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: Let Apple, Bombardier, Chat.sun and Doomsayer do 1/a ,1/b,1/c and 1/d parts of work per day respectively . Then , 5(1/a +1/b) +9(1/c +1/b) +4(1/c +1/d) =1 ; 7(1/a +1/b) +6(1/c +1/b) +5(1/c +1/d) =1
and 1/a +1/b + 1/c +1/d = 1/n (say)
=>12( 1/c +1/b) = 1-3/n which implies n>3 and 4( 1/a +1/b) = 1- 7/n which implies n>7 .
Consequently, 1/c +1/d = 1/n - ( 1/a +1/b) = 1/n -1/4(1-7/n) =(11-n)/4n
which implies n<11; So, from the given options n=11 is not possible
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.
Let divya work alone for a days Raveena work alone for b days and they work together for c days then a/20 + b/25 + 9c/80 = 1
=> 20 a + 16b + 45 c = 400
=> here when c=4 ; a= 7 and b = 5 it satisfies the above equation . 5+4+7 = 16 is possible .
when c=4 ; a= 3 and b = 10 it satisfies the above equation . 3+4+10 = 17 is possible
a+b+c = 10 or 14 is not possible.
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