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CBSE Class 10 Board Exam 2026 : Mathematics

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Govindu Srinivas
The Institution of Engineers (India) IEI, Kolkata

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SAMPLE QUESTION PAPER MARKING SCHEME SUBJECT: MATHEMATICS- STANDARD CLASS X SECTION - A 1 (c) 35 1 2 (b) x2 (p+1)x +p=0 1 3 (b) 2/3 1 4 (d) 2 1 5 (c) (2,-1) 1 6 (d) 2:3 1 7 (b) tan 30 1 8 (b) 2 1 9 (c) x= + 1 10 (c) 8cm 1 11 (d) 3 3cm 1 12 (d) 9 cm2 1 13 (c) 96 cm2 1 14 (b) 12 1 15 (d) 7000 1 16 (b) 25 1 17 (c) 11/36 1 18 (a) 1/3 1 19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) 1 20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) 1 1 SECTION B 21 Adding the two equations and dividing by 10, we get : x+y = 10 Subtracting the two equations and dividing by -2, we get : x-y =1 Solving these two new equations, we get, x = 11/2 y = 9/2 22 23 In ABC, 1 = 2 AB = BD (i) Given, AD/AE = AC/BD Using equation (i), we get AD/AE = AC/AB .(ii) In BAE and CAD, by equation (ii), AC/AB = AD/AE A= A (common) BAE ~ CAD [By SAS similarity criterion] PAO = PBO = 90 ( angle b/w radius and tangent) AOB = 105 (By angle sum property of a triangle) AQB = x105 = 52.5 (Angle at the remaining part of the circle is half the 1 angle subtended by the arc at the centre) 24 We know that, in 60 minutes, the tip of minute hand moves 360 In 1 minute, it will move =360 /60 = 6 From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 6 = 210 Area of swept by the minute hand in 35 min = Area of sector with sectorial angle of 210 and radius of 6 cm = 360x x 62 210 = 7 12 x 22 7 x6x6 =66cm2 OR Let the measure of A, B, C and D be , , and respectively Required area = Area of sector with centre A + Area of sector with centre B + Area of sector with centre C + Area of sector with centre D 2 = = 360 x x 72 + ( + + + ) 360 ( ) 360 x x 72 + 360 x x 72 + 360 x x 72 x x 72 = x x 7x 7 ( By angle sum property of a triangle) 360 7 2 = 154 cm 25 sin(A+B) =1 = sin 90, so A+B = 90 .(i) cos(A-B)= 3/2 = cos 30, so A-B= 30 (ii) From (i) & (ii) A = 60 And B = 30 OR cos sin 1 3 = 1+ 3 Dividing the numerator and denominator of LHS by cos , we get cos +sin 1 tan 1 3 = 1+ 3 Which on simplification (or comparison) gives tan = 3 Or = 60 1+tan 26 SECTION - C Let us assume 5 + 2 3 is rational, then it must be in the form of p/q where p and q are co-prime integers and q 0 i.e 5 + 2 3 = p/q So 3 = 5 2 1 (i) Since p, q, 5 and 2 are integers and q 0, HS of equation (i) is rational. But LHS of (i) is 3 which is irrational. This is not possible. This contradiction has arisen due to our wrong assumption that 5 + 2 3 is rational. So, 5 + 2 3 is irrational. 27 Let and be the zeros of the polynomial 2x2 -5x -3 Then + = 5/2 And = -3/2. Let 2 and 2 be the zeros x2 + px +q Then 2 + 2 = -p 2( + ) = -p 2 x 5/2 =-p So p = -5 And 2 x 2 = q 4 = q So q = 4 x-3/2 = -6 3 28 Let the actual speed of the train be x km/hr and let the actual time taken be y hours. Distance covered is xy km If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x+6)km/hr, time of journey is (y 4) hours. Distance covered =(x+6)(y 4) xy=(x+6)(y 4) 4x+6y 24=0 2x+3y 12=0 .(i) Similarly xy=(x 6)(y+6) 6x 6y 36=0 x y 6=0 (ii) Solving (i) and (ii) we get x=30 and y=24 1 Putting the values of x and y in equation (i), we obtain Distance =(30 24)km =720km. Hence, the length of the journey is 720km. OR Let the number of chocolates in lot A be x And let the number of chocolates in lot B be y total number of chocolates =x+y Price of 1 chocolate = 2/3 , so for x chocolates = x and price of y chocolates at the rate of 1 per chocolate =y. by the given condition x +y=400 2x+3y=1200 ..............(i) Similarly x+ y = 460 5x+4y=2300 ........ (ii) Solving (i) and (ii) we get x=300 and y=200 29 x+y=300+200=500 1 So, Anuj had 500 chocolates. sin3 / cos3 1+ sin2 /cos2 LHS : + cos3 / sin3 1+ cos2 / sin2 4 = sin3 / cos3 + (cos2 + sin2 )/cos2 cos3 / sin3 (sin2 + cos2 )/ sin2 = sin3 + cos3 cos sin = sin4 + cos4 cos sin = (sin2 + cos2 )2 2 sin2 cos2 cos sin = 1 - 2 sin2 cos2 cos sin = 1 - 2 sin2 cos2 cos sin cos sin = sec cosec 2sin cos = RHS 30 Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively. We know that the tangents drawn to a circle from an exterior point are equal in length. AP = AS .(1) BP = BQ (2) CR = CQ ...(3) DR = DS (4). Adding (1), (2), (3) and (4) we get AP+BP+CR+DR = AS+BQ+CQ+DS (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) AB+CD=AD+BC-----------(5) Since AB=DC and AD=BC (opposite sides of parallelogram ABCD) putting in (5) we get, 2AB=2AD or AB = AD. AB=BC=DC=AD Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a rhombus OR 5 1 1 Join OC In OPA and OCA OP = OC (radii of same circle) PA = CA (length of two tangents from an external point) 1 AO = AO (Common) Therefore, OPA OCA (By SSS congruency criterion) Hence, 1 = 2 (CPCT) Similarly 3 = 4 PAB + QBA =180 (co interior angles are supplementary as XY X Y ) 2 2 + 2 4 = 180 2 + 4 = 90 -------------------------(1) 2 + 4 + AOB = 180 (Angle sum property) Using (1), we get, AOB = 90 31 3 1 1 (i) P (At least one head) = (ii) P(At most one tail) = 4 (iii) P(A head and a tail) = 4 = 2 3 2 4 1 1 SECTION D 32 Let the time taken by larger pipe alone to fill the tank= x hours Therefore, the time taken by the smaller pipe = x+10 hours 4 Water filled by larger pipe running for 4 hours = litres 9 Water filled by smaller pipe running for 9 hours = +10 litres 6 We know that 4 9 1 + +10 = 2 Which on simplification gives: x2 16x 80=0 x2 20x + 4x 80=0 x(x-20) + 4(x-20)= 0 (x +4)(x-20)= 0 x=- 4, 20 x cannot be negative. Thus, x=20 x+10= 30 Larger pipe would alone fill the tank in 20 hours and smaller pipe would fill the tank alone in 30 hours. 1 1 1 OR Let the usual speed of plane be x km/hr and the reduced speed of the plane be (x-200) km/hr Distance =600 km [Given] According to the question, (time taken at reduced speed) - (Schedule time) = 30 minutes = 0.5 hours. 1 600 600 1 =2 Which on simplification gives: x2 - 200x 240000=0 x2 -600x + 400x 240000=0 x(x- 600) + 400( x-600) = 0 (x-600)(x+400) =0 x=600 or x= 400 But speed cannot be negative. The usual speed is 600 km/hr and 600 the scheduled duration of the flight is 600 =1hour 200 33 For the Theorem : Given, To prove, Construction and figure 1 1 1 Proof 1 7 Let ABCD be a trapezium DC AB and EF is a line parallel to AB and hence to DC. To prove : = Construction : Join AC, meeting EF in G. Proof : In ABC, we have GF AB CG/GA=CF/FB [By BPT] ......(1) In ADC, we have EG DC ( EF AB & AB DC) DE/EA= CG/GA [By BPT] .....(2) From (1) & (2), we get, = 34. Radius of the base of cylinder (r) = 2.8 m = Radius of the base of the cone (r) Height of the cylinder (h)=3.5 m Height of the cone (H)=2.1 m. Slant height of conical part (l)= r2+H2 = (2.8)2+(2.1)2 = 7.84+4.41 1 = 12.25 = 3.5 m 1 Area of canvas used to make tent = CSA of cylinder + CSA of cone 1 = 2 2.8 3.5 + 2.8 3.5 = 61.6+30.8 = 92.4m2 1 1 Cost of 1500 tents at 120 per sq.m = 1500 120 92.4 = 16,632,000 Share of each school to set up the tents = 16632000/50 = 332,640 OR 8 First Solid Second Solid (i) SA for first new solid (S ): 6 7 7 + 2 3.52 - 3.52 = 294 + 77 38.5 = 332.5cm2 SA for second new solid (S ): 6 7 7 + 2 3.52 - 3.52 = 294 + 77 38.5 = 332.5 cm2 So S : S = 1:1 2 (ii) Volume for first new solid (V )= 7 7 7 - 3 3.53 = 343 - 539 6 = 1519 2 6 1 1 1 cm3 1 Volume for second new solid (V )= 7 7 7 + 3 3.5 = 343 + 35 539 6 3 = 2597 6 cm3 Median = 525, so Median Class = 500 600 Class interval Frequency Cumulative Frequency 0 100 2 2 100 200 5 7 200 300 x 7+x 300 400 12 19+x 400 500 17 36+x 500 600 20 56+x 600 700 y 56+x+y 700 800 9 65+x +y 800 900 7 72+x+y 900 1000 4 76+x+y 9 1 1 76+x+y=100 x+y=24 .(i) 1 n cf 2 Median = l + f xh Since, l=500, h=100, f=20, cf=36+x and n=100 Therefore, putting the value in the Median formula, we get; 525 = 500 + 50 (36+x) 20 x 100 so x = 9 y = 24 x (from eq.i) y = 24 9 = 15 Therefore, the value of x = 9 and y = 15. 36 (i) B(1,2), F(-2,9) BF = ( -2-1) + ( 9-2) = ( -3) + ( 7) = 9 + 49 = 58 So, BF = 58 units 1 (ii) W(-6,2), X(-4,0), O(5,9), P(3,11) Clearly WXOP is a rectangle Point of intersection of diagonals of a rectangle is the mid point of the diagonals. So the required point is mid point of WO or XP 6+5 2+9 =( 2 , 2 ) 1 11 =(2, (iii) 2 ) A(-2,2), G(-4,7) Let the point on y-axis be Z(0,y) AZ = GZ 10 ( 0+2) + ( y-2) = ( 0+4) + ( y-7) ( 2) + y + 4 -4y= (4) + y + 49 -14y 8-4y = 65-14y 10y= 57 So, y= 5.7 i.e. the required point is (0, 5.7) OR A(-2,2), F(-2,9), G(-4,7), H(-4,4) Clearly GH = 7-4=3units AF = 9-2=7 units So, height of the trapezium AFGH = 2 units 1 So, area of AFGH = (AF + GH) x height 2 1 = 2(7+3) x 2 = 10 sq. units 37. (i) Since each row is increasing by 10 seats, so it is an AP with first term a= 30, and common difference d=10. So number of seats in 10th row = 10 = a+ 9d = 30 + 9 10 = 120 n (ii) Sn = 2( 2a + (n-1)d) n 1500 = 2( 2 30 + (n-1)10) 3000 = 50n + 10n2 n2 +5n -300 =0 n2 + 20n -15n 300 =0 (n+20) (n-15) =0 Rejecting the negative value, n= 15 OR No. of seats already put up to the 10th row = S10 10 S10 = 2 {2 30 + (10-1)10)} 11 = 5(60 + 90) = 750 So, the number of seats still required to be put are 1500 -750 = 750 (iii) If no. of rows =17 then the middle row is the 9th row 8 = a+ 8d = 30 + 80 = 110 seats 38 (i) 1 P and Q are the two positions of the plane flying at a height of 3000 3m. A is the point of observation. (ii) In PAB, tan60 =PB/AB Or 3 = 3000 3/ AB So AB=3000m tan30 = QC/AC 1/ 3= 3000 3 / AC AC = 9000m distance covered = 9000- 3000 = 6000 m. 1 OR In PAB, tan60 =PB/AB Or 3 = 3000 3/ AB So AB=3000m tan45 = RD/AD 1= 3000 3 / AD 12 AD = 3000 3 m distance covered = 3000 3 - 3000 = 3000( 3 -1)m. (iii) speed = 6000/ 30 = 200 m/s = 200 x 3600/1000 = 720km/hr Alternatively: speed = 3000( 3 1) 15( 3 1) = 200 m/s = 200 x 3600/1000 = 720km/hr 13

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