10th ICSE Guess Paper with sol

Transcription

L.K. Gupta (Mathematics Classes)
www.pioneer mathematics.com MOBILE: 9815527721, 4617721
BOARDS
(2010)
(10th ICSE GUESS PAPER)
TIME: 2:30 HOURS
MAX. MARKS: 80
GENERAL INSTRUCTIONS & MARKING SCHEME
1. Answers to this paper must be written on the paper provided separately.
2. You will not be allowed to write during the first 15 minutes.This time is to be spent
in reading the question paper.
3. The time given at the head of this paper is the time allowed for writing the answers.
4. Attempt all questions from Section A and any four questions from Section B.
All working, including rough work, must be clearly shown and must be done on the same
sheet as the rest of the answer. Omission of essential working will result in the loss of marks.
5. The intended marks for questions or parts of questions are given in brackets [ ].
6. Mathematical tables are provided.
NAME OF THE CANDIDATE
PHONE NUMBER
Pioneer Education (The Best Way To Success)
S.C.O. 320, Sector 40- D, Chandigarh
Ph: - 9815527721, 0172 – 4617721.
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com  NOTICES
1
SECTION – A
1.


(a) Determine the value of ‘m’ if x  m is a factor of the polynomial x3  m2  1 x  2
hence find the value of k if 3m  6  k .
Sol:
(a)
x  m  is a factor of x3  m2  1 x  2
 for x  m


[3]


m3  m2  1 m  2  0
or m3  m3  m  2  0
or m  2
Now 3m  6  k
 3 2   6  k
 k 0
 m  2 and k  0
(b) If 3a  8b :3c  8d ::3a  8b :3c  8d ,then show that a, ,b, c, d,are in Proportion
Sol:
3a  8b 3a  8b

3c  8d 3c  8d
3a  8b 3c  8d

Using alternendo
3a  8b 3c  8d
Using componendo and dividendo
3a  8b  3a  8b 3c  8d  3c  8d

3a  8b  3a  8b 3c  8d  3c  8d
6a
6a

or
16b 16d
a c
 
b d
 a, b,c and d are in proportion.
[3]
(c) On what sum will the difference between the simple interest and compound interest for 2
years at 5% per annum will be equal to Rs. 50?
[4]
Sol:
Let Principle = Rs. x
Now , C.I. –S.I =Rs. 50
n

 p r t
r 

1
 50
or P  1 


100
100




2
2

 x  5 2
5 

1
 50
or x   1 


100
100




  21 2  x
 50
or x     1  
20
10




 441  x
or x 
 1 
 50
 400  10
 441  400  x
x
 50

 400  10
41x  40x
or
 50
400
x  20,000
 principle , Rs. x = Rs. 20,000
2. (a) Mehak deposits Rs. 150 per month in a recurring deposit account for 8 months at
the rate of 8 % per annum. what amount will she get on maturity?
[4]
Sol:
8 8  1 
Total P for 1 month = Rs 150 
2
 150  36  Rs.5400
5400  8 
1
 Rs.36
100  12
Total amount paid in 8 months  150  8
 Rs.1,200
Amount received on maturity
 Rs. 1,200  36  Rs.1236
(b) Solve the inequation and represent the solution on the number
2
x
2
    1  , xR .
Line:
3
3
3
Sol:
2
x
2
   1
3
3
3
[4]
3
Multiplying through out by 3
2   x  3  2
or 2  3   x  2  3
or 5   x  1
or 5  x  1
1 x 5
(c) In the given figure PQ ||MN and
(i)
ar ΔLPQ 
LP 2
 calculate the value of :
PM 3
ar ΔLMN 
(ii)
area of trapeziumPMNQ
area of ΔLMN
[3]
Sol:
Given : PQ || MN and
L  L
LP 2

PM 3
Common
P  M
 ΔLPQ  ΔLMN
[Corresponding angles ]
[ AA similarity ]
Let LP = 2x then PM = 3x
LM = 2x+3x = 5x
ar ΔLPQ  LP2  LP 2  2x 2
(i)



ar ΔLMN  LM2  LM   5x 
4x2
4


 4 :25
2
25x 25
(ii) Let ar ΔLPQ   4y then ar ΔLMN   25y
ar trapeziumPMNQ   25y  4y  21y
4

ar trapizium PMNQ  21y 21


 21 :25
ar ΔLMN 
25y 25
3. (a)
(i) Point P(k, m) is reflected in the x-axis to P' 5, 2 .Write down the values of k and m .
(ii) P’’ is the reflection of P when reflected in the y-axis .Write down the coordinates of P’’.
(iii) Name a single transformation that maps P’ to P’’.
[4]
Sol:
(a)
(i) Mx k,m   5, 2
 k  5,m  2
Mx x,y   x, y 
 Co –ordinates of P are 5,2
(ii) My 5,2   5,2 
 M x,y   x,y 
y
 Co –ordinates of P’’ are 5,2
(iii) Reflection of p’ in the origin
M0 x, y   x, y 
(b) Find the mean, mode and median of the following data : 25 , 27 , 19, 29, 21 , 23 , 25 , 30 ,
28 , 20
[3]
Sol:
x 247
(i) mean   
 24.7
n
10
(ii) Ascending order : 19, 20, 21 , 23, 25, 27, 28, 29, 30
n  10(even)
T  Next term
Median  n/2
2
T  Next term T5  T6
 10/2

2
2
25  25 50


 25
2
2
(iii) Mode =25 as it occurs maximum number of times in the data.
(c) The area enclosed between two concentric circles is 770 cm2 . If the radius
22 

of the outer circle is 21 cm . Calculate the radius of the inner circle  Use π  
7 

Sol:
[3]
5
Radius of outer circle (R) = 21 cm
Let radius of inner circle = r cm
or πR 2  πr 2  770cm2
or π R 2  r 2  770

or
or
or
or
or
or




22
2
21  r2  770
7
770  7
441  r2 
22
2
441  r  245
r 2  441  245
r 2  196
r  14cm
Radius of the inner circle is r  14cm .
4.
(a) Prove that :
1
1

secθ  tanθ cosθ

1
1

cosθ secθ  tanθ
[3]
Sol:
(a)
1
1

secθ  tanθ cosθ
secθ  tanθ
1


 secθ
secθ  tanθ  secθ  tanθ
L.H.S =
secθ  tanθ
 secθ
sec2 θ  tan2 θ
 secθ  tanθ  secθ
 secθ  secθ  tanθ 

 secθ  secθ  tan θ
secθ  tan θ
secθ  tan θ
sec θ  tan θ 

 secθ 
2
2
secθ  tanθ
1
1


cosθ secθ  tanθ
 RHS.
Hence proved.
(b) If the mean of the distribution is 62.8 and sum of frequencies is 50, find P and Q.
[4]
6
Class
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
5
P
10
Q
7
8
Sol:
Class
0-20
20-40
40-60
60-80
80-100
100-120
X
10
30
50
70
90
110
F
5
P
10
Q
7
8
50
Fx
50
30p
500
70q
630
880
2060+30p+70q
Now, 5  p  10  q  7  8  50
or p  q  50  30
or p  q  20................(i)
fx
 x
f
2060  30p  70q
or 62.8 
50
10 206  3p  7q 
or 62.8 
50
or 314  206  3p  7q
or 3p  7q  108......................(ii)
(i) × 3  3p + 3q = 60 ………….(iii)
Subtracting (iii) from (ii)
 4q  48  q  12
 p  12  20  p  8 from (i)
 p  8 and q  12
(c) List Price of a washing machine is Rs. 17,658.The rate of sales tax is 8% .The customer
requests the shopkeeper to allow a discount in the Price of the washing machine to such an
extent that the Price remains Rs. 17,658 inclusive of sales tax. Find the discount in the price
of the washing machine.
[3]
Sol:
Let new marked price of the washing machine= Rs. x
7
No total amount to be paid = M.P +S.T% of M.P
17658 = x  8% of x
8
17658 = x 
x
100
2
17,658  x  x
25
27x
 17658 
25
17,658  25

x
27
 x  16,350
 Discount in the price of the washing machine
 17,658  16,350  Rs.1,308
SECTION-B
(Attempt any four question from this Section.}
5. (a) Solve the following equation and give your answer up to two decimal places:
7
3x  1 
[3]
x
Sol:
7
3x  1   0  3x 2  x  7  0
x
Comparing this to the equation ax 2  bx  c  0
we get a  3,b  1,c  7
b  b2  4ac
x
2a

 1  
1
2
 4 3 7 
2 3
1  1  84
6
1  85 1  9.219


6
6
1  9.219 1  9.219

,
6
6


10.219 8.219
,
6
6
8
 1.703,  1.369
 1.70, 1.37
(b) In the given figure , a circle touches all the four sides of a quadrilateral ABCD
whose sides are AB = 6cm , BC = 7cm and CD = 4 cm. Find AD .
[3]
Sol:
Given that :AB = 6cm , BC =7cm , CD = 4cm . let the circle touches the sides AB , BC, CD
and DA at points P,Q,R and S respectively.
then AP = AS , BP =BQ ,CR =CQ , DR = DS
(tangents from external point are of equal length)
Adding we get :
AP +BP+CR+DR =AS +DS+BQ+QC
AP  PB  CR  RD  BQ  QC  DS  SA 
AB  CD  BC  DA
6  4  7  AD
AD = (10-7) cm = 3 cm.
(c) Find the equation of the straight line that passes through the point (3,4) and
Perpendicular to the line 3x  2y  5  0 .
Sol:
[4]
Coefficient of x 3

Coefficient of y
2
1 1
Slope of CD m 2  

 2 /3
m1 3
2
slope of AB m1  
9
then equation of CD , y  y 1  m x  x 1 
 N 3, 4  is the passing point and slope  2 /3
2
 y  4  x  3  3y  12  2x  6
3
 2x  3y  6  0  2x  3y  6  0
6.
(a) Manan, Produces an item for Rs. 216, which he sells to Rohan , Rohan sells it to
Sohan and Sohan sells it to Mohan. The tax rate is 10 %. The Profit Rs. 20 at each
stage of the selling chain. Find the amount of VAT.
[4]
 2 2
(b) If A  
, then find  A 2  6A
[3]

 3 4 
(c) A bag contain 3 red, 5 black and 6 white balls. A ball is drawn at a random. Find the
Probability that the ball drawn is (i) black (ii) not red (iii) either red or white . [3]
Sol:
(a) The selling price for Manan = Rs. 216 +Rs. 20
= Rs. 236
236  10
 Rs.23.60
Tax charged = Rs.
100
So VAT = Rs. 23. 60
The value of invoice = Rs. 236 +Rs.23.60
 Rs. 259.60
cost price for Rohan = 236
The selling price for Rohan = Rs. 236+Rs, 20
= Rs. 256
256  10
Tax charged = Rs.
 Rs.25.60
100
So, VAT = Rs. 25.60 –Rs. 23.60
=Rs. 2.00
The value of invoice = Rs. 256 +Rs. 25.60
=Rs. 281.60
Cost Price for Sohan = Rs. 256
The selling price of Sohan = Rs. 256 + Rs. 20
= Rs. 276
276  10
Tax charged = Rs.
100
=Rs. 27.60
VAT = Rs. 27.60 – Rs. 25.60 = Rs. 2.00
The value of invoice = Rs. 276 + Rs. 27.60
=Rs. 303.60
10
Cost price for Mohan = Rs. 276
The selling price for Mohan = Rs. 276 + Rs. 20
=Rs. 296
296  10
Tax charged = Rs.
 Rs.29.60
100
VAT =Rs. 29.60 –Rs. 27.60 =Rs. 2.00
The value of invoice = Rs. 296+Rs. 29.60
=Rs. 325.60
So Total VAT =Rs. 23.60  3  2.00   Rs.29.60
(b)
 2  2
A

4 
 3
A2  A  A
 2  2  2  2 



4   3 4 
 3
2  2   2  3 2  2   2  4  


3  2   4  3  3  2   4  4 
 4  6  4  8   10  12



22 
 6  12 6  16   18
 10 12 
A2  

18  22
 2  2  12  12
6A  6 



3
4

  18 24 
 10 12   12  12
 A 2  6A  


18  22  18 24 
2 0


0 2 
(c)
No.of favaurable outcomes
P E  
Total no.of possible outcomes
Total numbers of balls  3  5  6  14
(i) No. of black balls = 5
11
P(black ball) =
5
14
B
w
(ii) No. of balls which are not red  5 6  11
11
P ( not red ) =
14
or
3 11
P(not red)  1  p(red)  1 

14 14
(iii) No. of favorable outcomes = 3  6  9
9
P(either red or while) 
14
7. (a) Anu sold Rs. 100 shares at 10% discount and invested in 15 % Rs . 50 shares
at Rs. 33 . If she sold her shares at 10 % premium instead of 10 % discount, she would have
earned Rs. 450 more. Find the number of shares sold by her.
[3]
Sol:
(a)
Let the number of shares sold = x
S.P. of x shares = Rs. 90x (Sold at 10% discount)
S.P. of x shares sold at 10% premium = Rs. 110x
90x
No. of shares (each of Rs. 50 ) that can be purchased wit h Rs. 90x =
33
90
Total face value = Rs.50  x
33
90
Annual income  15% of 50  x
33
15
90

 50  x
100
33
225
 Rs.
x.
11
110x
No. of shares (each of Rs. 50) that can be purchased with Rs. 110x 
33
110
Total face value = Rs. 50 
x
33
15
110
 50 
x
Annual income 
100
33
 15 110x 
 Rs.  
33 
 2
 Rs.25x
12
225x
 450
11
275  225x

 450
11
 50x  450  11
450  11
x
50
 x  90
Number of shares sold by Anu = 99

25x 
(b) From a window (60m high above the ground ) of a house in a street,the angles of
elevation and depression of the top and the foot of another house on the opposite side of
the street are 600 and 450 respectively . Show that the height of the opposite house
is 60 1  3 m .
[4]


Sol:
Let P denote the position of the window of a house and AB denote the opposite house .In
right triangle PQA
AQ
 tan450  1
PQ
Each angle of quadrilatral OPQA is 900

OPQA is a rec tan gle


 AQ  60m

 PQ  AQ
 PQ  60  AQ  OP  60m 
In right triangle PQB
BQ
 tan600
PQ
 BQ  3  60
 BQ  60 3
 The height of the opposite house  AB  AQ  BQ

 

 60  60 3  60 1  3 m
13
(c) In the given figure , find CEB and ADB , where E is the Point of intersection of chords
AC and BD of circle.
[3]
Sol:
(c)  BAC   BDC [ angle of same segment ]
 BAC  350
in AEB
EAB  AEB  EBA  1800
350  AEB  500  1800
AEB  1800  350  500  950
CEB  AEB  1800
[Linear pair angles]
0
0
CEB  95  180
CEB  1800  950  850
In ΔADB
ADB  DAB  ABD  1800
ADB  550  350  500  1800


ADB  1400  1800
14
ADB  180  140
ADB  400
0
0
8. (a) Prove that if the bisector of any angle of a triangle and the perpendicular bisector of
its apposite side intersect, they will intersect on the circumcircle of the triangle.
[3]
Sol:
(a)
Let ABC be the given triangle and AD be the angle bisector of A and PR be right bisector
of side BC, intersecting each other at P.
As P lies on right bisector of BC
PB  PC and PQB  900
with BP and PC as diameters draw circles
these circles will pass through Q.
 BQP  CQP  900


and will touch the sides AB and AC
Now ABL is a tangent and BQ is a chord.
 ABD  BPA [ angle in the alt. segment ]
Similarly,
 ACD  CPA
ABD  ACD  BPA  CPA  BPC
Adding A to the both sides, we get
 ABD  ACD  A   A   BPC
or 1800  A  BPC [ angle sum property]
Now, ABPC is a quadrilateral in which
A  P  1800
 A,B,P and C are concylic .
(b) The radii of the internal and external surface of a hollow spherical shell are 3cm and 5cm
15
2
respectively. If it is melted and recast in to a solid cylinder of height 2 cm, find the
3
diameter and the curved surface area of the cylinder.
[4]
Sol:
r  3cm
R  5cm
4
4 22 3 3
Volume of sphere (shell )  π R 3  r3  
5 3
3
3 7
4 22
  125  27 
3 7
4 22
88  14 3
   98 
cm
3 7
3
For cylinder
2
8
h  2 cm  cm
3
3
Volume of shell = volume of cylinder
88  14
 πr2h
3
88  14 22 2 8
 r 
3
7
3
88  14  7  3
r2 
3  22  8
2
r  49
r  7m
Diameter of the cylinder  2r  2  7  14cm
Surface area of cylinder  2πrh
22
8
 2 7 
7
3
2
 117.33cm




(c) From the given fig. find the value of x.
[3]
16
PA × PB = PC × PD
x×x=1×8
x² = 8
x = √8 = 2√2
9.
(a) Draw ΔABC , having A 2,0 ,B 6,0  and C 2,8 
(i) Draw the line of symmetry of ΔABC .
(ii) Find the coordinates of the point D, if the line (i) and BC are both lines of
symmetry of the quadrilateral ABCD.
(iii) Assign, special name to the quadrilateral ABCD.
Sol: (a) (i)
[3]
(ii) D(6,8) (iii) Square
(b) The Point A 5, 1  on reflection in x-axis is mapped as A’. Also A on reflection in y-axis is
mapped as A’’. Write the coordinates of A’ and A’’ also calculate the distance AA’:
Sol:
A ' 5,1 and A ''5, 1 
So, AA ' 

[3]
x2  x1   y 1  y 2 
2
2
5  5  1  1
2
2
17

10  2
2
2
 100  4
 104
 2 26
(c) A page from Pooja’s saving bank account is given below.
Date
Particulars
Withdrawals
Deposits
Balance
Rs. P
Rs, P
Rs, P.
01.01.2000
B/f
- - 2,800.00
08.01.2000
By cash
- 2,200.00
5,000.00
18.02.2000
To cheque
2,700.00
- 2,300.00
19.05.2000
By cash
1,800.00
4,100.00
Calculate the total interest earned by her upto 30 -06-2000, the rate of interest are as follows.
(i) 4.5% p.a. from 01. 10-99 to 31 -03. 2000.
(ii) 4 % p.a. from 01. 04.2000 to date .
[4]
Sol:
Month
Qualifying amount (in Rs.)
January
5,000
February
2,300
March
2,300 =9600
April
2,300
May
2,300
June
4,100 = 8700
P  R  T 9600  4.5  1

 Rs.36
100
100  12
8,700  4  1
 Rs.29
(ii) Interest earned by pooja at 4 % 
100  12
Total interest =Rs. 36  29   Rs.65 .
(i) Interest earned by pooja at 4.5% 
18
10. (a) Using a ruler and compass only construct :
(i) A triangle ABC in which AB = 9 cm, BC = 10 cm and ABC  450 .
(ii) Also, Construct a circle of radius 2 cm to touch the arms of  ABC .
[4]
Sol:
(a)
(b) The Marks of 200 students in an exam were recorded as follows:
Marks %
10-20
20-30
30-40
40-50
50-60
60-70
70-80
No. of
7
11
20
46
57
37
15
students
Draw a cumulative frequency table and hence, draw the Ogive and use it to find:
(i) The median, and
(ii) The number of students who score more than 40 % marks.
Sol:
(b)
Marks %
10-20
No. of
students(frequency)
7
Cumulative
(frequency)
7
20-30
11
18
30-40
20
38
40-50
46
84
50-60
57
141
60-70
37
178
70-80
15
193
80-90
7
200
80-90
7
[6]
19
n n

   1
2 2
 th observation
(i) Median 
2
200  200


 1
2  2
 th observation

2
100  101

the observation
2
 100.5 th observation = 52.5%
(ii) Number of students getting more than 40%
Marks = 200 – 38 = 162
11. (a) Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also
cyclic.
[3]
Sol:
(a)
PQRS is a quadrilateral
20
P  Q  R  S  3600
1  2  3  4  5  6  7  8  3600
21  24  25  28  3600
 1  2, 3  4, 5  6, 7  8 
2 1  4  5  8   3600
1  4  5  8  1800
1  8 4  5  1800
180
0


 9  1800  10  1800
3600  9  10   1800
So, 9  10  3600  1800
9  10  1800
Hence XUYW is a cyclic quadrilatral.
(b) Without using mathematical tables evaluate :
tan20 tan30 tan450 tan870 tan880

2 sec2 200  cot 2 700

[3]
Sol:
tan20 tan30 tan450 tan870 tan880

2 sec2 200  cot 2 700




 
90  20
tan20.tan30. tan450.tan 900  30 tan 90  20

2 sec2 200  cot 2

0
tan20 tan30 tan450 cot 30 cot 20

2 sec2 200  tan2 200

21
Using ,

tan 90  θ  cot θ & sec θ  tan θ  1
0
2
2
1
1

21 2
(c) A train covers a distance of 90km at a uniform speed. Had the speed been 15 km per hour
more, it would have taken 30 minute less for the journey . Find the original speed of the train.
[4]
Sol:
Total distance = 90km
Let original speed of the train be = x km/hr
Distence
 Time 
Speed
90
 hr
x
Now the increased speed of the train
 x  15 km / hr
Time 
Distance
speed
90
hr
x  15
According to the Problem
90
90
1


x x  15 2
90 x  15   90x 1


x x  15 
2
90x  1350  90x 1


x 2  15x
2
2
 2  1350  x  15x
 x 2  15x  2700  0
 x x  60   45 x  60   0

Either x  60  0 or x  45  0
 x  60 or x  45
Rejecting x  60 [as speed cannot be negative]
Hence original speed of the train = 45 km / hr
22

10th ICSE Guess Paper with sol

Transcription

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