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

►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.

1. Find per-pipe filling rates

   • Set X: 10 pipes fill 0.7 tank in 7 min ⇒ rate per pipe = 0.7⁄(7·10) = 0.01 tank/min

   • Set Y: 5 pipes fill 3/8 tank in 3 min ⇒ rate per pipe = (3/8)⁄(3·5) = 0.025 tank/min

2. New filling rates after adding one pipe each

   • X: 11 pipes × 0.01 = 0.11 tank/min

   • Y: 6 pipes × 0.025 = 0.15 tank/min

3. Set Z emptying rate increase

   • Original emptying: 8 pipes empty 0.5 tank in 10 min ⇒ total = 0.5⁄10 = 0.05 tank/min

   • +20% ⇒ 0.05×1.2 = 0.06 tank/min (outflow)

4.  Net Filling Rate
   0.11 + 0.15 - 0.06 = 0.20 tank / min

5. Time to fill 1 tank:
   1/0.20 min = 5 mins

6. Completion time:
   Started at 2: 58 p.m + 5 mins = 3: 03 p.m 

Here’s the same step-by-step solution written entirely in plain text (no LaTeX or special math notation):

1. Define daily work rates:
   – Ajit’s rate = A (work-units per day)
   – Baljit’s rate = B (work-units per day)
   – Diljit’s rate = D (work-units per day)

2. From “Ajit can do as much work in 2 days as Baljit can do in 3 days”:
   2 × A = 3 × B
   ⇒ A = (3/2) × B

3. From “Baljit can do as much in 4 days as Diljit in 5 days”:
   4 × B = 5 × D
   ⇒ D = (4/5) × B

4. Combined rate when all three work together = A + B + D
   = (3/2)B + B + (4/5)B
   = 1.5B + 1.0B + 0.8B
   = 3.3B

5. They finish the entire job (1 work-unit) in 20 days together, so:
   (combined rate) × 20 days = 1 work-unit
   ⇒ 3.3B × 20 = 1
   ⇒ 66B = 1
   ⇒ B = 1/66 (work-units per day)

6. Time for Baljit alone to do 1 work-unit = 1 ÷ B = 1 ÷ (1/66) = 66 days

Step 1: Define capacity

• Total tank capacity = 1 unit

Step 2: Define rates

• Inflow rate (filling pipe) = 1/x per hour

• Outflow rate (emptying pipe) = 1/y per hour

Step 3: Net rate when both pipes are open

Net rate = Inflow − Outflow
= 1/x − 1/y
= (y − x) / (xy) units per hour
Step 4: Remaining volume to be filled

Tank is already 1/3 full, so remaining = 2/3 units

Step 5: Time required

Time = (Remaining volume) ÷ (Net rate)
= (2/3) ÷ [(y − x)/xy]
= (2/3) × (xy / (y − x))
= 2xy / [3(y − x)]

Step 1: Find A’s work rate

A completes 6/7 of the work in 2z hours.

Work rate of A = (6/7) ÷ (2z) = 3/(7z).

Step 2: Find B’s work rate

B works twice as fast as A.

So, B’s rate = 2 × (3/(7z)) = 6/(7z).

Step 3: Find remaining work after A

Total work = 1.

Work completed by A = 6/7.

Remaining work = 1 − 6/7 = 1/7.

Step 4: Time taken by B

Time = Work ÷ Rate.

Time taken by B = (1/7) ÷ (6/(7z))
= (1/7) × (7z/6)
= z/6

z/6 = (3/18)z.

(Because dividing by 6 is the same as multiplying numerator and denominator by 3:
z/6 = (3z)/(18) = (3/18)z.)

= (3/18)z.

►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. 

Step 1: Effective digging rate

• 40 m in 4 days at 8 hrs/day

• ⇒ 10 m dug per day (before refill)

Step 2: Refill condition

• At the end of each day, one-fifth of the achieved depth refills.

• ⇒ Only 4/5 of the end-of-day depth remains for the next morning.

Step 3: Day-wise calculation

• Day 1
  Before refill: 10 m
  After refill: 10 × (4/5) = 8 m

• Day 2
  Before refill: 8 + 10 = 18 m
  After refill: 18 × (4/5) = 14.4 m

• Day 3
  Before refill: 14.4 + 10 = 24.4 m
  After refill: 24.4 × (4/5) = 19.52 m

Step 4: Depth at beginning of Day 4
= 19.52 m

​

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

Step 1: Define variables

• R = time to read one passage

• A = time to answer one question

Given: In time R, he answers 12 questions
→ A = R/12

Step 2: Questions per passage

• Passage 1 → 5 questions

• Passage 2 → 8 questions

• Passage 3 → 8 questions

• Passage 4 → 6 questions

Step 3: Time per passage

• T₁ = R + 5A = R + 5(R/12) = (17/12)R

• T₂ = R + 8A = R + 8(R/12) = (20/12)R

• T₃ = R + 8A = (20/12)R

• T₄ = R + 6A = R + 6(R/12) = (18/12)R

Step 4: Condition given
Time for first three passages exceeds time for last passage by 13 minutes:

T₁ + T₂ + T₃ − T₄ = 13
⇒ (17 + 20 + 20 − 18)/12 × R = (39/12)R = 13
⇒ R = 13 × (12/39) = 4 minutes

Step 5: Total time
T_total = T₁ + T₂ + T₃ + T₄
= (17 + 20 + 20 + 18)/12 × R
= (75/12)R
= (75/12) × 4 = 25 minutes

Step 6: Time on first passage
T1 = (17/12)R = (17/12) × 4 = 17/3 ≈ 5.67 minutes

Step 7: Percentage of total
(T1 / T_total) × 100
= (5.67 / 25) × 100 ≈ 22.67% ≈ 22.6%

Step 1: Define variables
Let T be the time Anoop takes to read one passage.
In this time T, he can answer 12 questions.

Step 2: Total questions
Passages = 4
Questions per passage = 5, 8, 8, 6
Total questions = 5 + 8 + 8 + 6 = 27

Step 3: Time for answering questions

• He answers 12 questions in time T → time per question = T/12

• For 27 questions → time = 27 × (T/12) = 27T/12

Step 4: Original total time

• Reading time = 4T

• Answering time = 27T/12

Total time = 4T + 27T/12
= (48T + 27T)/12 = 75T/12

Step 5: New target time
Target = 20% less than original
= 0.8 × (75T/12) = 60T/12 = 5T

Step 6: Allowed reading time
Answering time is fixed = 27T/12

So, allowed reading time = 5T − 27T/12
= (60T − 27T)/12 = 33T/12

Original reading time = 4T = 48T/12

Reduction = from 48T/12 to 33T/12 → factor = 33/48

Step 7: Required speed increase
Reading speed must increase by reciprocal factor:
= 48/33 = 16/11

Percentage increase = (16/11 − 1) × 100%
= (5/11) × 100% ≈ 45.45%

Let the total work = 1 unit.

Rates of work per day:

• P = 1/16

• Q = 1/20

• R = 1/30

Step 1: Work done in first 4 days (all together)
Work = 4 × (1/16 + 1/20 + 1/30)
= 4 × [(15 + 12 + 8) / 240]
= 4 × (35/240)
= 140/240 = 7/12

Step 2: Distribution of work duration

• P leaves after 4 days → works only 4 days in total.

• Q leaves 4 days before completion → works for (T – 4) days.

• R works throughout → works for T days.

Step 3: Total work equation
(P’s contribution) + (Q’s contribution) + (R’s contribution) = 1

(4/16) + (T – 4)/20 + T/30 = 1

Step 4: Simplify
4/16 = 1/4 = 5/20

So, (5/20 + (T – 4)/20) + T/30 = 1
⇒ (T + 1)/20 + T/30 = 1

LCM of 20 and 30 = 60

[3(T + 1) + 2T] / 60 = 1
⇒ 3T + 3 + 2T = 60
⇒ 5T = 57
⇒ T = 11.4 days