CAT Integer Solutions Bible Top 10 Must Solve Integer Solutions Questions covering 80% of Syllabus – Page 2

log_(1/4) of (n² − 7n + 11) > 0 is _____. Using DPAC Strategy: Cycle n² − 7n + 11 < (1/4)^0→ n² − 7n + 11 < 1 Now we need values of n which satisfy 0 < n² − 7n + 11 < 1 Compute E(n) = n² − 7n + 11 for nearby integers: n = 1 → E = 5 (not in (0,1)) n = 2 → E = 1 (equals 1, excluded) n = 3 → E = −1 (negative, excluded) n = 4 → E = −1 (negative, excluded) n = 5 → E = 1 (equals 1, excluded) For n ≤ 0 or n ≥ 6, E ≥ 5 (too large).

Given: (m + 2n)(2m + n) = 27 Since m and n are integers, (m + 2n) and (2m + n) must be integer factors of 27. Possible factor pairs (m + 2n, 2m + n) are: (1, 27), (3, 9), (9, 3), (27, 1), (−1, −27), (−3, −9), (−9, −3), (−27, −1) We now solve each pair as a system: m + 2n = A; 2m + n = B From these: Multiply first equation by 2: 2m + 4n = 2A Subtract from second: (2m + n) − (2m + 4n) = B − 2A ⇒ −3n = B − 2A ⇒ n = (2A − B) / 3 Then m = A − 2n m + 2n 2m + n n = (2A − B)/3 m = A − 2n 2m − 3n Valid / Reject 1 27 (2 − 27)/3 = −25/3 — — Reject 3 9 (6 − 9)/3 = −1 3 − 2(−1) = 5 13 Valid 9 3 (18 − 3)/3 = 5 9 − 10 = −1 −17 Valid 27 1 (54 − 1)/3 = 53/3 — — Reject −1 −27 (−2 + 27)/3 = 25/3 — — Reject −3 −9 (−6 + 9)/3 = 1 −3 − 2 = −5 −13 Valid −9 −3 (−18 + 3)/3 = −5 −9 + 10 = 1 17 Valid −27 −1 (−54 + 1)/3 = −53/3 — — Reject Maximum possible valid value of 2m − 3n = 17

5, which are 11 and 12. To satisfy the sum constraint a + b + c + d = 46, we can take values 11, 11, 12, 12.

Final Answer: 12 Given: 3ac = 8(a + b) ⇒ 8b = 3ac − 8a ⇒ b = (3ac/8) − a Since a, b, c are natural numbers, 3ac must be divisible by 8. Now try small values of a (since we want the minimum value of 3a + 2b + c).

Given Information: Physics students = 75, Mathematics students = 111 , Chemistry students = 40 Assume: x - number of students who chose both Physics and Mathematics , y - students who chose both Mathematics and Chemistry , z - student who chose both Physics and Chemistry , t - number of students who chose all three subjects And z = y , y = x/2 (Given) Using Set formula (the total number of students who chose at least one subject) |P ∪ M ∪ C| = |P| + |M| + |C| − |P ∩ M| − |M ∩ C| − |P ∩ C| + |P ∩ M ∩ C| Substitute the values: 150 = 75 + 111 + 40 - x - y - z + t Now, substitute z = y and y = x/2: 150 = 75 + 111 + 40 - x - (x/2) - (x/2) + t 150 = 226 - x - x + t 150 = 226 - 2x + t 2x - 76 = t (Equation 1) We want to maximize 75 - x , (students who chose only Physics but not mathematics). 30 , 75 - x = 30 , x = 45 , t = 14 Since 75 - x is maximum at option 2 thus option 2 is the answer