CAT 2023 Slot 3 QA
Q1–10 of 22For a real number $x$, if $\dfrac{1}{2}$, $\dfrac{\log_3(2x-9)}{\log_3 4}$, and $\dfrac{\log_5\left(2x+\dfrac{17}{2}\right)}{\log_5 4}$ are in an arithmetic progression, then the common difference is
By change of base: log_3(2x-9)/log_3 4 = log_4(2x-9) and log_5(2x+17/2)/log_5 4 = log_4(2x+17/2).
So the AP is: 1/2, log_4(2x-9), log_4(2x+17/2)
Let u=2x-9. Then 2x+17/2 = u+35/2.
AP condition: 2 log_4 u = 1/2 + log_4(u+35/2)
log_4(u^2) - log_4(u+35/2) = 1/2
u^2/(u+35/2) = 4^(1/2) = 2
u^2 = 2u+35 => u^2-2u-35=0 => (u-7)(u+5)=0
Since u=2x-9>0 (log defined), u=7 => x=8.
Terms: 1/2, log_4 7, log_4(49/2)
Common difference d = log_4 7 - 1/2 = log_4 7 - log_4 2 = log_4(7/2).
Let $n$ and $m$ be two positive integers such that there are exactly 41 integers greater than $8^m$ and less than $8^n$, which can be expressed as powers of 2. Then, the smallest possible value of $n+m$ is
Answer: (c) 16
Powers of 2 strictly between 8^m=2^(3m) and 8^n=2^(3n): these are 2^k with 3mCount = (3n-1)-(3m+1)+1 = 3(n-m)-1
Set equal to 41: 3(n-m)-1=41 => n-m=14
Minimize n+m with n=m+14, m>=1: n+m=2m+14, minimized at m=1 => n+m=16.
For some real numbers $a$ and $b$, the system of equations $x+y=4$ and $(a+5)x+(b^2-15)y=8b$ has infinitely many solutions for $x$ and $y$. Then, the maximum possible value of $ab$ is
Answer: (b) 33
For infinitely many solutions, the two equations must be proportional:
(a+5)/1 = (b^2-15)/1 = 8b/4 = 2b
So a+5=2b and b^2-15=2b
From b^2-2b-15=0: (b-5)(b+3)=0 => b=5 or b=-3
Case b=5: a=2(5)-5=5, so ab=25
Case b=-3: a=2(-3)-5=-11, so ab=33
Maximum ab=33.
If $x$ is a positive real number such that $x^8+\left(\dfrac{1}{x}\right)^8=47$, then the value of $x^9+\left(\dfrac{1}{x}\right)^9$ is
Answer: (d) 34*sqrt(5)
Let S_n = x^n + 1/x^n. Using S_n = S_1*S_(n-1) - S_(n-2) (since x+1/x=S_1):Let A=S_2=x^2+1/x^2. Then S_4=A^2-2, and S_8=(A^2-2)^2-2=47
(A^2-2)^2=49 => A^2-2=7 (taking positive root, since A>=2) => A^2=9 => A=3
So S_1^2-2=3 => S_1^2=5 => S_1=sqrt(5) (x positive real)
Build up using S_n = sqrt(5)*S_(n-1) - S_(n-2):
S_1=sqrt(5), S_2=3
S_3=sqrt(5)(3)-sqrt(5)=2*sqrt(5)
S_4=sqrt(5)(2*sqrt(5))-3=10-3=7
S_5=sqrt(5)(7)-2*sqrt(5)=5*sqrt(5)
S_6=sqrt(5)(5*sqrt(5))-7=25-7=18
S_7=sqrt(5)(18)-5*sqrt(5)=13*sqrt(5)
S_8=sqrt(5)(13*sqrt(5))-18=65-18=47 (matches given)
S_9=sqrt(5)(47)-13*sqrt(5)=34*sqrt(5)
A quadratic equation $x^2+bx+c=0$ has two real roots. If the difference between the reciprocals of the roots is $\dfrac{1}{3}$, and the sum of the reciprocals of the squares of the roots is $\dfrac{5}{9}$, then the largest possible value of $(b+c)$ is
Answer: 9
Let roots be p,q; s=p+q=-b, m=pq=c.
|1/p-1/q| = |q-p|/m = 1/3 => (q-p)^2 = m^2/9 ...(I)
1/p^2+1/q^2 = (s^2-2m)/m^2 = 5/9 ...(II)
Also (q-p)^2=s^2-4m, so from (I): s^2-4m = m^2/9
From (II): 9(s^2-2m)=5m^2 => 9s^2=5m^2+18m
Substituting s^2=4m+m^2/9:
9(4m+m^2/9)=5m^2+18m => 36m+m^2=5m^2+18m => 18m-4m^2=0 => m(9-2m)=0
m=0 (rejected) or m=9/2
Then s^2 = 4(9/2)+(9/2)^2/9 = 18+9/4 = 81/4 => s=+-9/2
c=9/2, b=-s. To maximize b+c, take s=-9/2 so b=9/2:
b+c = 9/2+9/2 = 9
The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is
Answer: 468
Number of divisors formula: if 15=(a+1)(b+1)..., factor combinations of 15: 15x1, 5x3, 3x5, 1x15.
So a number with 15 divisors has form p^14 or p^4*q^2 (two primes).
Smallest such numbers (using smallest primes, higher exponent on smaller prime):
- 2^4*3^2=16*9=144 (divisors: 5*3=15) -- smallest
- Next smallest: 2^2*3^4=4*81=324
- (2^14=16384 is much larger, not relevant)
Sum of first two such numbers = 144+324 = 468
Let $n$ be any natural number such that $5n-1<3n+1$. Then, the least integer value of $m$ that satisfies $3n+1<2n+m$ for each such $n$, is
Answer: 5
First inequality: 5^(n-1) < 3^(n+1)
Taking logs: n < (ln3+ln5)/(ln5-ln3) = ln15/ln(5/3) ~ 2.708/0.510 ~ 5.31
So valid natural numbers: n in {1,2,3,4,5} (check: n=5: 5^4=625<3^6=729 OK; n=6: 5^5=3125>3^7=2187 fails)
We need least integer m such that 3^(n+1) < 2^(n+m) holds for every n in {1,...,5}.
For each n: m > (n+1)*log_2(3) - n
n=1: m>2(1.585)-1=2.17 => m>=3
n=2: m>3(1.585)-2=2.755 => m>=3
n=3: m>4(1.585)-3=3.34 => m>=4
n=4: m>5(1.585)-4=3.925 => m>=4
n=5: m>6(1.585)-5=4.51 => m>=5
The strictest requirement (largest n) needs m=5.
Check m=5,n=5: 3^6=729<2^10=1024 OK
Check m=4,n=5: 3^6=729<2^9=512? False.
So least m=5.
Rahul, Rakshita and Gurmeet, working together, would have taken more than 7 days to finish a job. On the other hand, Rahul and Gurmeet, working together would have taken less than 15 days to finish the job. However, they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. If Rakshita had worked alone on the job then the number of days she would have taken to finish the job, cannot be
Answer: (b) 21
Let R,K,G = daily rates of Rahul, Rakshita, Gurmeet.
"R+K+G together take more than 7 days" => R+K+G < 1/7
"R+G together take less than 15 days" => R+G > 1/15
Given: 6(R+K+G)+3K=1 (6 days all three, then 3 more days Rakshita alone)
=> R+K+G = (1-3K)/6
Constraint 1: (1-3K)/6 < 1/7 => 7-21K<6 => K>1/21
Constraint 2: R+G = (R+K+G)-K = (1-3K)/6-K = (1-9K)/6 > 1/15
=> 15-135K>6 => K<1/15
So 1/21 < K < 1/15, i.e., Rakshita's alone-time 1/K satisfies 15<1/K<21
Among options 17, 21, 16, 20 -- the value 21 is excluded (boundary, not strictly less than 21).
Cannot be 21.
Anil mixes cocoa with sugar in the ratio $3:2$ to prepare mixture A, and coffee with sugar in the ratio $7:3$ to prepare mixture B. He combines mixtures A and B in the ratio $2:3$ to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percentage of sugar in this drink will be
Answer: (d) 17
Mixture A (cocoa:sugar = 3:2): sugar fraction = 2/5
Mixture B (coffee:sugar = 7:3): sugar fraction = 3/10
Mix A:B = 2:3 -> take 2 units A + 3 units B = 5 units total (mixture C)
Sugar in C = 2*(2/5)+3*(3/10) = 4/5+9/10 = 8/10+9/10 = 17/10
Mix C (5 units) with equal amount of milk (5 units) -> total drink = 10 units, sugar unchanged = 17/10
Sugar % = (17/10)/10 * 100 = 17%
The population of a town in 2020 was $100000$. The population decreased by $y\%$ from the year 2020 to 2021, and increased by $x\%$ from the year 2021 to 2022, where $x$ and $y$ are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between $x$ and $y$ is 10, then the lowest possible population of the town in 2021 was