CDS II 2026 Elementary Mathematics with Solutions
Q.1 [Trigonometry]
If cot θ + cos θ = m, cot θ − cos θ = n, where 0 < θ < π/2, then what is (m² − n²)/√(mn) equal to?
mn = cot²θ − cos²θ = cos²θ/sin²θ − cos²θ = cos²θ(1 − sin²θ)/sin²θ = cos⁴θ/sin²θ.
Since 0 < θ < π/2 everything is positive, so √(mn) = cos²θ/sin θ.
Therefore (m²−n²)/√(mn) = 4·(cos²θ/sin θ) ÷ (cos²θ/sin θ) = 4.
Trap: forgetting that cot²θ − cos²θ simplifies to cos⁴θ/sin²θ and not to something with a stray sin.
Q.2 [Trigonometry]
If 1/(cosec θ + cot θ) − 1/sin θ = p, then what is 1/sin θ − 1/(cosec θ − cot θ) equal to?
So 1/(cosec θ + cot θ) = cosec θ − cot θ, and 1/(cosec θ − cot θ) = cosec θ + cot θ.
First expression: p = (cosec θ − cot θ) − cosec θ = − cot θ.
Second expression: cosec θ − (cosec θ + cot θ) = − cot θ = p.
Both expressions equal −cot θ, so the answer is p.
Q.3 [Trigonometry]
If 8 sin θ − cos θ = 4, where π/6 < θ < π/3, then what is sin θ equal to?
s² + (8s − 4)² = 1 → 65s² − 64s + 15 = 0.
Discriminant = 64² − 4·65·15 = 4096 − 3900 = 196, √196 = 14.
s = (64 ± 14)/130 → s = 3/5 or s = 5/13.
Now use the range: π/6 < θ < π/3 means 1/2 < sin θ < √3/2, i.e. 0.5 < sin θ < 0.866.
3/5 = 0.6 ✓ 5/13 ≈ 0.385 ✗
Answer 3/5. The range clause is not decoration — it is the whole second half of the question.
Q.4 [Heights & Distances]
A person walking along a straight road observes that at two consecutive kilometre-stones the angles of elevation of a hill in front of him are 30° and 60° respectively. What is the height of the hill?
From the nearer stone: tan 60° = h/x → x = h/√3.
From the farther stone (1 km more): tan 30° = h/(x + 1) → x + 1 = h√3.
Subtract: h√3 − h/√3 = 1 → h(3 − 1)/√3 = 1 → 2h/√3 = 1 → h = √3/2 km.
Standard formula: h = d·tanA·tanB/(tanB − tanA) with d = 1.
Q.5 [Heights & Distances]
A tower subtends an angle α at a point P on the same level as the foot of the tower. Q is a point vertically above P and PQ = h. If the angle of depression of the foot of the tower measured from Q is β, then what is the height of the tower?
From Q, the angle of depression of the foot of the tower is β, and Q is h above P:
tan β = h/d → d = h cot β.
At P, the tower subtends α: tan α = H/d → H = d tan α.
Hence H = h cot β · tan α = h tan α cot β.
Trap: mixing up which angle goes with which vertical distance. The angle β is measured from Q, so it pairs with h; the angle α is measured from P, so it pairs with H.
Q.6 [Trigonometry]
If 5 sin θ + 12 cos θ = 13, where 0 < θ < π/2, then what is tan θ + cot θ equal to?
sin θ = 5/13, cos θ = 12/13.
tan θ + cot θ = sin/cos + cos/sin = 1/(sin θ cos θ) = 1/((5/13)(12/13)) = 169/60.
Answer 169/60.
Shortcut worth remembering: a sin θ + b cos θ = √(a²+b²) forces sin θ = a/√(a²+b²), cos θ = b/√(a²+b²).
Q.7 [Trigonometry]
What is the ratio of the greatest value of sin²x + 2 (where 0 ≤ x ≤ π/2) to its least value?
Greatest value of sin²x + 2 = 1 + 2 = 3 (at x = π/2).
Least value = 0 + 2 = 2 (at x = 0).
Ratio = 3/2.
Q.8 [Trigonometry]
If A + B + C = π (A, B, C > 0) and the angle C is obtuse, then which of the following is/are correct?
I. sin A · sin B < 1
II. tan A · tan B > 1
I. 0 < sin A < 1 and 0 < sin B < 1, so their product is less than 1. Correct.
II. From A + B < π/2 we get A < π/2 − B, and tan is increasing on (0, π/2), so
tan A < tan(π/2 − B) = cot B = 1/tan B → tan A · tan B < 1. Incorrect (the inequality runs the other way).
Answer: I only.
Q.9 [Trigonometry]
If (cos 35° + cos 55°) = p, then what is sin 35° · cos 35° equal to?
So p = cos 35° + sin 35°.
Square both sides: p² = cos²35 + sin²35 + 2 sin35 cos35 = 1 + 2 sin35 cos35.
Hence sin 35° cos 35° = (p² − 1)/2.
Q.10 [Trigonometry]
If p = sec θ + tan θ and q = cosec θ − cot θ, then what is (p − q − pq) equal to?
p − q − pq = p(1 − q) − q.
1 − q = (sin θ − 1 + cos θ)/sin θ.
p(1 − q) = (1 + sin θ)(sin θ + cos θ − 1)/(sin θ cos θ).
Expand the numerator: (1+s)(s + c − 1) = s + c − 1 + s² + sc − s = c − 1 + s² + sc = c − 1 + (1 − c²) + sc = c(1 − c + s).
So p(1 − q) = c(1 − c + s)/(sc) = (1 − c + s)/s = (1 − cos θ)/sin θ + 1 = q + 1.
Therefore p − q − pq = (q + 1) − q = 1.
Quick check with θ = 45°: p = √2 + 1 = 2.414, q = √2 − 1 = 0.414; 2.414 − 0.414 − 1.000 = 1 ✓
Q.11 [Sets]
Consider the following statements:
I. The set of all birds living on the Earth is an infinite set.
II. The set of all real numbers between 0 and 10 is a finite set.
Which of the statements given above is/are correct?
II. Between any two distinct real numbers there are infinitely many real numbers. The interval (0, 10) is an infinite — indeed uncountably infinite — set. Incorrect.
Answer: Neither I nor II.
The pairing is deliberate: one statement calls a finite set infinite, the other calls an infinite set finite.
Q.12 [Sets]
Consider the following sets:
I. The set of even prime numbers
II. {x ∈ ℝ : x³ + 1 = 0}
III. {n ∈ ℤ : n² < 1}
How many of the above are null sets?
II. x³ = −1 has the real solution x = −1, so the set is {−1} — not null.
III. n² < 1 with n an integer gives n = 0, so the set is {0} — not null.
Answer: None.
Trap in III: candidates read n² < 1 and conclude no integer works, forgetting 0² = 0 < 1.
Q.13 [Sets]
Consider the following statements:
I. If A is a subset of U and U is the Universal set, then the complement of the set A is also a subset of U.
II. The complement of a Universal set is a singleton set.
Which of the statements given above is/are correct?
II. U′ = U − U = ∅, the empty set. The empty set has no elements; a singleton has exactly one. Incorrect.
Answer: I only.
Q.14 [Algebra]
What should be added to (x² − 1)/(x² + 1) to get (2x³ + 3x² − 1)/((x + 2)(x² + 1))?
= [2x³+3x²−1 − (x²−1)(x+2)] / ((x+2)(x²+1)).
(x²−1)(x+2) = x³ + 2x² − x − 2.
Numerator = 2x³+3x²−1 − x³ − 2x² + x + 2 = x³ + x² + x + 1 = x²(x+1) + (x+1) = (x+1)(x²+1).
So the required expression = (x+1)(x²+1)/((x+2)(x²+1)) = (x+1)/(x+2).
Q.15 [Algebra]
If x + y + z = 0, then what is x(y + z)² + y(z + x)² + z(x + y)² equal to?
So the expression = x(−x)² + y(−y)² + z(−z)² = x³ + y³ + z³.
The standard identity x³ + y³ + z³ − 3xyz = (x+y+z)(x²+y²+z² − xy − yz − zx) gives, when x+y+z = 0,
x³ + y³ + z³ = 3xyz.
Q.16 [Trigonometry]
If tan 6θ · tan 3θ = 1, where 0 < θ < 30°, then what is θ equal to?
Here 6θ + 3θ = 90° → 9θ = 90° → θ = 10°.
Check the range: 0 < 10° < 30° ✓, and 6θ = 60°, 3θ = 30°, with tan60 · tan30 = √3 × (1/√3) = 1 ✓
Q.17 [Trigonometry]
If (1 − tan x)/(1 + tan x) = 1 − 2 tan x/(1 + tan²x), where 0 ≤ x < π/2, x ≠ π/4, then what is (sin x + cos x) equal to?
So RHS = 1 − 2 sin x cos x = (sin x − cos x)².
LHS: divide numerator and denominator by cos x → (cos x − sin x)/(cos x + sin x).
Let u = cos x − sin x. The equation becomes u/(cos x + sin x) = u².
Either u = 0, which gives x = π/4 — excluded by the question;
or u(cos x + sin x) = 1 → cos²x − sin²x = 1 → cos 2x = 1 → x = 0.
At x = 0: sin x + cos x = 0 + 1 = 1.
The exclusion x ≠ π/4 is the hint that you must find the other root.
Q.18 [Trigonometry]
If cos²x + cos⁴x = 1, where 0 < x < π/2, then what is (sin²x + sin³x) + (sin³x + sin⁴x) equal to?
Hence sin²x = c², and since both are positive, sin x = c = cos²x.
The expression = sin²x + 2sin³x + sin⁴x = sin²x(1 + sin x)² = [sin x (1 + sin x)]².
Substituting sin x = c: [c(1 + c)]² = [c + c²]² = [1]² = 1, using c + c² = 1.
Elegant, and no surds needed — the whole question is the single substitution sin x = cos²x.
Q.19 [Trigonometry]
If sin⁴x + cos⁴x = 1, where 0 ≤ x ≤ π/2, then what is sin x · cos x equal to?
Setting this equal to 1 gives 2 sin²x cos²x = 0 → sin x cos x = 0.
(This happens at x = 0 or x = π/2, both inside the stated closed interval.)
Q.20 [Trigonometry]
If 3 cos θ + 4 sin θ = 4, where 0 ≤ θ < π/2, then what is 4 cos θ + 3 sin θ equal to?
16(1 − s)²/9 + s² = 1 → 16(1 − 2s + s²) + 9s² = 9 → 25s² − 32s + 7 = 0.
Discriminant = 1024 − 700 = 324, √324 = 18 → s = (32 ± 18)/50 → s = 1 or s = 7/25.
s = 1 gives θ = π/2, which the range 0 ≤ θ < π/2 excludes. So s = 7/25 and c = 24/25.
4 cos θ + 3 sin θ = 96/25 + 21/25 = 117/25.
Note the asymmetry: swapping coefficients does not give 4 back — that is exactly what the question tests.
Q.21 [Algebra]
If 1/(x + k) + 1/(x + 2k) + 1/(x + 5k) = 1/k, then what is the solution of the equation?
1/(2k) + 1/(3k) + 1/(6k) = (3 + 2 + 1)/(6k) = 6/(6k) = 1/k ✓
So x = k.
Method note: clearing denominators gives a cubic, and testing the four options takes a fraction of the time. In a 100-question paper in 120 minutes, that arithmetic discipline is itself a skill.
Q.22 [Algebra]
Consider the polynomial p(x) = x⁵ + 2x⁴ + x³ − x² − 2x − 1:
I. x² + 2x + 1 is a factor of p(x).
II. x² + x + 1 divides p(x).
Which of the statements given above is/are correct?
p(x) = x³(x² + 2x + 1) − (x² + 2x + 1) = (x² + 2x + 1)(x³ − 1).
I. x² + 2x + 1 is a factor. Correct.
Further, x³ − 1 = (x − 1)(x² + x + 1), so
p(x) = (x + 1)²(x − 1)(x² + x + 1).
II. x² + x + 1 divides p(x). Correct.
Answer: Both I and II. Spotting the grouping x³(…) − (…) is the whole question.
Q.23 [Quadratic Equations]
If 5 is a root of the equation x² + px − 55 = 0 and the quadratic equation x² + px + k = 0 has equal roots, then what is the value of k?
The second equation x² + 6x + k = 0 has equal roots, so its discriminant is zero:
36 − 4k = 0 → k = 9.
Q.24 [Time, Speed & Distance]
Had I walked 0·5 km/hr faster, I would have taken 8 minutes less to walk 2 km. What is my original speed?
2/v − 2/(v + 0·5) = 2/15
2[(v + 0·5) − v] / (v(v + 0·5)) = 2/15
1/(v² + 0·5v) = 2/15 → v² + 0·5v = 7·5 → 2v² + v − 15 = 0
(2v − 5)(v + 3) = 0 → v = 2·5 (rejecting the negative root).
Answer 2·5 km/hr. Check: 2/2·5 = 48 min; 2/3 = 40 min; difference 8 min ✓
Q.25 [Quadratic Equations]
If the roots of the equation x² + mx + n = 0 are increased by the same quantity k, then they become the roots of the equation x² + nx + m = 0. What is the value of (m + n)?
The roots of the second are α + k, β + k:
Sum: (α + β) + 2k = −n → −m + 2k = −n → 2k = m − n.
Product: αβ + k(α + β) + k² = m → n − mk + k² = m.
Substituting k = (m − n)/2 and multiplying through by 4:
4n − 2m(m − n) + (m − n)² = 4m
4n − 2m² + 2mn + m² − 2mn + n² = 4m
n² − m² + 4n − 4m = 0 → (n − m)(n + m) + 4(n − m) = 0 → (n − m)(m + n + 4) = 0.
n = m gives k = 0, the trivial case where nothing is shifted. So m + n = −4.
Q.26 [Quadratic Equations]
If α and β are the roots of the equation 2x² − 2(n + 1)x + (n² + n + 1) = 0, then what is (2α³ + 2β³) equal to?
α³ + β³ = (α + β)³ − 3αβ(α + β).
2(α³ + β³) = 2(n+1)³ − 3(n² + n + 1)(n + 1).
2(n+1)³ = 2n³ + 6n² + 6n + 2.
3(n²+n+1)(n+1) = 3(n³ + 2n² + 2n + 1) = 3n³ + 6n² + 6n + 3.
Difference = (2n³ − 3n³) + (6n² − 6n²) + (6n − 6n) + (2 − 3) = −n³ − 1 = −(n³ + 1).
Q.27 [Number System]
The product of two numbers is 1050. The quotient when the larger number is divided by the smaller number is 4 and the remainder is 10. What is the sum of the two numbers?
s(4s + 10) = 1050 → 4s² + 10s − 1050 = 0 → 2s² + 5s − 525 = 0.
Discriminant = 25 + 4200 = 4225, √4225 = 65 → s = (−5 + 65)/4 = 15.
L = 4(15) + 10 = 70. Check: 15 × 70 = 1050 ✓ and remainder 10 < 15 ✓
Sum = 85.
Q.28 [Polynomials]
What is the HCF of x³ + x²y − 10xy² + 8y³ and x³ + x²y − 4xy² − 4y³?
x³ + x²y − 10xy² + 8y³ = (x − y)(x² + 2xy − 8y²) = (x − y)(x + 4y)(x − 2y).
Second polynomial: putting x = −y gives −1 + 1 + 4 − 4 = 0, so (x + y) is a factor.
x³ + x²y − 4xy² − 4y³ = (x + y)(x² − 4y²) = (x + y)(x − 2y)(x + 2y).
The only common factor is (x − 2y).
Q.29 [Surds]
What is (√2 + √3 + √5)(√2 + √3 − √5)(√2 − √3 + √5)(− √2 + √3 + √5) equal to?
First pair = (a+b)² − c² = (2 + 3 + 2√6) − 5 = 2√6.
Second pair = c² − (a−b)² = 5 − (2 + 3 − 2√6) = 2√6.
Product = 2√6 × 2√6 = 4 × 6 = 24.
This is Heron's formula in disguise — the same product appears as 16 × (area)² for a triangle with sides √2, √3, √5.
Q.30 [Polynomials]
The LCM and HCF of two polynomials p(x) and q(x) are (x + k)²(x² − 7x + 6) and (x + k) respectively. If [x³ + (2k − 1)x² + k(k − 2)x − k²] is one of the polynomials, then which of the following statements is/are correct?
I. The other polynomial is (x + k)(x − 6) where k ≠ − 6.
II. (x + k) cannot be the HCF if k = − 6.
x³ + (2k−1)x² + (k²−2k)x − k² = (x + k)(x² + (k−1)x − k) = (x + k)(x + k)(x − 1) = (x + k)²(x − 1).
Since LCM × HCF = p(x) × q(x):
(x+k)²(x−1)(x−6) × (x+k) = (x+k)²(x−1) × q(x)
q(x) = (x + k)(x − 6). I is correct — and the caveat k ≠ −6 matters, because…
II. If k = −6 then p(x) = (x−6)²(x−1) and q(x) = (x−6)², making the HCF (x−6)², not (x−6). So (x+k) could not be the HCF. Correct.
Answer: Both I and II.
Q.31 [Number System]
Two positive integers x and y are in the ratio 17 : 19. If the LCM of the two numbers is 1615, then what is (x + y) equal to?
LCM = 17 × 19 × a = 323a.
323a = 1615 → a = 5.
So x = 85, y = 95 and x + y = 180.
Check: HCF(85, 95) = 5, LCM = 85 × 95 / 5 = 1615 ✓
Q.32 [Ratio & Proportion]
If p/q = q/r = r/s = k, then what is (p³ + q³ + r³)/(q³ + r³ + s³) equal to?
p³ + q³ + r³ = k³q³ + k³r³ + k³s³ = k³(q³ + r³ + s³).
So the required ratio = k³.
Now express k³ in the options: k³ = (p/q)(q/r)(r/s) = p/s.
Q.33 [Polynomials]
If x² − 4x + 3 is a factor of x⁴ + px² + q, then what is (p − q) equal to?
x = 1: 1 + p + q = 0 → p + q = −1.
x = 3: 81 + 9p + q = 0 → 9p + q = −81.
Subtracting: 8p = −80 → p = −10, and then q = −1 − (−10) = 9.
p − q = −10 − 9 = −19.
Q.34 [Quadratic Equations]
Suppose p and q are positive integers where q is not a perfect square. If 1/(p + √q) is a root of a quadratic equation having integer coefficients, then which of the following statements is/are correct?
I. The sum of the roots is 2p times the product of the roots.
II. The sum of the reciprocals of the roots is equal to 2p.
α = 1/(p + √q) = (p − √q)/(p² − q) and β = 1/(p − √q) = (p + √q)/(p² − q).
Sum = 2p/(p² − q). Product = 1/((p+√q)(p−√q)) = 1/(p² − q).
I. 2p × product = 2p/(p² − q) = sum ✓ Correct.
II. 1/α + 1/β = (p + √q) + (p − √q) = 2p ✓ Correct.
Answer: Both I and II.
Q.35 [Algebra]
If 1/(2b + 2c) + 1/(2c + 2a) = 1/(a + b), then what is a² + b² + c² equal to?
→ [(c+a) + (b+c)] (a+b) = 2(b+c)(c+a)
→ (a + b + 2c)(a + b) = 2(b+c)(c+a)
LHS = a² + 2ab + b² + 2ac + 2bc.
RHS = 2(bc + ab + c² + ac) = 2ab + 2bc + 2c² + 2ac.
Subtracting: a² + b² − 2c² = 0 → a² + b² = 2c².
Hence a² + b² + c² = 2c² + c² = 3c².
Q.36 [Time, Speed & Distance]
A train completely crossed two persons X and Y moving in the same direction as the train in 8 seconds and 8·1 seconds respectively. X is walking at a uniform speed of 3 km/hr and Y is walking at a uniform speed of 4 km/hr. What is the speed of the train?
L = (v − 3) × 8 and L = (v − 4) × 8·1 (up to the same conversion factor).
8(v − 3) = 8·1(v − 4)
8v − 24 = 8·1v − 32·4
0·1v = 8·4 → v = 84 km/hr.
Sense check: Y walks faster, so the relative speed is lower and the crossing takes longer — consistent with 8·1 s > 8 s ✓
Q.37 [Number System]
A number is formed by three digits, each less by unity than the digit that follows it. If 15 is added to the number, then the sum is 30 times the sum of the digits of the number. What is the product of the digits of the number?
The number N = 100a + 10(a+1) + (a+2) = 111a + 12. Sum of digits = 3a + 3.
N + 15 = 30(3a + 3)
111a + 27 = 90a + 90 → 21a = 63 → a = 3.
Digits are 3, 4, 5; the number is 345. Check: 345 + 15 = 360 and 30 × 12 = 360 ✓
Product = 3 × 4 × 5 = 60.
Q.38 [Logarithms]
Consider the following statements in respect of common logarithms:
I. The logarithm of a number greater than 100 but less than 1000 lies between 2 and 3.
II. The logarithm of a positive number less than unity is negative.
II. For 0 < N < 1, log₁₀N < log₁₀1 = 0. Correct.
Answer: Both I and II.
Q.39 [Logarithms]
What is the number of zeros immediately after the decimal point in (0·5)¹⁰⁰⁰? (Given that log₁₀2 = 0·30103)
log₁₀(2⁻¹⁰⁰⁰) = −1000 × 0·30103 = −301·03.
Write this in standard form: −301·03 = −302 + 0·97, so the number is 10^0·97 × 10⁻³⁰².
10^0·97 ≈ 9·3, a single non-zero digit, and it sits at the 302nd decimal place.
Therefore the digits in places 1 to 301 after the decimal point are zeros: 301 zeros.
Rule: if the characteristic of log N is −(k+1), the number of zeros immediately after the decimal point is k.
Q.40 [Surds]
If x = 11 + 2√30, then what is the value of x^(1/2) + x^(−1/2)?
x = (√6 + √5)² → √x = √6 + √5.
1/√x = 1/(√6 + √5) = (√6 − √5)/((√6)² − (√5)²) = √6 − √5.
Sum = (√6 + √5) + (√6 − √5) = 2√6.
Q.41 [Profit & Loss]
A shopkeeper sold his goods at the cost price. By using false weights, he gained 11 1/9 %. What weight did he use for 1 kg?
Gain % = (1000 − w)/w × 100.
11 1/9 % = 100/9 %, so
(1000 − w)/w = 1/9 → 9000 − 9w = w → 10w = 9000 → w = 900 gm.
Trap: dividing the shortfall by 1000 instead of by w. The denominator in gain % is always the cost to the seller — here the 900 g he actually parts with.
Q.42 [Quadratic Equations]
If the roots of the equation px² − 40x + 96 = 0 are even integers, then what is the value of p?
Sum: 2m + 2n = 40/p → m + n = 20/p.
Product: 4mn = 96/p → mn = 24/p.
Test p = 4: m + n = 5, mn = 6 → m, n are roots of t² − 5t + 6 = 0, i.e. t = 2 and 3.
So the roots are 4 and 6, both even ✓
Verify: 4x² − 40x + 96 = 0 → x² − 10x + 24 = 0 → (x − 4)(x − 6) = 0 ✓
Answer 4. (For p = 1, 2 and 3 the discriminant is not a perfect square, or the sum is not an integer.)
Q.43 [Number System]
A number N consists of two digits. The digit in the tens place is 3 times the digit in the units place. The digits are reversed and the resulting number is denoted by R. If N × R = 3627, then what is the product of the digits of the number N?
u = 1: N = 31, R = 13, N×R = 403 ✗
u = 2: N = 62, R = 26, N×R = 1612 ✗
u = 3: N = 93, R = 39, N×R = 3627 ✓
The digits are 9 and 3, so the product is 27.
Q.44 [Boats & Streams]
A boatman can row to a place (Y) at a distance of 24 km from the starting point (X) and back in 14 hours. If he can row 4 km with the stream in the same time as 3 km against it, what is the speed of the stream?
Equal times for 4 km downstream and 3 km upstream: 4/d = 3/u → u = 3d/4.
Total time: 24/d + 24/u = 14
24/d + 24 × 4/(3d) = 14 → 24/d + 32/d = 14 → 56/d = 14 → d = 4, hence u = 3.
Speed of stream = (d − u)/2 = (4 − 3)/2 = 0·5 km/hr.
(Speed of boat in still water = (d + u)/2 = 3·5 km/hr.)
Q.45 [Alligation]
An alloy (X) of gold and silver is taken in the ratio 1 : 2 and another alloy (Y) of gold and silver is taken in the ratio 2 : 3. How many parts of X and Y must be taken to obtain a new alloy Z consisting of gold and silver in the ratio 4 : 7?
X: 1/3 Y: 2/5 Z (required): 4/11.
By alligation, X : Y = (2/5 − 4/11) : (4/11 − 1/3)
= (22 − 20)/55 : (12 − 11)/33 = (2/55) : (1/33)
= 2 × 33 : 55 × 1 = 66 : 55 = 6 : 5.
Check: 6 parts of X → gold 2, silver 4. 5 parts of Y → gold 2, silver 3. Total gold 4, silver 7 ✓
Q.46 [Number System]
N is a 3-digit number. The middle digit of N is equal to the sum of the other two digits and the sum of the digits is 10. How many such numbers (N) can be possible?
Given b = a + c and a + b + c = 10.
Substituting: b + b = 10 → b = 5, so a + c = 5.
Since a ≥ 1 (it is a three-digit number) and c ≥ 0:
(a, c) = (1,4), (2,3), (3,2), (4,1), (5,0) — five possibilities.
The numbers are 154, 253, 352, 451, 550 → 5.
Q.47 [Profit & Loss]
The ratio of the cost price of two articles X and Y is 1 : 2. A businessman earns a profit of p% by selling X and a loss of 3p% by selling Y. If he loses 20% in this transaction, then what is the value of p?
Gain on X = p/100 × 1 = p/100.
Loss on Y = 3p/100 × 2 = 6p/100.
Net loss = 6p/100 − p/100 = 5p/100.
Overall loss % = (5p/100) / 3 × 100 = 5p/3.
Setting 5p/3 = 20 gives p = 12%.
Trap: averaging the percentages (p and 3p) directly. Percentages must be applied to their own bases before combining.
Q.48 [Compound Interest]
A sum of money at the rate of 5% per annum compounded annually becomes n times in 100 years. What is the value of n? (Given log₁₀2 = 0·301, log₁₀3 = 0·477 and log₁₀7 = 0·845)
log 105 = log(3 × 5 × 7) = log 3 + log 5 + log 7.
log 5 = log(10/2) = 1 − 0·301 = 0·699.
log 105 = 0·477 + 0·699 + 0·845 = 2·021.
log n = 100 × (2·021 − 2) = 100 × 0·021 = 2·1.
n = 10^2·1 = 10² × 10^0·1 ≈ 100 × 1·26 ≈ 126.
So n is more than 100. The three logarithm values given are the clue that you are meant to take logs rather than attempt the power directly.
Q.49 [Variation]
A variable y is directly proportional to a variable quantity xⁿ. Given that when x = 2, y = 20·8 and when x = 3, y = 105·3. Which one of the following is the constant of proportionality if it is greater than 1?
105·3/20·8 = (3/2)ⁿ → 5·0625 = 1·5ⁿ.
Now 1·5² = 2·25, 1·5³ = 3·375, 1·5⁴ = 5·0625 ✓ so n = 4.
Then k = 20·8/2⁴ = 20·8/16 = 1·3.
Check with the second data point: 1·3 × 3⁴ = 1·3 × 81 = 105·3 ✓
Q.50 [Surds]
If x = 3 − 3^(1/3) − 3^(2/3), then what is x(x − 3)(x − 6) equal to?
x(x − 3)(x − 6) = (3 − t)(−t)(−3 − t) = t(3 − t)(3 + t) = t(9 − t²).
Now t² = (a + a²)² = a² + 2a³ + a⁴ = a² + 6 + 3a (using a³ = 3, a⁴ = 3a).
So 9 − t² = 3 − 3a − a².
t(9 − t²) = (a + a²)(3 − 3a − a²)
= 3a − 3a² − a³ + 3a² − 3a³ − a⁴
= 3a − 3a² − 3 + 3a² − 9 − 3a = −12.
Numerical check: a ≈ 1·4422, x ≈ −0·5223; (−0·5223)(−3·5223)(−6·5223) ≈ −12·00 ✓
Q.51 [Statistics]
If the mean of a set of 7 observations is 10 and the mean of a set of 3 observations is 5, then what is the combined mean?
Total = 7 × 10 + 3 × 5 = 70 + 15 = 85. Count = 10.
Combined mean = 85/10 = 8·5.
Trap: averaging 10 and 5 to get 7·5. The means must be weighted by the number of observations.
Q.52 [Alligation]
The average weekly wages of male employees in a company is ₹4,200 and that of females is ₹3,200. If the average weekly wage of all employees is ₹4,000, then what is the ratio of male to female employees?
Male : Female = (4000 − 3200) : (4200 − 4000) = 800 : 200 = 4 : 1.
Check: 4 males and 1 female → total wage = 4(4200) + 3200 = 20,000 over 5 employees = ₹4,000 ✓
Note the mean 4,000 sits closer to the male figure, so there must be more males — a useful sanity test before you compute.
Q.53 [Statistics]
Consider the observations x, x + 4, 30, 33, 74, 78, 49, 52, 98, 85. If the median of the data is 67, then what is the value of x?
The eight known values sorted: 30, 33, 49, 52, 74, 78, 85, 98.
If both x and x+4 were below 49, the middle pair would be 49 and 52 (median 50·5) ✗
No pair of the known values sums to 134 (52+74 = 126, 74+78 = 152), so x and x+4 must themselves occupy the middle two positions:
x + (x + 4) = 134 → 2x = 130 → x = 65.
Verify: sorted data 30, 33, 49, 52, 65, 69, 74, 78, 85, 98 → median = (65+69)/2 = 67 ✓
Q.54 [Averages]
A cyclist pedals from her house to her office at a speed of 3 kmph and back from the office to her house at 6 kmph. What is the average speed?
Average = 2uv/(u + v) = 2 × 3 × 6/(3 + 6) = 36/9 = 4 kmph.
Why not 4·5? She spends twice as long on the slow leg, so the slow speed gets more weight. The average speed is always less than the arithmetic mean of the two speeds.
Q.55 [Statistics]
If the heights (in cm) of 9 students of a class are 150, 165, 145, 149, 150, 147, 152, 144 and 148, then what is the algebraic sum of the heights measured from their arithmetic mean?
Σ(xᵢ − x̄) = Σxᵢ − n·x̄ = n·x̄ − n·x̄ = 0.
The sum of deviations of any data set taken about its own arithmetic mean is always zero — this is precisely why the mean is described as the centre of gravity of the data. (For the record, the mean here is 150, but you never needed it.)
Q.56 [Statistics]
Consider the grouped data — Class 0–10: 4, 10–20: 8, 20–30: 15, 30–40: 10, 40–50: 3. What is the mode of the above distribution?
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
= 20 + [(15 − 8)/(30 − 8 − 10)] × 10
= 20 + (7/12) × 10 = 20 + 5·833 = 25·83.
Q.57 [Statistics]
If the mean of 15 observations, namely x₁, x₂, x₃, …, x₁₅ is 2·5, then what is the value of Σ(i=1 to 15) [5(4xᵢ + 1)]?
Σ[5(4xᵢ + 1)] = 5 Σ(4xᵢ + 1) = 5 [4Σxᵢ + Σ1]
= 5 [4(37·5) + 15] = 5 [150 + 15] = 5 × 165 = 825.
Trap: forgetting that Σ1 taken 15 times is 15, not 1.
Q.58 [Statistics]
Consider the discrete grouped data — x: 12, 14, 10, 16, 18 with f: 6, 8, 10, 9, 3. What is the arithmetic mean of the above distribution?
Σfx = 12(6) + 14(8) + 10(10) + 16(9) + 18(3)
= 72 + 112 + 100 + 144 + 54 = 482.
Mean = Σfx/Σf = 482/36 = 13·39 (to two decimals).
Sense check: the largest frequency sits at x = 10, pulling the mean below the midpoint of the x-values — consistent with 13·39.
Q.59 [Statistics]
Consider the following incomplete frequency distribution with a total frequency of 100 — Class 0–10: 10, 10–20: x, 20–30: 20, 30–40: y. If the mean of the frequency distribution is 25, then what are the missing frequencies?
Equation 2 (mean): class midpoints are 5, 15, 25, 35.
Σfx = 10(5) + 15x + 20(25) + 35y = 550 + 15x + 35y.
Mean 25 → Σfx = 25 × 100 = 2500, so 15x + 35y = 1950 → 3x + 7y = 390.
Substituting x = 70 − y: 3(70 − y) + 7y = 390 → 210 + 4y = 390 → y = 45, x = 25.
Answer: x = 25, y = 45.
Q.60 [Data Interpretation]
The data gives investment (₹ in crores) by a governmental organization — Science, Technology and Environment: 7,200; Transport and Communication: 3,000; Rural Development: 1,500; Industry and Minerals: 3,800; Agricultural Services: 2,500.
I. The percentage investment in the 'Science, Technology and Environment' category is 40%.
II. The angle (in degrees) for 'Agricultural Services' in a pie diagram is 50°.
I. 7200/18000 × 100 = 40% ✓ Correct.
II. Angle = (2500/18000) × 360° = 2500/50 = 50° ✓ Correct.
Answer: Both I and II.
Shortcut for pie angles: when the total is 18,000, each 50 crore is 1°, so you can read the angle by dividing the component by 50.
Q.61 [Geometry — Circles]
[Items 61–63] The centre O of a circle inside a triangle ABC is at a distance of 13 cm from each of the vertices of the triangle. The diameter of the circle is 10 cm and the circle touches only two sides of the triangle, AB and AC.
If x is the perimeter, in cm, of the triangle, then which one of the following is correct?
Since the circle touches AB, the perpendicular distance from O to AB is 5. For a chord at distance d from the centre of a circle of radius R, half the chord = √(R² − d²):
AB/2 = √(13² − 5²) = √(169 − 25) = √144 = 12 → AB = 24. Similarly AC = 24.
For BC, place O at the origin with A = (0, 13). By symmetry B and C are mirror images. From AB = 24:
x² + y² = 169 and x² + (y − 13)² = 576 → −26y + 169 = 576 − 169 → y = −119/13.
x² = 169 − (119/13)² = 14400/169 → x = 120/13, so BC = 240/13 ≈ 18·46.
Perimeter = 24 + 24 + 18·46 = 66·46 cm → lies between 65 and 67.
Q.62 [Geometry — Triangles]
[Item 62 of the same set] If y is the area in cm² of the triangle, then which one of the following is correct?
Height from A to BC = 13 + 119/13 = (169 + 119)/13 = 288/13.
Area = ½ × base × height = ½ × (240/13) × (288/13)
= ½ × 69120/169 = 34560/169 ≈ 204·5 cm²
which lies between 200 and 210 cm².
Cross-check with the circumradius formula: Area = abc/(4R) = (24 × 24 × 240/13)/(4 × 13) = 204·5 ✓
Q.63 [Geometry — Triangles]
[Item 63 of the same set] Consider the following statements:
I. ∠ABC lies between 60° and 90°.
II. If z is the distance in cm from the centre O of the circle to the midpoint of BC, then 7 cm < z < 8 cm.
cos B = BM/AB = (120/13)/24 = 5/13 → B = cos⁻¹(0·3846) ≈ 67·4°, which lies between 60° and 90°. Correct.
II. The midpoint of BC is at (0, −119/13) and O is at the origin, so
z = 119/13 ≈ 9·15 cm, which is not between 7 and 8. Incorrect.
Answer: I only.
Note: z ≠ 5 confirms the circle does not touch BC, exactly as the question stated.
Q.64 [Geometry — Circles]
[Items 64–66] A circle with centre O passes through the vertex A of an equilateral triangle ABC and touches BC at its midpoint M. The circle cuts AB at D and AC at E.
What is AD : DB equal to?
Now apply the tangent–secant (power of a point) theorem at B, where BM is the tangent and BDA is the secant:
BM² = BD × BA
(a/2)² = BD × a → BD = a/4.
Therefore AD = a − a/4 = 3a/4, and
AD : DB = (3a/4) : (a/4) = 3 : 1.
Q.65 [Geometry — Circles]
[Item 65 of the same set] If x is the area of the circle and y is the area of the triangle such that z = (x/y)², then which one of the following is correct?
Area of circle x = πr² = π(3a²/16) = 3πa²/16.
Area of triangle y = (√3/4)a².
x/y = (3πa²/16) ÷ (√3a²/4) = (3π/16)(4/√3) = 3π/(4√3) = π√3/4 ≈ 3·1416 × 1·7321/4 ≈ 1·3603.
z = (1·3603)² ≈ 1·851, which lies between 1·5 and 2.
Q.66 [Geometry — Triangles]
[Item 66 of the same set] What is the ratio of the area of the triangle ADO to the area of the quadrilateral BDOM?
From Q64, BD = a/4, so D divides BA one-quarter of the way from B: D = (a/8, a√3/8).
Area of ΔADO by the shoelace formula with A(a/2, a√3/2), D(a/8, a√3/8), O(a/2, a√3/4):
= ½ |(a/2)(a√3/8 − a√3/4) + (a/8)(a√3/4 − a√3/2) + (a/2)(a√3/2 − a√3/8)|
= ½ a²√3 |−1/16 − 1/32 + 3/16| = ½ a²√3 (3/32) = 3a²√3/64.
Area of ΔABM = ½ × (a/2) × (a√3/2) = a²√3/8 = 8a²√3/64.
Quadrilateral BDOM = ABM − ADO = (8 − 3)a²√3/64 = 5a²√3/64.
Ratio = 3 : 5.
Q.67 [Geometry — Triangles]
[Items 67–68] ABC is an equilateral triangle. BP is perpendicular to AC, PR is perpendicular to AB, and PQ is perpendicular to BC.
What is the ratio of AB² : BP² : PR²?
In right triangle APR (right angle at R), ∠A = 60° and AP = a/2:
PR = AP · sin 60° = (a/2)(√3/2) = a√3/4.
Now form the squares:
AB² = a²
BP² = 3a²/4
PR² = 3a²/16
Ratio = a² : 3a²/4 : 3a²/16 = 1 : 3/4 : 3/16 = multiply by 16 → 16 : 12 : 3.
Q.68 [Geometry — Triangles]
[Item 68 of the same set] What is the ratio of the area of Δ PRB to the area of Δ PQC?
Area = ½ × PR × RB = ½ × (a√3/4)(3a/4) = 3a²√3/32.
ΔPQC (right angle at Q): PC = a/2 and ∠C = 60°, so
PQ = PC sin 60° = a√3/4 and QC = PC cos 60° = a/4.
Area = ½ × (a√3/4)(a/4) = a²√3/32.
Ratio = 3a²√3/32 : a²√3/32 = 3 : 1.
Note PR = PQ; the whole ratio comes from RB being three times QC.
Q.69 [Geometry — Triangles]
[Items 69–70] PQR is a triangle such that QP = QR = 15 cm and PR = 18 cm. PN, QM and RT are the altitudes of the triangle which intersect at O.
What is QT : QO equal to?
RT is the altitude to PQ, so RT = 2 × Area/PQ = 216/15 = 14·4.
In right triangle QTR: QT = √(QR² − RT²) = √(225 − 207·36) = √17·64 = 4·2.
Now locate the orthocentre O. Put P(−9, 0), R(9, 0), Q(0, 12); O lies on the y-axis.
The altitude from P is perpendicular to QR, whose direction is (9, −12) ∝ (3, −4); so the altitude direction is (4, 3). Parametrising from P: (−9 + 4t, 3t); setting x = 0 gives t = 9/4 and y = 27/4 = 6·75.
So O = (0, 6·75) and QO = 12 − 6·75 = 5·25.
QT : QO = 4·2 : 5·25 = 4 : 5.
Q.70 [Geometry — Triangles]
[Item 70 of the same set] What is the ratio of QO to OM?
QO = 12 − 6·75 = 5·25
OM = 6·75 − 0 = 6·75
QO : OM = 5·25 : 6·75 = 525 : 675 = 7 : 9.
Check: QO + OM = 12 = QM ✓ and 7 + 9 = 16 parts, each of 0·75 ✓
Q.71 [Data Sufficiency]
[Data sufficiency] Question: ABCD is a rhombus of side 10 cm. What is the sum of the squares of the diagonals of the rhombus?
Statement I: AC : BD = 3 : 4.
Statement II: AC = 6 cm.
AC² + BD² = (2p)² + (2q)² = 4(p² + q²) = 4 × side².
Here side = 10, so AC² + BD² = 4 × 100 = 400 cm² — determined by the side alone.
Neither statement is needed. Answer: (d).
Note: Statement II is in fact inconsistent with a rhombus of side 10 and Statement I (it would force the other diagonal to 8, giving side 5, not 10) — a reminder to test consistency, not just sufficiency.
Q.72 [Data Sufficiency]
[Data sufficiency] Question: The diagonal of a rectangle ABCD (AB > BC) is 5√2 cm. What is its perimeter?
Statement I: Length AB is an integer.
Statement II: Length BC is an integer.
Statement I alone: AB an integer. AB = 7 → BC = 1 ✓; AB = 6 → BC = √14 ≈ 3·74 ✓ (still AB > BC). Two different perimeters — insufficient.
Statement II alone: BC an integer. BC = 1 → AB = 7 ✓; BC = 3 → AB = √41 ≈ 6·4 ✓. Two different perimeters — insufficient.
Both together: both integers with AB² + BC² = 50 gives (7,1) and (5,5); the second fails AB > BC. So AB = 7, BC = 1 and the perimeter = 2(7 + 1) = 16 cm.
Answer: (c).
Q.73 [Data Sufficiency]
[Data sufficiency] Question: ABC is a triangle right-angled at A and AD is perpendicular to BC. If AD = 7·2 cm, then what is BD × CD equal to?
Statement I: AB : AC = 3 : 4.
Statement II: BC = 15 cm.
AD² = BD × CD.
This follows from the similarity ΔBDA ~ ΔADC.
So BD × CD = (7·2)² = 51·84 cm², using nothing but the given AD.
Neither statement is required. Answer: (d).
Q.74 [Data Sufficiency]
[Data sufficiency] Question: ABCD is a quadrilateral with AB = 6 cm, BC = 8 cm, CD = 4√5 cm, DA = 2√5 cm and inscribed in a circle. What is the diameter of the circle?
Statement I: The area of the quadrilateral is 44 cm².
Statement II: ∠ABC = 90°.
Area (Brahmagupta): s = (6 + 8 + 4√5 + 2√5)/2 = 7 + 3√5 ≈ 13·708.
K = √((s−a)(s−b)(s−c)(s−d)) = √(7·708 × 5·708 × 4·764 × 9·236) = 44 — so Statement I is merely a consequence of the data, adding nothing.
Circumradius of a cyclic quadrilateral: R = (1/4K)·√((ab+cd)(ac+bd)(ad+bc)).
(ab+cd) = 48 + 40 = 88; (ac+bd) = 24√5 + 16√5 = 40√5; (ad+bc) = 12√5 + 32√5 = 44√5.
Product = 88 × 40√5 × 44√5 = 774400, √774400 = 880.
R = 880/(4 × 44) = 5 → diameter = 10 cm.
Neither statement is needed. Answer: (d). (Statement II is also a consequence: 6² + 8² = 10² confirms ∠ABC = 90°.)
Q.75 [Data Sufficiency]
[Data sufficiency] Question: ABC is a triangle inscribed in a semi-circle such that AC coincides with the diameter of the semi-circle. Let P be any point on the arc of the semi-circle. What is AP² + CP² + AC² equal to?
Statement I: The radius of the circle is 5 cm.
Statement II: AB = 6 cm and BC = 8 cm.
AP² + CP² + AC² = 2AC².
So all we need is AC.
Statement I: radius 5 → AC = 10 → answer = 200 cm². Sufficient.
Statement II: B is also on the semicircle, so ∠ABC = 90° and AC = √(6² + 8²) = 10 → answer = 200 cm². Sufficient.
Either alone works. Answer: (b).
Q.76 [Mensuration — Solids]
A thin metallic sheet, 1·92 m² in area, is cut into two equal pieces. One piece is used to make a hollow cube of volume P (in m³) and the other is used to make a hollow cuboid of volume Q (in m³) with dimensions in the ratio 4 : 2 : 1. If no sheet is wasted, which one of the following is correct?
Cube: 6s² = A → s = √(A/6), so P = s³ = (A/6)^(3/2).
Cuboid with sides 4k, 2k, k: surface area = 2(8k² + 2k² + 4k²) = 28k² = A → k = √(A/28).
Q = 4k·2k·k = 8k³ = 8(A/28)^(3/2).
Q/P = 8 (A/28)^(3/2) ÷ (A/6)^(3/2) = 8 (6/28)^(3/2) = 8 (3/14)^(3/2).
(Q/P)² = 64 × (3/14)³ = 64 × 27/2744 = 1728/2744 = 216/343.
So 343Q² = 216P².
Note the area cancels entirely — the relation holds for any sheet size.
Q.77 [Geometry — Triangles]
ABC is a triangle such that ∠ABC = 120°. If BD is the bisector of ∠B that meets AC at D, then what is the ratio of the area of Δ ABD to the area of Δ CBD?
[ABD] : [CBD] = AD : DC.
By the angle bisector theorem, AD : DC = AB : BC.
Therefore the ratio = AB : BC.
The 120° is a distractor — the result holds for any angle at B, because the bisector property does not depend on the angle's size.
Q.78 [Geometry — Similarity]
Consider the following statements:
I. The areas of two similar triangles are in the ratio of the squares of the corresponding medians.
II. The areas of two similar triangles are in the ratio of the squares of the corresponding heights.
I. Correct — medians are corresponding linear elements.
II. Correct — heights are corresponding linear elements.
Answer: Both I and II.
Q.79 [Geometry — Similarity]
If two similar scalene triangles ABC and PQR have the same area, then which of the following statements is/are correct?
I. Both the triangles are congruent.
II. Both the triangles have equal perimeters.
k = 1 means every pair of corresponding sides is equal, so the triangles are congruent — Statement I is correct.
Congruent triangles necessarily have equal perimeters — Statement II is correct.
Answer: Both I and II. (The word 'scalene' rules out the symmetry cases that sometimes confuse this argument.)
Q.80 [Geometry — Similarity]
In a triangle ABC, AB = 8 cm and D and E are points on AB and AC respectively such that DE is parallel to BC. If BC = 5DE, then what is AD × BD equal to?
AD/AB = DE/BC = 1/5 (since BC = 5DE).
AD = 8/5 = 1·6 cm, and BD = AB − AD = 8 − 1·6 = 6·4 cm.
AD × BD = 1·6 × 6·4 = 10·24 cm².
Trap: reading AD : DB as 1 : 5 instead of AD : AB = 1 : 5. The ratio from similarity is to the whole side, not to the remaining part.
Q.81 [Geometry — Circles]
In a cyclic quadrilateral ABCD, ∠A = (4x + 3)°, ∠B = (3y + 9)°, ∠C = (4y − 3)°, ∠D = (5x − 10)°. What is ∠A + ∠B equal to?
∠A + ∠C = 180: (4x + 3) + (4y − 3) = 180 → 4x + 4y = 180 → x + y = 45.
∠B + ∠D = 180: (3y + 9) + (5x − 10) = 180 → 5x + 3y = 181.
Substituting y = 45 − x: 5x + 135 − 3x = 181 → 2x = 46 → x = 23, y = 22.
∠A = 4(23) + 3 = 95°, ∠B = 3(22) + 9 = 75°.
∠A + ∠B = 170°.
Note ∠A + ∠B need not be 180° — that holds for opposite angles, not adjacent ones.
Q.82 [Mensuration — Areas]
The area of a rhombus is 720 cm² and the sum of its diagonals is 98 cm. What is the perimeter of the rhombus?
Area = ½d₁d₂ = 720 → d₁d₂ = 1440. Also d₁ + d₂ = 98.
(d₁ − d₂)² = (d₁ + d₂)² − 4d₁d₂ = 9604 − 5760 = 3844 → d₁ − d₂ = 62.
Solving: d₁ = 80, d₂ = 18.
Side = ½√(d₁² + d₂²) = ½√(6400 + 324) = ½√6724 = ½ × 82 = 41 cm.
Perimeter = 4 × 41 = 164 cm.
Q.83 [Mensuration — Circles]
A chord of a circle makes an angle of 90° at the centre of the circle. If the area of the major segment is k times the area of the minor segment, then what is the value of k? (Take π = 22/7)
Minor segment = area of quadrant − area of the right triangle at the centre
= (π/4)(1) − ½(1)(1) = π/4 − 1/2.
With π = 22/7: = 22/28 − 14/28 = 11/14 − 7/14 = 4/14 = 2/7.
Area of the whole circle = π = 22/7.
Major segment = 22/7 − 2/7 = 20/7.
k = (20/7) ÷ (2/7) = 10.
The value π = 22/7 is given precisely because the answer is a clean integer only with that approximation.
Q.84 [Mensuration — Areas]
ABC is a triangle in which (AB + BC) exceeds CA by 10 cm, (BC + CA) exceeds AB by 8 cm and (CA + AB) exceeds BC by 72 cm. What is the area of the triangle?
c + a − b = 10, a + b − c = 8, b + c − a = 72.
Adding all three: a + b + c = 90, so the semi-perimeter s = 45.
Now notice each condition is exactly twice one of Heron's terms:
s − b = (a + c − b)/2 = 5
s − c = (a + b − c)/2 = 4
s − a = (b + c − a)/2 = 36
Area = √(s(s−a)(s−b)(s−c)) = √(45 × 36 × 5 × 4) = √32400 = 180 cm².
Recognising that each 'exceeds' clause is 2(s − side) turns a messy system into one line.
Q.85 [Mensuration — Solids]
A reservoir is in the form of a cuboid. Its length is 30 m. If 24 kL of water is removed from the reservoir, the water level goes down by 20 cm. What is the width of the reservoir?
Volume removed = length × width × drop in level
24 = 30 × w × 0·2
24 = 6w → w = 4 m.
Unit conversion is the whole difficulty here — a candidate who leaves the depth in centimetres gets an answer 100 times too small.
Q.86 [Mensuration — Solids]
A well is dug with a diameter of 3·5 m and 16 m depth. The earth so excavated is spread in the form of a right circular cone of radius 7 m. What is the height of the cone?
Volume of the cone = ⅓πR²H = ⅓π(49)H.
Equating (no earth is lost):
⅓ × 49 × H = 49 → H = 3 m.
Both π values cancel, so you never need 22/7 here.
Q.87 [Mensuration — Circles]
The chord of a circle is 20 cm and the height of the minor segment is 5 cm. What is the diameter of the circle?
r² = (r − 5)² + 10²
r² = r² − 10r + 25 + 100
10r = 125 → r = 12·5 cm.
Diameter = 25 cm.
General sagitta relation: (half-chord)² = h(2r − h) → 100 = 5(2r − 5) ✓
Q.88 [Mensuration — Circles]
On a circular metal plate of uniform thickness, 16 holes, each of diameter 2 cm, are made. If the plate thereby has lost one-ninth of its original weight, then what is the diameter of the plate?
Area removed = 16 × π(1)² = 16π (each hole has radius 1 cm).
This is one-ninth of the original area πR²:
16π/(πR²) = 1/9 → R² = 144 → R = 12 cm.
Diameter = 24 cm.
Trap: using the hole diameter 2 cm as the radius, which would inflate the area four-fold.
Q.89 [Mensuration — Areas]
A square of side length 4 cm has its corners cut away in such a manner so as to form a regular octagon. What is the length of the side of the octagon? (Take √2 = 1·41)
For a regular octagon these must be equal:
4 − 2x = x√2 → 4 = x(2 + √2) → x = 4/(2 + √2) = 4(2 − √2)/2 = 4 − 2√2.
Octagon side = x√2 = (4 − 2√2)√2 = 4√2 − 4 = 4(√2 − 1)
= 4(1·41 − 1) = 4 × 0·41 = 1·64 cm.
Q.90 [Mensuration — Solids]
A tall cylindrical reservoir kept vertically is 20 m in diameter. Water is poured into it at the rate of 264 m³ per hour. What is the rate at which the water level rises in the reservoir per minute? (Take π = 22/7)
Rise per hour = volume rate ÷ base area = 264 ÷ (2200/7) = 264 × 7/2200 = 1848/2200 = 0·84 m per hour.
Rise per minute = 0·84/60 = 0·014 m = 1·4 cm.
The question asks per minute but gives the rate per hour — the division by 60 is exactly what option (d), 14 cm, is designed to catch.
Q.91 [Mensuration — Solids]
How many square metres of canvas (approximately) will be required to make a conical tent 3 m high so that a man 2 m tall may stand anywhere within a radius of 1 m from its centre without stooping?
height at distance d = 3(1 − d/R).
The man must clear 2 m at d = 1 m:
3(1 − 1/R) = 2 → 1 − 1/R = 2/3 → R = 3 m.
Slant height l = √(h² + R²) = √(9 + 9) = 3√2.
Canvas = curved surface area = πRl = π × 3 × 3√2 = 9√2π ≈ 9 × 1·414 × 3·1416 ≈ 40 m².
Only the curved surface is canvas — the base of a tent is the ground.
Q.92 [Coordinate Geometry]
PQRS is a square. M is a point on PS such that PM : MS = 2 : 1 and N is a point on SR such that SN : NR = 1 : 2. If the area of triangle MQN is 10 square units, then what is the perimeter of the square?
M on PS with PM : MS = 2 : 1 → M is two-thirds of the way from P to S → M = (0, a/3).
N on SR with SN : NR = 1 : 2 → N = (a/3, 0).
Shoelace with M(0, a/3), Q(a, a), N(a/3, 0):
Area = ½|0(a − 0) + a(0 − a/3) + (a/3)(a/3 − a)|
= ½|−a²/3 − 2a²/9| = ½ × 5a²/9 = 5a²/18.
Setting 5a²/18 = 10 → a² = 36 → a = 6.
Perimeter = 4 × 6 = 24 units.
Q.93 [Mensuration — Solids]
A cubical wooden block of length 10 cm is sliced to produce a cross-section of maximum area. What is the area of the cross-section?
Area = 10 × 10√2 = 100√2 cm² ≈ 141·4 cm².
Compare: a face gives 100 cm²; the regular hexagonal cross-section through the centre gives (3√3/4)(10√2)² /… ≈ 129·9 cm². The diagonal rectangle wins.
Q.94 [Geometry — Triangles]
In a triangle ABC, AD is the bisector of angle A that meets BC at D. If AB = 21 cm, DC = 20 cm and AB : AC = 3 : 4, then what is AC² − BD² equal to?
By the angle bisector theorem, BD : DC = AB : AC = 3 : 4, so
BD = (3/4) × DC = (3/4)(20) = 15 cm.
AC² − BD² = 28² − 15² = 784 − 225 = 559.
Q.95 [Mensuration — Areas]
A square of maximum area is inscribed in an equilateral triangle. If the side of the triangle is equal to (6 + 4√3) cm, then what is the area of the square?
s = a(h − s)/h → s(h + a) = ah → s = ah/(h + a).
Substituting h = a√3/2:
s = a(a√3/2) ÷ (a√3/2 + a) = a√3/(√3 + 2) = a√3(2 − √3) = a(2√3 − 3).
With a = 6 + 4√3:
s = (6 + 4√3)(2√3 − 3) = 12√3 − 18 + 24 − 12√3 = 6 cm.
Area = 6² = 36 cm².
The unusual side length is chosen precisely so that the surds cancel and s comes out a whole number.
Q.96 [Mensuration — Solids]
A room is L m long, B m wide and H m high. Further, L > B > H and L, B and H are integers. The length of the longest pole that can be placed in the room is 17 m and the length of the longest pole that can be placed on the floor is 15 m. What is the volume of the room?
The longest pole on the floor is the floor diagonal: √(L² + B²) = 15 → L² + B² = 225.
Subtracting: H² = 289 − 225 = 64 → H = 8.
Now find integers L > B > 8 with L² + B² = 225. The Pythagorean triple (9, 12, 15) gives L = 12, B = 9, and indeed 12 > 9 > 8 ✓
Volume = 12 × 9 × 8 = 864 m³.
Q.97 [Mensuration — Solids]
A spherical metal ball is molten and made into n smaller identical spheres. In this process, the surface area of the smaller balls increases by 900%. What is the value of n?
Volume is conserved: n·(4/3)πr³ = (4/3)πR³ → r = R/n^(1/3).
Total new surface area = n · 4πr² = 4πn·R²/n^(2/3) = 4πR² · n^(1/3).
So the surface area multiplies by n^(1/3):
n^(1/3) = 10 → n = 1000.
Trap: reading 900% as 'becomes 9 times', which gives n = 729 — option (a), placed there for exactly that error.
Q.98 [Mensuration — Areas]
ABCD is a parallelogram with ∠ABC = 150° and the sides have integer values. If the area of the parallelogram is 17·5 cm², then consider the following statements:
I. It is possible to have a perimeter equal to 24 cm.
II. It is possible to have a perimeter equal to 72 cm.
sin 150° = ½, so ab × ½ = 17·5 → ab = 35.
Integer factor pairs of 35: (1, 35) and (5, 7).
I. a = 5, b = 7 → perimeter = 2(5 + 7) = 24 cm ✓ Correct.
II. a = 1, b = 35 → perimeter = 2(1 + 35) = 72 cm ✓ Correct.
Answer: Both I and II. (A parallelogram with sides 1 and 35 at 150° is thin but perfectly valid.)
Q.99 [Mensuration — Solids]
N number of cubes each of side length equal to 10 cm are joined end to end in a row. If the total surface area of the resulting cuboid is 3800 cm², then what is the value of N?
Total surface area = 2(lb + bh + hl)
= 2(10N × 10 + 10 × 10 + 10 × 10N)
= 2(100N + 100 + 100N) = 400N + 200.
400N + 200 = 3800 → 400N = 3600 → N = 9.
Check: a 90 × 10 × 10 cuboid has surface area 2(900 + 100 + 900) = 3800 ✓
Q.100 [Geometry — Triangles]
In a quadrilateral ABCD, ∠ABC = 90° and ∠ACD = 90°. Let AB = p units, BC = q units, AC = r units, CD = s units, AD = t units, where p < q < r < s < t < 15. If p, q, r, s and t are integers, then what is the area of the quadrilateral?
ΔABC: p² + q² = r² ΔACD: r² + s² = t²
So r must be the hypotenuse of one triple and a leg of another, with all five values integers, strictly increasing and t < 15.
Taking (p, q, r) = (3, 4, 5), we need (5, s, t) with 5 < s < t < 15 — the triple (5, 12, 13) fits.
Check the ordering: 3 < 4 < 5 < 12 < 13 < 15 ✓
Area = ½(p)(q) + ½(r)(s) = ½(3)(4) + ½(5)(12) = 6 + 30 = 36 square units.
The constraint t < 15 is what makes the answer unique — it eliminates (6,8,10) with (10,24,26) and similar scalings.