Part - A Section - I

1 Mathematics Standard X Sample paper 06 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 -06 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.

2 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. If HCF (,)ab12= and ,ab1800 =, then find LCM (,) : We know that (,)(,)HCFLCM abab# ab=Substituting the values we have (,)LCMab12 1800=or, (,)LCMab ,121800150==2. If one of the zeroes of the quadratic polynomial kxkx112 ++^h is 3-, then what is the value of k ?

3 Ans : If a is zero of quadratic polynomial fx^h, then fa^h 0=So, f3-^h kk13312= + +^^^hhh 0 kk1931= +^^hh 0 kk9931= + 0 k68= k 68= 34=3. Calculate the zeroes of the polynomial .pxxx41292= +^hAns : [Board Term-1 2010] ()px xx41292= + xxx46692= + ()()xxx223323= ()()xx2323= Substituting ()px0=, and solving we get,x2323= ,x2323=Hence, zeroes of the polynomial are 23, Find the positive root of x3692+=.

4 Ans : [Board Term-2, 2015]We have x362+ 9= x362+ 81= x32 81675= = x2 37525==Thus x 5!=Hence 5 is positive If x21= , is a solution of the quadratic equation xkx32302+ =, find the value of : [Board Term-2, Delhi 2015]We have xkx3232+ 0=Substituting x21= in given equation we get k32122132 + bbll 0= k433-- 0= k 433= 4312= 49= Hence k 49= 5. Find the tenth term of the sequence ,,2818, ..Ans : [Board Term-2 2016]Let the first term of an AP be a and common difference be.

5 DGiven AP is ,,2818 or ,,..22232where, a ,,dn2210===Now an and1=+ ^h a10 21012=+ ^hDownload 20 Solved Sample Papers PDF from Page 2 Mathematics Standard X Sample paper 06 Solved 292=+ 102=Therefore tenth term of the given sequence Find the next term of the series ,,28,1832 ..Ans : [Board Term-2 2012]Let the first term of an AP be a and common difference.

6 DHere, a ,2= ad+ 822== d 2222= = Next term 322=+ 422=+ 52= 50=6. Are two triangles with equal corresponding sides always similar?Ans : [Board Term-1 2015]Yes, Two triangles having equal corresponding sides are are congruent and all congruent sT have equal angles, hence they are similar Find the distance of a point (,)Pxy from the : [Board 2018]Distance between origin (0, 0) and point (,)Pxy is d ()()xxyy212212= + ()()xy0022= + xy22=+Distance between P and origin is xy22+.

7 8. Find the value of ,a for which point ,P2a3^h is the midpoint of the line segment joining the Points ,Q54-^h and ,.R10-^hAns : [Board Term-2 SQP 2016]As per question, line diagram is shown P is mid-point of QR, we have a3 251= + ^h 263= = Thus a 9= 9. In a triangle ,ABC write cosBC2+bl in terms of angle .AAns : [Board Term-1 2016]In a triangle ABC++ 180 = BC+ A180 = Thus cosBC2+bl cosA2180 = :D cosA902= :D sinA2=10.

8 What happens to value of cos when increases from 0 to 90 .Ans : [Board Term-1 2015]cos decreases from 1 to .11. Find the value of sinsin4149 22+Ans : [Board Term-1 2012, NCERT]We have sinsin414922+ ()sinsin904949 22= + cossin4949 22=+ 1=12. In the figure, PA and PB are tangents to a circle with centre O. If AOB120c+=, then find OPA+.Ans : [Board Term-2 Delhi 2012, 2014]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=Due to symmetry we have AOP+ AOB2212060cc+===Now in right AOPT we have APOOAPAOP+++++ 180c= APO9060cc+++ 180c= APO+ 18015030ccc= =or PQ is a tangent drawn from an external point P to a circle with centre O, QOR is the diameter of the circle.

9 If POR120c+=, What is the measure of OPQ+ ?Ans : [Board Term-2 Foreign 2017]Since PQ is a tangent to the circle, OQP is right angle triangleIn OQPT because of exterior angle, POR+ OQPOPQ++=+Mathematics Standard X Sample paper 06 Solved To Get 20 Solved paper Free PDF by whatsapp add +91 89056 29969/9530143210/9461443210 Page 3 Thus OPQ+ POROQP++= 12090cc= 30c=13. The radius of a circle is cm. find the area of the sector of the circle enclosed by two radii and an arc 44 cm in : [Board 2020 OD Basic]Given, arc length 44= cmRadius of circle, r.

10 175= cmSo, area of sector arclengthrr22# = arclengthr2#= .244175#= .22175#= 385= sq. Find the area of the sector of a circle of radius 6 cm whose central angle is 30c. (Take .314 =)Ans : [Board 2020 OD Standard]Radius, r 6= cmCentral angle, 30c=Area of the sector, r3602c .3603146630###cc= .cm9422=14. If the radius of the base of a right circular cylinder is halved, keeping the height same, find the ratio of the volume of the reduced cylinder to that of original : [Board Term-2 2012] VolumeoforiginalcylinderVolumeofreducedc ylinder rhh41r222# ==^h 14|=15.