Net rate = 1/10 + 1/15 - 1/20 = (6+4-3)/60 = 7/60. Time = 60/7 ≈ 8.57. Rechecking: (6+4-3)/60 = 7/60, so 60/7. Let me verify: LCD(10,15,20)=60. 1/10=6/60, 1/15=4/60, 1/20=3/60. Net = 7/60. Time = 60/7 ≈ 8.57. But checking options, 9.6 = 48/5 = 9.6. Verifying: 7/60 gives 8.57, not 9.6. Re-examining: rates sum to 7/60, time = 60/7 ≠ options. This Q needs revision.
Rate A = 1/12, Rate B = 1/15. Combined = 9/60 per hour. In 2 hours: 18/60 filled. Remaining = 42/60. New rate = 9/60 - 1/10 = 3/60. Time = (42/60)/(3/60) = 14 hours. Total = 2 + 7.5 = 9.5 hours
# Solution: Speed of Boat in Still Water
Step 1: Define variables
Let b = speed of boat in still water (km/h)
Let c = speed of current (km/h)
Step 2: Set up equations using Speed = Distance ÷ Time
For upstream (boat moves against current):
- Speed upstream = b - c
- 90 ÷ (b - c) = 6
- Therefore: b - c = 15 ... (equation 1)
For downstream (boat moves with current):
- Speed downstream = b + c
- 110 ÷ (b + c) = 5
- Therefore: b + c = 22 ... (equation 2)
Step 3: Solve simultaneously
Add equations 1 and 2:
- (b - c) + (b + c) = 15 + 22
- 2b = 37
- b = 18.5 km/h
Wait—let me recalculate. From equation 2: 110 ÷ 5 = 22 ✓
Actually: 2b = 37 gives 18.5, but let me verify the answer differently.
Step 3 (Revised): Check if answer C works
If b = 20 km/h:
- From b - c = 15: c = 5 km/h
- From b + c = 22: c = 2 km/h ✗ (inconsistent)
Let me recalculate: 110 ÷ 5 = 22 ✓ and 90 ÷ 6 = 15 ✓
- Adding: 2b = 37, so b = 18.5 km/h
However, if the answer key states C: 20 km/h, there may be a typo in the problem values. Based on the given data (90 km/6 hrs upstream, 110 km/5 hrs downstream), the mathematically correct answer is 18.5 km/h.
Conclusion: If this is truly answer C (20 km/h), verify the problem statement. Using the given numbers, the boat speed should be 18.5 km/h. Answer C would only be correct if the problem distances or times were different.
1/A + 1/B = 1/12, 1/B + 1/C = 1/15, 1/A + 1/C = 1/20. Adding: 2(1/A + 1/B + 1/C) = 1/12 + 1/15 + 1/20 = 12/60. 1/A + 1/B + 1/C = 1/10. Time = 10 days
Net rate = 1/6 + 1/8 - 1/12 = 4/24 + 3/24 - 2/24 = 5/24. Time = 24/5 = 4.8 hours [Recalc needed for precision].
Speed = 400/20 = 20 m/s = 20 × 18/5 = 72 km/h.
We need to find what fraction of the total work A completes when working together with B for 6 days, given that A's 3-day work equals B's 4-day work.
Step 1: Set up the work rate relationship
Let A's work rate be \(a\) (fraction of work per day) and B's work rate be \(b\).
Given: A's 3-day work = B's 4-day work
Step 2: Express combined work for 6 days
When A and B work together for 6 days, the total work completed is:
Step 3: Substitute the rate relationship
Substitute \(a = \frac{4b}{3}\):
Therefore: \(a = \frac{4}{3} \cdot \frac{1}{14} = \frac{4}{42} = \frac{2}{21}\)
Step 4: Calculate A's fraction of total work
In 6 days, A completes:
Let A’s 1-day work = a
Let B’s 1-day work = b
Given:
3a=4b
a=
3
4b
So, ratio of efficiencies:
A:B=4:3
Assume:
A’s 1-day work = 4 units
B’s 1-day work = 3 units
Together, 1-day work:
4+3=7 units
In 6 days, total work done:
6×7=42 units
A alone does in 6 days:
6×4=24 units
Required fraction:
42
24
=
7
4
✅ Answer:
7
4
Answer: A completes \(\frac{4}{7}\) of the total work (Option C)
When a shopkeeper offers a discount on the marked price but still earns profit, we relate cost price (CP), selling price (SP), and marked price (MP) using profit and discount percentages.
Step 1: Calculate Selling Price from Profit
Given that the shopkeeper makes 20% profit on CP = Rs. 200:
Step 2: Relate Selling Price to Marked Price using Discount
The shopkeeper gives 15% discount on the marked price, so:
Step 3: Solve for Marked Price
Substitute \(SP = 240\) into the discount equation:
Step 4: Calculate the Final Answer
Cost Price (CP) = ₹200
Profit = 20%
So, Selling Price (SP):
SP=200×
100
120
=₹240
A discount of 15% means the selling price is 85% of the marked price (MP).
240=85% of MP
MP=
85
240×100
=
85
24000
=₹282.35 (approx)
Therefore, the marked price is approximately ₹282.35.
Answer: Marked Price = Rs. 282.35 (Option B)
When two trains cross each other, the total distance covered equals the sum of their lengths, and we use the relative speed to find this distance.
Step 1: Convert speed to m/s
The moving train travels at \(80\text{ km/h}\). Convert to metres per second:
Step 2: Find distance covered in 18 seconds
Distance = Speed × Time
Step 3: Apply the crossing condition
When the moving train completely crosses the stationary train, the distance covered equals the sum of both train lengths:
where \(L_{\text{moving}}\) is the length we need to find and \(L_{\text{stationary}} = 200\text{ m}\).
Step 4: Calculate the length of the moving train
Answer: The length of the moving train is \(200\text{ m}\) (Option A)
14641/10000 = 1.4641. (1+r)^4 = 1.4641. 1+r = 1.11. r = 11%.
