CMAT questions on Algebra - Number Theory

Key idea: if the HCF is 21, the numbers are 21a and 21b with a, b co-prime, and LCM = 21ab.

21ab = 3003, so ab = 143 = 11 × 13.

Both numbers must exceed 21, so we cannot take a = 1. That leaves a = 11, b = 13.

The numbers are 231 and 273.

Sum = 231 + 273 = 504 (or 21 × 24 = 504).

Answer: 504

Set-up: let the number be 10t + u (tens digit t, units digit u).

Double the tens digit and halve the units digit. The new number is 10(2t) + u/2.

Interchanging the digits gives 10u + t.

20t + u/2 = 10u + t

19t = 9.5u, so u = 2t.

The units digit is twice the tens digit (numbers like 12, 24, 36, 48).

Why the others fail: option 1 fails for 12, whose digit sum is 3. Options 2 and 3 contradict u = 2t.

Answer: Digit in the unit's place is twice the digit in ten's place (option 4)

Given: abc > 0. Either all three are positive, or exactly two are negative. None of them is 0.

Option 2: \( \frac{ab}{c} = \frac{abc}{c^2} \). Here abc > 0 and c² > 0, so ab/c > 0 always. Must be true.

Others fail:

• Option 1: a = b = c = 1 gives a/b = 1 > 0.
• Option 3: a = b = c = 1 gives bc = 1 > 0.
• Option 4: a = 1, b = c = 2 gives a = 1 < bc = 4.

Answer: ab/c > 0 (option 2)

Given: 1/x is positive, so x > 0. 1/y is negative, so y < 0.

• (A) 1/x + 1/y: a positive plus a negative can have either sign. For example, x = 1, y = −2 gives +0.5, while x = 2, y = −1 gives −0.5. Not necessarily true.
• (B) 1/x − 1/y = positive − (negative) = positive + positive. Always true.
• (C) \( \frac{x-y}{xy} \): x − y > 0 (positive minus negative), and xy < 0. Positive ÷ negative is negative. Always true. (In fact (x − y)/xy = 1/y − 1/x, which is the negative of (B).)
• (D) \( \frac{1}{x^2} - \frac{1}{y^2} \) depends on whether |x| or |y| is bigger. Not necessarily true.

Answer: Only (B) and (C) are true (option 2)

Statement I

• There are 8 numbers with mean 9, so their sum is 72.
• Known numbers: 7 + 8 + 9 + 10 + 8 + 7 = 49, so a + b = 23.
• 7 and 8 each already appear twice. For 8 to be the mode, 8 must occur more often than 7, so one of a, b must be 8.
• Then the other is 23 − 8 = 15, and a × b = 8 × 15 = 120.

Statement I is correct.

Statement II

• Add 1 to both sides: a + b + ab + 1 = 85, which factorises as (a + 1)(b + 1) = 85.
• 85 = 5 × 17 (the factorisation 1 × 85 would make a = 0, which is not positive). So a = 4 and b = 16.
• a + b = 20. Check: 4 + 16 + 64 = 84. ✓

Statement II is correct.

Answer: Both Statement I and Statement II are correct (option 1)

Identity: \( x^3 + y^3 = (x+y)^3 - 3xy(x+y) \).

\( = 3^3 - 3 \times 2 \times 3 = 27 - 18 = 9 \).

Check: x + y = 3 and xy = 2 give x = 1, y = 2. Then 1 + 8 = 9. ✓

Answer: 9