10th ICSE Guess Paper with sol - Pioneer Mathematics

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

L.K. Gupta (Mathematics Classes) 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

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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     PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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]

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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 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


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

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: PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

[3]

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www.pioneer mathematics.com MOBILE: 9815527721, 4617721 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. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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Class 0-20 20-40 40-60 60-80 80-100 100-120

www.pioneer mathematics.com MOBILE: 9815527721, 4617721 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 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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www.pioneer mathematics.com MOBILE: 9815527721, 4617721 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


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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 . 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

www.pioneer mathematics.com MOBILE: 9815527721, 4617721 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 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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


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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 ]  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 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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

www.pioneer mathematics.com MOBILE: 9815527721, 4617721 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] PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH SOLUTIONS ON www.pioneermathematics.com  NOTICES

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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. 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

www.pioneer mathematics.com MOBILE: 9815527721, 4617721 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 - Pioneer Mathematics

10th ICSE Guess Paper with sol - Pioneer Mathematics

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