Directions: Read the following information to answer the question.
Solution:
If the amount of work done per day by Sara, Kyna, Riddhi per day are S/6, K/9, R/15 then K/9 + R/15 = S/n, R/15 + S/6 = K/n and S/6 + K/9 = R/n.
Eliminating S, K and R we get n^3 + 15n^2 = 405. How? Can you use determinants here and get this in just 1 step?
Let f(n) = n^3 + 15n^2 - 405 -> and is an increasing function for n > 0, f(4.5) < 0 and f(5) > 0.
Directions: Read the following information to answer the question.
Solution: Let Aravind complete 1/a portion of the work in a day
and Swarnali 1/b portion of the work in a day
Aravind on holiday for x days implies loss of (1/a)*x fraction of the work This loss is compensated by arvind and swarnali in the coming extra y days so
((1/a)+(1/b))*y =(1/a)*x
Similarly, for Swarnali's holiday of x days
((1/a)+(1/b))*z= (1/b)*x
Adding the two equations
((1/a)+(1/b))*(y+z)=x*((1/a)+(1/b)) implying y+z=x
x/2=(y+z)/2
Hence y,x/2,z are in AP
Directions: Read the following information to answer the question.
(a) If Akshay and John do the work without absence and Akshay digs 6 km, then the work gets completed in a little over 7 days
(b) n = 4/3
(c) If the work gets completed in 7.5 days, when Akshay and John dug without any absence, then Akshay dug 5 km
(d) all of the foregoing
(e) none of the foregoing
Solution: Let W be the amount of work done in digging. Let Akshay, John do W/a and W/b parts of work per day respectively.
Let N, p, q be the number of days in which the work can be done if Akshay, John do the work together without absence, with Akshay absent for 3 days, with John absent for 4 days respectively.
=> NW/a + NW/b = W, (p-3)W/a + pW/b = W and qW/a + (q-4)W/b = W.
Also, pW/b - NW/b = W/10 and qW/a - NW/a = nW/10.
=> p = b/10 + N, q = an/10 + N
=> (b/10 + N-3)/a + (b/10 + N)/b = 1; (na/10 + N)/a + (na/10 + N-4)/b = 1
=> (b/10 - 3)/a + 1/10 + N(1/a + 1/b) = 1 and n/10 + (na/10 - 4)/b + N(1/a+1/b) = 1, but N(1/a + 1/b) = 1
=> b/10 - 3 + a/10 = 0; and n/10(1/a + 1/b) = 4 => n = 4/3 and a+b = 30
Directions: Read the following information to answer the question.
A fill pipe can fill a tank in 20 hours, a drain pipe can drain a tank in 30 hours. If a system of n pipes (fill pipes and drain pipes put together) can fill the tank in exactly 5 hours, which of the following are possible values of n (More than one option could be correct)?
3 fill pipes cancel out 2 drain pipes. Plus, you need an additional 4 fill pipes fill the tank in 5 hours. so the answer has to be 5k + 4.
Both 54 and 29 are possible.
Directions: Read the following information to answer the question.
Let us say A and B split their share of the task and started doing their respective shares simultaneously.
Let’s say A takes A days to finish the task. Therefore, B takes A + 12 days to finish the entire task.
A has to finish 40% of the task, since B is doing the rest. So A will only take 2A/5 number of days.
B only has to finish 60% of the task, so B will take (3(A+12)/5) number of days.
But as we know, B starts working along with A and finishes 12 days after A stops working.
So, (3(A+12)/5) = ((2A)/5+12)
3A + 36 = 2A + 60
A = 24; B = 36 days.