Directions: Read the following information to answer the question.
There are four varieties of pipes Pipe A, Pipe B, Pipe C and Pipe D. Each pipe can be either an inlet pipe or an outlet pipe but cannot be both. there are 5 tanks of equal volume.
Tank P is filled by Pipe A and Pipe B
Tank Q is filled by Pipe A and Pipe C
Tank R is filled by Pipe A and Pipe D
Tank S is filled by Pipe B and Pipe C
Tank T is filled by Pipe C and Pipe D
Time taken for the first 3 tanks (P, Q and R) to get filled are in the ratio 1:2:4 and the time taken for the S and T tanks to be filled are in the ratio 7:10.
Find the outlet pipes among the 4 varieties.
Solution:
Let A,B,C and D do a,b,c and d with of work in an hour.
Let A and B fill the tank in 1 hour.
Then A and C would fill the tank in 2 hours while A and D in 4 hours.
a+b=1 ---------- (i)
a+c=1/2 ----------- (ii)
a+d=1/4 ------------ (iii)
Let B and C take 7k hours while c and D take 10k hours to fill the tank
⇒ b+c=1/7k --------------- (iv)
c+d=1/10k -------------- (v)
a=[(i)+(ii)−(iv)]/2 = [(ii)+(iii)−(v)]/2
⇒ a={1+1/2−1/7k}/2={1/2+1/4−1/10k}/2 ---------- (vi)
⇒ k=4/70
On substituting value of k in equation (vi) we get a<0
⇒ a<0,b>0,c>0 and d>0
Hence only A is the outlet pipe.
Directions: Read the following information to answer the question.
Solution:
When A and B work together, they will take √5×15=√225= 15 days.
Where 5 and 45 are the extra time that A and B take to complete the job if they work alone compared to the time that they will take if they worked together.
Directions: Read the following information to answer the question.
Solution:
Let us take the complete work to be 20 units (LCM of 5 and 4) to make things easier.
3a + 4b = 4 ---(1)
4 < 4a + 2b < 5 ---(2)
Double (1) and add it to (2) to get:
12 < 10 (a+b) < 13 -----(3)
Now,
Multiply (1) by 4 and subtract (3) from it
12a + 16b - 10 (a+b) [= 2a+6b]
= 4(4) - (12,13)
= (3,4) [ since 16 - 12 = 4, 16 - 13 = 3 =>the value will be between 12 and 3 and 4 ]
=> they will complete the work between 20/3 and 20/4 days (inclusive)
Thus, the work can get completed in 5th, 6th, 7th day.
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