What is the HCF of 84 and 132?
Answer & suggested solution
B. 12
- 84=2²×3×7 and 132=2²×3×11.
- Their common prime product is 2²×3=12.
Loading content...
AMIT SHRIVASTAVACONCEPTS TO CONFIDENCE
BOARD EXAM 2027 · DIGITAL PRACTICE
CBSE CLASS 10 · MATHEMATICS STANDARD (041)
SAMPLE
PAPERS
Practise the paper. Review every step.
Original questions · suggested solutions
A · B · C · D · E OBJECTIVE → CASE STUDIES
1 COMPLETE PAPER FREEAttempt first · solutions included
FULL PACK · NOTIFY MENo payment now₹199Planned price
AMIT SIR · ORIGINAL BOARD-STYLE PRACTICE
Class 10 Maths Standard Sample Papers
Attempt a complete paper before checking the solutions. Practise managing your time, choosing a method and presenting clear working across all 14 chapters.
80marks3 hourssuggested time38questions
Start the free paper ↓An original practice paper by Amit Sir, using the section structure of the CBSE 2026–27 Maths Standard sample paper. This is not an official CBSE paper. The worked solutions include a suggested mark split for self-review.
See the official CBSE paper pattern ↗FREE COMPLETE PAPER · 2026–27 SESSION
Original Practice Paper 01
Mathematics Standard (041) · Class 10
Maximum marks: 80Time: 3 hours
Before you begin
- Attempt all 38 numbered questions; internal alternatives are choices within the same question, not extra questions.
- Use a separate answer sheet. Give reasoning and draw labelled diagrams wherever helpful.
- Section A: 20×1; B: 5×2; C: 6×3; D: 4×5; E: 3×4.
- Attempt either alternative in Q21,Q23,Q27,Q29,Q32,Q33, and in part (iii) of each case study.
- Use π=22/7 where required. Calculators are not allowed.
- For a mock attempt, set aside 3 hours and keep solutions closed until you finish.
For a blank paper, leave solutions closed before printing. Solutions show a suggested marking breakdown for self-review.
Section A · Objective questions
Questions 1–18 are MCQs; 19–20 are Assertion–Reason questions. Choose one option for each.
The sum of the zeros of 3x²−10x+3 is:
Answer & suggested solution
C. 10/3
- For ax²+bx+c, the sum is −b/a.
- Here −(−10)/3=10/3.
The pair x+2y=7 and 2x+4y=14 has:
Answer & suggested solution
D. Infinitely many solutions
- The second equation is twice the first.
- The lines coincide, giving infinitely many common solutions.
The discriminant of x²−6x+5=0 is:
Answer & suggested solution
A. 16
- D=b²−4ac=(−6)²−4(1)(5).
- D=36−20=16.
The 12th term of the AP 7,11,15,… is:
Answer & suggested solution
B. 51
- a=7 and d=4.
- a₁₂=7+(12−1)×4=51.
A monic quadratic with zeros 2 and −5 is:
Answer & suggested solution
C. x²+3x−10
- Sum=−3 and product=−10.
- The monic polynomial is x²−(−3)x−10=x²+3x−10.
If x+y=9 and x−y=3, then x equals:
Answer & suggested solution
B. 6
- Adding the equations gives 2x=12.
- Thus x=6.
The distance between (−2,3) and (4,11) is:
Answer & suggested solution
D. 10 units
- Distance=√[(4+2)²+(11−3)²].
- This is √(36+64)=10 units.
The midpoint of (2,−4) and (6,8) is:
Answer & suggested solution
A. (4,2)
- Average the corresponding coordinates.
- ((2+6)/2,(−4+8)/2)=(4,2).
The areas of two similar triangles are in the ratio 9:25. Their corresponding sides are in the ratio:
Answer & suggested solution
C. 3:5
- Area ratio is the square of the corresponding side ratio.
- Taking the positive square root gives 3:5.
A tangent from P touches a circle at T. OP=13 cm and radius OT=5 cm. PT is:
Answer & suggested solution
B. 12 cm
- OT is perpendicular to PT, so triangle OTP is right-angled.
- PT=√(13²−5²)=√144=12 cm.
sin30°+cos60° equals:
Answer & suggested solution
C. 1
- sin30°=1/2 and cos60°=1/2.
- Their sum is 1.
For an acute angle θ, sec²θ−tan²θ equals:
Answer & suggested solution
B. 1
- The identity sec²θ=1+tan²θ holds where defined.
- Subtract tan²θ to obtain 1.
A vertical pole casts a 10 m shadow when the Sun’s angle of elevation is 45°. The pole’s height is:
Answer & suggested solution
D. 10 m
- tan45°=height/shadow=1.
- Height=10×1=10 m.
If sinθ=3/5 for acute θ, then tanθ is:
Answer & suggested solution
A. 3/4
- Take opposite side 3 and hypotenuse 5; adjacent side is 4 by Pythagoras.
- tanθ=opposite/adjacent=3/4.
For acute θ, sinθ=cosθ implies:
Answer & suggested solution
C. θ=45°
- Divide by the positive cosθ: tanθ=1.
- Within the acute interval, θ=45°.
A sector has radius 7 cm and central angle 90°. Using π=22/7, its area is:
Answer & suggested solution
B. 38.5 cm²
- Area=(90/360)×π×7².
- This equals (1/4)×(22/7)×49=38.5 cm².
A cylinder has radius 3 cm and height 7 cm. Its volume, using π=22/7, is:
Answer & suggested solution
C. 198 cm³
- Volume=πr²h=π×9×7=63π.
- Using π=22/7 gives 198 cm³.
Assertion (A): The arithmetic mean of 2,4,6 is 4. Reason (R): The arithmetic mean equals the sum of observations divided by their number.
Answer & suggested solution
A. Both A and R are true, and R correctly explains A.
- The sum is 12 and there are 3 observations.
- Mean=12/3=4. The stated rule directly explains A.
Assertion (A): An event can have probability 1.2. Reason (R): For every event E, 0≤P(E)≤1.
Answer & suggested solution
D. A is false, but R is true.
- A probability cannot exceed 1, so A is false.
- R states the valid probability bounds, so R is true.
Section B · Very short answers
Five questions of 2 marks each. Attempt either alternative in Q21 and Q23.
Find the HCF of 72 and 120 by prime factorisation.
Answer & suggested solution
24.
- 72=2³×3² and 120=2³×3×5.
- HCF=2³×3=24. [1+1 marks]
OR · attempt one alternative
Two bells ring at intervals of 12 and 18 minutes. If both ring now, after how many minutes will they next ring together?
Answer & suggested solution
36 minutes.
- 12=2²×3 and 18=2×3².
- The next common interval is their LCM=2²×3²=36 minutes. [1+1 marks]
Find the zeros of x²−7x+10 and verify their product.
Answer & suggested solution
Zeros 2 and 5; product 10.
- x²−7x+10=(x−2)(x−5), so the zeros are 2 and 5. [1 mark]
- Their product is 2×5=10=c/a. [1 mark]
Tangents PA and PB from the same external point P have lengths (7x−3) cm and (4x+9) cm. Find x and the common tangent length.
Answer & suggested solution
x=4; each tangent is 25 cm.
- Tangents from an external point are equal: 7x−3=4x+9, so x=4. [1 mark]
- The length is 7(4)−3=25 cm, also 4(4)+9=25 cm. [1 mark]
OR · attempt one alternative
Two similar triangles have perimeters 18 cm and 30 cm. A side of the smaller triangle is 9 cm. Find its corresponding side in the larger triangle.
Answer & suggested solution
15 cm.
- The corresponding side ratio equals the perimeter ratio, 18:30=3:5. [1 mark]
- The required side is 9×5/3=15 cm. [1 mark]
If tanθ=5/12 for acute θ, find sinθ+cosθ.
Answer & suggested solution
17/13.
- Use opposite=5 and adjacent=12; the hypotenuse is √(25+144)=13. [1 mark]
- sinθ+cosθ=5/13+12/13=17/13. [1 mark]
Two fair dice are rolled. Find the probability that their sum is 9.
Answer & suggested solution
1/9.
- The favourable ordered pairs are (3,6),(4,5),(5,4),(6,3): four outcomes. [1 mark]
- All 36 ordered outcomes are equally likely; probability=4/36=1/9. [1 mark]
Section C · Short answers
Six questions of 3 marks each. Attempt either alternative in Q27 and Q29.
Given that √2 is irrational, prove that 7−3√2 is irrational.
Answer & suggested solution
7−3√2 is irrational.
- Suppose 7−3√2=r is rational. [1 mark]
- Rearrange to obtain √2=(7−r)/3, which would be rational because r is rational. [1 mark]
- This contradicts the given irrationality of √2, so 7−3√2 is irrational. [1 mark]
Three notebooks and two pens cost ₹94. Two notebooks and five pens cost ₹103. Find the price of each item.
Answer & suggested solution
Notebook ₹24; pen ₹11.
- Let notebook and pen prices be n and p: 3n+2p=94 and 2n+5p=103. [1 mark]
- Multiplying by 5 and 2 gives 15n+10p=470 and 4n+10p=206, so 11n=264 and n=24. [1 mark]
- Then 72+2p=94 gives p=11. Check: 48+55=103. [1 mark]
OR · attempt one alternative
Solve 2x²−7x+3=0 by factorisation and check the product of roots.
Answer & suggested solution
Roots 3 and 1/2; product 3/2.
- 2x²−7x+3=(2x−1)(x−3). [1 mark]
- The zero-product rule gives x=1/2 or 3. [1 mark]
- Their product is 3/2, equal to c/a=3/2. [1 mark]
In triangle ABC, D lies on AB and E on AC, with DE∥BC. AD=3 cm, DB=2 cm and AE=4.5 cm. Find EC and the ratio AE:AC.
Answer & suggested solution
EC=3 cm; AE:AC=3:5.
- By the Basic Proportionality Theorem, AD/DB=AE/EC. [1 mark]
- 3/2=4.5/EC gives EC=3 cm. [1 mark]
- AC=4.5+3=7.5 cm, so AE:AC=4.5:7.5=3:5. [1 mark]
A tangent PT touches a circle of centre O at T. If OP=17 cm and OT=8 cm, find PT and justify the right angle used.
Answer & suggested solution
PT=15 cm.
- The radius through the point of contact is perpendicular to the tangent, so ∠OTP=90°. [1 mark]
- By Pythagoras, PT²=OP²−OT²=17²−8²=225. [1 mark]
- Thus PT=15 cm (positive length). [1 mark]
OR · attempt one alternative
Tangents PA and PB touch a circle at A and B. If ∠APB=60°, find ∠AOB, where O is the centre, and explain the theorem used.
Answer & suggested solution
∠AOB=120°.
- OA⊥PA and OB⊥PB because radii are perpendicular to tangents at contact. [1 mark]
- In quadrilateral OAPB, angles sum to 360°. [1 mark]
- ∠AOB=360°−90°−90°−60°=120°. [1 mark]
A 90° sector of radius 14 cm contains the triangle formed by its two radii and chord. Find the area between the chord and arc. Use π=22/7.
Answer & suggested solution
56 cm².
- Sector area=(90/360)π(14²)=154 cm². [1 mark]
- The included triangle is right-angled, with area=(1/2)×14×14=98 cm². [1 mark]
- Segment area=154−98=56 cm². [1 mark]
Find the mean of this grouped data: class intervals 0–10,10–20,20–30 with frequencies 3,5,2 respectively.
Answer & suggested solution
14.
- The class marks are 5,15,25. [1 mark]
- Σf=10 and Σfx=3×5+5×15+2×25=140. [1 mark]
- Grouped mean=Σfx/Σf=140/10=14. [1 mark]
Section D · Long answers
Four questions of 5 marks each. Attempt either alternative in Q32 and Q33.
A rectangular playground has area 180 m². Its length is 3 m more than its width. Form a quadratic equation and find its dimensions, explaining which root is valid.
Answer & suggested solution
Width 12 m; length 15 m.
- Let width be x>0 metres and length x+3 metres. [1 mark]
- x(x+3)=180 gives x²+3x−180=0. [1 mark]
- Factor: (x+15)(x−12)=0. [1 mark]
- The algebraic roots are −15 and 12; reject −15 because a width must be positive. [1 mark]
- Width=12 m and length=15 m; check 12×15=180. [1 mark]
OR · attempt one alternative
An auditorium has 12 seats in its first row and 2 more seats in each successive row. If it has 390 seats in total, find the number of rows and seats in its last row.
Answer & suggested solution
15 rows; last row 40 seats.
- This is an AP with a=12 and d=2; let n be the positive number of rows. [1 mark]
- Sₙ=n/2[24+2(n−1)]=n(n+11)=390. [1 mark]
- n²+11n−390=(n+26)(n−15)=0. [1 mark]
- Reject n=−26; hence n=15 rows. [1 mark]
- Last row has 12+14×2=40 seats. [1 mark]
In triangle ABC, D lies on AB and E on AC, with DE∥BC. AD=4 cm, DB=6 cm and AE=6 cm. Prove that triangles ADE and ABC are similar, find EC, and find area(ABC) if area(ADE)=24 cm².
Answer & suggested solution
Triangles ADE and ABC are similar; EC=9 cm; area(ABC)=150 cm².
- ∠DAE=∠BAC (common angle) and ∠ADE=∠ABC (DE∥BC); hence ΔADE∼ΔABC by AA. [2 marks]
- AB=4+6=10, so AD/AB=2/5=AE/AC. This gives AC=15 and EC=9 cm. [1 mark]
- The area ratio ADE:ABC is (2/5)²=4/25. [1 mark]
- Therefore area(ABC)=24×25/4=150 cm². [1 mark]
OR · attempt one alternative
Triangles PQR and XYZ are similar in that order. PQ=6 cm, QR=8 cm, PR=10 cm and XY=9 cm. Find YZ and XZ, the ratio of their areas, and the perimeter of the larger triangle.
Answer & suggested solution
YZ=12 cm; XZ=15 cm; areas PQR:XYZ=4:9; larger perimeter=36 cm.
- Similarity gives scale factor XY/PQ=9/6=3/2. [1 mark]
- YZ=(3/2)×8=12 cm and XZ=(3/2)×10=15 cm. [2 marks]
- The smaller-to-larger area ratio is (2/3)²=4/9. [1 mark]
- The larger perimeter is 9+12+15=36 cm. [1 mark]
From point A on level ground, the angle of elevation of the top of a vertical tower is 30°. After walking 20 m towards its foot to B, the angle becomes 60°. Find the tower height and the distance from B to its foot.
Answer & suggested solution
Height 10√3 m; distance from B to the foot 10 m.
- Let AF=x metres and tower height be h; then BF=x−20. [1 mark]
- tan30°=h/x gives h=x/√3. [1 mark]
- tan60°=h/(x−20) gives h=√3(x−20). [1 mark]
- Equating: x=3(x−20), so x=30 and BF=10 m. [1 mark]
- Then h=30/√3=10√3 m. [1 mark]
A solid toy consists of a right circular cylinder of radius 7 cm and height 10 cm, topped by a hemisphere of the same radius. Find its volume and total exposed surface area, including the flat circular base. Use π=22/7.
Answer & suggested solution
Volume=6776/3 cm³; exposed surface area=902 cm².
- Volume is cylinder plus hemisphere: V=πr²h+(2/3)πr³. [1 mark]
- V=490π+(686/3)π=(2156/3)π=6776/3 cm³. [1 mark]
- The circular joining face is internal and not exposed. [1 mark]
- Exposed area=2πrh+2πr²+πr²=140π+98π+49π=287π. [1 mark]
- Using π=22/7 gives 902 cm². [1 mark]
Section E · Case studies
Three questions of 4 marks each, split 1+1+2. In each case choose either option for part (iii).
A school sets out chairs for a celebration. The first row has 18 chairs, and each successive row has 2 more chairs than the previous row. Answer parts (i)–(iii).
(i) [1] How many chairs are in the fourth row?
Answer & suggested solution
24.
- a₄=18+3×2=24. [1 mark]
(ii) [1] Write the number of chairs in the nth row.
Answer & suggested solution
2n+16.
- aₙ=18+2(n−1)=2n+16. [1 mark]
(iii) [2] Find the total number of chairs in the first 10 rows.
Answer & suggested solution
270.
- S₁₀=10/2[2(18)+9(2)]. [1 mark]
- S₁₀=5×54=270. [1 mark]
OR · attempt one alternative for this part
Find the total number of chairs in the first 15 rows.
Answer & suggested solution
480.
- S₁₅=15/2[36+14×2]. [1 mark]
- S₁₅=(15/2)×64=480. [1 mark]
On a park plan, A(0,0), B(6,0), C(6,8) and D(0,8) form a rectangular walking track. Coordinates are in metres. Answer parts (i)–(iii).
(i) [1] Find AB.
Answer & suggested solution
6 m.
- AB=√[(6−0)²+(0−0)²]=6 m. [1 mark]
(ii) [1] Find the diagonal AC.
Answer & suggested solution
10 m.
- AC=√(6²+8²)=10 m. [1 mark]
(iii) [2] Find the midpoint of AC and show that it is also the midpoint of BD.
Answer & suggested solution
Both midpoints are (3,4).
- Midpoint of AC=((0+6)/2,(0+8)/2)=(3,4). [1 mark]
- Midpoint of BD=((6+0)/2,(0+8)/2)=(3,4), so they coincide. [1 mark]
OR · attempt one alternative for this part
P divides AB internally in the ratio AP:PB=1:2. Find P and its distance from C.
Answer & suggested solution
P=(2,0); PC=4√5 m.
- Section formula gives P=((1×6+2×0)/3,0)=(2,0). [1 mark]
- PC=√[(6−2)²+(8−0)²]=√80=4√5 m. [1 mark]
A teacher groups 20 students by minutes spent on a revision task: 10–20:4 students; 20–30:8; 30–40:6; 40–50:2. Treat classes as continuous with width 10. Answer parts (i)–(iii).
(i) [1] State the total frequency N.
Answer & suggested solution
20.
- N=4+8+6+2=20. [1 mark]
(ii) [1] Identify the median class.
Answer & suggested solution
20–30.
- N/2=10 falls in the class with cumulative frequency rising from 4 to 12, so the median class is 20–30. [1 mark]
(iii) [2] Estimate the median time.
Answer & suggested solution
27.5 minutes.
- Median=l+[(N/2−cf)/f]h=20+[(10−4)/8]×10. [1 mark]
- Median=20+7.5=27.5 minutes. [1 mark]
OR · attempt one alternative for this part
Estimate the mode of the grouped data.
Answer & suggested solution
80/3 minutes, or approximately 26.67 minutes.
- The modal class is 20–30; l=20, f₁=8, f₀=4, f₂=6 and h=10. [1 mark]
- Mode=20+[(8−4)/(16−4−6)]×10=20+20/3=80/3 minutes. [1 mark]
Finished your attempt?
Review your work against the suggested solutions. Note the marks lost to concepts, calculation, presentation or time, then revise those topics before your next paper.
Get updates about the full pack ↑