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ICSE Class X Board Exam 2018 : Chemistry

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Kanishka Srivastava
R. V. S. Academy (RVS), Jamshedpur

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X - ICSE Board Mathematics - Question Paper Solutions Date: 27.02.2018 SECTION - A (40 Marks) Attempt all questions from this Section Question 1 (a) Ans. Find the value of x and y if : x 2 9 7 6 7 10 7 y 5 4 5 22 15 x 2 9 7 6 7 10 7 y 5 4 5 22 15 [3] 14 6 7 10 7 2x 18 2 y 10 4 5 22 15 14 7 10 7 2x 6 18 4 2 y 10 5 22 15 Using equality of matrix 2 x 6 10 and 2 y 10 5 15 2 x 4 2 y 5 15 x 2 2 y 20 y = 10 (b)Sonia had a recurring deposit account in a bank and deposited Rs. 600 per month for 21/2 years. If the rate of interest was 10% p.a., find the maturity value of this account. [3] Ans. Recurring deposite = 600 per month Period = 21/2 yrs = 30 months R.O.I = (r) = 10% Total principal per 1 month www.vedantu.com 1 n n 1 600 300 30 31 Rs.2, 79, 000 2 PRT 279000 10 1 Rs.2325 100 100 12 Interest = Maturity vale = 600 30+2325=Rs.20325 (c) Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag A card is drawn at random from the bag. Find the probability of getting a card which is : [4] (i) a prime number (ii) a number divisible by 4 (iii) a number that is a multiple of 6 (iv) an odd number 2 Ans. 14 4 6 8 10 12 16 18 20 n S 10 C1 10 (i) A = a prime number = {2} P A 1 10 (ii) B = Number divisible by 4 = {4, 8, 12, 16, 20} P B 5 1 10 2 (iii) C = a number that is multiple of 6 = {6, 12, 18} P C 3 10 (iv) D = an odd number = { } P D 0 0 10 www.vedantu.com 2 Question 2 (a) The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the [3] (i) radius of the cylinder 22 (ii) volume of cylinder Use 7 Ans. (i) Given circumference 2 r 132 2 22 r 7 r 3 7 21cm Radius = 21 cm (ii) Volume of cylinder r 2 h 25 22 21 21 25 7 21 22 21 25 3 = 34, 650 cm3 (b) If (k 3), (2k + 1) and (4k + 3) are three consecutive terms of an A.P., find the value of k. Ans. k 3 , 2k 1 , 4k 3 are consecutive numbers in AP.. [3] 2 2k 1 k 3 4k 3 4k 2 k 3 4k 3 k =2 (c)PQRS is a cyclic quadrilateral. Given QPS =73 , PQS =55 and PSR =82 , calculate :[4] S 82 73 R 55 P Q (i) QRS (ii) RQS (iii) PRQ www.vedantu.com 3 Ans. From diagram (i) SPQ QRS 180 (Opposite angles are supplimentary) 73 QRS 180 QRS 180 73 107 (ii) PSR PQR 180 82 PQR 180 PQR 180 82 PQR 98 But PQR PQS RQS 98 55 RQS 98 55 RQS 43 (iii) S R o 55 o o o 73 o 107 - 55 = 52o 82 o 55 P Q PRQ =52o Question 3 (a) If (x + 2) and (x + 3) are factors x3 + ax + b, find the values of a and b . Ans. x + 2 is factor of x3 + ax + b [3] 3 2 a 2 b 0 8 2a b 0 2a b 8 ...(i) x 3 is factor of x 3 ax b 3 3 a 3 b 0 27 3a b 0 www.vedantu.com 4 ...(ii) 3a b 27 (ii) (i) 3a b 2a b 27 8 3a b 2a b 27 8 a 19 Put a = 19 in (i) 2( 19) b = 8 38 b = 8 38 + 8 = b = 30 (b) Prove that sec 2 cosec2 tan cot Ans. L.H.S [3] sec2 cosec2 tan 2 1 cot 2 1 tan 2 2 cot 2 tan 2 2 tan cot cot 2 tan cot 1 tan cot 2 tan + cot (c) Using graph paper draw a histogram for the given distibution showing the number of runs scored by 50 batsman. Estimate the mode of the data : [4] Runs 1000- 4000- 5000- 6000- 7000- 8000- 9000Scored 4000 5000 6000 7000 8000 9000 10000 No. of 4 18 9 6 7 2 4 batsman Ans. www.vedantu.com 5 B A 18 Scale : On x axis 1 cm = 1000 units On y axis 16 1 cm = 2 units 14 K 12 10 8 6 4 2 3000 4000 5000 6000 7000 8000 9000 10000 MODE approx 4550 or 4600 Question 4 (a)Solve the following inequation, write down the solution set and represent it on the real number line :[3] Ans. 12 3x and 3x 14 4 x and x 14 x Z 3 -4 -3 -2 -1 -0 1 2 3 4 x = {-4,-3,-2,-1,0,1,2,3,4} (b)If the straight lines 3x 5 y =7 and 4x+ay+9=0 are perpendicular to one another, find the value of a. [3] www.vedantu.com 6 Ans. Slope of 3 x 5 y 7 is m1 3 3 5 5 Slope of 4 x ay 9 0 is m2 4 a lines are m1 m2 1 3 4 1 5 a 12 a 5 a 12 5 (c)Solve x 2 +7x =7 and give your answer correct to two decimal places. Ans. x 2 +7x =7 x2 7 x [4] 49 49 7 4 4 2 7 77 x 2 4 x 7 77 2 4 x 77 7 4 2 x 77 7 4 2 x 77 7 2 x = 0.88 or x = 7.88 www.vedantu.com 7 SECTION - B (40 Marks) Attempt any four questions from this Section Question 5 (a) The 4th term of a G.P. is 16 and the 7th terms is 128. Find the first term and common ratio of the series. [3] Ans. Let the first term of a G.P. a and common ratio r a4 ar 3 16 ...(i) a7 ar 6 128 ...(ii) ii i ar 6 128 ar 3 16 r3 8 r 2 Put r 2 in equation (ii) 3 a 2 16 a 2 First term (a) = 2 Common ratio = 2 (b) A man inversts Rs.22,500 in Rs.50 shares available at 10% discount. If the dividend paid by the company is 12%, calculate : [3] (i) The number of shares purchased (ii) The annual dividend received (iii) The rate of return he gets on his investment. Give your answer correct to the nearest whole number. Ans. Actual price = 50 Rs./Share Price after discount = 50 10% of 50 = Rs.45 (i) Total shares bought (Purchased) 22500 500 45 (ii) Annual dividend received 500 50 12% www.vedantu.com 8 500 50 12 100 5 600 Rs. 3000 (iii) ROR = 25500 22500 100 22500 3000 100 13.33% 22500 aprrox. 13% (c) Use graph paper for this question (Take 2cm = 1 unit along both x and y axis). ABCD is a quadrilateral whose vertices are A(2, 2), B(2, 2), C(0, 1) and D(0, 1). (i) Reflect quadrilateral ABCD on the y-axis and name it as A ' B ' CD . (ii) Write down the coordinaes of A 'and B ' . (iii) Name two points which are invariant under the above reflection. (iv) Name the polygon A ' B ' CD . Ans. A'( 2, 2) A(2,2) 2 D 1 (0,1) 2 0 1 1 2 1 C(0, 1) B'( 2, 2) 2 (ii) A ' =( 2, 2) B ' =( 2, 2 ) (iii) C and D (iv) A' B 'CD is trapezium B(2, 2) www.vedantu.com 9 Question 6 (a) Using properties of proportion, solve for x. Given that x is positive : 2 x 4x2 1 2x 4x2 1 Ans. 2x 4x2 1 2 [3] 4 2x 4 x 1 4 1 Applying componendo and dividendo 2x 2x 4 x 1 2 x 4 1 4 x 1 4 1 4x2 1 2x 4x2 1 2 4x 2 4x2 1 2 5 3 Squaring on both sides, we get 4x2 25 2 4x 1 9 36 x 2 100 x 2 25 64 x 2 25 x2 x 25 64 5 8 (b) 2 3 If A , 5 7 0 4 1 0 B and C , find AC B 2 10C. 1 7 1 4 Ans. 2 3 0 4 1 0 A , B and C 5 7 1 7 1 4 [3] 2 3 1 0 0 4 0 4 1 0 AC B 2 10C 10 5 7 1 4 1 7 1 7 1 4 0 2 3 12 4 28 10 AC B 2 10C 5 7 28 7 45 10 40 www.vedantu.com 10 12 28 1 4 10 AC B 2 10C 2 7 10 28 45 40 15 40 AC B 2 10C 1 33 (c) Prove that Ans. Taking LHS : [4] 1 cot cosec 1 tan sec 1 sin 1 cos 1 1 sin sin cos cos sin cos 1 sin cos 1 sin sin cos cos 2 12 sin cos sin 2 cos2 2sin cos 1 sin cos 1 2 sin cos 1 sin cos 2sin cos sin cos =2 Question 7 (a) Find the value of k for which the following equation has equal roots. [3] Sol. following equation having equal roots b 2 4ac 0 here a 1, b 4k, 4k 2 c k2 k 2 4 1 k 2 k 2 0 www.vedantu.com 11 16k 2 4k 2 4k 8 0 12k 2 4k 8 0 3k 2 k 2 0 3k 2 3k 2k 2 0 3k k 1 2 k 1 0 k 1 3k 2 0 k 1 or (b) k 2 3 On a map drawn to a scale of 1 : 50,000, a rectangular plot of land ABCD has the following dimensions. AB = 6cm, BC = 8 cm and all angles are right angles. Find : (i) the actual length of the diagonal distance AC of the plot in km. (ii) the actual area of the plot in sq km. [3] Sol. D C 8cm A 6cm B Using pythagoras theory AC 2 AB 2 BC 2 AC 2 62 82 AC 2 100 AC 10 cm 1) Actual length 10 50000 500000 cm 500000 5 km 1000 100 2) Area of ABCD 6 8 48 cm 2 (c) 48 50000 50000 48 12 sq. km 100000 100000 4 A (2, 5), B(-1, 2) and C(5, 8) are the vertices of a triangle ABC, M is a point on AB such that AM : MB = 1 : 2. Find the co-ordinates of M . Hence find the equation of the line passing through the points C and M. [4] www.vedantu.com 12 Sol. A 2,5 , B 1, 2 and C 5,8 A (2, 5) 1 M 2 B (-1, 2) C (5, 8) Let the co-ordinates of M is (x, y) x 2 2 1 1 4 1 1 2 1 3 y 2 5 1 2 12 4 2 1 3 point M = (1, 4) Equation of line passing through C (5, 8) and M (1, 4). y 8 4 8 x 5 1 5 y 8 4 x 5 4 y 8 1 x 5 y 8 x 5 x y 3 0 Question 8 (a) Rs. 7500 were divided equally among a certain number of children. Had there been 20 less children, each woule have received Rs. 100 more. Find the original number of chlidren.[3] Sol. Let the original number of person be x, then 7500 divided equally between x person, 7500 x 7500 divided equally between x - 20 children each one get each one get 75 7500 x 20 www.vedantu.com 13 According to the question 7500 7500 100 x 20 x 1 7500 7500 100x x 20 x 7500x x 20 7500 100x 7 5 x x 2 0 7 5 x 75x 75x x 2 1500 20x x 2 20x 1500 0 x 20 400 4 1500 2 x 20 400 6000 2 x 20 80 2 20 80 2 or x 30 (not possible) x 50 original number of children = 50 x (b) 20 80 2 x If the mean of the following distribution of 24, find the value of a Marks Number of students Sol. or [3] 0 - 10 10 - 2020 - 30 30 - 40 40 - 50 7 a 8 10 5 Mean = 24 Class Class mark (x)i 5 fi xi 0 - 10 Frequency (fi) 7 10 - 20 a 15 15a 20 - 30 8 15 200 30 - 40 10 35 350 40 - 50 5 45 225 Total 30 + a 35 810 + 15a www.vedantu.com 14 Mean f x f i i 24 i 810 15a 24 30 a 810 15a 720 24a 90 9a a 10 (c) Using ruler and compass only, construct a ABC such that BC = 5 cm and AB = 6.5 cm and [4] ABC 120o (i) Construct a circm - circle of ABC (ii) Construct a cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Sol. Step of construction : D (i) Draw BC = 5 cm (ii) At B, draw X A O XBC 120 6.5cm (iii) From BX, cut off AB = 6.5 cm 120 5cm (iv) Join AC to get ABC C B (v) Draw the perpendicular bisector of BC and AB. These bisectors meet at O. With O as centre and radius equal to OA, draw a circle, which passes through A, B and C. This is the required circumcircle of ABC (vi) Produce the perpendicular bisector of BC so that it meets the circle at D. Join CD and AD to get the required cyclic quadrilateral ABCD. o o Question 9 (a) Priyanka has a recurring deposit account of Rs. 1000 per month at 10% per annum. If she gets Rs. 5550 as interest at the time of maturity, find dthe total time for which the account was held. [3] Sol. Amount of recurring deposit per month = Rs. 1000 Rate of interest = 10 % p.a. let period = n months Amount of interest = 5550 ......(1) Total principal for one month Interest 1000 n n 1 2 1000n n 1 10 1 2 100 12 25 n n 1 6 From (1) nd (2) , we get ......(2) 25 n n 1 5550 6 www.vedantu.com 15 25n 2 25n 33300 25n 2 25n 33300 0 n 2 n 1332 0 n 2 37 n 36n 1332 0 n n 37 36 n 37 0 n 36 n 37 0 n 36 (b) PM 2 In PQR, MN in parallel to QR and MQ 3 (i) Find P M N O MN QR R (ii) Prove that OMN and ORQ are similar.. (iii) Find. Area of OMN : Area of ORQ Sol. [3] In PQR , MN || QR is such a way that PM : MQ 2 : 3 (i) In PQR , MN || QR PM PN 2 MQ 3 MQ NR 3 PM 2 Adding 1 on both sides, 1 MQ 3 1 PM 2 PM MQ 3 2 PM 2 PQ 5 PM 2 PM 2 PQ 5 Now in PMN and PQR , PMN PQR (corresponding angles) P P (Common) PMN PQR (AA postulates) PM MN PN PQ QR NR www.vedantu.com 16 But PM 2 PQ 5 MN 2 QR 5 (ii) In OMN and ORQ (a) MON QOR (Vertically opposite angles) Since MN | | QR, (b) MNO OQR (Alternate angles) (c) NMO ORQ (Alternate angles) By AAA postulates, OMN ORQ Ar OMN (iii) Ar ORQ Ar OMN Ar ORQ (c) MN 2 QR 2 4 25 The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. This common radius is 7 cm. The height of the cylinder and cone are each of 4 cm. Find the volume of the solid. [4] 4 cm 4 cm 7 cm Sol. Volume = Volume of cone + Volume of cylinder + Volume of hemishpere 1 2 r 2h r 2 H r 3 3 3 1 r 2 h 3 H 2r 3 1 22 7 7 4 4 3 2 7 3 7 1 22 7 30 3 =22 7 10 =1540 cm3 www.vedantu.com 17 Question 10 (a) Use Remainder theorem to factorize the following polynomial : 3 [3] 2 2 x 3x 9 x 10 Sol. p x 2x 3 3x 2 9x 10 p 1 2 1 3 1 9 1 10 0 x 1 is a factor of p x Now, dividing p x by x 1, we get 2x 2 x 10 x 1 2x 3 3x 2 9x 10 2x 3 2x 2 x 2 9x 10 x2 x 10x 10 10x 10 0 3 2 2 2x 3x 9x 10 x 1 2x x 10 x 1 2x 2 5x 4x 10 x 1 x 2x 5 1 2x 5 x 1 x 2 2x 5 (b) In the figure given below O is the center of the circle. If QR = OP and ORP 20o . Find the value of x giving reasons. [3] T x0 P O S 20 0 Q www.vedantu.com R 18 Sol. T 60 o x O 0 100 0 20 40 0 P 40 0 140 0 S 200 Q R QR OP QR OP OQ Hence OQR is isoscleles QRO 180 20 20 140 OQP 40o linear pair of OQR OPQ 40o as OPQ is isosceles POQ 180o 40o 40o 100o POT =x =180o 100o 20o =60o (c) The angle of elevation from a point P of the top of a tower QR, 50 m high is 60o and that of the tower PT freom a point Q is 30o. Find the height of the tower PT, correct to the nearest metre. [4] R T P 60 o 30 o Q R Sol. 50 T P 60o x 30 o Q www.vedantu.com 19 Let the height of the tower PT is h. and PQ is x In PQT PT PQ tan 30o 1 H 3 x ......(1) x 3h In PQR tan 60o 50 x ......(2) 3 x 50 3 3 h 50 3h 50 h 50 3 h 50 3 Question 11 (a) The 4th term of an A. P. is 22 and 15th term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 terms. [4] Sol. Let the first term at a A.P. is a and common difference is d. a4 a 3d 22 ......(1) a15 a 14d 66 ......(2) ______________________________ 11d 44 a=4 put d = 4 in equation (1) a 3 4 22 a 12 22 a 10 Sn n 2a n 1 d 2 S8 8 20 7 4 2 S8 4 20 8 4 www.vedantu.com 20 S8 =4[ 48 = ] 192 (b) Use graph paper for htis questin. A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded : [6] Height in cm 135-140 140-145 145-150 150-155 155-160 160-165 165-170 No. of boys 4 8 20 14 7 6 1 Taking 2cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following : (i) the medium (ii) lower Quarile (iii) if above 158 cm is considered as the tall boys of the class. Find the number of boys in the class who are tall. Sol. Height in cm No.of boys C. f . 135-140 4 4 140-145 8 12 145-150 20 32 150-155 14 46 155-160 7 53 160-165 6 59 165-170 1 60 Median = 60 30th item 2 www.vedantu.com 21 60 55 50 45 40 35 30 A 25 20 15 10 7.5 C 5 B 0 135 140 145 150 155 160 165 170 (i) Meadian = 150.5 (ii) Lower quartile = 142.5 (iii) 10 boys are above 158 cm www.vedantu.com 22

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