CBSE · Class 10 Mathematics — Teacher Copy
Full Solutions
Each question opens with the formula you’ll need, then short numbered steps with the working shown on its own line — built so you can glance down and pick up the thread instantly while teaching.
1Real Numbers
Question 1
Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576.
Formula: a = bq + r, where 0 ≤ r < b — repeat until r = 0; the last divisor is the HCF
1
Divide the larger number by the smaller
12576 = 4052 × 3 + 420
2
Divide 4052 by 420
4052 = 420 × 9 + 272
3
Divide 420 by 272
420 = 272 × 1 + 148
4
Divide 272 by 148
272 = 148 × 1 + 124
5
Divide 148 by 124
148 = 124 × 1 + 24
6
Divide 124 by 24
124 = 24 × 5 + 4
7
Divide 24 by 4 — remainder becomes 0, so stop here
24 = 4 × 6 + 0
✓HCF(4052, 12576) = 4
Teaching tip: Say the pattern out loud each time — “big ÷ small → the remainder becomes the new small” — so the student sees the repeating rule before you name it “algorithm.”
Question 2
Prove that √3 is an irrational number.
Method: proof by contradiction — assume the opposite, then show it leads to a contradiction
1
Assume √3 IS rational — then it can be written as a fraction in lowest terms
√3 = p/q, where p and q are co-prime, q ≠ 0
2
Square both sides
3 = p²/q² → p² = 3q² …(i)
3
3 divides p² → since 3 is prime, 3 must divide p too. So write p = 3c
p = 3c, for some integer c
4
Substitute p = 3c into (i) and simplify
9c² = 3q² → q² = 3c²
5
Same logic: 3 divides q² → 3 divides q too
so 3 | p AND 3 | q
6
But p and q were supposed to share no common factor — contradiction!
✓√3 is irrational (proved by contradiction)
Teaching tip: This exact structure (assume → chase it to a contradiction) reappears for √2, √5, and similar proofs — teach the shape of the argument, not just this one case.
2Quadratic Equations
Question 3
Solve the quadratic equation 2x² − 5x + 3 = 0 by factorisation.
Rule: split the middle term — find two numbers that multiply to a×c and add to b
1
a = 2, b = −5, c = 3, so a × c = 6. Find two numbers that multiply to 6 and add to −5
−2 and −3 (−2 × −3 = 6, −2 + −3 = −5)
2
Split the middle term using these numbers
2x² − 2x − 3x + 3 = 0
3
Group in pairs and factor each pair
2x(x − 1) − 3(x − 1) = 0
4
Factor out the common bracket
(x − 1)(2x − 3) = 0
5
Set each bracket to zero and solve
x = 1 or x = 3/2
✓x = 1 or x = 3/2
Teaching tip: Make the student say “which two numbers multiply to ac and add to b” out loud every time — it prevents the most common sign mistake.
Question 4
For 3x² − 4x + 1 = 0: (a) find the discriminant and nature of roots, (b) solve using the quadratic formula.
Formulas: D = b² − 4ac | x = (−b ± √D) / 2a
1
a = 3, b = −4, c = 1 → calculate the discriminant
D = (−4)² − 4(3)(1) = 16 − 12 = 4
2
D > 0 and D is a perfect square →
roots are real, unequal, rational
3
Apply the quadratic formula
x = (4 ± √4) / 6 = (4 ± 2) / 6
4
Work out both roots
x = 6/6 = 1 or x = 2/6 = 1/3
✓D = 4 (real, unequal roots); x = 1 or x = 1/3
Teaching tip: Sketch all three discriminant cases (D>0 two crossings, D=0 one touch, D<0 no crossing) quickly on the board — it turns D into something visual, not just a formula to memorise.
3Introduction to Trigonometry
Question 5
In a right triangle, if sin A = 3/5, find cos A and tan A.
Formulas: sin = opp/hyp | cos = adj/hyp | tan = opp/adj | Pythagoras: a² + b² = c²
1
sin A = opposite/hypotenuse = 3/5 → take
opposite = 3, hypotenuse = 5
2
Find the adjacent side using Pythagoras’ theorem
adjacent² = 5² − 3² = 25 − 9 = 16 → adjacent = 4
3
Now read off the two remaining ratios
cos A = 4/5 tan A = 3/4
✓cos A = 4/5, tan A = 3/4
Teaching tip: Always draw the triangle and label all three sides before calculating anything — most opposite/adjacent mix-ups happen when students skip the sketch.
Question 6
Evaluate: 2 tan²45° + cos²30° − sin²60°
Standard values: tan 45° = 1 | cos 30° = √3/2 | sin 60° = √3/2
1
Substitute the standard values
2(1)² + (√3/2)² − (√3/2)²
2
Simplify each term
2(1) + 3/4 − 3/4
3
The last two terms cancel out
2 + 0 = 2
✓Answer = 2
Teaching tip: Point out cos²30° and sin²60° are identical terms before substituting — spotting shortcuts like this is exactly what CBSE rewards, and it’s a nice “aha” moment.
4Pair of Linear Equations in Two Variables
Question 7
Solve using substitution: x + y = 14 and x − y = 4.
Method: make one variable the subject, substitute into the other equation
1
From equation 1, make x the subject
x = 14 − y
2
Substitute into equation 2
(14 − y) − y = 4 → 14 − 2y = 4
3
Solve for y
−2y = −10 → y = 5
4
Substitute back to find x
x = 14 − 5 = 9
✓x = 9, y = 5
Teaching tip: Finish by checking both values in the OTHER original equation (9 − 5 = 4 ✓) — build this habit from the first example onward.
Question 8
Solve using elimination: 3x + 2y = 11 and 2x + 3y = 4.
Method: match one variable’s coefficient in both equations, then subtract
1
Multiply eqn 1 by 3 and eqn 2 by 2, to match the y-coefficients (6)
9x + 6y = 33 and 4x + 6y = 8
2
Subtract to eliminate y
5x = 25 → x = 5
3
Substitute x = 5 into the original equation 1
15 + 2y = 11 → 2y = −4 → y = −2
✓x = 5, y = −2
Teaching tip: Ask the student to pick which variable to eliminate BEFORE multiplying anything — spotting the smallest common multiple of the coefficients saves messy arithmetic.
5Arithmetic Progressions
Question 9
Find the 10th term of the AP: 2, 7, 12, 17, …
Formula: aₙ = a + (n − 1)d
1
Identify a and d
a = 2, d = 7 − 2 = 5
2
Substitute n = 10 into the formula
a₁₀ = 2 + (10 − 1) × 5 = 2 + 45
✓a₁₀ = 47
Teaching tip: Have the student double-check d against a second pair of terms (12 − 7 = 5) before using the formula — catches a very common miscalculation.
Question 10
Find the sum of the first 15 terms of the AP: 3, 7, 11, …
Formula: Sₙ = n/2 [2a + (n − 1)d]
1
Identify a and d
a = 3, d = 7 − 3 = 4
2
Substitute n = 15 into the formula
S₁₅ = 15/2 [2(3) + 14(4)] = 15/2 [6 + 56]
3
Simplify
S₁₅ = 15/2 × 62 = 15 × 31
✓S₁₅ = 465
Teaching tip: Show the alternative Sₙ = n/2 (a + l) using the 15th term as l, and let the student cross-check via a totally different route.
6Coordinate Geometry
Question 11
Find the distance between (3, 4) and (7, 1).
Formula: d = √[(x₂−x₁)² + (y₂−y₁)²]
1
Substitute the coordinates
d = √[(7−3)² + (1−4)²]
2
Simplify inside the root
d = √[16 + 9] = √25
✓d = 5 units
Teaching tip: Point out this is Pythagoras’ theorem in disguise — the coordinate differences are the two legs, distance is the hypotenuse. Far easier to remember than rote-learning it.
Question 12
Find the point dividing (2, 3) and (7, 8) internally in the ratio 2 : 3.
Formula: x = (m₁x₂+m₂x₁)/(m₁+m₂), y = (m₁y₂+m₂y₁)/(m₁+m₂)
1
(x₁,y₁)=(2,3), (x₂,y₂)=(7,8), m₁=2, m₂=3 → find x
x = (2×7 + 3×2)/5 = 20/5 = 4
2
Now find y using the same pattern
y = (2×8 + 3×3)/5 = 25/5 = 5
✓Point = (4, 5)
Teaching tip: Sketch the two points first — since the ratio is 2:3, the dividing point should sit closer to (2,3). A quick glance at the sketch catches most sign/order mistakes.