47
n = 7 [ am = an m = n] 7 terms must be added [3]
(b) Diameter of base of cylindrical container = 42 cm
Radius (r) =422
= 21 cmLet rise in level of water = h cm
Size of rectangular solid = 22 cm × 14 cm × 10.5 cm Volume of solid = 3234 cm Now, Volume of rise in level of water = volume of solid r2h = 22 × 14 × 10.5
227
× 21 × 21 × h = 22 × 14 × 10.5
h =22 14 10.5 7
22 21 21
h =73
cm
h = 2.33 cm Rise in level of water is 2.33 cm. [3]
(c) 9 22
1+xx
– 91
+xx
– 52 = 0
Let x +1x
= y
x2 + 2
1x
+ 2 = y2 [Squaring both sides]
x2 + 2
1x
= y2 – 2
Given equation reduces to :9(y2 – 2) – 9 y – 52 = 0
9y2 – 18 – 9y – 52 = 0 9y2 – 9y – 70 = 0 9y2 – 30y + 21y – 70 = 0 3y(3y – 10) + 7(3y – 10) = 0 (3y – 10)(3y + 7) = 0
... 17 ...
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Set A... 18 ...
3y – 10 = 0 or 37 + 7 = 0
y =103
or y = –73
When y =103
,
x +1x
=103
2 1xx
=103
3x2 + 3 = 10x 3x2 – 10x + 3 = 0 3x2 – 9x – x + 3 = 0 3x(x – 3) –1(x – 3) = 0 (x – 3)(3x – 1) = 0
x = 3 or x =13
Similarly when y = –73
,
x +1x
= –73
,
2 1xx
= –73
3x2 + 3 = – 7
3x² + 7x + 3 = 0
a = 3, b = 7, c = 3
b² – 4ac = 49 – 4 × 3 × 3= 49 – 36= 13
Using formula x =2– ± – 4
2b b ac
a
By solving, we get x =13– 7 ±
6
The solution is 3,13 ,
– 7±6
13[4]