CAT 2024 Question Paper (Slot 1, QA) Practice Questions & Answers

Let the number of apples be $5k$5k and mangoes be $2k$2k. Oranges = $187 - 7k$187−7k. After selling, unsold apples are $5k-75$5k−75 and unsold oranges are $(187-7k)/2$(187−7k)/2. Given the ratio: $(5k-75) / ((187-7k)/2) = 3/2$(5k−75)/((187−7k)/2)=3/2. Solving this gives $2(5k-75) = 3(187-7k)/2 \Rightarrow 20k-300 = 561-21k \Rightarrow 41k = 861 \Rightarrow k=21$2(5k−75)=3(187−7k)/2⇒20k−300=561−21k⇒41k=861⇒k=21. Apples=105, Mangoes=42, Oranges=40. Unsold fruits = $(105-75) + (42-26) + (40/2) = 30+16+20=66$(105−75)+(42−26)+(40/2)=30+16+20=66.

We need to find $\sqrt{29 - 12\sqrt{5}}$29−125​​. We can write this as $\sqrt{29 - 2\sqrt{180}}$29−2180​​. We look for two numbers whose sum is 29 and product is 180. The numbers are 20 and 9. So, the square root is $\sqrt{20} - \sqrt{9} = 2\sqrt{5} - 3$20​−9​=25​−3. To match the form $a+b\sqrt{n}$a+bn​, we can write this as $-3 + 2\sqrt{5}$−3+25​. This gives $a=-3, b=2, n=5$a=−3,b=2,n=5. Then $a+b+n = 4$a+b+n=4. Alternatively, we can write $2\sqrt{5}$25​ as $\sqrt{20}$20​. This gives $-3+\sqrt{20}$−3+20​, so $a=-3, b=1, n=20$a=−3,b=1,n=20. Then $a+b+n = -3+1+20 = 18$a+b+n=−3+1+20=18. The maximum value is 18.

The equation can be written in terms of powers of 2. Since $8=2^3$8=23 and $32768=2^{15}$32768=215, the equation is $(2^{-3})^k \times (2^{-15})^{1/3} = 2^{-3} \times (2^{-15})^{1/k}$(2−3)k×(2−15)1/3=2−3×(2−15)1/k. This simplifies to $2^{-3k} \times 2^{-5} = 2^{-3} \times 2^{-15/k}$2−3k×2−5=2−3×2−15/k, so $2^{-3k-5} = 2^{-3-15/k}$2−3k−5=2−3−15/k. Equating exponents gives $-3k-5 = -3-15/k$−3k−5=−3−15/k. Multiplying by ${k}$k yields the quadratic equation $3k^2+2k-15=0$3k2+2k−15=0. For a quadratic equation $ax^2+bx+c=0$ax2+bx+c=0, the sum of the roots is $-b/a$−b/a. Here, the sum of values for ${k}$k is $-2/3$−2/3.

The second inequality can be rewritten by completing the square for x and y: $4 \leq (x+2)^2 + (y-2)^2 \leq 8$4≤(x+2)2+(y−2)2≤8. This represents the area between two concentric circles centered at $(-2, 2)$(−2,2), with inner radius $r_1=2$r1​=2 and outer radius $r_2=\sqrt{8}$r2​=8​. The line $y = x+4$y=x+4 passes through the center of these circles $(-2, 2)$(−2,2). The inequality $y \geq x+4$y≥x+4 represents the half-plane above this line. Therefore, the required area is exactly half the area of the annulus, which is $\frac{1}{2} \pi (r_2^2 - r_1^2) = \frac{1}{2} \pi (8 - 4) = 2\pi$21​π(r22​−r12​)=21​π(8−4)=2π.

Total work for Renu is $15 \times 4 = 60$15×4=60 hours. Total work for Seema is $8 \times 5 = 40$8×5=40 hours. Let the total work be 120 units. Renu's rate is 2 units/hr, Seema's is 3 units/hr. Renu works 2 hrs/day ($h_R=2$hR​=2). Seema works double, so $h_S=4$hS​=4 hrs/day. Renu works double the days of Seema ($d_R = 2d_S$dR​=2dS​). Total work: $(2 \text{ units/hr} \times 2 \text{ hr/day} \times 2d_S \text{ days}) + (3 \text{ units/hr} \times 4 \text{ hr/day} \times d_S \text{ days}) = 120$(2 units/hr×2 hr/day×2dS​ days)+(3 units/hr×4 hr/day×dS​ days)=120. $8d_S + 12d_S = 120 \Rightarrow 20d_S=120 \Rightarrow d_S=6$8dS​+12dS​=120⇒20dS​=120⇒dS​=6.

Let incomes in September be 800, 600, 500. Total rent = $0.15(800) + 0.12(600) + 0.18(500) = 120 + 72 + 90 = 282$0.15(800)+0.12(600)+0.18(500)=120+72+90=282. New incomes in October are $800(1.1)=880$800(1.1)=880, $600(1.12)=672$600(1.12)=672, and $500(1.15)=575$500(1.15)=575. New total income = $880+672+575 = 2127$880+672+575=2127. Rent is still 282. Percentage is $(282/2127) \times 100 \approx 13.26\%$(282/2127)×100≈13.26%.

This is the number of surjective (onto) functions from a set with $|A|=6$∣A∣=6 elements to a set with $|B|=3$∣B∣=3 elements. Using the Principle of Inclusion-Exclusion, the number is $\sum_{k=0}^{3} (-1)^k \binom{3}{k} (3-k)^6 = \binom{3}{0}3^6 - \binom{3}{1}2^6 + \binom{3}{2}1^6 = 1(729) - 3(64) + 3(1) = 729 - 192 + 3 = 540$∑k=03​(−1)k(k3​)(3−k)6=(03​)36−(13​)26+(23​)16=1(729)−3(64)+3(1)=729−192+3=540.

Substitute ${a}$a from the second equation into the first: $4(x^2+y^2+z^2) = 4(x-y-z)-3$4(x2+y2+z2)=4(x−y−z)−3. Rearrange this into $4x^2-4x+4y^2+4y+4z^2+4z+3=0$4x2−4x+4y2+4y+4z2+4z+3=0. Completing the square gives $(2x-1)^2 + (2y+1)^2 + (2z+1)^2 = 0$(2x−1)2+(2y+1)2+(2z+1)2=0. Since $x, y, z$x,y,z are real, each squared term must be zero. This gives $x=1/2, y=-1/2, z=-1/2$x=1/2,y=−1/2,z=−1/2. Substituting these back into the first equation gives $a = 4((1/2)^2 + (-1/2)^2 + (-1/2)^2) = 4(1/4+1/4+1/4) = 3$a=4((1/2)2+(−1/2)2+(−1/2)2)=4(1/4+1/4+1/4)=3.

The sum of all numbers formed by ${n}$n distinct digits is $(n-1)! \times (\text{sum of digits}) \times (111...1 \text{ n times})$(n−1)!×(sum of digits)×(111...1 n times). Here, $n=4$n=4. The sum is $(4-1)! \times (a+b+c+d) \times 1111 = 6 \times (a+b+c+d) \times 1111 = 6666(a+b+c+d)$(4−1)!×(a+b+c+d)×1111=6×(a+b+c+d)×1111=6666(a+b+c+d). Let $S=a+b+c+d$S=a+b+c+d. Then $6666S = 153310+n$6666S=153310+n. Dividing gives $S = (153310+n)/6666$S=(153310+n)/6666. $153310/6666 \approx 23$153310/6666≈23. So we test $S=23$S=23. $6666 \times 23 = 153318$6666×23=153318. This means $153318=153310+n$153318=153310+n, which gives $n=8$n=8. The required value is $S+n = 23+8=31$S+n=23+8=31.

We need to find $10^{100} \pmod 7$10100(mod7). Since $10 \equiv 3 \pmod 7$10≡3(mod7), this is equivalent to finding $3^{100} \pmod 7$3100(mod7). By Fermat's Little Theorem, $3^6 \equiv 1 \pmod 7$36≡1(mod7). We write $100$100 as $100 = 6 \times 16 + 4$100=6×16+4. So, $3^{100} = (3^6)^{16} \times 3^4 \equiv 1^{16} \times 3^4 \equiv 81 \pmod 7$3100=(36)16×34≡116×34≡81(mod7). Since $81 = 11 \times 7 + 4$81=11×7+4, the remainder is 4.