Problems on Numbers Level - 3 (CAT PYQ)
Q1–10 of 50Find the number of non-negative integer values of k for which the quadratic equation x² − 5x + k = 0 has only integer roots.(CAT 2025 Slot 1)
Let the roots be $\alpha$ and $\beta$.
Sum of roots: $\alpha + \beta = 5$
Product of roots: $\alpha \beta = k$
Since $k \ge 0$, both roots must be non-negative integers (as their sum is $+5$).
Possible integer pairs $(\alpha, \beta)$ summing to $5$:
$(0, 5) \implies k = 0 \times 5 = 0$
$(1, 4) \implies k = 1 \times 4 = 4$
$(2, 3) \implies k = 2 \times 3 = 6$
Number of non-negative integer values of $k$: $3$.
Correct Answer: (B) 3
For what range of values of c is the minimum value of f(x) = x² − 4cx + 8c strictly greater than the maximum value of g(x) = −x² + 3cx − 2c?(CAT 2025 Slot 1)
min f = 8c−4c² (at x=2c). max g = 9c²/4−2c (at x=3c/2). Setting 8c−4c² > 9c²/4−2c and simplifying gives 25c² < 40c, i.e. 0 < c < 8/5.
If m and n are integers such that (m + 2n)(2m + n) = 27, then the maximum possible value of 2m − 3n is:(CAT 2025 Slot 2)
Let $A = m + 2n$ and $B = 2m + n$. Then $A \times B = 27$.
Solving for $m$ and $n$:
Target expression: $2m - 3n = 2\left(\frac{2B - A}{3}\right) - 3\left(\frac{2A - B}{3}\right) = \frac{7B - 8A}{3}$.
Test integer factors of 27 $(A, B)$:
If $A = -1, B = -27$: $2m - 3n = \frac{7(-27) - 8(-1)}{3} = \frac{-189 + 8}{3} = -60.33$ (Not integer)
If $A = 1, B = 27$: $m = \frac{54 - 1}{3}$ (not integer)
If $A = 3, B = 9$: $m = \frac{18 - 3}{3} = 5$, $n = \frac{6 - 9}{3} = -1$.
Check condition: $(5 + 2(-1))(2(5) + (-1)) = 3 \times 9 = 27$.
Value: $2(5) - 3(-1) = 10 + 3 = 13$.
If $A = -9, B = -3$: $m = \frac{-6 + 9}{3} = 1$, $n = \frac{-18 + 3}{3} = -5$.
Check condition: $(1 - 10)(2 - 5) = -9 \times -3 = 27$.
Value: $2(1) - 3(-5) = 2 + 15 = 17$.
Maximum value: $17$.
If a, b, c, and d are integers such that their sum is 46, then the minimum possible value of (a−b)² + (a−c)² + (a−d)² is:(CAT 2025 Slot 2)
To minimize squares of differences, $a, b, c, d$ should be as close to each other as possible.
$46 / 4 = 11.5 \implies$ set the values to $11, 11, 12, 12$ (sum $= 46$).
Let $a = 12$, $b = 12$, $c = 11$, $d = 11$:
Correct Answer: (C) 2
Suppose a, b, c are three distinct natural numbers such that 3ac = 8(a + b). The smallest possible value of 3a + 2b + c is:(CAT 2025 Slot 2)
Rearranging: $3ac - 8a = 8b \implies a(3c - 8) = 8b \implies b = \frac{a(3c-8)}{8}$.
Since $b > 0$, we must have $3c - 8 > 0 \implies c \ge 3$.
Testing small values for distinct natural numbers $a, b, c$:
If $c = 4 \implies b = \frac{a(12-8)}{8} = \frac{a}{2}$.
For $a = 2 \implies b = 1$. Distinct numbers: $a=2, b=1, c=4$.
Expression value: $3(2) + 2(1) + 4 = 6 + 2 + 4 = 12$.
Correct Answer: (B) 12
Let p, q, and r be three natural numbers such that their sum is 900, and r is a perfect square whose value lies between 150 and 500. If p is not less than 0.3q and not more than 0.7q, find the sum of the maximum and minimum possible values of p.(CAT 2025 Slot 3)
$p + q = 900 - r$.
Since $0.3q \le p \le 0.7q$, we have $1.3q \le p + q \le 1.7q \implies 1.3q \le 900-r \le 1.7q$.
Thus, $p = \frac{p}{p+q}(900-r)$.
Minimum fraction for $p$: $\frac{0.3}{1.3} = \frac{3}{13} \implies p_{\text{min}} \approx \frac{3}{13}(900-r)$
Maximum fraction for $p$: $\frac{0.7}{1.7} = \frac{7}{17} \implies p_{\text{max}} \approx \frac{7}{17}(900-r)$
Evaluating perfect squares $r \in (150, 500)$ to optimize integer values gives the sum of limits $= 397$.
Correct Answer: (C) 397
If (a + b√n) is the positive square root of (29 − 12√5), where a and b are integers and n is a natural number, the maximum possible value of (a + b + n) is:(CAT 2024 Slot 1)
Writing 12√5 as 2ab√n for different n=5k² forms: n=5 gives (a,b)=(−3,2), sum=4. n=20 (since √20=2√5) gives (a,b)=(−3,1): check (−3+√20)² = 9−12√5+20 = 29−12√5 ✓, and −3+√20≈1.47>0 (valid positive root). This gives a+b+n = −3+1+20 = 18, larger than the n=5 case.
If x is a positive real number such that 4log₁₀x + 4log₁₀₀x + 8log₁₀₀₀x = 13, then the greatest integer not exceeding x is:(CAT 2024 Slot 1)
Convert to base 10:
$$\log_{100} x = \frac{1}{2}\log_{10} x, \quad \log_{1000} x = \frac{1}{3}\log_{10} x$$Substitute:
$$4\log_{10} x + 4\left(\frac{1}{2}\log_{10} x\right) + 8\left(\frac{1}{3}\log_{10} x\right) = 13$$$$\left(4 + 2 + \frac{8}{3}\right)\log_{10} x = 13 \implies \frac{26}{3}\log_{10} x = 13$$$$\log_{10} x = \frac{13 \times 3}{26} = \frac{3}{2} = 1.5$$$x = 10^{1.5} = 10\sqrt{10} \approx 10 \times 3.162 = 31.62$.
$\lfloor x \rfloor = 31$.
Correct Answer: (B) 31
If m and n are natural numbers such that n > 1, and mⁿ = 2²⁵ × 3⁴⁰, then m − n equals:(CAT 2024 Slot 2)
Since $m$ and $n$ are natural numbers and $m^n = 2^{25} \times 3^{40}$, $m$ can be written as:
For $m$ to be an integer, $n$ must be a common positive divisor of the exponents $25$ and $40$.
Find the common divisors:
$$\gcd(25, 40) = 5$$The positive divisors of $5$ are $1$ and $5$.
Determine $n$:
Since $n > 1$, $n$ must be $5$.
Calculate $m$:
$$m = 2^{25/5} \times 3^{40/5} = 2^5 \times 3^8 = 32 \times 6561 = 209,952$$Calculate $m - n$:
$$m - n = 209,952 - 5 = 209,947$$
If (a + b√3)² = 52 + 30√3, where a and b are natural numbers, then a + b equals: (CAT 2024 Slot 3)
Expand LHS: $a^2 + 3b^2 + 2ab\sqrt{3} = 52 + 30\sqrt{3}$.
Equating parts:
$2ab = 30 \implies ab = 15$
$a^2 + 3b^2 = 52$
Factors of 15:
If $a = 5, b = 3$: $5^2 + 3(3^2) = 25 + 27 = 52$ (Satisfied!).
$a + b = 5 + 3 = 8$.
Correct Answer: (B) 8