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At the bus stop 3/4 of commuters leave the bus and 7 new come in. This process repeats 3 times. What is the minimal number of people initially in the bus.
0.25(0.25 (0.25 x + 7) + 7) = x/64 + 7/16 + 7/4 is an integer, thus the minimum x is 52. Any quicker solution?
Yup, there is a quicker solution. First note that if x is the answer, then the number of people on the bus is always a linear function of x => we are looking for the smallest possible number of people in the end. If the number of people in the end is y, then the number in the beginning is 4(4(4(y-7)-7)-7)=64y-588. Hence y is at least 10, and in that case the minimum x is 52. It remains to check that 52 works, and it does. This isn't a radical change from your solution, but multiplication is usually easier than division, especially considering that this is probably a first-round problem, where no pen/paper are allowed. Another method: First step subtracts 7 from y, so y>6. Second step requires 4y>35, or equivalently y>8. Finally, y=9 doesn't work and y=10 does, so x=52.
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