3. Start P₀=2, P₁=5. Compute P₂=23, P₃=110, P₄=527 and P₅=2525.
4. P₆=5×2525−527=12098. P₇=5×12098−2525=57965.
5. Independent identity check: P₃P₄=P₇+P₁, so P₇=110×527−5=57965.
Why this method works: The identity (x+1/x)²=x²+1/x²+2 determines P₁=5 because x is positive. To derive the shortcut, expand (x³+x⁻³)(x⁴+x⁻⁴): it equals x⁷+x⁻⁷+x+x⁻¹, hence P₇=P₃P₄−P₁.
Independent check and common trap: P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
G Strategy shortcut
Spot: Algebraic Identities. Apply: P₃=5(23−1)=110; P₄=23²−2=527. Then P₇=P₃×P₄−P₁=110×527−5=57,965. This skips P₅ and P₆. Check: P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
G Strategy: Spot: Algebraic Identities.
Apply: P₃=5(23−1)=110; P₄=23²−2=527. Then P₇=P₃×P₄−P₁=110×527−5=57,965. This skips P₅ and P₆.
Check: P₃=P₁³−3P₁=125−15=110 and P₄=P₂²−2=529−2=527. Thus 110×527−5=57965. This uses only two smaller powers and provides a separate check on the recurrence arithmetic.
Solve log base (x-4) of (x^2-16) = log base (x-4) of (x+2) + 2. Find the sum of all valid values of x.
Answer and solution
Correct answer: A
Solution
Answer: A — (3 + √57)/2
1. A logarithm needs a positive base other than 1 and positive arguments. Here x>4 and x≠5 ensure all these conditions.
2. Combine logarithms: x²−16=(x+2)(x−4)². Factor the left as (x−4)(x+4).
3. Since x−4≠0, divide by it: x+4=(x+2)(x−4)=x²−2x−8.
4. Solve x²−3x−12=0: x=(3±√57)/2.
5. Only (3+√57)/2≈5.275 satisfies the domain; it is not 5. Sum of valid roots=(3+√57)/2, option A.
Why this method works: The base x−4 must be positive and unequal to 1. This gives x>4 with x≠5; both log arguments are then positive. Replace the constant 2 by log base(x−4) of (x−4)² before combining the right side into a single logarithm.
Independent check and common trap: Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
G Strategy shortcut
Spot: Logarithms. Apply: Write the domain first: x>4, x≠5. Cancel x−4 only after this check. Reject the negative quadratic root; the sum is the one surviving root. Check: Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
G Strategy: Spot: Logarithms.
Apply: Write the domain first: x>4, x≠5. Cancel x−4 only after this check. Reject the negative quadratic root; the sum is the one surviving root.
Check: Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
If 6^(4x) × 9^(3x+6) × 25^y = 27^(2z) × 10^(4x) × 125^(x-2), where x, y and z are natural numbers, find the minimum possible value of x+y+z.
Answer and solution
Correct answer: 36
Solution
Answer: 36
1. Rewrite every base using primes 2,3,5. Left side is 2⁴ˣ×3¹⁰ˣ⁺¹²×5²ʸ. Right side is 2⁴ˣ×3⁶ᶻ×5⁷ˣ⁻⁶.
2. Equate exponents: z=(5x+6)/3 and y=(7x−6)/2.
3. For z to be an integer, x is divisible by 3. For y to be an integer, x is even. Thus x=6k.
4. Then y=21k−3 and z=10k+2. Nonnegative x=0 would give y=−3, so every natural-number convention requires k≥1.
5. x+y+z=37k−1 increases with k. At k=1: (x, y, z)=(6,18,12), total=36. The word minimum makes the answer unique.
Why this method works: Unique prime factorization permits comparison of the exponents of 2,3 and 5 separately. The powers of 2 already match, leaving two linear relations. Integrality then supplies the crucial divisibility conditions that ordinary real-variable solving would miss.
Independent check and common trap: At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
G Strategy shortcut
Spot: Indices and Prime Factorisation. Apply: Ignore matching powers of 2. Powers of 3 force x to be a multiple of 3; powers of 5 force x even. Smallest feasible x=6, yielding y=18, z=12. Check: At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
G Strategy: Spot: Indices and Prime Factorisation.
Apply: Ignore matching powers of 2. Powers of 3 force x to be a multiple of 3; powers of 5 force x even. Smallest feasible x=6, yielding y=18, z=12.
Check: At the smallest feasible x=6, y=(42−6)/2=18 and z=(30+6)/3=12. Check prime exponents:2 has 24 on both sides,3 has 72 and 5 has 36. Larger x=6k increases the sum by 37 each time, proving minimality.
Let f(x)=(x^2+5x)(x^2+5x+4). Solve √(f(x)+4)=2501 and find the sum of all real roots.
Answer and solution
Correct answer: B
Solution
Answer: B — -5
1. Let u=x²+5x. Then f(x)+4=u(u+4)+4=(u+2)².
2. The nonnegative square root is
u+2
, so
u+2
=2501 gives u=2499 or u=−2503.
3. For u=2499, x²+5x−2499=0 has discriminant 10021>0, hence two real roots.
4. For u=−2503, x²+5x+2503=0 has discriminant 25−10012<0, hence no real roots.
5. By Vieta, the two valid roots sum to −5, option B. Do not include the complex-root branch.
Why this method works: The square root of a square is an absolute value, not automatically the original expression. Therefore both signs of u+2 must be examined after completing the square; ignoring the negative branch without explanation is incomplete.
Independent check and common trap: Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.
G Strategy shortcut
Spot: Quadratic Equations and Perfect Squares.
Apply: Set u=x²+5x. The radical becomes
u+2
. Of u=2499 and −2503, only the first is possible since x²+5x≥−25/4. Vieta gives −5.
Check: Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.
G Strategy: Spot: Quadratic Equations and Perfect Squares.
Apply: Set u=x²+5x. The radical becomes |u+2|. Of u=2499 and −2503, only the first is possible since x²+5x≥−25/4. Vieta gives −5.
Check: Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.
Three natural numbers p, q and r add up to 1000. The number r is a perfect square between 200 and 600, and p is at least 25% of q but at most 75% of q. Find the sum of the maximum and minimum possible values of p.
Answer and solution
Correct answer: 417
Solution
Answer: 417
1. Allowed squares are 225,256,289,324,361,400,441,484,529,576. Set S=p+q=1000−r.
2. From p≥q/4 and q=S−p, obtain p≥S/5. From p≤3q/4 obtain p≤3S/7.
3. To maximize p choose largest S=775, at r=225. Then p≤2325/7=332.142…, so max p=332. q=443 satisfies both ratio bounds.
4. To minimize p choose smallest S=424, at r=576. Then p≥424/5=84.8, so min p=85. q=339 satisfies both bounds.
5. Both extremes are attainable natural numbers. Required sum=332+85=417.
Why this method works: The condition on p is relative to q, not to the combined sum S. Substituting q=S−p changes p≥q/4 into 5p≥S, and p≤3q/4 into 7p≤3S. This distinction explains the denominators 5 and 7.
Independent check and common trap: For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
G Strategy shortcut
Spot: Number Properties. Apply: Write S=1000−r. The ratio bounds give S/5≤p≤3S/7. Use largest square 576 for min p and smallest square 225 for max p; round inward. Check: For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
G Strategy: Spot: Number Properties.
Apply: Write S=1000−r. The ratio bounds give S/5≤p≤3S/7. Use largest square 576 for min p and smallest square 225 for max p; round inward.
Check: For the minimum, ceil(424/5)=85 givesq=339 and 85≥84.75. For the maximum, floor(3×775/7)=332 givesq=443 and 332≤332.25. Check the opposite inequality too; both are satisfied, so both endpoint constructions are feasible.
For real x, find the range of f(x)=(x-1)/(x^2+2x-2).
Answer and solution
Correct answer: A
Solution
Answer: A — (-∞, 1/6] ∪ [1/2, ∞)
1. The function is undefined at x=−1±√3. Set y=(x−1)/(x²+2x−2) and rearrange to yx²+(2y−1)x+1−2y=0.
2. For y≠0, real x requires Δ=(2y−1)²−4y(1−2y)=12y²−8y+1=(2y−1)(6y−1)≥0.
3. Thus y≤1/6 or y≥1/2. For y=0 the equation is linear: x=1, whose denominator is 1, so 0 is included.
4. No solution of the rearranged equation can have denominator zero: that would also require x−1=0, but the denominator at x=1 is 1.
5. Endpoints are attained: f(2)=1/6 and f(0)=1/2. Range=(−∞,1/6] ∪ [1/2,∞), option A.
Why this method works: Solve for x at a fixed proposed outputy. A y belongs to the range exactly when the resulting equation has a real, permitted x. The discriminant gives the range boundary, but y=0 changes the equation from quadratic to linear and needs its own check.
Independent check and common trap: At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
G Strategy shortcut
Spot: Functions and Range. Apply: Treat x as the unknown in yx²+(2y−1)x+1−2y=0. Its discriminant gives (2y−1)(6y−1)≥0. Separately check y=0 and denominator exclusions. Check: At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
G Strategy: Spot: Functions and Range.
Apply: Treat x as the unknown in yx²+(2y−1)x+1−2y=0. Its discriminant gives (2y−1)(6y−1)≥0. Separately check y=0 and denominator exclusions.
Check: At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
Let n be the least positive integer such that 180 is a factor of 270^n. If m is the least positive integer such that 270^n is a factor of 180^m, find m+n.
Answer and solution
Correct answer: B
Solution
Answer: B — 5
1. Prime factorizations: 180=2²×3²×5 and 270=2×3³×5.
2. For 180 to divide 270ⁿ, require n≥2,3n≥2, n≥1. Therefore the least n is 2.
3. Now 270²=2²×3⁶×5² must divide 180ᵐ=2²ᵐ×3²ᵐ×5ᵐ.
4. Need 2m≥2,2m≥6, m≥2. Least m=3. Hence m+n=5, option B.
Why this method works: For divisibility, each prime exponent in the divisor must be no larger than the corresponding exponent in the dividend. The strongest of these inequalities determines the smallest exponent; checking only one arbitrary prime can miss the constraint.
Independent check and common trap: For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
G Strategy shortcut
Spot: HCF and LCM / Prime Factorisation. Apply: 180=2²3²5 and 270=2·3³5. First the scarce 2 forces n=2; then the required 3⁶ forces m=3. Sum=5. Check: For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
G Strategy: Spot: HCF and LCM / Prime Factorisation.
Apply: 180=2²3²5 and 270=2·3³5. First the scarce 2 forces n=2; then the required 3⁶ forces m=3. Sum=5.
Check: For n=1 the power of 2 is too small, while n=2 works. Then m=2 gives only 3⁴, too small for the required 3⁶; m=3 supplies 2⁶,3⁶,5³ and works for every prime. Thus both minima are independently established.
3. For t≠0, solve 3t²−7t+6=0. Its discriminant is 49−72=−23, so there are no real t.
4. Thus the original equation has only the real solution x=0, which is an integer. Count=1.
Why this method works:
Replacing
x
by t does not discard negative x because x²=t² for both signs. Factor outt before dividing; otherwise the valid zero solution would be lost. A positive t would correspond to two opposite x values, but none exists here.
Independent check and common trap: The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.
G Strategy shortcut
Spot: Absolute Value and Quadratic Equations.
Apply: Put t=
x
. Then t[3t²−7t+6]=0. The quadratic has Δ=−23, so only t=0 and x=0; answer 1.
Check: The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.
G Strategy: Spot: Absolute Value and Quadratic Equations.
Apply: Put t=|x|. Then t[3t²−7t+6]=0. The quadratic has Δ=−23, so only t=0 and x=0; answer 1.
Check: The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.
A trader sells grain to three buyers. To the first buyer, the trader gives half the stock plus an extra 3 kg. To the second buyer, the trader gives half of the remaining stock plus an extra 3 kg. To the third buyer, the trader again gives half of the remaining stock plus an extra 3 kg, after which no grain is left. Find the initial quantity of grain.
Answer and solution
Correct answer: 42
Solution
Answer: 42
1. After each sale, remaining grain = previous stock −(half of it+3)=previous stock/2−3.
2. Invert the rule: previous stock=2(remaining stock+3).
3. Before buyer 3: 2(0+3)=6 kg. Before buyer 2: 2(6+3)=18 kg.
4. Before buyer 1: 2(18+3)=42 kg.
5. Forward check: sells 24, then 12, then 6 kg; total 42 and nothing remains.
Why this method works: Work backwards because the final remainder is known exactly. The forward map takes half the current stock and then subtracts 3; its inverse must add 3 first and then double. Doubling before adding 3 would invert a different operation.
Independent check and common trap: Forward verification from 42 kg: the first buyer receives 21+3=24, leaving 18; the second gets 9+3=12, leaving 6; the last gets 3+3=6, leaving 0. The three sales add to 42, so no stock is missing.
G Strategy shortcut
Spot: Linear Equations and Word Problems. Apply: Forward remainder=x/2−3, so reverse is x=2(remainder+3). Start from 0: 0→6→18→42. Check: Forward verification from 42 kg: the first buyer receives 21+3=24, leaving 18; the second gets 9+3=12, leaving 6; the last gets 3+3=6, leaving 0. The three sales add to 42, so no stock is missing.
G Strategy: Spot: Linear Equations and Word Problems.
Apply: Forward remainder=x/2−3, so reverse is x=2(remainder+3). Start from 0: 0→6→18→42.
Check: Forward verification from 42 kg: the first buyer receives 21+3=24, leaving 18; the second gets 9+3=12, leaving 6; the last gets 3+3=6, leaving 0. The three sales add to 42, so no stock is missing.
An arithmetic progression has the property that the 3rd, 6th and 9th terms add up to 84, while the sum of its first 12 terms is 402. Find the sum of its first 4 terms.
Answer and solution
Correct answer: A
Solution
Answer: A — -42
1. Write tn=a+(n−1)d. The three given terms sum to 3a+15d=84, so a+5d=28.
2. S₁₂=6(2a+11d)=402, hence 2a+11d=67.
3. Double the first equation to get₂a+10d=56. Subtract to obtain d=11.
4. Then a=28−55=−27. S₄=2(2a+3d)=2(−54+33)=−42.
5. Check the first four terms: −27,−16,−5,6 sum to −42, option A.
Why this method works: Terms 3,6,9 are equally spaced in an AP, so their sum is three times the middle termt₆. The mean of 12 terms lies halfway between t₆ andt₇; these two observations determine d without listing the sequence.
Independent check and common trap: The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
G Strategy shortcut
Spot: Arithmetic Progression. Apply: t₃+t₆+t₉=3t₆, so t₆=28. S₁₂/12=33.5=(t₆+t₇)/2, giving d=11. Then a=−27 and S₄=−42. Check: The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
G Strategy: Spot: Arithmetic Progression.
Apply: t₃+t₆+t₉=3t₆, so t₆=28. S₁₂/12=33.5=(t₆+t₇)/2, giving d=11. Then a=−27 and S₄=−42.
Check: The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
The full CAT 2026 Slot 3 mock on the real exam screen: VARC, DILR and Quant, 40 minutes each (120 minutes). It always opens the whole paper, whichever page you start from.