Conducting IIM: IIM Calcutta
The evening slot paper, often considered the 'decider' slot. This dataset allows for a comprehensive review of the hardest questions asked in 2024.
Highlights:
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The number of distinct integer solutions (x,y) of the equation ∣x+y∣+∣x−y∣=2, is:
8
The equation represents a square with vertices at (1,0), (0,1), (-1,0), and (0,-1). The integer solutions are the vertices and the points on the sides, which total to 8.
For some constant real numbers p,k and a, consider the following system of linear equations in x and y :\npx−4y=23x+ky=a\nA necessary condition for the system to have no solution for (x,y), is:
2a+k=0
For a system of linear equations to have no solution, the lines must be parallel and distinct. This means the ratio of coefficients of x and y must be equal, but not equal to the ratio of the constant terms. The condition is 3p=k−4=a2. From the inequality, we get ak=−12 and ap=6. From the equality we get pk=−12. The condition 2a+k=0 is a necessary condition for no solution.
If (a+b3)2=52+303, where a and b are natural numbers, then a+b equals:
8
Expanding the left side gives (a2+3b2)+2ab3. Comparing with the right side, we have a2+3b2=52 and 2ab=30, so ab=15. By testing integer pairs for ab=15 (e.g., a=5, b=3), we find 52+3(32)=25+27=52. So a=5 and b=3. Thus, a+b=8.
A circular plot of land is divided into two regions by a chord of length 103 meters such that the chord subtends an angle of 120∘ at the center. Then, the area, in square meters, of the smaller region is:
25(34π−3)
Let r be the radius. Using the law of cosines, (103)2=r2+r2−2r2cos(120∘). This gives 300=3r2, so r=10. The area of the sector is 360120πr2=3100π. The area of the triangle formed by the chord and radii is 21r2sin(120∘)=253. The area of the smaller region is the area of the sector minus the area of the triangle.
In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are:
72 and 80, respectively
The number of students who opted for both is given by (Number swimming) + (Number running) - (Total students). This must be calculated for the minimum and maximum number of girls (110 and 150) to find the range.
If 1068 is divided by 13 , the remainder is:
9
Using Fermat's Little Theorem, ap−1≡1(modp). So, 1012≡1(mod13). We can write 1068=(1012)5⋅108. The remainder is the same as 108(mod13). 102=100≡9(mod13), so 108=(102)4≡94(mod13)≡(−4)4(mod13)≡256(mod13)≡9(mod13).
The midpoints of sides AB, BC, and AC in △ ABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of △ ABC is 1440 sq cm, then the area, in sq cm, of △ XYZ is:
90
△ MNP is the medial triangle of △ ABC, so its area is 1/4 of Area(ABC). △ XYZ is the medial triangle of △ MNP. Thus, Area(XYZ) = (1/4) * Area(MNP) = (1/4) * (1/4) * Area(ABC) = 1440/16 = 90.
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was:
25
Let the first increment be x%. The new salary is S(1+x/100)(1+2x/100)=1.875S. Solving (1+x/100)(1+2x/100)=1.875 for x gives x=25.
Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayna in the same job. They undertake a job with an arrangement where Sam and Mohit work together on the first day, Sam and Ayna on the second day, Mohit and Ayna on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is:
3/10
Let Sam's rate be R. Mohit's is 2R, Ayna's is 2R/3. The work done in 3 days is (R+2R) + (R+2R/3) + (2R+2R/3) = 9R. Sam's work in this period is R+R = 2R. The fraction of work done by Sam is 2R/9R = 2/9. Whoops, rechecking... Let Sam's work rate be 3 units/day. Mohit's rate is 6 units/day. Ayna's rate is 2 units/day. Total work is 20*3=60 units. Day 1: S+M=9. Day 2: S+A=5. Day 3: M+A=8. Cycle of 3 days does 22 units. 2 cycles = 44 units in 6 days. Remaining 16 units. Day 7: S+M=9. Rem: 7. Day 8: S+A=5. Rem: 2. Day 9: M+A does 2 units in 2/8=1/4 day. Sam works on Day 1,2,4,5,7,8. Sam's total work: 3+3+3+3+3+3=18. Fraction = 18/60 = 3/10.
A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was:
30
Let initial water be 'w' litres. Initial milk is 300-w. 2w litres of solution is removed. Milk removed = 2w∗(300−w)/300. Remaining milk is (300−w)−2w(300−w)/300. This is equal to 72% of 300, which is 216. Solving the equation gives w=30.
