Mathematics Question Paper 2000 Section – A Question1) For what value of k will the following system of linear equations have an infinite number of solutions : 2x + 3y = 2; (k + 2)x + (2k + 1)y = 2(k – 1) ? (Marks 2)Question 2) Find whether the number 6, 10, 14 and 22 are in proportional or not. If not what number be added to each number so that they become proportional ? (Marks 2)Question 3) Reduce the following relation expression into lowest form: (x4 – 10×2 + 9)/(x3 + 42 + 3x) (Marks 2)Question 4) Solve for x and y ax + by = a – b – (i) bx – ay = a + b – (ii) (Marks 2)Question 5) Find the value of k such that sum of the roots of the quadratic equation 3×2 + (2k + 1)x – (k + 5) = 0 is equal to the product of its roots. (Marks 2)Question 6) Find two consecutive numbers, whose square have sum 85.Question 7) Without using trignometric table, show that: tan 7o . tan 23o . tan 60o . tan 67o . tan 83o = Ö3 (Marks 2)Question 8) Show that (sinq – 2sin3q )/(2cos3q – cosq ) = tanq . (Marks 2)Question 9) In the given figure chord PQ and RS of a circle intersect at T. If RS = 18 cm, ST = 6 cm and PT = 18 cm, find the length of TQ. (Marks 2)Question 10) In the given figure DE || BC and AD : DB = 5 : 4 Find ar(DFE)/ar(CFB). (Marks 2)Question 11) In the figure BAC = 30o. Show that BC is equal to the radius of the circum-circle of ABC whose centre is O. (Marks 2)Question 12) Rita purchased a car, with a market price of Rs. 2,10,000 at a discount of 5%. If the sales tax charged at 10%, find the amount Rita had to pay for purchasing the car. (Marks 2)Question 13) The mean weight of 21 students of a class is 52 kg. If the mean weight of first 11 students of the class is 50 kg and that of last 11 students is 54 kg, find the weight of the 11th student. (Marks 2)Question 14) The following data has been arranged in ascending order: 12, 14, 17, 21, x, 26, 28, 32, 36. If the median of the data is 23, find x. If 32 is changed to 23, find the new median. (Marks 2)Question 15) For what value of x, in the mode of the following data 5 ? 2, 4, 3, 5, 4, 5, 6, 4, x, 7, 5. (Marks 2) no Answer (sorry) Section – B Question 16) Determine graphically the co-ordinates of the vertices of the triangle, the equation of whose sides are: y = x, 3y = x, x + y = 8. (Marks 4) Sorry, no Answer because I do not have graphical tool to display the solution.Question 17) A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs. 1000 as hostel charges where as a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day. (Marks 4)Question 18) Find the value of a and b so that the polynomials p(x) and q(x) have (x + 1)(x – 2) as their HCF. p(x) = (x2 + 3x + 2)(x2 + x + a) q(x) = (x2 – 3x + 2)(x2 – 3x + b) (Marks 4)Question 19) A page of pass book of Ved is given below :- Date Particulars Amt. withdrawn Rs. Amt. deposited Rs. Balance 8/3/98 B/F – - – - – - – - 4500 12/3/98 To cheque 600 – - – 3900 18/4/98 By cheque 1600- – - 5500 26/4/98 By cash – 3500- – - 9000 12/8/98 By cash – 500 – - – 9500 16/10/98 To cheque 4500- – - 5000 12/11/98 By cheque – 1650- – 6650 3/12/98 By cash – 1350- – - 8000 Find the interest Ved gets for the period March, 98 to Dec.’ 98 at 5% per annum simple interest. (Marks 4) click for AnswerQuestion 20) The annual income of Seema (excluding HRA) is Rs. 1,60,000. She contributes Rs. 5000 per month to her provident fund and pays a half yearly insurance premium of Rs. 5000. Calculate the income tax along with surcharge Seema has to pay in the last month of the year if her earlier deductions as income tax for the first 11 months were at the rate of Rs. 400 per month. (Marks 4)Assume the following for calculating income tax : (a) Standard Deduction 1/3 of the total income subject to a maximum of Rs. 20,000 (Rs. 25,000 if income is less than Rs. 1 lac) (b) Rates of Income tax Slab (i) Upto Rs. 50,000 (ii) From Rs. 50,001 to 60,000 (iii) From Rs. 60,001 to Rs. 1,50,000 (iv) From Rs. 1,50,001 onwards Income Tax Nil 10% of the amount exceeds Rs. 50,000 Rs. 1000 + 20% of the amount exceeding Rs. 60,000 Rs. 19,000 + 30% of the amount exceeding Rs. 1,50,000 (c) Rebate in Tax 20% of the total savings subject to a maximum of Rs. 12,000 (d) Surcharge 10% of the tax payable ANSWERS Answer 1) For what value of k will the following system of linear equations have an infinite number of solutions : 2x + 3y = 2; (k + 2)x + (2k + 1)y = 2(k – 1) ? (Marks 2) 2x + 3y = 2 ……………………… (i) (k + 2)x + (2k + 1)y = 2(k – 1) ……. (ii) (ii) cab represented as follow: ((k + 2)/(k – 1))x + ((2k + 1)/(k – 1))y = 2 ……. (iii) (i) and (iii) can have infinite number of solution if ==> ((k + 2)/(k – 1))/2 = ((2k + 1)/(k – 1))/3 ==> (k + 2)/(2(k – 1)) = (2k + 1)/(3(k – 1)) ==> 3(k + 2) = 2(2k + 1) ==> 3k + 6 = 4k + 2 ==> k = 4Answer 2) Find whether the number 6, 10, 14 and 22 are in proportional or not. If not what number be added to each number so that they become proportional ?6, 10, 14 and 22 are in proportional ==> if 6/10 is equal to 14/22 ==> 6 x 22 = 132 ==> 10 x 14 = 140 ==> As 132 and 140 are not equal so 6, 10, 14 and 22 are NOT in proportional Assume that if number “n” is added to each number, then 6, 10, 14 and 22 become proportional. ==> (6 + n)/(10 + n) = (14 + n)/(22 + n) ==> (6 + n)(22 + n) = (14 + n)(10 + n) ==> n2 + 28n + 132 = n2 + 24n + 140 ==> 4n = 8 ==> n = 2 ==> So number 2 can be added to each number so that they become proportional.Answer 3) Reduce the following relation expression into lowest form: (x4 – 10×2 + 9)/(x3 + 4×2 + 3x) ==> (x4 – 9×2 – x2 + 9)/(x3 + 3×2 + x2 + 3x) ==> ((x4 – 9×2 – x2 + 9)) / ((x3 + 3×2 + x2 + 3x)) ==> (x2(x2 – 9) – (x2 – 9)) / ( x(x2 + 3x) + (x2 + 3x) ) ==> (x2(x2 – 9) – (x2 – 9)) / ( x(x2 + 3x) + (x2 + 3x) ) ( (x2 – 1) (x2 – 9) ) / ( (x + 1)(x2 + 3x) ) ==> ( (x + 1)(x – 1)(x + 3)(x – 3) ) / ( x(x + 1)(x + 3) ) ==> ( (x – 1)(x – 3) ) / (x) ==> (x2 – 4x + 3) / (x) Answer 4) Solve for x and y ax + by = a – b – (i) bx – ay = a + b – (ii) Multiply (i) by b and multiply (ii) by a ==> abx + b2y = ab – b2 …………(iii) ==> abx – a2y = a2 + ab …………(iv) ==> Subtract (iii) and (iv) ==> b2y + a2y = – b2 – a2 ==> y(b2 + a2) = -(b2 + a2) ==> y = -1 ==> Substitue the vale of y in (i) ==> ax – b = a – b ==> ax = a ==> x = 1Answer 5) Find the value of k such that sum of the roots of the quadratic equation 3×2 + (2k + 1)x – (k + 5) = 0 is equal to the product of its roots. (Marks 2)Sum of the roots = -(2k + 1)/3 Product of the roots = -(k + 5)/3 ==> So -(2k + 1)/3 = -(k + 5)/3 => -2k – 1 = -k – 5 => k = 4Answer 6) Find two consecutive numbers, whose square have sum 85.Assume that the two consecutive numbers are n and n+1 ==> n2 + (n+1)2 = 85 ==> n2 + n2 + 2n + 1 = 85 ==> 2n2 + 2n -84 = 0 ==> n2 + n – 42 = 0 ==> n2 + 7n – 6n – 42 = 0 ==> n(n + 7) -6(n + 7) = 0 ==> (n – 6)(n + 7) = 0 ==> n = 6 or n = -7 ==> So the two consecutive numbers are 6,7 OR -7, -6Answer 7) Without using trignometric table, show that: tan 7o . tan 23o . tan 60o . tan 67o . tan 83o = Ö3(Left hand side) (LHS) = tan 7o . tan 23o . tan 60o . tan 67o . tan 83o (Right hand side) (RHS) = 3 LHS = tan 7o . tan 23o . tan 60o . tan 67o . tan 83o ==> tan(90o – 83o) . tan(90o – 67o) . tan 60o . tan 67o . tan 83o ==> cot83o . cot67o . tan 60o . tan 67o . tan 83o ==> 1/(tan83o) . 1/(tan67o) . tan 60o . tan 67o . tan 83o ==> tan 60o ==> Ö3 ==> Therefore LHS = RHS (proved)Answer 8) Show that (sinq – 2sin3q )/(2cos3q – cosq ) = tanq . (Marks 2)L.H.S = (sinq – 2sin3q )/(2cos3q – cosq ) R.H.S = tanq ==> L.H.S = (sinq – 2sin3q )/(2cos3q – cosq ) ==> L.H.S = ((sinq)(1 – 2sin2q )/((cosq)(2cos2q – 1 ) ==> L.H.S = ((sinq)(1 – sin2q – sin2q)/((cosq)(cos2q + cos2q – 1 ) ==> L.H.S = ((sinq)(cos2q – sin2q)/((cosq)(cos2q – sin2q ) ==> L.H.S = ((sinq)(cos2q – sin2q)/((cosq)(cos2q – sin2q ) ==> L.H.S = (sinq)/(cosq) ==> L.H.S = tanq = R.H.SAnswer 12) Rita purchased a car, with a market price of Rs. 2,10,000 at a discount of 5%. If the sales tax charged at 10%, find the amount Rita had to pay for purchasing the car. (Marks 2)Assume that Rita had to pay amount y for purchasing the car Market price of the car = 210000 Discount (@5%) = (5/100)x210000 = Rs 10500 Price of car before sales tax = 210000 – 10500 = Rs 199500 Sales tax @10%) = (10/100)x199500 = Rs 19950 Price of car after sales tax = 199500 + 19950 = Rs 219450Answer 13) The mean weight of 21 students of a class is 52 kg. If the mean weight of first 11 students of the class is 50 kg and that of last 11 students is 54 kg, find the weight of the 11th student.Assume the weight of 11th student is n. Total weight of first 11 student = 11 x 50 = 550 Total weight of first 11 student = 11 x 54 = 594 Total weight of first 21 student = 21 x 52 = 1092 ==> 550 + 594 = 1092 + n ==> n = 52 ==> So weight of the 11th student is 52Answer 14) The following data has been arranged in ascending order: 12, 14, 17, 21, x, 26, 28, 32, 36. If the median of the data is 23, find x. If 32 is changed to 23, find the new median. (Marks 2)Number of observations n = 10 Median = (5th observation + 6th observation)/2 ==> 23 = (22 + x)/2 ==> x = 24 If 32 is changed to 23 then median = (22 + 23)/2 ==> median = 45/2 = 22.5Answer 17) A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs. 1000 as hostel charges where as a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day. (Marks 4) Assume the fixed charges of hostel is = p Assume cost of food per day is = v So charges for student A = p + 20v = 1000 ……………..(i) Charges for student B = p + 26v = 1180 ……………..(i) Subtract (ii) – (i) 6v = 180 ==> v = 30 Substitute the value of v in equation (i) ==> p + 600 = 1000 ==> p = 400 ==> So Fixed charge = p = Rs 400.00 , and cost of food per day = v = Rs 30.00Answer 18) Find the value of a and b so that the polynomials p(x) and q(x) have (x + 1)(x – 2) as their HCF. p(x) = (x2 + 3x + 2)(x2 + x + a) q(x) = (x2 – 3x + 2)(x2 – 3x + b) (Marks 4)p(x) = (x2 + 3x + 2)(x2 + x + a) q(x) = (x2 – 3x + 2)(x2 – 3x + b) ==> x – 2 is a factor of x2 + x + a => 4 + 2 + a = 0 ==> a = -6 x + 1 is a factor of x2 – 3x + b => 1 + 3 + b = 0 ==> b = -4 ==> a = -6, b = -4Question 19) A page of pass book of Ved is given below :- Date Particulars Amt. withdrawn Rs. Amt. deposited Rs. Balance 8/3/98 B/F - - 4500 12/3/98 To cheque 600 - 3900 18/4/98 By cheque - 1600 5500 26/4/98 By cash - 3500 9000 12/8/98 By cash - 500 9500 16/10/98 To cheque 4500 - 5000 12/11/98 By cheque - 1650 6650 3/12/98 By cash - 1350 8000 Find the interest Ved gets for the period March, 98 to Dec.’ 98 at 5% per annum simple interest. (Marks 4)Answer 19) Month March April May June July Aug Sept Oct Nov Dec Principal (Rs.) 3900 3900 9000 9000 9000 9000 9500 5000 5000 8000 Total Principal = Rs. 71,300 Interest = Rs. (71,300 x 5 x 1)/(100 x 12) = Rs. 297.08Answer 20 ) Annual Income = Rs. 1,60,000, Standard Deduction = Rs. 20,000 Taxable Income = Rs. 1,60,000 – Rs. 20,000 = Rs. 1,40,000 Income Tax = Rs. 1,000 + Rs. 16000 = Rs. 17,000 Total Saving = 5000 x 12 + 5000 x 2 = 60,000 + 10,000 = Rs. 70,000 Rebate = Rs. 12,000 Income Tax payable = Rs. 17,000 – Rs. 12,000 = Rs. 5000 Total Income tax, including surcharge = 5000 + 10/100 x 5000 = Rs. 5500 Advance tax paid = 11 x 400 = Rs. 4400 Income Tax payable in the last month = Rs. 1100Read more at:http://www.icbse.com/papers/cbse-class-10-mathematics-solved-question-paper-2000

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