ICSE Pariksha Mathematics-X 1 RSPL/1 I ICSE Pariksha-2010 MathematicsX SECTION ‘A’ Q1. (a) Salman Khan invests Rs. 25,600 for three years at the rate of 10% per annum compound interest. Find : (i) The sum due to Salman at the end of the first year. (ii) The interest he earns for the second year. (iii) The total amount due to him at the end of the third year. [3] (b) Using properties of proportion, solve for x : 2 2 2 2 + - - -xx = 3 [3] (c) The polynomial 2x3 + mx2 + nx – 2 when divided by (2x – 3), leaves remainder 7 and has (x + 2) as its factor. Find m and n. [4] Sol. (a) Given : P = Rs. 25,600, R = 10% (i) Interest for the first year = P × R× T 100 = 25,600 ×10 ×1 100 = Rs. 2,560. \ Sum due to Salman at the end of the first year = Rs. 25,600 + Rs. 2,560 = Rs. 28,160. (ii) Principal for second year = Rs. 28,160 \ Interest earned for the second year = 28,160 ×10 ×1 100 = Rs. 2,816. (iii) Principal for third year = Rs. 28,160 + Rs. 2,816 = Rs. 30,976 Interest earned for third year = 30,976 ×10 ×1 100 = Rs. 3,097.60. \ Total amount due to him at the end of the third year = Rs. 30,976 + Rs. 3,097.60 = Rs. 34,073.60 (b) Given : 2 2 2 2 + - - -xx = 3 It can be written as, 2 2 2 2 + - - -xx = 31 Using Componendo and Dividendo, we get 2 2 2 2 2 2 2 2 + - + - - + - - + - x x x x = 3 1 3 1 +- Þ 2 2 2 2-x = 42 Þ 2 2-x = 2 On squaring both sides, 2 2 - x = 42 ICSE Pariksha Mathematics-X or 2 = 8 – 4x Þ 4x = 8 – 2 Þ 4x = 6 Þ x = 6 3 = 4 2 (c) The given polynomial is 2x3 + mx2 + nx – 2 When divided by (2x – 3), it leaves remainder 7 i.e., 2x – 3 = 0 Þ 2x = 3 Þ x = 32 \ 3 2 3 3 3 2 2 2 2 2 æ ö æ ö æ ö + + - ç ÷ ç ÷ ç ÷ è ø è ø è ø m n = 7 Þ 27 9 3 2× 2 8 4 2 + + - m n = 7 Þ 27 9 3 2 4 4 2 m n + + - = 7 Þ 27 9 6 8 4 + + - m n = 7 Þ 9m + 6n + 19 = 28 Þ 3m + 2n = 3 ...(i) Also, (x + 2) is a factor, So, x + 2 = 0 Þ x = – 2 \ 2(–2)3 + m(–2)2 + n(–2) – 2 = 0 or –16 + 4m – 2n – 2 = 0 Þ 4m – 2n = 18 ...(ii) (i) + (ii) Þ 3m + 2n = 3 + 4m – 2n = 18 7m = 21 Þ m = 3 By equation (i) Þ 3(3) + 2n = 3 or 9 + 2n = 3 or 2n = – 6 Þ n = – 3 \ m = 3 and n = – 3. Q2. (a) Solve the given inequation and graph the solution on the number line. [3] 2 1 2 3 , . 3 3 2 - £ + < Î 3 y y Z (b) Kareena purchases an article for Rs. 21,384 which includes 10% rebate on the marked price and 8% sales tax on the remaining price. Find the marked price of the article. [3] (c) In the given figure ABCD is a rectangle. Two quadrants with centres C and D are drawn. Find the area and perimeter of the shaded region. Given that, AB = 42 cm and BC = 28 cm [4] D A CBICSE Pariksha Mathematics-X 3 Sol. (a) 2 2 1 2 3 , 3 3 3 y y Z - £ + < Î Þ 8 2 10 3 3 3 - £ + < y Þ 8 2 10 2 3 3 3 3 y - - £ < - Þ 10 8 3 3 - £ £ y Þ – 3.3 £ y £ 2.67 Þ y Î {– 3, – 2, – 1, 0, 1, 2} –4 –3 –2 –1 0 1 2 3 4 (b) Let the marked price of the article = Rs. x Total amount paid = D% S.T.% M.P. 1 1 100 100 æ öæ ö - + ç ÷ç ÷ è øè ø Þ Rs. 21,384 = 10 8 1 1 100 100 xæ öæ ö - + ç ÷ç ÷ è øè ø Þ Rs. 21,384 = 9 27 10 25 xæ öæ ö ç ÷ç ÷ è øè ø Þ x = 21,384 × 25×10 27 × 9 Rs. Þ x = Rs. 22,000 \ Marked price of the article = Rs. 22,000 (c) For bigger quadrant R = 28 cm For smaller quadrant r = 42 – 28 = 14 cm Shaded area = Area of rectangle – (Area of bigger quadrant + Area of smaller quadrant) = 42 × 28 – 2 2 4 4 R r p p æ ö + ç ÷ è ø = 1176 – [ ] 2 2 4R r p + = 1176 – 2 2 22 (28 14 ) 7× 4 + = 1176 – 22 ×980 7× 4 = 1176 – 770 = 406 cm2 Perimeter of the shaded region = R 14 42 2 2 p p æ ö + + + ç ÷ è ø r cm = ( R) 56 p + + 2 r = 22 (14 28) 56 14´ + + = 22 × 42 56 14 + = 66 + 56 = 122 cm Q3. (a) Let A × 1 1 0 2 é ù ê ú ë û = [1 2], where A is a matrix : (i) State the order of matrix A (ii) Find the matrix A. [3] (b) Aerika opened a recurring deposit account in a bank and deposited Rs. 150 per month for 8 months. If she received Rs. 1,236 at the time of maturity, find the rate of interest per annum. [3]4 ICSE Pariksha Mathematics-X (c) Use a graph paper for this question. Plot the points A(3, 2) and B(–3, –2). From A and B draw perpendiculars AM and BN on the x-axis (Take 1 cm = 1 unit on both the axes). [4] (i) Name the image of A on reflection in the origin (ii) Name the figure AMBN and find its area (iii) Write the co-ordinates of the point to which M is mapped on reflection in x-axis and y-axis. Sol. (a) (i) 1 1 0 2 é ù ê ú ë û A × = [1 2] 1 × 2 2 × 2 1 × 2 Order of A is (1 × 2) ìïíïî For multiplication of two matrices number of columns of Ist matrix should be same as the number of rows of the IInd matrix then the resulting matrix has order 1 × 2 (ii) Let matrix A = [x y] then [x y] 1 1 0 2 é ù ê ú ë û = [1 2] Þ [x × 1 + y × 0 x × 1 + y × 2] = [1 2] Þ [x x + 2y] = [1 2] Þ x = 1 and x + 2y = 2 or 1 + 2y = 2 Þ 2y = 1 Þ y = 12 martix A = 1 1 2 é ù ê ú ë û (b) Monthly instalment = Rs. 150 Time = 8 months Amount deposited in 8 months = Rs. 150 × 8 = Rs. 1200 Equivalent principal for 1 month = 150 × 8(8 1) 2+ = 150 × 8 × 9 2 = Rs. 5400 Let rate of interest be r % p.a. Then interest on Rs. 5400 for 1 month = Rs. 5400 ×1× 12×100 r = Rs. 92 r Amount received on maturity = Rs. 1236 \ 1200 + 92 r = 1236 Þ 92 r = 1236 – 1200 Þ 92 r = 36 Þ r = 8 Hence rate of interest = 8% p.a. ìïíïîICSE Pariksha Mathematics-X 5 (c) (i) Image of A on reflection in origin is (– 3, – 2) i.e., the point B. (ii) Figure AMBN is a parallelogram. ar (AMBN) = 2 ar (DAMN) = 2 × 1 (MN × AM) 2 = 6 × 2 = 12 sq. units –5 –4 –3 –2 –1 0 1 2 3 4 5 X X¢ 1234 Y–1 –2 –3 –4 Y¢ A (3, 2) B (–3, –2) M (3,0) N (–3,0) (iii) The co-ordinates of the points to which M is mapped on reflection in x-axis are (3, 0) and in y-axis are (–3, 0). Q4. (a) Without using tables, evaluate : 2 2 2 tan 35 sin 20 sec 70 sin 50 sin 40 cot 55° æ ö- ° ° + ° + ° ç ÷ è ø° [3] (b) Kamalpreet has an account in the State Bank of India. A page from her pass book is given below : Date Particulars Withdrawals Deposits Balance (in Rs.) (in Rs. ) (in Rs.) 01-04-2008 BF — — 1400.00 10-05-2008 By cash — 700.00 2100.00 02-06-2008 To self 1000.00 — 1100.00 11-07-2008 To cheque 300.00 — 800.00 21-08-2008 By cash — 1700.00 2500.00 03-10-2008 By cash — 2400.00 4900.00 If the interest earned by Kamalpreet at the end of December, 2008 was Rs. 87.75, find the rate of interest per annum. [4]6 ICSE Pariksha Mathematics-X (c) In the given figure ABCD is a quadrilateral, inscribed in a circle with centre O. [3] Side CD produced to E. If ÐADE = 95° and ÐOBA = 30°, find : (i) ÐOBC (ii) ÐOAC Sol. (a) 2 tan 35 cot 55° æ ö ç ÷ è ø° – sin 20° sec 70° + sin2 50° + sin2 40° = 2 tan 35 tan (90 55 ) ° ì ü í ý ° - ° î þ – sin 20° cosec (90° – 70°) + cos2 (90° – 50°) + sin2 40° = 2 2 2 tan 35 sin 20 cosec 20 (cos 40 sin 40 ) tan 35° æ ö- ° ° + ° + ° ç ÷ è ø° = (1)2 – sin 20° × 1 1 sin 20 + ° = 1 – 1 + 1 = 1 (b) Minimum balance for the month of April = Rs. 1400 Minimum balance for the month of May = Rs. 2100 Minimum balance for the month of June = Rs. 1100 Minimum balance for the month of July = Rs. 800 Minimum balance for the month of August = Rs. 800 Minimum balance for the month of September = Rs. 2500 Minimum balance for the month of October = Rs. 4900 Minimum balance for the month of November = Rs. 4900 Minimum balance for the month of December = Rs. 4900 Total = Rs. 23400 Now, I = P×R ×T 100 Þ 87.75 = 23400 × R ×1 12 ×100 Þ R = 87.75×12 234 R = 4.5% (c) (i) 95° 30° A O D C B E ÐADE = ÐABC [Exterior Ð of a cyclic quadrilateral] Þ 95° = ÐOBA + ÐOBC Þ 95° = 30° + ÐOBC Þ ÐOBC = 65° (ii) ÐADC + ÐADE = 180° [Linear pair Тs] ÞÐADC + 95° = 180° A B CD E OICSE Pariksha Mathematics-X 7 Þ ÐADC = 85° Now ÐAOC = 2 ÐADC subtended by an arc at the centre is double the angle subtended by it at any other point on the remaining part of the circle Ð ìïíïî= 2 × 85° ÐAOC = 170° Now OA = OC [Radii] \ ÐOAC = ÐOCA = x [say] \ In DAOC, x + x + ÐAOC = 180° Þ 2x + 170° = 180° Þ 2x = 10° Þ x = 5° \ ÐOAC = 5° SECTION ‘B’ Q5. (a) If x : y : z = 1 : 2 : 3, Prove that : 2 2 2 12 13 4 5 + + = x y z y [3] (b) Given : sin A + sin2 A = 1, Prove that : cos2 A + cos4 A = 1. [3] (c) If a line segment with end points (3, 4) and (14, –3) meets the x-axis at P, in what ratio does P divide the segment ? Also, find the co-ordinates of point P. [4] Sol. (a) Let 1x = = = 2 3 y z k Then, x = k, y = 2k, z = 3k L.H.S. 2 2 2 12 13 4 + + x y z = 2 2 2 12( ) 13(2 ) 4(3 ) + + k k k = 2 2 2 12 52 36 + + k k k = 2 100k = 10 k = 5 × 2 k = 5y = R.H.S. (b) Given, sin A + sin2A = 1 Þ sin A = 1 – sin2 A Þ sin A = cos2 A Þ sin2 A = cos4 A [By squaring both sides] Þ 1 – cos2 A = cos4 A Þ 1 = cos2 A + cos4 A Hence proved. (c) A P B (3 , 4 ) ( , ) x y (1 4, – 3 ) k 1 Let the join of A(3, 4) and B(14, – 3) be divided by the x-axis in the ratio k : 1 at P. then, x = 14 3 3 4 = 1 1 + - + + + k k y k k Since P lies on x-axis, So y = 0. Þ 3 41 - ++ kk = 0 Þ k = 43 So, ratio = 4 : 38 ICSE Pariksha Mathematics-X Hence the co-ordinates of point P = 4 14 × 3 3 , 0 4 1 3 æ ö + ç ÷ ç ÷ + ç ÷ è ø= 65 ,0 7 æ ö ç ÷ è ø Q6. (a) Solve for x : 2x2 + x – 4 = 0 (By using quadratic formula) [3] (b) In the given figure, XY is diameter of the circle, PQ is a tangent to the circle at Y. If ÐAXB = 50° and ÐABX = 70°, calculate ÐBAY and ÐAPY. [3] P YO B Q C 7 0 ° 50 ° A X (c) Mr. Negi sold a certain number of shares of Rs. 20 paying 8% dividend at Rs. 18 and invested the proceeds in Rs. 10 shares, paying 12% dividend at 50% premium. If the change in his annual income is Rs. 120. Find the number of shares sold by Mr. Negi. [4] Sol. (a) 2x2 + x – 4 = 0 a = 2, b = 1, c = – 4 By using formula, x = 2 4 2 - ± - b b ac a = 2 1 (1) 4 2 4 2 2 - ± - ´ ´ - ´ = 1 1 32 4 - ± + = 1 33 4 - ± = 1 5.74 4 - ± = 1 5.74 1 5.74 , 4 4 - + - - = 4.74 6.74 , 4 4 - =2.37 3.37 , 2 2 - = 1.18, – 1.68 (b) In DAXB, ÐXAB + ÐAXB + ÐABX = 180° [Triangle property] Þ ÐXAB + 50° + 70° = 180° Þ ÐXAB = 180° – 120° = 60° Þ ÐXAY = 90° [Angle of semi-circle] \ ÐBAY = ÐXAY – ÐXAB = 90° – 60° = 30°ICSE Pariksha Mathematics-X 9 and ÐBXY = ÐBAY = 30° [Angle of same segment] \ ÐACX = ÐBXY + ÐABX = 30° + 70° = 100° also, ÐXYP = 90° [\ Diameter ^ tangent] ÐAPY = ÐACX – ÐCYP ÐAPY = 100° – 90° ÐAPY = 10°. (c) Let the number of shares sold be x Annual income = Rs. 8 × ×20 100 æ ö ç ÷ è ø x = Rs. 85x S.P. of x shares = Rs. 18x Number of new shares bought = 18 15x \ Annual income = Rs. 12 18 10 100 15 æ ö ´ ´ ç ÷ è ø x = Rs. 36 25x So, 8 36 5 25 - x x = 120 Þ 40 36 25 - x x = 120 Þ 4x = 120 × 25 Þ x = 120 × 25 4 Þ x = 750 Hence, number of shares sold by Mr. Negi = 750. Q7. (a) The following table shows the age distribution of cases of certain disease admitted during a year in particular hospital. [6] Age (in years) No. of cases 5 – 14 6 15 – 24 11 25 – 34 21 35 – 44 23 45 – 54 14 55 – 64 5 Total 80 Draw a histogram, frequency polygon and hence find the mode from the graph. (b) A man from the top of a vertical tower observe a car moving at a uniform speed coming directly towards him. If it takes 12 minutes to change the angle of depression from 30° to 45°, how soon after it, will the car reach the tower. [4]10 ICSE Pariksha Mathematics-X Sol. (a) Age (in years) No. of cases 4 .5 14.5 24.5 34.5 44.5 54.5 64.5 C.I (age in y ears) 2468 10 24 12 14 16 18 20 22 No. of cases Histogram 4.5–14.5 6 14.5 – 24.5 11 24.5 – 34.5 21 34.5 – 44.5 23 44.5 – 54.5 14 54.5 – 64.5 5 Hence, mode from the graph = 36.3 (b) Let AB be the tower of height h metres, let the speed of the car be x m/min. Distance = Speed × Time So, CD = 12x Let, car takes t minutes to reach the tower AB, then AD = tx In DABD, tan q = AB AD Þ tan 45° = h tx 3 0° 4 5° C D A 1 2x t x B 45 ° 30° h Þ 1 = htx Þ h = tx In DABC, tan q = AB AC Þ tan 30° = 12 + hx tx Þ 13 = (12 ) + tx x t Þ 13 = 12 + t t Þ 3 t = 12 + t Þ 3 t – t = 12 Þ ( ) 3 1 t - = 12 Þ t = 12 12 12 12000 = = = 1.732 1 0.732 732 3 1 - - = 16.39 min Thus, the car will reach the tower in = 16.39 min. Q8. (a) In the given figure, ABCD is a parallelogram and E is a point on BC. The diagonal BD intersect AE at F. Prove that : DF × FE = FB × FA. [3] D A C E F BICSE Pariksha Mathematics-X 11 (b) Construct a DABC, in which BC = 7.5 cm, ÐB = 60°, altitude AD = 3 cm. Construct a circle to touch BC at its mid-point and to pass through A. [4] (c) P, Q and R have co-ordinates (–2, 1), (2, 2) and (6, –2) respectively, write down : [3] (i) The gradient of QR. (ii) The equation of the line passing through P and perpendicular to QR. Sol. (a) Draw GF || AD || BC cutting AB in G. In DABE, GF || BE D A C E F G B \ AG GB = AF FE ...(i) In DBAD, GF || AD \ AG GB = DF FB ...(ii) From equation (i) and (ii), we get AF FE = DF FB Þ AF × FB = FE × DF (b) Draw BC = 7.5 cm. Construct ÐZBC = 60°. Draw LN. parallel to BC at a distance of 3 cm, cutting BZ at A. Draw AD perpendicular to BC. Then AD = 3 cm. Join A to C. Then ABC is the required triangle. Bisect BC at M by drawing the right bisector XY of BC. Draw the right bisector PQ of AM to meet XY at O. With O as centre and OA or OM as radius draw the required circle which will touch BC at its mid-point M and also pass through A. (c) (i) P(–2, 1), Q(2, 2) and R(6, – 2) Gradient of QRm1 = 2 1 2 1 -- y y x x m1 = 2 2 6 2 - -- = 4 4 - = –1 (ii) Slope of the line ^ to QR (m2) = 11 11 m- - =- = 1 So, equation of line ^ to QR and passing through P. y – y1 = m2 (x – x1) Þ y – 1 = 1(x + 2) Þ x – y + 3 = 0 Q9. (a) If X = 4 1 1 2 é ù ê ú -ë û, show that 6X – X2 = 9I, where I is the unit matrix. [3] (b) Prove that : 2 2 2 2 1 sin .sec 1 cos .cosec + q q + q q = tan q [3] (c) Find the number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm. [4] Sol. (a) X= 4 1 , 1 2 é ù ê ú -ë û I = 1 0 0 1 é ù ê ú ë û P D MYX CN Q O Z L B12 ICSE Pariksha Mathematics-X L.H.S., 6X – X2 = 6 4 1 4 1 4 1 1 2 1 2 1 2 é ù é ùé ù - ê ú ê úê ú - - - ë û ë ûë û = 24 6 15 6 6 12 6 3 é ù é ù - ê ú ê ú - - ë û ë û = 9 0 0 9 é ù ê ú ë û = 1 0 9 0 1 é ù ê ú ë û = 9I, R.H.S. (b) L.H.S. 2 2 2 2 1 sin sec 1 cos cosec + q× q + q× q = 2 2 2 21 1 sin .cos1 1 cos .sin + q q + q q = 2 2 2 2 2 2 cos sin sin cos cos sin /q+ q q+ q æ ö æ ö ç ÷ ç ÷ è ø è ø q q = 22 sin cos qq = sin cosqq = tan q = R.H.S. (c) Since each coin will be in cylindrical form. So, r = 0.75 cm and h = 0.2 cm \ Volume of each coin = pr2h = p × (0.75)2 × 0.2 cm3 For the required cylinder, R = 2.25 cm, and H = 10 cm \ Volume of new cylinder = pR2H = p × (2.25)2 × 10 cm3 \ Number of coins = Volume of new cylinder Volume of each coin = 2 2 × (2.25) ×10 × (0.75) × 0.2 p p = 450. Q10. (a) A two digit number is such that the product of the digits is fourteen. If 45 is added to the number then the digits interchange their places. Find the number. [3] (b) The bisectors of ÐB and ÐC of a quadrilateral ABCD, intersect at P. Show that P is equidistant from the opposite sides AB and CD. [4] (c) What point (or points) on the x-axis are at a distance of 5 units from the point (5, – 4). [3] Sol. (a) Let the unit's digit be = x Then, ten's digit according the question be = 14 x the original number = 14 10 x x ´ + = 140 x x + The number obtained by interchanging the places of digit = 14 10 x x ´ + = 14 10x x +ICSE Pariksha Mathematics-X 13 Now, according the question 140 45 x x + + = 14 10x x + Þ 2 140 45 x x x + + = 2 10 14 x x+ Þ 140 + x2 + 45x = 10x2 + 14 Þ 9x2 – 45x – 126 = 0 Þ x2 – 5x – 14 = 0 Þ x2 – 7x + 2x – 14 = 0 x(x – 7) + 2(x – 7) = 0 Þ (x – 7) (x + 2) = 0 Þ So, x – 7 = 0 or x + 2 = 0 x = 7 or x = – 2 (neglected) Hence, original number = 140 x x + = 140 7 7 + = 20 + 7 = 27 (b) Draw PS ^ BC P lies on the bisector of ÐB PM ^ AB [By construction] and PS ^ BC So, PM = PS ...(i) B S C D A N P M Now, P lies on the bisector of ÐC PS ^ BC and PN ^ CD So, PN = PS ...(ii) From equation (i) and (ii), we get PM = PN (c) Let (k, 0) be any point on x-axis, then according the question, 2 2 (5 ) ( 4 0) - +- - k = 5 Þ 25 + k2 – 10k + 16 = 25 Þ k2 – 10k + 16 = 0 Þ (k – 8) (k – 2) = 0 Þ k – 8 = 0 or k – 2 = 0 Þ k = 8 or k = 2 \ The required points are (8, 0) and (2, 0). Q11. (a) Compute mean for the following data. (By Step Deviation Method). Marks No. of students less than 10 12 less than 20 19 less than 30 35 less than 40 47 less than 50 58 less than 60 65 less than 70 84 less than 80 100 [4]14 ICSE Pariksha Mathematics-X (b) Draw an ogive from the following frequency table. Class interval Frequency 8 – 12 10 12 – 16 16 16 – 20 22 20 – 24 18 24 – 28 12 28 – 32 4 Total 82 from this ogive, calculate : (i) The median. (ii) The upper and lower quartile. (iii) The inter quartile range. [6] Sol. (a) Marks c.f. f x d = x – A u = d/h fu 0 – 10 12 12 5 – 30 – 3 – 36 10 – 20 19 7 15 – 20 – 2 – 14 20 – 30 35 16 25 – 10 – 1 – 16 30 – 40 47 12 35 0 0 0 40 – 50 58 11 45 10 1 11 50 – 60 65 7 55 20 2 14 60 – 70 84 19 65 30 3 57 70 – 80 100 16 75 40 4 64 Sf = 100 Sfu = 80 Here, A = 35, h = 10 Mean = A + × SSfu h f = 80 35 ×10 100 + = 35 + 8 = 43ICSE Pariksha Mathematics-X 15 (b) Class interval frequency (f) c.f. 8 – 12 10 10 12 – 16 16 26 16 – 20 22 48 20 – 24 18 66 24 – 28 12 78 28 – 32 4 82 N = 82 (i) N = 82 (even) The position of median is given by = N 82 41 2 2 = = 0 8 12 15 16 20 24 28 32 36 10 20 30 40 50 60 70 80 90 Y X (1 2, 1 0) (1 6, 2 6) (2 0, 4 8) 23 19 (2 4, 6 6) (3 2, 8 2) (2 8, 7 8) Cumulative Frequency Cla ss-In terv a l So, from the graph, median = 19 (ii) The position of Upper Quartile is given by = 3N 3× 82 246 = = 4 4 4 = 61.5 So, from the graph, Upper Quartile = 23 The position of lower Quartile is given by = N 82 20.5 4 4 = = So, from the graph, lower Quartile = 15 (iii) Inter quartile range = Q3 – Q1 = 23 – 15 = 8