MBA MAHCET Preparation 2026 | Percentage , Ratio, Time & Work, Distance Time & Speed | MBA Wallah

A classic mixture and alligation problem is solved, where two varieties of tea (costing ₹60/kg and ₹65/kg) are mixed. The mixture is sold at ₹68.20/kg, yielding a 10% profit. The key here is to convert the selling price of the mixture back to its cost price, using the profit percentage. With an SP of 68.20 and 10% profit, the CP of the mixture is calculated as ₹62. Using alligation, the diagonal differences (62-60=2 and 65-62=3) give the ratio of the two varieties as 3:2. The concept of factors being uniform (all CP or all SP) in alligation is reinforced.

This segment deals with average production. A company produces an average of 4000 items/month for the first 3 months. The question asks for the average production per month over the next 9 months to achieve an overall average of 4375 items/month over a 12-month period. The combined average formula (n1x1 + n2x2) / (n1+n2) is applied. Alternatively, the problem can be solved using mixture and alligation concepts, where the average of the two periods is weighted by their duration. Both methods yield an answer of 4500 items/month for the next 9 months.

The discussion shifts to percentage-based problems, starting with compound interest. The problem requires finding the compound interest on ₹15,625 for 9 months at 16% per annum, compounded quarterly. The first step is to adjust the annual interest rate to a quarterly rate (16%/4 = 4%). For 9 months, there are three quarters. The amount is calculated using successive multiplication: Principal * (1 + 4/100)^3. After canceling out common factors and performing calculations, the amount is ₹17,576. The compound interest is then Amount - Principal, which is ₹1951. The importance of understanding compounding frequency (quarterly, half-yearly) is highlighted.

A profit and loss problem is explored: by selling 45 lemons for ₹40, a loss of 20% is incurred. The question asks how many lemons should be sold for ₹24 to gain 20%. The instructor first calculates the cost price (CP) of 45 lemons using the selling price (SP) and loss percentage. CP = SP / (1 - Loss Percent), which comes out to ₹50. So, one lemon's CP is 50/45. To gain 20%, the new SP of one lemon is (50/45) * (1 + 20/100) = 4/3. Finally, to find how many lemons can be sold for ₹24 at this new SP, 24 / (4/3) results in 18 lemons. The unitary method is also explained for clarity.

This problem involves Cost Price (CP), Marked Price (MP), and Profit Percent. The CP of an article is 64% of its MP. A discount of 12% is allowed on the MP. The goal is to calculate the profit percentage. Assuming MP = ₹100, the CP is ₹64. A 12% discount on MP means the selling price (SP) is 100 - 12 = ₹88. With CP = ₹64 and SP = ₹88, the profit is ₹24. The profit percentage is (Profit / CP) * 100 = (24/64) * 100 = 37.5%. This demonstrates the relationship between these key pricing terms.

A scenario involving a bulk purchase and segmented sales at different profits is discussed. 20 dozen notebooks are purchased at ₹48/dozen, making the total CP = ₹960. 8 dozen are sold at 10% profit (profit of 4.8/dozen * 8 dozen = 38.4). The remaining 12 dozen are sold at 20% profit (profit of 9.6/dozen * 12 dozen = 115.2). The total profit is 38.4 + 115.2 = 153.6. The overall profit percentage is (Total Profit / Total CP) * 100 = (153.6 / 960) * 100 = 16%. The method covers calculating profits for different quantities and then finding the overall profit percentage.

This problem illustrates a purchase using various payment methods and deferred payment with interest. A mobile's price is paid 1/6 via UPI, 1/3 via cash. The balance is paid a year later with 10% interest. If the interest paid was ₹6000, what was the original mobile price? The instructor recommends assuming a total price that is a multiple of 6 and 3, e.g., ₹60. UPI payment = 1/6 of 60 = ₹10. Cash = 1/3 of 60 = ₹20. Total paid = ₹30. Balance (loan) = 60 - 30 = ₹30. Interest on balance (₹30) at 10% = ₹3. Using the unitary method, if ₹3 interest (assumed) equals ₹6000 (actual), then the original price of ₹60 (assumed) equals ₹120,000. This method simplifies complex payment structures.

The session transitions to Time and Work problems. This question involves 8 pipes connected to a water tank: some fill (water pipes) and some empty (waste pipes). Each water pipe fills the tank in 12 hours, and each waste pipe empties it in 36 hours. When all 8 pipes are open, an empty tank fills in 3 hours. The approach is to assume the tank's capacity as the LCM of 12, 36, and 3, which is 36 liters. This gives the efficiency of each pipe type: water pipe = 3 L/hr, waste pipe = 1 L/hr, and all 8 pipes combined = 12 L/hr. Let x be the number of water pipes, so (8-x) are waste pipes. The equation is (x * 3) - ((8-x) * 1) = 12. Solving for x gives x=5. Thus, there are 5 water pipes and 3 waste pipes. The question asked for waste pipes, so the answer is 3.

Another tank problem: an inlet pipe fills an empty tank in 8 hours when the outlet pipe is closed. When both are open (inlet filling, outlet emptying), the tank fills in 10 hours. The capacity of the tank is assumed as the LCM of 8 and 10, which is 40 liters. Inlet pipe's efficiency is 40/8 = 5 L/hr. Combined efficiency of inlet and outlet is 40/10 = 4 L/hr. Since the inlet alone fills 5 L/hr but with the outlet, only 4 L/hr is net filled, the outlet pipe must be emptying 1 L/hr (5-4=1). The question asks how long it will take for the full tank (40 liters) to become half-empty (i.e., 20 liters remaining). So, 20 liters need to be emptied. At 1 L/hr, it will take 20 hours.

This question involves two individuals, Sunil and Nilesh, checking quality. Sunil checks 1000 items in 5 hours (efficiency = 200 items/hr). Nilesh checks 75% of 1000 items (750 items) in 3 hours (efficiency = 250 items/hr). They need to check 1300 items together, but Nilesh stops after 2 hours. In the first 2 hours, both work, checking (200+250)*2 = 900 items. The remaining items are 1300-900 = 400. Sunil alone checks these remaining 400 items at his rate of 200 items/hr, taking 2 more hours. The total time taken to check 1300 items is 2 hours (both) + 2 hours (Sunil alone) = 4 hours.

A complex alternating work problem is presented. B finishes a job in 40 days. A is twice as fast as B, and thrice as fast as C. Assuming a total work of 120 units (LCM of 40 and other implicit timings), B's efficiency is 3 units/day. A's efficiency is 6 units/day (twice of B). C's efficiency is 2 units/day (A is thrice of C). The work proceeds in a 3-day roster: Day 1: A+B (9 units), Day 2: B+C (5 units), Day 3: A+C (8 units). A cycle of 3 days produces 22 units (9+5+8). The question asks for the total days A worked until the job is finished. To complete 120 units, 5 full cycles (5*22 = 110 units) take 15 days (5*3). In these 15 days, A worked 2 days per cycle (Day 1 and Day 3), so 5*2=10 days. The remaining work is 10 units (120-110). On Day 16 (first day of the next cycle), A+B work, producing 9 units. A works on this day, so A's total is 10+1 = 11 days. The remaining 1 unit will be completed on Day 17 by B+C, where A is not involved. So, A worked for a total of 11 days.

The last section focuses on Time, Speed, Distance. A bus covers 50 km in 40 minutes, then next 50 km at 2 km/minute, and next 30 km at 1 km/minute. The goal is to find the average speed. The formula for average speed is Total Distance / Total Time. First segment: Distance = 50 km, Time = 40 min. Second segment: Distance = 50 km, Speed = 2 km/min, so Time = 50/2 = 25 min. Third segment: Distance = 30 km, Speed = 1 km/min, so Time = 30/1 = 30 min. Total Distance = 50+50+30 = 130 km. Total Time = 40+25+30 = 95 minutes. To convert average speed to km/hr, convert total time to hours: 95/60 hours. Average Speed = 130 / (95/60) = 130 * 60 / 95 = 82.1 km/hr (approx). The session concludes with a homework problem on upstream-downstream, which will be discussed in the next class based on student feedback.

The session begins by welcoming students to MBA Wallah's ' अंतिम युद्ध' (final battle) for the MBA CET exam. The instructor, Vineet Kankariya, highlights the significance of the CET as the last major MBA entrance exam for many aspirants, especially for prestigious institutions like JBIMS due to its high ROI. He emphasizes that while the syllabus for Quant in CET and other OMETs is similar, the level of difficulty varies. The session will focus on Arithmetic, specifically Ratio, Percentage, Time & Work, and Time, Speed & Distance, and encourages revisiting foundational concepts.

The first problem presented is a partnership question. A, B, and C invest varying amounts over a 3-year period, with A adding to his investment and B withdrawing from his. The goal is to determine the profit-sharing ratio at the end of 3 years. The instructor explains that profit is distributed based on the product of investment and the period of investment. The calculations show A's investment is (25*1 + 35*2) = 95, B's is (35*2 + 25*1) = 95, and C's is (30*3) = 90. Simplifying the ratio 95:95:90 gives 19:19:18, demonstrating how to handle changing investments over time in a partnership.

This section tackles a problem involving successful and unsuccessful candidates in an exam. Out of 132 examinees, the ratio of successful to unsuccessful is 9:2. The instructor calculates that there are 108 successful and 24 unsuccessful candidates. The twist comes when 4 more students pass after rechecking. The common mistake of only adding to successful count is addressed, emphasizing the need to subtract from the unsuccessful count. Therefore, the new successful count is 112 (108+4) and unsuccessful is 20 (24-4), leading to a new ratio of 28:5.