Quant with RR  Arithmetic

A and B do a piece of work in 6 days and 9 days. They work on alternate days, starting with A on the first day. In how many days will the work be completed?
There are many ways to solve this kind of problem, but I will try to expalin throughly
Lcm of (6,9) = 18 units
A = 18/6 = 3 units in one day
B = 18/9 = 2 units in one day
So when they work together alternatively
In two days = 3+2 = 5 units
In other two day = 3+2 = 5 units
Again in two day = 3+2 = 5 units
Total
In 6 days = 15 units
Remaining = 18  15 = 3 units
Now A turn will come to finish this 3 units of work and he will do it in one day onlySo total days they took to complete the work = 6+1 = 7 days
If 1 litre of pop contains 85% water and 15% fruit juice, how much more water would need to be added to make it 99% water with 1% fruit juice?
total =1000ml
Water =850ml
Juice = 150ml
Now we are adding water that means juice is still untouched and it's given that now juice is 1 % of the quantity
Clearly
Total = 15000ml
Juice = 150 ml
So water = 14850ml
Water added = 14850850 = 14000ml I.e 14ltrPerson A can do a piece of work in 20 days. He works at it for 5 days and then person B finishes it in 10 more days. How many days will they need, in order to finish the work together?
A work for 5 days , so he work 5/20 = 1/4 work
Remaining 3/4 is done by B in 10 days
So B take 10*4/3 = 40/3 daysTogether
1/20 + 3/40
= 5/40
= 1/8
= 8 daysRam and Shayam run a race of 2000m. First, Ram gives Shayam a start of 200m and beats him by 30 seconds. Next Ram gives Shayam a lead of 3 minutes and gets beaten by 1000m. What is the time in minutes in which Ram and Shayam can run the race separately?
Let the speed of Ram be "R" & Shyam be "S"
Condition 1:
When shyam gets head start of 200m and beaten by 30 sec
2000/R  1800/S = 30 > eq (1)Condition 2:
When shyam gets a head start of 3 min and won by 1000m
1000/R  2000/S = 180 > eq (2)Solving this we will get
R = 25/3 m/s
S = 20/3 m/sR = 2000 * 3 * /25 = 4 min
S = 2000 * 3/20 = 5 minThus time taken by Ram & Shyam to complete the race is 4 min & 5 min
A cistern is filled in 6h but take 4h longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in how much time?
With a leak it takes 10 hr to fill the tank
1/6  1/x = 1/10
1/x = 1/6  1/10
1/x = 4/60
1/x = 1/15
X = 15 hrsA take 5 days more than B to do a job & 9 days more than C; A & B together can do the job in the same time as C. How many days A would take to do it?
Clearly
1/A + 1/B = 1/C
1/A + 1/(A5) = 1/(A9)
Solve this quadratic
You will get
A = 15 DaysThe vessels contain water and milk in the ratio 1:2 and 2:5 are mixed in the ration is 1:4 then what will be the resulting ratio?
Quantity of water in the resultant mixture
1 * 1/3 + 4 * 2/7 = 1/3 + 8/7 = 31/21Quantity of milk in the resultant mixture
1 * 2/3 + 4 * 5/7 = 2/3 + 20/7 = 74/21Required ratio
31/21 : 74/21
= 31 : 74Ram and Shyam start at same end of a swimming competition.The length of pool is 50m.The race is for completing 1000m. If Ram beats Shyam and meets him 17 times during the race what could be the speed of Shyam if speed of Ram is 5m/s?
Length of pool is 50m
So 100m (2x of pool)
Since they meet 17 times
So ram covers 17 * 100 = 1700 m
But race is of 1000m
When ram covers the race I.e 1000m
So shyam covers 17001000 = 700m
Ram  shyam
1000> 700
5 > ?
5 * 700/1000
= 3.5 m/sA and B can finish a work together in 12 days, and B and C together in 16 days. If A alone works for 5 days and then B alone continues for 7 days, then remaining work is done by C in 13 days. In how many days can C alone finish the complete work?
Lcm of (12,16) = 48 units <  total work
So A+B = 48/12 = 4 units per day
B+C = 48/16 = 3 units per day
Now,
5A+7B+13C = 48 units
5 (A+B)+2 (B+C)+11C = 48
5 (4)+2 (3)+11C = 48
20+6+11C = 48
11C = 22
C = 2 units
So C does 2 units of work daily
So total time required to complete work alone is 48/2 = 24 daysAround a circular path of 360km, three men start cycling at 48, 60 and 72 km per day. After how many days will they meet?
The time each takes to complete a round of circular field,
360/48 = 7.5
360/60 = 6
360/72 = 5 days.
They'll meet at the time equal to the LCM of (7.5,6,2)
which is 30 days.