Algebra – MBA CET questions with solutions

Approach: x + 1/x = 2 forces x = 1 (for real x).

Solution:
x + 1/x = 2 → x² − 2x + 1 = 0 → (x − 1)² = 0 → x = 1.
(2x² + 2)/(3x² + 5x + 3) = (2 + 2)/(3 + 5 + 3) = 4/11.

Cetking shortcut: Divide numerator and denominator by x: 2(x + 1/x) / (3(x + 1/x) + 5) = 4/(6 + 5) = 4/11.

Answer: A

Approach: Remove brackets from the innermost outward, flipping signs after a minus.

Solution:
{2a³ − 7a³} = −5a³ ⇒ −{…} = +5a³.
Inside [ ]: 3a³ + 4a² + 5a³ + 5a³ − 7a² = 13a³ − 3a².
2a³ − [13a³ − 3a²] = 2a³ − 13a³ + 3a² = −11a³ + 3a².

Why the other options are wrong:
C 11a³ − 3a²: whole sign reversed.
D −11a³ − 3a²: forgets to flip the a² sign.
B, E: wrong coefficients.

Cetking shortcut: Put a = 1: 2 − [3 + 4 + 5 + 5 − 7] = 2 − 10 = −8; only A gives −11 + 3 = −8.

Answer: A

Approach: Match with (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Solution:
Square terms give 2a, 3b, c (signs to fix).
+12ab = 2(2a)(3b) ⇒ 2a and 3b have the same sign.
−4ac = 2(2a)(−c) and −6bc = 2(3b)(−c) ⇒ c has the opposite sign.
So the expression = (2a + 3b − c)² ⇒ square root = 2a + 3b − c.

Why the other options are wrong:
B / C 2a − 3b − c: would make the ab term −12ab.
A 2a + b − 3c, E 2a + b − 2c: squares give b², not 9b².

Cetking shortcut: Only the sign pattern matters: ab positive, ac and bc negative ⇒ c alone is negative ⇒ D.

Answer: D

Approach: Use x + y + z = 0 to replace each pair sum by the negative of the third.

Solution:
x + y = −z, y + z = −x, x + z = −y.
(x + y)(y + z)(x + z) = (−z)(−x)(−y) = −xyz.

Why the other options are wrong:
D xyz: misses one minus sign (three negatives give a negative).
A, B, E: B and E are 0 here; A is unrelated.

Cetking shortcut: Take x = 1, y = 1, z = −2: (2)(−1)(−1) = 2 and −xyz = 2 ✓.

Answer: C

Approach: Factor theorem: (x − a) is a factor if f(a) = 0.

Solution:
f(x) = x³ − 19x + 30
x + 2 → f(−2) = −8 + 38 + 30 = 60 ≠ 0
x + 1 → f(−1) = −1 + 19 + 30 = 48 ≠ 0
x − 2 → f(2) = 8 − 38 + 30 = 0 ✓
(Full factorisation: (x − 2)(x − 3)(x + 5))

Why the other options are wrong:
A: f(−2) = 60
B: f(−1) = 48
D: f(1) = 12
E: f(7) = 343 − 133 + 30 = 240

Answer: C

Approach: Factor theorem gives one equation; combine with the given linear equation.

Solution:
x + 1 is a factor → f(−1) = 0
2(−1)³ + a(1) + 2b(−1) + 1 = 0 → a − 2b − 1 = 0 → a = 2b + 1
Substitute in 2a − 3b = 4: 4b + 2 − 3b = 4 → b = 2
a = 5 → (a, b) = (5, 2)

Cetking shortcut: Test options in 2a − 3b = 4: only (5, 2) gives 10 − 6 = 4; confirm a − 2b = 1 ✓.

Answer: B

Approach: Identity a² − b² = (a + b)(a − b).

Solution: a + b = 20, a − b = 8.
a² − b² = 20 × 8 = 160.
(Check: numbers are 14 and 6 → 196 − 36 = 160 ✔)

Answer: C

Approach: a² − b² = (a + b)(a − b).

Solution:
a + b = 20, a − b = 8.
a² − b² = 20 × 8 = 160.
(Numbers are 14 and 6: 196 − 36 = 160 ✓.)

Answer: B

Approach: Equate pairs of line sums to get linear relations among x, y, z.

Solution:
Grid: [3x+4y, 2x, 2x+y+z] / [2x², 4y, y²+z] / [y+z, 3x+2z, z−1].
Row 1 = 7x + 5y + z; Column 2 = 5x + 4y + 2z → z = 2x + y.
Row 3 = 3x + y + 4z − 1; equate with Row 1 → 4x + 4y − 3z + 1 = 0 → with z = 2x + y: y = 2x − 1, z = 4x − 1.
Anti-diagonal = 2x + 6y + 2z; equate with Row 1 → 5x = y + z = 6x − 2 → x = 2, y = 3, z = 7.
N = 7(2) + 5(3) + 7 = 36. Check Row 2: 8 + 12 + 16 = 36 ✓; Column 1: 18 + 8 + 10 = 36 ✓; Main diagonal: 18 + 12 + 6 = 36 ✓

Cetking shortcut: Use only linear lines (Row 1, Column 2, Row 3, anti-diagonal) — avoid the x² and y² cells until the check.

Answer: B

Approach: Match with (4a − q)².

Solution:
16a² = (4a)², so p = 4a
2 × 4a × q = 12a → q = 3/2
Add q² = 9/4 → 16a² − 12a + 9/4 = (4a − 3/2)²

Cetking shortcut: Number to add = (b²)/(4a) = 144/64 = 9/4.

Answer: A