Test: Arun Sharma Based Level 1: Time & Work - CUET Commerce MCQ

It is given that pipes of set x can fill 70% of the tank in 7 min.

• set x = 70/100 th in 7 min  

It means that, 1/10 th tank is filled in 1 min.

Similarly,  

It is given that pipes of set y can fill 3/8 th of the tank in 3 min.

So, we write as,

• set y = 3/8 th of tank in 3 min  

It means that, 1/8 th tank is filled in 1 min.

And,  pipes of the set z can empty 5/10 th  of the tank in 10 min.

• set z = 5/10 th  in 10 min  

Therefore, 1/20 th  tank is emptied in 1 min.

So,  we consider all three sets for one minute. Then,

1/10 + 1/8 - 1/20 = 7/40 th tank is filled in one minute.

Thus, the time required to fill the tank completely = 40/7 minutes

►Set X will do 5% per minute and Set Y will do 6.25% per minute, while set Z will do 5% per minute (negative work).

►Hence, Net work will be 6.25% per minute. To fill 49% it will take slightly less than eight minutes and the value will be a fraction.

►None of the first three options matches this requirement. 

►If 4 of the taps of set Z are closed, the net work done by Set Z would be -2.5% while the work done by Sets X and Y would remain 10% and 12.5% respectively.

►Thus, the total work per minute would be 20% and hence the tank would take 5 minutes to fill up.

►Again if we close 4 taps of set Z, the rate of emptying by set Z would be 2.5% per minute.

►A half filled tank would contain 50% of the capacity and hence would take 50 / 2.5 = 20 minutes to empty.

►The rate per minute with the given changes (in percentage terms) would be :

►Set X = 11%, Set Y = 15% and Set Z = -6%.

►Hence, the net rate = 11 + 15 - 6 = 20%, per minute and it would take 5 minutes for the tank to fill. If all pipes are opened at 2:58, the tank would get filled at 3:03.

►Let Ajit's rate of work be 100/2 = 50 work units per day.

►Baljit would do 100/3 = 33.33 work units per day and Diljit does 133.33/5 = 26.66 units of work per day.

►Their 1 day work = 50 + 33.33 + 26.66 = 110 units of work per day.

►In 20 days, the total work done would be 2200 units of work and hence for Baljit to do it alone it would take : 2200 / 33.33 = 66 days to complete the same work.

• We will approach this question by hit or trial
• So let x=24 minutes
• Mini (time/question)=6,
• So Mini(questions/24m)=4
• Vinay (time/question)=8,
• So Vinay(questions/24m)=3
• y=4-3=1
• Now let us reduce time taken per question by 2 minutes for both:
• Mini (time/question)=4, So Mini(questions/24m)=6
• Vinay (time/question)=6, So Vinay(questions/24m)=4
• y=6-4=2(2y)
• Now, let us plug in the values of x(24) and y(1) in the answer choices and see which gives us the answer of 4 (because in 24 minutes, Mini solves 24/6=4 questions.
• 1) 1/4 * [2*(24+1) -√(2*24*24 + 4*1*1)] = 1/4 * [50 -√(1156)] = 1/4 * [50 - √(2*2*17*17)] = 1/4 * [50-34] = 1/4*16 = 4
• We got our answer on the first try, no need to try other options

►To solve this question first assume the values of x and y (such that x < y).

►If you take x as 10 hours and y as 15 hours, you will get a net work of 3.33% per hour.

►At this rate it will take 20 hours to fill the tank from one third full. Using this condition try to put these values of x and y into the options to check the values.

►For instance option (a) gives the value as 3 x 10 x 15 /10 = 45 which is not equal to 20.

Since A finishes 6 / 7^th of the work in 2z hours.
B would finish 12 / 7 of the work in 2z hours.
Thus, to do 1/7^th of the work (which represents the remaining work), B would require
     ► 2z /12 = z / 6 hours.

►The per day digging of all three combined is 54 metres. Hence, their average should be 18.

►This means that the first should be 18 - x, the second, 18 & the third 18 + x.

►The required conditions are met if we take the values as 15, 18 and 21 metres for the first, second and third diggers, respectively. 

First day — 8
Second day — 18 x (4 / 5) = 14 .4
Third day — 14.4 + 10 = 24.4 x (4 / 5) = 19.52

►Let us assume that time to read one passage = x

►So, according to the question, 3x + 21( x / 12) - 13 ⇒ x = 4

►Therefore, total time taken for answering all questions = 4 x 4 + (4 x 27 / 12) = 25 min

►Time spent on first passage = 4 + (4 x 5 / 12) = 17 / 3 min

►Required percentage = 22 .6%

►As we have found in the previous question that total time spent = 25 min

►If total time spent is cut down by 20%, then time = 20 min

►Now, let us assume increased reading speed = x

►So, 20 = 4 x + (4 x 27 / 12) ⇒ x =11 / 4 = Multiplication.

►Hence, percentage increase = 45.45%

P—16—15
Q—20—12
R—30—8
LCM = 240
 ► 35 x 4 + 8 x 4 = 140 + 32 = 172
 ► 240 - 172 = 68
 ► 68 / 20 = 3.4
Total time = 4 + 4 + 3.4 = 11.4 days