(10th icse guess paper) - pioneermathematics.com icse guess paper...... find the mean, mode and...
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L.K. Gupta (Mathematics Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
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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
L.K. Gupta (Mathematics Classes) Pioneer Education (The Best Way To Success)
S.C.O. 320, Sector 40- D, Chandigarh
Ph: - 9815527721, 0172 – 4617721.
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SECTION – A 1. (a) Determine the value of ‘m’ if x m is a factor of the polynomial 3 2x m 1 x 2
hence find the value of k if 3m 6 k . [3] Sol: (a)
x m is a factor of 3 2x m 1 x 2
for x m
3 2m m 1 m 2 0
or 3 3m m 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 [3] Sol: 3a 8b 3a 8b3c 8d 3c 8d
Using alternendo 3a 8b 3c 8d3a 8b 3c 8d
Using componendo and dividendo 3a 8b 3a 8b 3c 8d 3c 8d3a 8b 3a 8b 3c 8d 3c 8d
or 6a 6a
16b 16d
a cb d
a,b,c and d are in proportion. (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
or n
r p r tP 1 1 50
100 100
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or 2
5 x 5 2x 1 1 50
100 100
or 2
21 xx 1 50
20 10
or 441 x
x 1 50400 10
441 400 xx 50
400 10
or 41x 40x
50400
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:
Total P for 1 month = Rs 150 8 8 1
2
150 36 Rs.5400 5400 8
1 Rs.36100 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
Line: 2 x 2
1 ,x R3 3 3
. [4]
Sol: 2 x 2
13 3 3
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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 LP 2
PQ||MN andPM 3
calculate the value of :
(i)
ar ΔLPQar ΔLMN
(ii) area of trapeziumPMNQ
area of ΔLMN [3]
Sol:
Given : PQ || MN and LP 2PM 3
L L Common
P M [Corresponding angles ] ΔLPQ ΔLMN [ AA similarity ]
Let LP = 2x then PM = 3x LM = 2x+3x = 5x
(i)
2 22
2
ar ΔLPQ LP LP 2xar ΔLMN LM 5xLM
2
2
4x 44 :25
2525x
(ii) Let ar ΔLPQ 4y then ar ΔLMN 25y
ar trapeziumPMNQ 25y 4y 21y
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ar trapizium PMNQ 21y 2121 :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) xM k,m 5, 2
k 5,m 2 xM x,y x, y
Co –ordinates of P are 5,2
(ii) y yM 5,2 5,2 M x,y x,y
Co –ordinates of P’’ are 5,2
(iii) Reflection of p’ in the origin
0M 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:
(i) mean x 247
24.7n 10
(ii) Ascending order : 19, 20, 21 , 23, 25, 27, 28, 29, 30 n 10(even)
Median n/2T Next term
2
10/2 5 6T Next term T T
2 2
25 25 5025
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 2cm . If the radius
of the outer circle is 21 cm . Calculate the radius of the inner circle 22
Use π7
[3]
Sol:
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Radius of outer circle (R) = 21 cm Let radius of inner circle = r cm or 2 2 2πR πr 770cm or 2 2π R r 770
or 2 22221 r 770
7
or 2 770 7441 r
22
or 2441 r 245 or 2r 441 245 or 2r 196 or 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)
L.H.S =1 1
secθ tanθ cosθ
secθ tanθ1
secθsecθ tanθ secθ tanθ
2 2
secθ tanθsecθ
sec θ tan θ
secθ tanθ secθ secθ secθ tanθ
secθ tan θ
secθ secθ tan θsecθ tan θ
2 2sec θ tan θsecθ
secθ tanθ
1 1cosθ 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]
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Sol:
Class X F Fx 0-20 10 5 50 20-40 30 P 30p 40-60 50 10 500 60-80 70 Q 70q 80-100 90 7 630 100-120 110 8 880 50 2060+30p+70q
Now, 5 p 10 q 7 8 50 or p q 50 30 or p q 20................(i)
fxx
f
or 2060 30p 70q
62.850
or 10 206 3p 7q
62.850
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 andq 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
Class Frequency 0-20 5 20-40 P 40-60 10 60-80 Q 80-100 7 100-120 8
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No total amount to be paid = M.P +S.T% of M.P 17658 = x 8% of x
17658 = 8
x x100
217,658 x x
25
27x17658
25
17,658 25x
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 1x
[3]
Sol: 27
3x 1 0 3x x 7 0x
Comparing this to the equation 2ax bx c 0 we get a 3,b 1,c 7
2b b 4acx
2a
21 1 4 3 7
2 3
1 1 846
1 85 1 9.2196 6
1 9.219 1 9.219,
6 6
10.219 8.219,
6 6
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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 . [4] Sol:
slope of 1Coefficient of x 3
AB mCoefficient of y 2
Slope of CD 21
1 1m 2/3
3m2
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then equation of CD , 1 1y y m x x
N 3,4 is the passing point and slope 2/3
2y 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]
(b) If 2 2
A3 4
, then find 2A 6A [3]
(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
Tax charged = Rs. 236 10
Rs.23.60100
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
Tax charged = Rs. 256 10
Rs.25.60100
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
Tax charged = Rs. 276 10
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
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Cost price for Mohan = Rs. 276 The selling price for Mohan = Rs. 276 + Rs. 20 =Rs. 296
Tax charged = Rs. 296 10
Rs.29.60100
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 2A
3 4
2A A A 2 2 2 2
3 4 3 4
2 2 2 3 2 2 2 4
3 2 4 3 3 2 4 4
4 6 4 8 10 12
6 12 6 16 18 22
210 12
A18 22
2 2 12 126A 6
3 4 18 24
210 12 12 12
A 6A18 22 18 24
2 0
0 2
(c)
No.of favaurable outcomesP E
Total no.of possible outcomes
Total numbers of balls 3 5 6 14 (i) No. of black balls = 5
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P(black ball) = 5
14
(ii) No. of balls which are not red B w
5 6 11
P ( not red ) = 1114
or
P(not red) 3 11
1 p(red) 114 14
(iii) No. of favorable outcomes = 3 6 9
P(either red or while) 9
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
No. of shares (each of Rs. 50 ) that can be purchased wit h Rs. 90x = 90x33
Total face value =90
Rs.50 x33
Annual income 90
15% of 50 x33
15 9050 x
100 33
225Rs. x
11 .
No. of shares (each of Rs. 50) that can be purchased with Rs. 110x 110x
33
Total face value = Rs. 110
50 x33
Annual income 15 110
50 x100 33
15 110xRs.
2 33
Rs.25x
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225x25x 450
11
275 225x450
11
50x 450 11 450 11
x50
x 90 Number of shares sold by Anu = 99 (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 060 and 045 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
0AQtan45 1
PQ
0Each angle of quadrilatral OPQA is 90
OPQA is a rec tan gle
AQ 60m
PQ AQ
PQ 60 AQ OP 60m
In right triangle PQB 0BQ
tan60PQ
BQ 3 60
BQ 60 3 The height of the opposite house AB AQ BQ
60 60 3 60 1 3 m
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(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 ]
0BAC 35 in AEB
0EAB AEB EBA 180 0 0 035 AEB 50 180
0 0 0 0AEB 180 35 50 95 0CEB AEB 180
[Linear pair angles] 0 0CEB 95 180
0 0 0CEB 180 95 85 In ΔADB
0ADB DAB ABD 180
0 0 0 0ADB 55 35 50 180 0 0ADB 140 180
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0 0ADB 180 140 0ADB 40
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
0PB PC and PQB 90 with BP and PC as diameters draw circles these circles will pass through Q.
0BQP CQP 90
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 0180 A BPC [ angle sum property] Now, ABPC is a quadrilateral in which
0A P 180 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
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respectively. If it is melted and recast in to a solid cylinder of height 2
23
cm, find the
diameter and the curved surface area of the cylinder. [4] Sol: r 3cm R 5cm
Volume of sphere (shell ) 3 3 3 34 4 22π R r 5 33 3 7
4 22125 27
3 7
4 2298
3 7 388 14
cm3
For cylinder 2 8
h 2 cm cm3 3
Volume of shell = volume of cylinder 288 14 πr h
3
288 14 22 8r
3 7 3
2 88 14 7 3r
3 22 8
2r 49 r 7m Diameter of the cylinder 2r 2 7 14cm Surface area of cylinder 2πrh
22 82 7
7 3
2117.33cm (c) From the given fig. find the value of x.
[3]
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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. [3] Sol: (a) (i)
(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’: [3] Sol: A' 5,1 and A'' 5, 1
So, 2 22 1 1 2AA' x x y y
2 25 5 1 1
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2 210 2
100 4 104
2 26 (c) A page from Pooja’s saving bank account is given below.
Date Particulars Withdrawals Rs. P
Deposits Rs, P
Balance 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
(i) Interest earned by pooja at 4.5% P R T 9600 4.5 1
Rs.36100 100 12
(ii) Interest earned by pooja at 4 % 8,700 4 1
Rs.29100 12
Total interest =Rs. 36 29 Rs.65 .
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10. (a) Using a ruler and compass only construct : (i) A triangle ABC in which AB = 9 cm, BC = 10 cm and 0ABC 45 . (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 80-90
No. of students
7 11 20 46 57 37 15 7
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. [6] Sol: (b)
Marks % No. of students(frequency)
Cumulative (frequency)
10-20 7 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
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(i) Median
n n1
2 22
th observation
200 2001
2 22
th observation
100 1012
the observation
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
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0P Q R S 360
01 2 3 4 5 6 7 8 360 02 1 2 4 2 5 2 8 360
1 2, 3 4, 5 6, 7 8
02 1 4 5 8 360 01 4 5 8 180
01 8 4 5 180
0 0 0180 9 180 10 180
0 0360 9 10 180
So, 0 09 10 360 180 09 10 180
Hence XUYW is a cyclic quadrilatral.
(b) Without using mathematical tables evaluate :
0 0 0 0 0
2 0 2 0
tan2 tan3 tan45 tan87 tan88
2 sec 20 cot 70 [3]
Sol:
0 0 0 0 0
2 0 2 0
tan2 tan3 tan45 tan87 tan88
2 sec 20 cot 70
0 0 0 0 0 0
2 0 2 0
tan2 .tan3 .tan45 .tan 90 3 tan 90 2
2 sec 20 cot 90 20
0 0 0 0 0
2 0 2 0
tan2 tan3 tan45 cot 3 cot 2
2 sec 20 tan 20
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22
Using , 0 2 2tan 90 θ cot θ & sec θ tan θ 1
1 12 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
DistenceTime
Speed
90hr
x
Now the increased speed of the train x 15 km / hr
DistanceTime
speed
90hr
x 15
According to the Problem 90 90 1x x 15 2
90 x 15 90x 1x x 15 2
2
90x 1350 90x 12x 15x
22 1350 x 15x 2x 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