Part - A Section - I

1 Mathematics Standard X Sample Paper 04 Solved To Get 20 Solved Paper Free PDF by whatsapp add +91 89056 29969/9530143210/9461443210 Page 1 CLASS X (2020-21)MATHEMATICS STANDARD (041)SAMPLE PAPER-04 Time : 3 Hours Maximum Marks : 80 General Instructions :1. This question paper contains two parts A and Both Part A and Part B have internal A :1. It consists of two sections- I and Section I has 16 questions. Internal choice is provided in 5 Section II has four case study-based questions. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 B :1. Question no. 21 to 26 are very short answer type questions of 2 mark Question no. 27 to 33 are short answer type questions of 3 marks Question no. 34 to 36 are long answer type questions of 5 marks Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 - ASection - I1.

2 A and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then calculate the least prime factor of ()ab+.Ans : [Board Term-1 2014]Here a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. The least prime factor of ()ab+ would be What is the HCF of the smallest composite number and the smallest prime number?Ans : [Board Term-1 OD 2018]The smallest prime number is 2 and the smallest composite number is 422=.Hence, required HCF is (,)2222=.2. The 2 digit number which becomes 65th of itself when its digits are reversed. If the difference in the digits of the number being 1, what is the two digits number?Ans : If the two digits are x and y, then the number is xy10+.Now ()xy6510+ yx10=+Solving, we get xy4455+ yx 45=Also xy1 =.

3 Solving them, we get x5= and .y4= Therefore, number is What are the values of k for which the quadratic equation xkxk202 += has equal roots?Ans : We have xkxk22 + 0=Comparing with axbxc2++ 0= we a 2=, bk= and ck=.For equal roots, the discriminant must be bac42- 0= kk422--^^hh 0= kk82- 0= kk8-^h 0= & k ,08=Hence, the required values of k are 0 and 8. or If one root of the quadratic equation axbxc02++= is the reciprocal of the other, then show that ac=.Ans : If one root is , then the other 1 .Product of roots, 1$aa ac= 1 ac= & a c=4. Show that ()ab2-, ()ab22+ and ()ab2+ are in : [Board 2020 Delhi Standard]Given, (),()abab222 + and ()ab2+.Common difference, d1 ()()abab222=+ ()()ababab22222=+ + ababab22222=+ + ab2=and d2 ()()abab222=+ + ababab22222=++ ab2=Since, d1 d2=, thus, ()ab2-, ()ab22+ and ()ab2+ are in Find the sum of all 11 terms of an AP whose middle term is Standard X Sample Paper 04 Solved Download 20 Solved Sample Papers PDF from Page 2 Ans : [Board 2020 OD Standard]In an AP with 11 terms, the middle term is 62111th=+ , a6 ad5=+ 30=Thus, S11 []ad211210=+ ()ad115=+ 1130#= 330=5.

4 In ABCT, cmAB63=, cmAC12= and cmBC6=, then B+= ..Ans : [Board 2020 OD Standard]We have AB 63= cm, AC 12= cm and BC 6= cmNow AB2 363108#== AC2 144=and BC2 36=In can be easily observed that above values satisfy Pythagoras theorem, ABBC22+ AC2= 10836+ cm144=Thus B+ 90c=6. In given figure ||.DEBC If ,ADcDBc34== cm and AE6= cm then find .ECAns : [Board Term-1 2016]In the given figure DE ||BC, thus BDAD ECAE= 43 EC6= EC 8= cm7. The value of the ()tansin604522cc+ is ..Ans : [Board 2020 OD Basic] tansin604522cc+ ()32122=+cm 321=+ 27=8. Evaluate sintancos602453022--cccAns : [Board 2019 OD]()sintancos60245302321234324322222ccc = = = ccmm 9. In the given figure, AB is a 6 m high pole and DC is a ladder inclined at an angle of 60c to the horizontal and reaches up to point D of pole.

5 If .AD254= m, find the length of ladder. ( use .3173=)Ans : [Board Term-2 Delhi 2016]We have AD .254= m DB ..6254346= = mIn BCDT, B+ 90c= sin60c DCBD= 23 .DC346= DC ..334621733464#===Thus length of ladder is 4 A ladder, leaning against a wall, makes an angle of 60c with the horizontal. If the foot of the ladder is m away from the wall, find the length of the : [Board Term-2 2011]As per given in question we have drawn figure ACBT with C60c+=, we get cos60c .AC25= 21 .AC25=Mathematics Standard X Sample Paper 04 Solved To Get 20 Solved Paper Free PDF by whatsapp add +91 89056 29969/9530143210/9461443210 Page 3 AC .2255#== m10. In figure, PA and PB are tangents to the circle with centre O such that .APB50+= Write the measure of .OAB+Ans : [[Board Term-2 Delhi 2015]We have APB+ 50 = PAB+ PBA21805065 +== =Here OA is radius and AP is tangent at A, since radius is always perpendicular to tangent at point of contact, we have OAP+ 90=Now OAB+ OAPPAB++= 906525 = =11.]

6 In the given figure, PQ and PR are tangents to the circle with centre O such that ,QPR50 += Then find .OQR+Ans : [Board Term-2 Delhi 2012, 2015]We have QPR+ 50 += (Given)Since QOR+ and QPR+ are supplementary angles QORQPR+++ 180 = QOR+ QPR180+= 18050130 = =From OQRT we have OQR+ ORQ2180130 +== 25025 ==12. If the circumference of a circle increases from 4 to 8 , then what about its area ?Ans : [Board Term-2 Delhi 2013]Circumference of the circle r2 4 = cm or r2= circumference R2 8 = cm or R4= of the 1st circle r2 242# ==^h cmArea of the new circle R2 416442# ===^hArea of the new circle 4= times the area of first If the radius of the circle is 6 cm and the length of an arc 12 cm.

7 Find the area of the : [Board Term-2 2014]Area of the sector 21#=(length of the correspondingarc)# radius lr2121126####== 36= cm213. A rectangular sheet paper 40 cm 22# cm is rolled to form a hollow cylinder of height 40 cm. Find the radius of the : [Board Term-2 Foreign 2014]Here, h40= cm, circumference 22= cm r2 22= r 222227##= 27= .35= cm14. The radius of sphere is r cm. It is divided into two equal parts. Find the whole surface of two : [Board Term-2 2012]Whole surface of each part rrr23222 =+=Total surface of two parts rr23622# ==15. Consider the following distribution :Marks Obtained0 or more10 or more20 or more30 or more40 or more50 or moreNumber of students635855514842(i) Calculate the frequency of the class 30 - 40.

8 (ii) Calculate the class mark of the class 10 - : [Board Term-1, 2014]Class (i) Frequency of the class 30 - 40 is Standard X Sample Paper 04 Solved Download 20 Solved Sample Papers PDF from Page 4(ii) Class mark of the class : 1025- 21025=+ .235175==16. For finding the popular size of readymade garments, which central tendency is used?Ans : For finding the popular size of ready made garments, mode is the best measure of central IICase study-based questions are compulsory. Attempt any 4 sub parts from each question. Each question carries 1 Lavanya throws a ball upwards, from a rooftop, which is 20 m above from ground. It will reach a maximum height and then fall back to the ground.

9 The height of the ball from the ground at time t is h, which is given by htt416202= ++.(i) What is the height reached by the ball after 1 second?(a) 64 m (b) 128 m(c) 32 m (d) 20 m(ii) What is the maximum height reached by the ball?(a) 54 m (b) 44 m(c) 36 m (d) 18 m(iii) How long will the ball take to hit the ground?(a) 4 seconds (b) 3 seconds(c) 5 seconds (d) 6 seconds(iv) What are the two possible times to reach the ball at the same height of 32 m?(a) 1 and 3 seconds (b) 1 and 4 seconds(c) 1 and 2 seconds (d) 1 and 5 seconds(v) Where is the ball after 5 seconds ?(a) at the ground (b) rebounds(c) at highest point (d) fall backAns : (i) Height is given by, h tt416202= ++At t 1= second, h ()()41161202= ++ 32= mThus (c) is correct option.(ii) Rearranging the given equation, by completing the square, h ()tt4452= ()tt444452= + [()]t4292= ()t42362= +Height is maximum, at t 2=, thus hmax 03636=+= mThus (c) is correct option.

10 (iii) When ball hits the ground, h 0=, thus tt416202 ++ 0= tt452-- 0= ()()tt51 + 0=Thus t 5= or t1= . Since, time cannot be negative, the t 5= seconds is correct (c) is correct option.(iv) Since, h tt4162022= ++ 32 tt4162022= ++ 8 tt4522= ++ tt432 + 0= ttt332+ + 0= ()()tt13-- 0= & t ,31=Thus (a) is correct option.(v) From (iii) at t5= we have h0=. Thus it will hit ground, then after that ball will (b) is correct To conduct sports day activities, in a rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AB, as shown in figure.