CAT 2024 Question Paper (Slot 2, QA) Practice Questions & Answers

By Fermat's Little Theorem, $3^{10} \equiv 1 \pmod{11}$310≡1(mod11). We can write $333 = 10 \times 33 + 3$333=10×33+3. Thus, $3^{333} = (3^{10})^{33} \times 3^3 \equiv 1^{33} \times 27 \equiv 27 \pmod{11}$3333=(310)33×33≡133×27≡27(mod11). The remainder of 27 divided by 11 is 5.

If $x<0$x<0, $|x|+x=0$∣x∣+x=0, so $y=15$y=15. The second eq becomes $x+|15|-15=20$x+∣15∣−15=20, so $x=20$x=20, which contradicts $x<0$x<0. Thus $x \ge 0$x≥0, and $|x|=x$∣x∣=x. The first eq becomes $2x+y=15$2x+y=15. If $y<0$y<0, $|y|=-y$∣y∣=−y, so the second eq is $x-2y=20$x−2y=20. Solving gives $x=10, y=-5$x=10,y=−5. Then $x-y=15$x−y=15. If $y \ge 0$y≥0, the second eq is $x=20$x=20. Then $2(20)+y=15$2(20)+y=15 gives $y=-25$y=−25, a contradiction. So $x=10, y=-5$x=10,y=−5 is the only solution. $x-y = 10 - (-5) = 15$x−y=10−(−5)=15.

We are given $m^n = 2^{25} \times 3^{40}$mn=225×340. We can write $m = 2^{25/n} \times 3^{40/n}$m=225/n×340/n. Since m is a natural number, n must be a common divisor of 25 and 40. The divisors of 25 are 1, 5, 25. The divisors of 40 are 1, 2, 4, 5, 8, 10, 20, 40. The common divisors are 1 and 5. Since $n>1$n>1, ${n}$n must be 5. Then $m=2^5 \times 3^8 = 32 \times 6561 = 209952$m=25×38=32×6561=209952. So $m-n = 209952 - 5 = 209947$m−n=209952−5=209947.

Let $g(x) = f(x)+1$g(x)=f(x)+1. Then $g(xy)-1 = (g(x)-1)(g(y)-1) + (g(x)-1) + (g(y)-1) = g(x)g(y) - 1$g(xy)−1=(g(x)−1)(g(y)−1)+(g(x)−1)+(g(y)−1)=g(x)g(y)−1. So, $g(xy) = g(x)g(y)$g(xy)=g(x)g(y). For any prime ${p}$p, $g(p)=f(p)+1=2$g(p)=f(p)+1=2. $160000 = 16 \times 10^4 = 2^4 \times (2\times5)^4 = 2^8 \times 5^4$160000=16×104=24×(2×5)4=28×54. $g(160000) = g(2^8)g(5^4) = (g(2))^8(g(5))^4 = 2^8 \times 2^4 = 2^{12} = 4096$g(160000)=g(28)g(54)=(g(2))8(g(5))4=28×24=212=4096. So $f(160000)=g(160000)-1=4095$f(160000)=g(160000)−1=4095.

Since $AB || CD$AB∣∣CD, $\triangle EDC \sim \triangle EAB$△EDC∼△EAB. The ratio of corresponding sides is $CD/AB = 1/2$CD/AB=1/2. Let $ED=x, EC=y$ED=x,EC=y. Then $EA=2x, EB=2y$EA=2x,EB=2y. So $AD=x, BC=y$AD=x,BC=y. Perimeter of $ABCD = AB+BC+CD+DA = 2+y+1+x=6$ABCD=AB+BC+CD+DA=2+y+1+x=6, so $x+y=3$x+y=3. The perimeter of $\triangle AEB = EA+EB+AB = 2x+2y+2 = 2(x+y)+2 = 2(3)+2 = 8$△AEB=EA+EB+AB=2x+2y+2=2(x+y)+2=2(3)+2=8.

Let the rates of work be a, v, s. Total work = 300 units. $a+v = 2$a+v=2, $v+s = 3$v+s=3. From the work sequence: $75a + 135v + 45s = 300$75a+135v+45s=300. We can write this as $75a + 75v + 60v + 45s = 300$75a+75v+60v+45s=300, so $75(a+v) + 15v + 45(v+s) - 45v = 300$75(a+v)+15v+45(v+s)−45v=300. This doesn't simplify well. Try: $75a + 135v + 45s = 75(a+v) + 60v + 45s = 75(2) + 60v + 45s = 150 + 60v + 45s = 300$75a+135v+45s=75(a+v)+60v+45s=75(2)+60v+45s=150+60v+45s=300. $60v+45s = 150 \implies 4v+3s=10$60v+45s=150⟹4v+3s=10. With $v+s=3$v+s=3, we get $v=1, s=2$v=1,s=2. Then $a=1$a=1. The combined rate for the new scenario on day ${k}$k is ${a}$a if $k \pmod 2 eq 0$k(mod2)eq0 and $k \pmod 3 eq 0$k(mod3)eq0, $a+v$a+v if $k \pmod 2 = 0$k(mod2)=0 and $k \pmod 3 eq 0$k(mod3)eq0, etc. Total rate over 6 days: $6a+3v+2s = 6(1)+3(1)+2(2) = 13$6a+3v+2s=6(1)+3(1)+2(2)=13. $300/13 approx 23.07$300/13approx23.07. So $23 \times 6 = 138$23×6=138 days completes $23 \times 13 = 299$23×13=299 units. Day 139 is a day 1 type: Amal works. Rate is 1. One more unit takes 1 day. Total = 139 days.

Let Bina's cost price be C. Selling price to Shyam = Rs. 4860. This is a 19% loss, so $0.81 \times C = 4860$0.81×C=4860, which means $C = 4860/0.81 = 6000$C=4860/0.81=6000. Let Hari's purchase price from Shyam be H. If Bina sold at H, she'd make a 17% profit. So $H = 1.17 \times C = 1.17 \times 6000 = 7020$H=1.17×C=1.17×6000=7020. Shyam's cost price is 4860 and his selling price is 7020. Shyam's profit is $7020 - 4860 = 2160$7020−4860=2160.

The series can be rewritten as $\sum_{n=1}^{\infty} (\frac{1}{5})^n [(\frac{1}{5})^n - (\frac{1}{7})^n] = \sum (\frac{1}{25})^n - \sum (\frac{1}{35})^n$∑n=1∞​(51​)n[(51​)n−(71​)n]=∑(251​)n−∑(351​)n. Both are infinite geometric series. The sum of the first is $\frac{1/25}{1-1/25} = \frac{1/25}{24/25} = 1/24$1−1/251/25​=24/251/25​=1/24. The sum of the second is $\frac{1/35}{1-1/35} = \frac{1/35}{34/35} = 1/34$1−1/351/35​=34/351/35​=1/34. The total sum is $\frac{1}{24} - \frac{1}{34} = \frac{34-24}{24 \times 34} = \frac{10}{816} = \frac{5}{408}$241​−341​=24×3434−24​=81610​=4085​.

Using the change of base formula, $\log_x y = \frac{\log_z y}{\log_z x}$logx​y=logz​xlogz​y​. So, $\frac{\log_c(a+b)/\log_c 8}{1} + \frac{\log_c(a-b)/\log_c 27}{1} = \frac{2}{3}$1logc​(a+b)/logc​8​+1logc​(a−b)/logc​27​=32​. This is not quite right. It should be $\log_c (a+b)^{1/3} + \log_c (a-b)^{1/3} = 2/3$logc​(a+b)1/3+logc​(a−b)1/3=2/3. So $\log_c((a^2-b^2)^{1/3}) = 2/3$logc​((a2−b2)1/3)=2/3. $(a^2-b^2)^{1/3} = c^{2/3}$(a2−b2)1/3=c2/3, so $a^2-b^2=c^2 \implies a^2=b^2+c^2$a2−b2=c2⟹a2=b2+c2. Since $10 \ge b \ge c$10≥b≥c, the maximum value for $b^2+c^2$b2+c2 is $10^2+10^2=200$102+102=200. So $a^2 < 200$a2<200 (since $a>10$a>10 and we must have $a-b>0$a−b>0). $a < \sqrt{200} \approx 14.14$a<200​≈14.14. The greatest integer value of ${a}$a is 14.

Let the groups be $E_1$E1​ (first 10), $E_2$E2​ (middle 20), $E_3$E3​ (last 10). Let sums be $S_1, S_2, S_3$S1​,S2​,S3​. $S_1+S_2 = 30*40k = 1200k$S1​+S2​=30∗40k=1200k. $S_2+S_3 = 30*60k=1800k$S2​+S3​=30∗60k=1800k. $(S_1+S_3)/20 = 50k \implies S_1+S_3=1000k$(S1​+S3​)/20=50k⟹S1​+S3​=1000k. Solving gives $S_1=200k, S_3=800k, S_2=1000k$S1​=200k,S3​=800k,S2​=1000k. Total 2022 bonus = $S_1+S_2+S_3 = 2000k$S1​+S2​+S3​=2000k. Average 2022 = $2000k/40=50k$2000k/40=50k. In 2023, $S'_1=2S_1=400k$S1′​=2S1​=400k, $S'_3=3S_3=2400k$S3′​=3S3​=2400k, $S'_2=S_2=1000k$S2′​=S2​=1000k. New total = $400k+1000k+2400k=3800k$400k+1000k+2400k=3800k. Since $k=1000$k=1000, new total = 3,800,000. New average = $3800k/40 = 95k = 95000$3800k/40=95k=95000.