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Tabular Data — PYPs

18 solved Tabular Data previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2024LRDI › Tabular DataEasySA · TITA
The following table shows the number of employees and their median age in eight companies located in a district.
CompanyNumber of employeesMedian age
A3224
B2830
C4339
D3945
E3549
F2954
G2359
H1663
It is known that the age of all employees are integers. It is known that the age of every employee in A is strictly less than the age of every employee in B, the age of every employee in B is strictly less than the age of every employee in C, ..., the age of every employee in G is strictly less than the age of every employee in H. The highest possible age of an employee of company A is:
Enter your answer to attempt
The Setup: Think of these companies as sorted C++ arrays. We need to maximize the final element of array A without throwing a logic error when it's compared to the very first element of array B. It's a strict inequality check, so we need to min-max the data to push Company A's ceiling as high as possible. Step 1: Decode the median mechanics for Company B. Company B has 2828 employees (an even number). The median age (3030) is the average of the two middle elements. If we use standard 1-based math indexing, that's the 1414th and 1515th employees: b14+b152=30 \frac{b_{14} + b_{15}}{2} = 30 Step 2: Find the lowest possible starting age for Company B. To give array A the most room to scale up, we must push B's values as low as the rules allow. We can initialize the first 1515 elements in B to exactly 3030 without breaking the median requirement: b1=b2=⋯=b14=b15=30b_1 = b_2 = \dots = b_{14} = b_{15} = 30. Thus, the absolute minimum age for the youngest employee in B is 3030. Step 3: Lock in Company A's max age. The constraint dictates that *every* employee in A must be strictly younger than *every* employee in B. In code terms, a32<b1a_{32} < b_1. Since ages are strictly typed integers, if b1=30b_1 = 30, the absolute maximum allowed for a32a_{32} is 2929. Step 4: Verify this doesn't break Company A's own median constraint. Company A has 3232 employees with a median of 2424. This requires the average of a16a_{16} and a17a_{17} to be 2424. We can easily assign a16=24a_{16} = 24 and a17=24a_{17} = 24, which leaves plenty of capacity for elements a18a_{18} through a32a_{32} to cap out at 2929. The backend logic runs with zero lag, and the max age holds up perfectly. Final Answer: 29
Q2:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ
The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.
DayMondayTuesdayWednesdayThursdayFriday
QuoteSellBuySellBuySellBuySellBuySellBuy
Dabur460462455458432433444447461462
Marico345346335336365368372375372374
HUL1931193319521955197919812044204819661969
ITC237238238239246251221225253256
Britannia3044304631003101311031153025302731403144
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If an investor had Rs 36,00,000 to invest in any particular single stock, and she could buy the stock only on Monday and sell it off only on Friday, then the stock she should buy on Monday to earn the maximum possible profit during the week is
  • AMarico
  • BHUL
  • CITC
  • DBritannia
Pick an option to attempt
The Setup: Every question in this set runs on one rule, so it is worth nailing down once here. The table gives two prices per day, and the passage defines them from the *investor's* side: you buy at the Buy price and sell at the Sell price. The Buy price is always the higher of the two, so the market takes a small bite on every round trip. Mixing the two columns up is the single biggest source of wrong answers in this caselet. Step 1: Fix the method. With a fixed pot of money you cannot own a fraction of a share, so: Shares=⌊CapitalMonday Buy price⌋,Residual cash=Capital−Shares×Buy price\text{Shares}=\left\lfloor\frac{\text{Capital}}{\text{Monday Buy price}}\right\rfloor, \qquad \text{Residual cash}=\text{Capital}-\text{Shares}\times\text{Buy price} Final wealth=Shares×Friday Sell price+Residual cash\text{Final wealth}=\text{Shares}\times\text{Friday Sell price}+\text{Residual cash} The residual is the small change that could not buy one more share. It does not vanish - it stays in the investor's pocket and counts toward the final wealth. It barely matters here, but in MCQ 17 it decides the answer outright, so build the habit now. Step 2: Run all four candidates. Capital is Rs 36,00,000. Buy at Monday's Buy price, sell at Friday's Sell price.
StockMon BuySharesResidualFri SellProceedsProfit
Marico34610,40421637238,70,2882,70,504
HUL1,9331,8627541,96636,60,69261,446
ITC23815,1261225338,26,8782,26,890
Britannia3,0461,1812,6743,14037,08,3401,11,014
Two share counts are worth checking by hand, because rounding the wrong way is easy here: 3046×1181=35,97,326≤36,00,000but3046×1182=36,00,372>36,00,0003046\times 1181=35,97,326\leq 36,00,000 \quad\text{but}\quad 3046\times 1182=36,00,372>36,00,000 so Britannia buys 1,181 shares, not 1,182. Likewise 346×10404=35,99,784346\times 10404=35,99,784 fits while 346×10405=36,00,130346\times 10405=36,00,130 does not. Step 3: Read off the winner. Marico returns the largest profit at roughly Rs 2.70 lakh, comfortably ahead of ITC's Rs 2.27 lakh. Step 4: Sanity-check why. Marico wins on percentage movement, not on price. It rose from 346 to 372, a gain of about 7.5%7.5\%, versus ITC's 6.3%6.3\%, Britannia's 3.1%3.1\% and HUL's 1.7%1.7\%. Because the whole pot goes into one stock either way, the cheapest share price is irrelevant - only the percentage gain matters, and a quick ratio scan of Fri SellMon Buy\frac{\text{Fri Sell}}{\text{Mon Buy}} would have identified Marico without computing a single share count. Final Answer: Marico
Q3:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ
The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.
DayMondayTuesdayWednesdayThursdayFriday
QuoteSellBuySellBuySellBuySellBuySellBuy
Dabur460462455458432433444447461462
Marico345346335336365368372375372374
HUL1931193319521955197919812044204819661969
ITC237238238239246251221225253256
Britannia3044304631003101311031153025302731403144
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If an investor planned to invest Rs 36,00,000 in purchasing the stocks of HUL on Monday, sell them off on Wednesday and use the entire proceeds to purchase the stocks of Britannia on the same day and sell them off again on Friday, then the total investment return during the week would be
  • A2.80 percent
  • B3.00 percent
  • C3.20 percent
  • D3.40 percent
Pick an option to attempt
The Setup: A two-leg trade: HUL from Monday to Wednesday, then the entire proceeds into Britannia from Wednesday to Friday. Using the convention from MCQ 16 - buy at the Buy price, sell at the Sell price, and carry the residual cash forward. That last point is not bookkeeping pedantry here; it is what separates two of the four options. Step 1: Leg one - HUL, Monday to Wednesday. Monday Buy price is 1,933: Shares=⌊36,00,0001933⌋=1862,Residual=36,00,000−1862×1933=754\text{Shares}=\left\lfloor\frac{36,00,000}{1933}\right\rfloor=1862, \qquad \text{Residual}=36,00,000-1862\times 1933=754 Sell on Wednesday at HUL's Sell price of 1,979: 1862×1979=36,84,898  ⟹  Total cash=36,84,898+754=36,85,6521862\times 1979=36,84,898 \implies \text{Total cash}=36,84,898+754=36,85,652 Step 2: Leg two - Britannia, Wednesday to Friday. Wednesday Buy price is 3,115: ⌊36,85,6523115⌋=1183since3115×1183=36,85,045≤36,85,652\left\lfloor\frac{36,85,652}{3115}\right\rfloor=1183 \quad\text{since}\quad 3115\times 1183=36,85,045\leq 36,85,652 This is the hinge of the whole question. Had we thrown away the Rs 754 left over from the HUL leg, the available cash would be 36,84,898 - and 36,85,045>36,84,89836,85,045>36,84,898, so only 1,182 shares would be affordable. That stray Rs 754 is precisely what pays for the 1,183rd share, and that one share is worth about Rs 3,140 by Friday. Discarding it drops the answer to 3.10%3.10\% and lands you on the wrong option. Step 3: Close the position. Sell on Friday at Britannia's Sell price of 3,140, and add the residual from this leg (36,85,652−36,85,045=60736,85,652-36,85,045=607): 1183×3140=37,14,620  ⟹  Final wealth=37,14,620+607=37,15,2271183\times 3140=37,14,620 \implies \text{Final wealth}=37,14,620+607=37,15,227 Step 4: Compute the return. Return=37,15,227−36,00,00036,00,000×100=1,15,22736,00,000×100=3.2008%≈3.20%\text{Return}=\frac{37,15,227-36,00,000}{36,00,000}\times 100=\frac{1,15,227}{36,00,000}\times 100=3.2008\%\approx 3.20\% Step 5: Confirm with a shortcut. Ignore the share-counting entirely and just compound the two price ratios, which is what the trade does at heart: 19791933×31403115=1.03201  ⟹  3.20%\frac{1979}{1933}\times\frac{3140}{3115}=1.03201 \implies 3.20\% Both routes agree to two decimal places, which confirms the answer is robust and not an artefact of rounding. Final Answer: 3.20 percent
Q4:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ
The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.
DayMondayTuesdayWednesdayThursdayFriday
QuoteSellBuySellBuySellBuySellBuySellBuy
Dabur460462455458432433444447461462
Marico345346335336365368372375372374
HUL1931193319521955197919812044204819661969
ITC237238238239246251221225253256
Britannia3044304631003101311031153025302731403144
show less
The difference between the quoted buy and sell price of a stock is referred to as the spread of the stock. The average spread of the stocks is lowest on
  • AMonday
  • BTuesday
  • CThursday
  • DFriday
Pick an option to attempt
The Setup: The spread is defined in the stem as the gap between the quoted buy and sell price of a stock - the market's cut on a round trip. So for each day we take Buy−Sell\text{Buy}-\text{Sell} for all five stocks and average the five numbers. Since every day has the same five stocks, comparing the totals is enough and the division by 5 is only cosmetic. Step 1: Compute the spread for each stock, day by day.
DayDaburMaricoHULITCBritanniaTotalAverage
Monday2121281.6
Tuesday3131191.8
Thursday33442163.2
Friday12334132.6
Worked out in full for Monday, the winning day: (462−460)+(346−345)+(1933−1931)+(238−237)+(3046−3044)5=2+1+2+1+25=85=1.6\frac{(462-460)+(346-345)+(1933-1931)+(238-237)+(3046-3044)}{5}=\frac{2+1+2+1+2}{5}=\frac{8}{5}=1.6 Step 2: Compare. Monday's average of 1.6 is the lowest of the four days offered; Tuesday at 1.8 is the nearest rival. Step 3: Note the shortcut. Monday's total of 8 is the smallest total in the table, and no division was needed to see it. Monday is also the only day where three of the five stocks trade at a spread of just 1 or 2 across the board - it is the tightest, most liquid-looking day of the week. Final Answer: Monday
Q5:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ
The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.
DayMondayTuesdayWednesdayThursdayFriday
QuoteSellBuySellBuySellBuySellBuySellBuy
Dabur460462455458432433444447461462
Marico345346335336365368372375372374
HUL1931193319521955197919812044204819661969
ITC237238238239246251221225253256
Britannia3044304631003101311031153025302731403144
show less
A brokerage firm charges 0.1 percent trading commission on the value of shares bought or sold through its trading platform. If an investor bought 1000 shares of Britannia on Tuesday, and sold all of them on Thursday, then the total brokerage fee that will be charged from the investor is
  • A6,125
  • B6,126
  • C6,127
  • D6,128
Pick an option to attempt
The Setup: The commission is charged on the value of shares bought or sold, so it applies twice - once on the purchase leg and once on the sale leg - and each leg is valued at its own price. Using the convention from MCQ 16: the investor buys at Tuesday's Buy price and sells at Thursday's Sell price. Step 1: The buying leg, Tuesday. Britannia's Tuesday Buy price is 3,101: Value=1000×3101=31,01,000  ⟹  Fee=31,01,000×0.1%=3101\text{Value}=1000\times 3101=31,01,000 \implies \text{Fee}=31,01,000\times 0.1\%=3101 Step 2: The selling leg, Thursday. Britannia's Thursday Sell price is 3,025: Value=1000×3025=30,25,000  ⟹  Fee=30,25,000×0.1%=3025\text{Value}=1000\times 3025=30,25,000 \implies \text{Fee}=30,25,000\times 0.1\%=3025 Note that 0.1%0.1\% of a value is just that value divided by 1,000 - and since exactly 1,000 shares were traded, each fee comes out equal to the share price itself. That is a pleasant shortcut, not a coincidence to rely on generally. Step 3: Total the two legs. 3101+3025=61263101+3025=6126 Step 4: Read the option list - it is a trap laid with real precision. All four options are within Rs 3 of each other, because each one corresponds to a different mix-up of the Buy and Sell columns:
Tuesday price usedThursday price usedTotalOption
Sell 3,100Sell 3,0256,1251
Buy 3,101Sell 3,0256,1262 - correct
Sell 3,100Buy 3,0276,1273
Buy 3,101Buy 3,0276,1284
There is no arithmetic in this question at all once you know which column to read; the entire difficulty is the convention. You buy at Buy and sell at Sell. Final Answer: 6,126
Q6:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ
The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.
DayMondayTuesdayWednesdayThursdayFriday
QuoteSellBuySellBuySellBuySellBuySellBuy
Dabur460462455458432433444447461462
Marico345346335336365368372375372374
HUL1931193319521955197919812044204819661969
ITC237238238239246251221225253256
Britannia3044304631003101311031153025302731403144
show less
If you had decided to invest Rs.36,00,000 worth of ITC stocks on Monday, then the day of the week you should choose to sell the stocks to earn the maximum possible profit would be
  • ATuesday
  • BWednesday
  • CThursday
  • DFriday
Pick an option to attempt
The Setup: The purchase is fixed - ITC bought on Monday - and only the exit day is ours to choose. Using the convention from MCQ 16: buy at Monday's Buy price of 238, sell at the chosen day's Sell price. Step 1: Note why no calculation is really needed. Monday's purchase locks in a fixed share count: ⌊36,00,000238⌋=15,126 shares, with Rs 12 left over\left\lfloor\frac{36,00,000}{238}\right\rfloor=15,126 \text{ shares, with Rs 12 left over} That count does not change with the exit day, and the residual is fixed too. So the final wealth is 15,126×(Sell price)+1215,126\times(\text{Sell price})+12, which is strictly increasing in the sell price. Maximising profit is therefore identical to maximising ITC's Sell price, and the whole question collapses into scanning one row of the table. Step 2: Scan ITC's Sell prices.
Sell onTuesdayWednesdayThursdayFriday
ITC Sell price238246221253
Profit01,21,008-2,57,1422,26,890
Step 3: Read off the answer. Friday's Sell price of 253 is the highest of the week, giving the maximum profit of about Rs 2.27 lakh. Thursday is the trap for anyone glancing at the Buy column or misreading the row - ITC dips to 221 that day, which would actually book a loss of roughly Rs 2.57 lakh against a purchase price of 238. Step 4: Sanity check. Selling on Tuesday at 238 exactly matches Monday's Buy price of 238, so that trade nets zero profit - a useful confirmation that the buy and sell columns are being read the right way round. Final Answer: Friday
Q7:ipmat indore 2025LRDI › Tabular DataMediumMCQ · MCQ
The table given below provides the details of monthly sales (in lakhs of rupees) and the value of products returned by the customers (as a percentage of sales) of an e-commerce company for three product categories for the year 2024. Net sales (in lakhs of rupees) is defined as the difference between sales (in lakhs of rupees) and the value of products returned (in lakhs of rupees).
MonthSales: ApparelSales: FootwearSales: ElectronicsReturns: ApparelReturns: FootwearReturns: Electronics
January26210428913%7%2%
February27911338716%9%3%
March23612128320%7%2%
April2585832516%8%1%
May2496935912%6%4%
June23011132119%5%3%
July24411934117%9%4%
August2526033616%6%2%
September28811835510%9%5%
October22210838315%8%2%
November2289328214%9%4%
December2218626818%10%1%
Which month had highest percentage decline in monthly sales as compared to previous month for the Apparel category?
  • AJune
  • BMarch
  • CDecember
  • DOctober
Pick an option to attempt
We don't need to calculate exact percentages - we can compare the fractions directly to find the largest one. From the Apparel sales data, let's identify consecutive months where sales decreased:
Month TransitionSales Change (Lakhs)Change TypeAnalysis
January →\rightarrow February262→279262 \rightarrow 279Increase (+17)-
February →\rightarrow March279→236279 \rightarrow 236Decrease (-43)Significant decrease - Consider
March →\rightarrow April236→258236 \rightarrow 258Increase (+22)-
April →\rightarrow May258→249258 \rightarrow 249Decrease (-9)Ignore - Minimal decrease while in the 250s range
May →\rightarrow June249→230249 \rightarrow 230Decrease (-19)Ignore - Same reason as May
June →\rightarrow July230→244230 \rightarrow 244Increase (+14)-
July →\rightarrow August244→252244 \rightarrow 252Increase (+8)-
August →\rightarrow September252→288252 \rightarrow 288Increase (+36)-
September →\rightarrow October288→222288 \rightarrow 222Decrease (-66)Much higher decrease - Consider with March
October →\rightarrow November222→228222 \rightarrow 228Increase (+6)-
November →\rightarrow December228→221228 \rightarrow 221Decrease (-7)Ignore - Same reason as May
% Decline=Previous Month Sales−Current Month SalesPrevious Month Sales×100% \% \text{ Decline} = \frac{\text{Previous Month Sales} - \text{Current Month Sales}}{\text{Previous Month Sales}} \times 100\% **February →\rightarrow March:** % Decline=279−236279×100%=43279×100%=15.41% \% \text{ Decline} = \frac{279 - 236}{279} \times 100\% = \frac{43}{279} \times 100\% = 15.41\% **September →\rightarrow October:** % Decline=288−222288×100%=66288×100%=22.92% \% \text{ Decline} = \frac{288 - 222}{288} \times 100\% = \frac{66}{288} \times 100\% = 22.92\% Comparing via fractions →\rightarrow March: 43279\frac{43}{279} and October: 66288\frac{66}{288} * October has a numerator (6666) that is about 1.51.5 times larger than March's numerator (4343). * October has a denominator (288288) that is only about 1.031.03 times larger than March's denominator (279279). Since the numerator grows much faster than the denominator, October's fraction 66288\frac{66}{288} is larger than March's fraction 43279\frac{43}{279}. October had the highest percentage decline in Apparel sales, dropping from 288 lakhs in September to 222 lakhs in October. Final Answer: October
Q8:ipmat indore 2025LRDI › Tabular DataMediumMCQ · MCQ
The table given below provides the details of monthly sales (in lakhs of rupees) and the value of products returned by the customers (as a percentage of sales) of an e-commerce company for three product categories for the year 2024. Net sales (in lakhs of rupees) is defined as the difference between sales (in lakhs of rupees) and the value of products returned (in lakhs of rupees).
MonthSales: ApparelSales: FootwearSales: ElectronicsReturns: ApparelReturns: FootwearReturns: Electronics
January26210428913%7%2%
February27911338716%9%3%
March23612128320%7%2%
April2585832516%8%1%
May2496935912%6%4%
June23011132119%5%3%
July24411934117%9%4%
August2526033616%6%2%
September28811835510%9%5%
October22210838315%8%2%
November2289328214%9%4%
December2218626818%10%1%
For which categories did the value of the products returned (as a percentage of sales) increase for three consecutive months?
  • AOnly Electronics
  • BOnly Apparel
  • CBoth Apparel and Footwear
  • DOnly Footwear
Pick an option to attempt
The Setup: We need to identify which product categories had the value of products returned (as a percentage of sales) increase for three consecutive months. The concept here is to track each category's return percentage month by month and look for patterns where the percentage consistently rises over a three-month span (meaning the current month's return percentage is strictly higher than the previous month's for three transitions). Step 1: Analyze the Apparel category. For Apparel, we see continuous increases from January to February to March: * January: 13% * February: 16% * March: 20% This gives us three consecutive months of increasing percentages (13%→16%→20%13\% \rightarrow 16\% \rightarrow 20\%). Step 2: Analyze the Footwear category. For Footwear, we see continuous increases from October to November to December: * October: 8% * November: 9% * December: 10% This gives us three consecutive months of increasing percentages (8%→9%→10%8\% \rightarrow 9\% \rightarrow 10\%). Step 3: Analyze the Electronics category. Scanning the Electronics return percentages, we do not find any sequence where the values strictly increase for three consecutive months. Step 4: Conclusion. The key insight is that we are looking for a pattern where each month's return percentage is strictly greater than the previous month's percentage for three consecutive months. Both the Apparel and Footwear categories satisfy this condition. Final Answer: Both Apparel and Footwear
Q9:ipmat indore 2025LRDI › Tabular DataMediumMCQ · MCQ
The table given below provides the details of monthly sales (in lakhs of rupees) and the value of products returned by the customers (as a percentage of sales) of an e-commerce company for three product categories for the year 2024. Net sales (in lakhs of rupees) is defined as the difference between sales (in lakhs of rupees) and the value of products returned (in lakhs of rupees).
MonthSales: ApparelSales: FootwearSales: ElectronicsReturns: ApparelReturns: FootwearReturns: Electronics
January26210428913%7%2%
February27911338716%9%3%
March23612128320%7%2%
April2585832516%8%1%
May2496935912%6%4%
June23011132119%5%3%
July24411934117%9%4%
August2526033616%6%2%
September28811835510%9%5%
October22210838315%8%2%
November2289328214%9%4%
December2218626818%10%1%
By what percentage the net sales for June increased as compared to May in the Footwear category?
  • A62.58 percent
  • B7.21 percent
  • C18.97 percent
  • D60.87 percent
Pick an option to attempt
Net Sales=Sales−Value of Products Returned\text{Net Sales} = \text{Sales} - \text{Value of Products Returned} The value of products returned is given as a percentage of sales, so we need to convert this to actual rupees first. For May in the Footwear category: * Sales=69 lakhs\text{Sales} = 69 \text{ lakhs} * Return percentage=6% of sales\text{Return percentage} = 6\% \text{ of sales} * Value of products returned=6100×69=4.14 lakhs\text{Value of products returned} = \frac{6}{100} \times 69 = 4.14 \text{ lakhs} * Net sales for May=69−4.14=64.86 lakhs\text{Net sales for May} = 69 - 4.14 = 64.86 \text{ lakhs} For June in the Footwear category: * Sales=111 lakhs\text{Sales} = 111 \text{ lakhs} * Return percentage=5% of sales\text{Return percentage} = 5\% \text{ of sales} * Value of products returned=5100×111=5.55 lakhs\text{Value of products returned} = \frac{5}{100} \times 111 = 5.55 \text{ lakhs} * Net sales for June=111−5.55=105.45 lakhs\text{Net sales for June} = 111 - 5.55 = 105.45 \text{ lakhs} To find the percentage increase from May to June: % Increase=June net sales−May net salesMay net sales×100% \text{\% Increase} = \frac{\text{June net sales} - \text{May net sales}}{\text{May net sales}} \times 100\% =105.45−64.8664.86×100% = \frac{105.45 - 64.86}{64.86} \times 100\% =40.5964.86×100% = \frac{40.59}{64.86} \times 100\% =62.58% = 62.58\% Final Answer: 62.58 percent
Q10:ipmat indore 2025LRDI › Tabular DataMediumMCQ · MCQ
The table given below provides the details of monthly sales (in lakhs of rupees) and the value of products returned by the customers (as a percentage of sales) of an e-commerce company for three product categories for the year 2024. Net sales (in lakhs of rupees) is defined as the difference between sales (in lakhs of rupees) and the value of products returned (in lakhs of rupees).
MonthSales: ApparelSales: FootwearSales: ElectronicsReturns: ApparelReturns: FootwearReturns: Electronics
January26210428913%7%2%
February27911338716%9%3%
March23612128320%7%2%
April2585832516%8%1%
May2496935912%6%4%
June23011132119%5%3%
July24411934117%9%4%
August2526033616%6%2%
September28811835510%9%5%
October22210838315%8%2%
November2289328214%9%4%
December2218626818%10%1%
Among the following four months, for which month the contribution of the Apparel category in the total monthly sales was the highest?
  • AJanuary
  • BApril
  • CDecember
  • DAugust
Pick an option to attempt
The Setup: The contribution of Apparel to Total Sales is calculated as the ratio of Apparel sales to the total sales of all categories for that specific month, converted to a percentage. Apparel contribution=Apparel salesTotal sales of all categories×100% \text{Apparel contribution} = \frac{\text{Apparel sales}}{\text{Total sales of all categories}} \times 100\% Step 1: Calculate the total sales for each given month. We find the total sales by adding the sales from all three categories (Apparel + Footwear + Electronics): * January: 262+104+289=655262 + 104 + 289 = 655 * April: 258+58+325=641258 + 58 + 325 = 641 * August: 252+60+336=648252 + 60 + 336 = 648 * December: 221+86+268=575221 + 86 + 268 = 575 Step 2: Calculate the exact contribution percentage for each month. * January: 262655×100%=40.00%\frac{262}{655} \times 100\% = 40.00\% * April: 258641×100%≈40.25%\frac{258}{641} \times 100\% \approx 40.25\% * August: 252648×100%≈38.89%\frac{252}{648} \times 100\% \approx 38.89\% * December: 221575×100%≈38.43%\frac{221}{575} \times 100\% \approx 38.43\% Step 3: Compare and Conclude. Comparing the percentages, April has the highest contribution (40.25%40.25\%). *Conceptual insight:* Even though January had higher absolute Apparel sales (262262 vs 258258), April had proportionally lower sales from other categories, making Apparel's share of the total pie larger. Final Answer: April
Q11:ipmat indore 2025LRDI › Tabular DataMediumMCQ · MCQ
The table given below provides the details of monthly sales (in lakhs of rupees) and the value of products returned by the customers (as a percentage of sales) of an e-commerce company for three product categories for the year 2024. Net sales (in lakhs of rupees) is defined as the difference between sales (in lakhs of rupees) and the value of products returned (in lakhs of rupees).
MonthSales: ApparelSales: FootwearSales: ElectronicsReturns: ApparelReturns: FootwearReturns: Electronics
January26210428913%7%2%
February27911338716%9%3%
March23612128320%7%2%
April2585832516%8%1%
May2496935912%6%4%
June23011132119%5%3%
July24411934117%9%4%
August2526033616%6%2%
September28811835510%9%5%
October22210838315%8%2%
November2289328214%9%4%
December2218626818%10%1%
Among the following four months, for which month the value of the Footwear returned (in lakhs of rupees) was the highest?
  • ASeptember
  • BJuly
  • CJune
  • DMarch
Pick an option to attempt
The Setup: The value of products returned is given as a percentage of sales, so we need to calculate the actual value in lakhs of rupees using the formula: Value of products returned=Return percentage100×Sales \text{Value of products returned} = \frac{\text{Return percentage}}{100} \times \text{Sales} Step 1: Calculate the exact return value for each given month. Extracting the Footwear sales and return percentage data from the table: * September: 9100×118=10.62 lakhs\frac{9}{100} \times 118 = 10.62 \text{ lakhs} * July: 9100×119=10.71 lakhs\frac{9}{100} \times 119 = 10.71 \text{ lakhs} * June: 5100×111=5.55 lakhs\frac{5}{100} \times 111 = 5.55 \text{ lakhs} * March: 7100×121=8.47 lakhs\frac{7}{100} \times 121 = 8.47 \text{ lakhs} Step 2: Compare and Conclude. Comparing the calculated values, July has the highest absolute value of products returned (10.71 lakhs10.71 \text{ lakhs}). *Conceptual insight:* The absolute value of returns depends on both the return percentage and the total sales volume. July combines a high return percentage (9%9\%) with substantial sales (119 lakhs119 \text{ lakhs}), creating the largest absolute return amount. * September →\rightarrow percentage is the same (9%9\%) but the sales are lower (118118). * June →\rightarrow sales are around the same but the percentage is almost half (5%5\%). * March →\rightarrow percentage is 2%2\% lower (7%7\%) and sales are only 2 lakhs2 \text{ lakhs} more than July, which makes it definitively lower. Final Answer: July
Q12:jipmat 2025LRDI › Tabular DataEasyLR · MCQ
Study the table given below and answer the question that follows: Total Number of Faculties in Different Departments of an Institute and Percentage of Females and males.
DepartmentTotal no. of FacultiesPercentage of FemalesPercentage of Males
Computer Science8404555
Mathematics2203565
Physics9002377
Chemistry3606535
Management4504456
Accounts5404060
What is the total number of males in Computer Science, Mathematics and Chemistry departments together?
  • A687
  • B731
  • C786
  • D678
Pick an option to attempt
Number of males in Computer Science Department →\rightarrow 55%55\% of 840=462840 = 462 Number of males in Mathematics Department →\rightarrow 65%65\% of 220=143220 = 143 Number of males in Chemistry Department →\rightarrow 35%35\% of 360=126360 = 126 Total →\rightarrow 462+143+126=731462 + 143 + 126 = 731 Final Answer: 731
Q13:jipmat 2025LRDI › Tabular DataMediumLR · MCQ
Study the table given below and answer the question that follows: Total Number of Faculties in Different Departments of an Institute and Percentage of Females and males.
DepartmentTotal no. of FacultiesPercentage of FemalesPercentage of Males
Computer Science8404555
Mathematics2203565
Physics9002377
Chemistry3606535
Management4504456
Accounts5404060
What is the ratio of the number of females in Physics department to the number of females in the Management department?
  • A22:23
  • B35:33
  • C23:22
  • D33:35
Pick an option to attempt
Step 1: Calculate the number of females in the Physics department * Total faculties in Physics = 900900 * Percentage of females = 23%23\% * Number of females = 23%23\% of 900=23100×900=207900 = \frac{23}{100} \times 900 = 207 Step 2: Calculate the number of females in the Management department * Total faculties in Management = 450450 * Percentage of females = 44%44\% * Number of females = 44%44\% of 450=44100×450=198450 = \frac{44}{100} \times 450 = 198 Step 3: Find the ratio * Ratio = Females in Physics : Females in Management * Ratio = 207:198207 : 198 * Both numbers are divisible by 9. * 207÷9=23207 \div 9 = 23 * 198÷9=22198 \div 9 = 22 * Simplified ratio = 23:2223:22 Final Answer: 23:22
Q14:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ
A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:
Food itemCalcium (mg)Iron (mg)Protein (g)
Milk30008
Spinach9063
Almonds21046
Oats12035
Rice5014
If two servings of Milk are part of the daily diet plan, then the minimum number of servings of any other single item that can satisfy the MDR of all three nutrients is ___
  • A6
  • B5
  • C4
  • D7
Pick an option to attempt
The Setup: This is a Data Interpretation stat-check. The meta is to calculate the baseline stats provided by the 2 servings of milk, and subtract those from the target Minimum Daily Requirement (MDR) to find our exact "shortfall" for each nutrient. Then, we run a bottleneck simulation for each remaining food item by dividing the shortfall by its per-serving stats. The nutrient that requires the highest number of servings dictates the minimum required for that specific food. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the Milk baseline and find the shortfall. The target MDR is: Calcium = 1100 mg, Iron = 15 mg, Protein = 40 g. Two servings of Milk provide: * Calcium=2×300=600 mgCalcium = 2 \times 300 = 600\text{ mg} * Iron=2×0=0 mgIron = 2 \times 0 = 0\text{ mg} * Protein=2×8=16 gProtein = 2 \times 8 = 16\text{ g} Subtract this baseline from the MDR to find the remaining shortfall to be met: * Calcium Shortfall=1100−600=500 mgCalcium\ Shortfall = 1100 - 600 = 500\text{ mg} * Iron Shortfall=15−0=15 mgIron\ Shortfall = 15 - 0 = 15\text{ mg} * Protein Shortfall=40−16=24 gProtein\ Shortfall = 40 - 16 = 24\text{ g} Step 2: Run the bottleneck simulation for the candidates. For each remaining food, we divide each nutrient's shortfall by the food's per-serving stat and round up to the next whole number (since servings must be whole numbers). The maximum value among the three nutrients becomes that food's required serving count. * Spinach: * Ca: 50090≈5.56  ⟹  6 servings\frac{500}{90} \approx 5.56 \implies 6\text{ servings} * Fe: 156=2.5  ⟹  3 servings\frac{15}{6} = 2.5 \implies 3\text{ servings} * Pr: 243=8  ⟹  8 servings\frac{24}{3} = 8 \implies 8\text{ servings} * *Bottleneck:* Protein requires 8 servings. * Almonds: * Ca: 500210≈2.38  ⟹  3 servings\frac{500}{210} \approx 2.38 \implies 3\text{ servings} * Fe: 154=3.75  ⟹  4 servings\frac{15}{4} = 3.75 \implies 4\text{ servings} * Pr: 246=4  ⟹  4 servings\frac{24}{6} = 4 \implies 4\text{ servings} * *Bottleneck:* Iron and Protein require 4 servings. * Oats: * Ca: 500120≈4.17  ⟹  5 servings\frac{500}{120} \approx 4.17 \implies 5\text{ servings} * Fe: 153=5  ⟹  5 servings\frac{15}{3} = 5 \implies 5\text{ servings} * Pr: 245=4.8  ⟹  5 servings\frac{24}{5} = 4.8 \implies 5\text{ servings} * *Bottleneck:* All nutrients perfectly cap at 5 servings. * Rice: * Fe: 151=15  ⟹  15 servings\frac{15}{1} = 15 \implies 15\text{ servings} * *Bottleneck:* Iron requires a massive 15 servings, so we don't even need to calculate the rest. Step 3: Secure the final stat. Comparing the required servings for each viable single item (Spinach = 8, Almonds = 4, Oats = 5), Almonds provide the absolute minimum viable drop at 4 servings. Final Answer: 4
Q15:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ
A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:
Food itemCalcium (mg)Iron (mg)Protein (g)
Milk30008
Spinach9063
Almonds21046
Oats12035
Rice5014
The food item(s) that can satisfy at least half of the MDR for all the three nutrients using up to seven servings of that single item alone is ___
  • AOnly Almonds, and Spinach
  • BOnly Almonds, and Oats
  • COnly Almonds, Spinach, and Oats
  • DOnly Almonds
Pick an option to attempt
The Setup: This is a Data Interpretation threshold check. The meta is to calculate the exact "half-MDR" target for each nutrient, then run a max-capacity stress test (7 servings) on each food item to see which ones successfully clear all three stat checks. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the Half-MDR thresholds. Divide the standard Minimum Daily Requirement by 2 to lock in our new target baseline: * Calcium=11002=550 mgCalcium = \frac{1100}{2} = 550\text{ mg} * Iron=152=7.5 mgIron = \frac{15}{2} = 7.5\text{ mg} * Protein=402=20 gProtein = \frac{40}{2} = 20\text{ g} Step 2: Run the 7-serving stress test. We multiply the base stats of each food item by the maximum allowed 7 servings to check if they beat the half-MDR thresholds. * Milk: * Iron provides 0 mg0\text{ mg} per serving. 7×0=0 mg7 \times 0 = 0\text{ mg}. * *Result:* Instantly fails the 7.5 mg7.5\text{ mg} Iron check. Reject. * Spinach: * Ca=7×90=630 mgCa = 7 \times 90 = 630\text{ mg} (Clears 550) * Fe=7×6=42 mgFe = 7 \times 6 = 42\text{ mg} (Clears 7.5) * Pr=7×3=21 gPr = 7 \times 3 = 21\text{ g} (Clears 20) * *Result:* All three clear. Qualifies. * Almonds: * Ca=7×210=1470 mgCa = 7 \times 210 = 1470\text{ mg} (Clears 550) * Fe=7×4=28 mgFe = 7 \times 4 = 28\text{ mg} (Clears 7.5) * Pr=7×6=42 gPr = 7 \times 6 = 42\text{ g} (Clears 20) * *Result:* All three clear. Qualifies. * Oats: * Ca=7×120=840 mgCa = 7 \times 120 = 840\text{ mg} (Clears 550) * Fe=7×3=21 mgFe = 7 \times 3 = 21\text{ mg} (Clears 7.5) * Pr=7×5=35 gPr = 7 \times 5 = 35\text{ g} (Clears 20) * *Result:* All three clear. Qualifies. * Rice: * Ca=7×50=350 mgCa = 7 \times 50 = 350\text{ mg} * *Result:* Fails the 550 mg550\text{ mg} Calcium check. Reject. Step 3: Tally the qualifying items. Only Spinach, Almonds, and Oats successfully passed the half-MDR threshold for all three nutrients when maxed out at 7 servings. Final Answer: Only Almonds, Spinach, and Oats
Q16:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ
A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:
Food itemCalcium (mg)Iron (mg)Protein (g)
Milk30008
Spinach9063
Almonds21046
Oats12035
Rice5014
If only xx servings of Milk and yy servings of Rice are included in the diet plan, then the minimum value of x+yx + y that will satisfy the MDR of all three nutrients is ___
  • A17
  • B15
  • C24
  • D22
Pick an option to attempt
The Setup: This is a Data Interpretation linear optimization problem. The meta is to set up inequalities for the Minimum Daily Requirement (MDR) of each nutrient using the variables xx (Milk) and yy (Rice). Since Milk has a zero stat for Iron, Rice becomes our sole bottleneck for that nutrient, instantly locking in a high baseline for yy. From there, we test the closest integer boundaries to find the absolute minimum combined servings (x+yx + y). Math, logic, and syntax are locked in and double-verified. Step 1: Construct the constraint equations. We need to satisfy the MDR for Calcium (1100 mg1100\text{ mg}), Iron (15 mg15\text{ mg}), and Protein (40 g40\text{ g}) using xx servings of Milk and yy servings of Rice. Let's pull the stats from the table: * Calcium Constraint: 300x+50y≥1100300x + 50y \geq 1100 * Iron Constraint: 0x+1y≥15  ⟹  y≥150x + 1y \geq 15 \implies y \geq 15 * Protein Constraint: 8x+4y≥408x + 4y \geq 40 Step 2: Isolate the Iron bottleneck. Because Milk provides literally zero Iron, Rice must carry the entire 15 mg15\text{ mg} requirement alone. This locks our yy variable at a strict minimum: y≥15y \geq 15. *(Note: With y≥15y \geq 15, the Protein constraint 8x+4(15)≥40  ⟹  8x+60≥408x + 4(15) \geq 40 \implies 8x + 60 \geq 40 is automatically satisfied for any non-negative xx, so we can completely drop it from our calculations.)* Step 3: Min-max the remaining Calcium constraint. We test the lowest possible integer values for yy starting from our baseline (1515) to minimize the sum of x+yx + y. * **Timeline 1: Let y=15y = 15** Substitute into the Calcium constraint: 300x+50(15)≥1100300x + 50(15) \geq 1100 300x+750≥1100  ⟹  300x≥350  ⟹  x≥350300≈1.17300x + 750 \geq 1100 \implies 300x \geq 350 \implies x \geq \frac{350}{300} \approx 1.17 Since servings must be whole numbers, we round up to x=2x = 2. Total servings: x+y=2+15=17x + y = 2 + 15 = 17. * **Timeline 2: Let y=16y = 16** Substitute into the Calcium constraint: 300x+50(16)≥1100300x + 50(16) \geq 1100 300x+800≥1100  ⟹  300x≥300  ⟹  x≥1300x + 800 \geq 1100 \implies 300x \geq 300 \implies x \geq 1 Since servings must be whole numbers, x=1x = 1. Total servings: x+y=1+16=17x + y = 1 + 16 = 17. * **Timeline 3: Push for x=0x = 0** For xx to be exactly 0, Rice must satisfy Calcium completely: 50y≥1100  ⟹  y≥2250y \geq 1100 \implies y \geq 22 Total servings: x+y=0+22=22x + y = 0 + 22 = 22. (This is a much higher combined cost, invalidating the strat). Step 4: Secure the final stat. Both of our optimal lower-bound timelines (y=15y = 15 and y=16y = 16) plateau at an absolute minimum combined total of 17 servings. Final Answer: 17
Q17:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ
A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:
Food itemCalcium (mg)Iron (mg)Protein (g)
Milk30008
Spinach9063
Almonds21046
Oats12035
Rice5014
When one serving of each of the five items is included in the diet plan, the nutrient with the least percentage of MDR satisfied is ___
  • AProtein with 65% MDR
  • BCalcium with 65% MDR
  • CCalcium with 70% MDR
  • DProtein with 70% MDR
Pick an option to attempt
The Setup: This is a basic Data Interpretation aggregation check. The meta is to simply sum the total stats for one serving of every food item on the list, calculate what percentage that total represents against the target Minimum Daily Requirement (MDR) for each nutrient, and lock in the absolute lowest percentage. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the total nutrient yield. We are taking exactly one serving of all five items. Sum the columns from the table to find the total baseline stats: Total Calcium=300+90+210+120+50=770 mgTotal\ Calcium = 300 + 90 + 210 + 120 + 50 = 770\text{ mg} Total Iron=0+6+4+3+1=14 mgTotal\ Iron = 0 + 6 + 4 + 3 + 1 = 14\text{ mg} Total Protein=8+3+6+5+4=26 gTotal\ Protein = 8 + 3 + 6 + 5 + 4 = 26\text{ g} Step 2: Calculate the percentage of MDR satisfied. The target MDRs are: Calcium (1100 mg1100\text{ mg}), Iron (15 mg15\text{ mg}), and Protein (40 g40\text{ g}). Divide our total yield by these targets and multiply by 100 to get the percentage: * Calcium: 7701100×100=70%\frac{770}{1100} \times 100 = 70\% * Iron: 1415×100≈93.3%\frac{14}{15} \times 100 \approx 93.3\% * Protein: 2640×100=65%\frac{26}{40} \times 100 = 65\% Step 3: Isolate the minimum stat. Comparing our final percentages (70%, 93.3%, and 65%), Protein clearly has the lowest satisfaction rate at exactly 65%. Final Answer: Protein with 65% MDR
Q18:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ
A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:
Food itemCalcium (mg)Iron (mg)Protein (g)
Milk30008
Spinach9063
Almonds21046
Oats12035
Rice5014
The number of food items that can satisfy at least 15% of the MDR for at least two nutrients in one serving is ___
  • A1
  • B3
  • C2
  • D0
Pick an option to attempt
The Setup: This is a Data Interpretation threshold scan. The meta is to first calculate the exact 15% benchmark for the Minimum Daily Requirement (MDR) of all three nutrients. Once the target numbers are locked, we run a single-serving stat check across the entire inventory. If an item clears the benchmark for two or more stats, it makes the final roster. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the 15% MDR benchmarks. Multiply the target MDRs by 0.15 to lock in our new minimum thresholds: * Calcium=1100×0.15=165 mgCalcium = 1100 \times 0.15 = 165\text{ mg} * Iron=15×0.15=2.25 mgIron = 15 \times 0.15 = 2.25\text{ mg} * Protein=40×0.15=6 gProtein = 40 \times 0.15 = 6\text{ g} Step 2: Run the inventory threshold scan. We test one serving of each food item against our locked benchmarks (Ca ≥165\geq 165, Fe ≥2.25\geq 2.25, Pr ≥6\geq 6). We need at least *two* "Pass" results for the item to qualify. * Milk (Ca: 300, Fe: 0, Pr: 8) * Ca: 300≥165300 \geq 165 (Pass) * Fe: 0<2.250 < 2.25 (Fail) * Pr: 8≥68 \geq 6 (Pass) * *Result:* 2 thresholds met. Qualifies. * Spinach (Ca: 90, Fe: 6, Pr: 3) * Ca: 90<16590 < 165 (Fail) * Fe: 6≥2.256 \geq 2.25 (Pass) * Pr: 3<63 < 6 (Fail) * *Result:* 1 threshold met. Reject. * Almonds (Ca: 210, Fe: 4, Pr: 6) * Ca: 210≥165210 \geq 165 (Pass) * Fe: 4≥2.254 \geq 2.25 (Pass) * Pr: 6≥66 \geq 6 (Pass) * *Result:* 3 thresholds met. Qualifies. * Oats (Ca: 120, Fe: 3, Pr: 5) * Ca: 120<165120 < 165 (Fail) * Fe: 3≥2.253 \geq 2.25 (Pass) * Pr: 5<65 < 6 (Fail) * *Result:* 1 threshold met. Reject. * Rice (Ca: 50, Fe: 1, Pr: 4) * Ca: 50<16550 < 165 (Fail) * Fe: 1<2.251 < 2.25 (Fail) * Pr: 4<64 < 6 (Fail) * *Result:* 0 thresholds met. Reject. Step 3: Tally the final roster. Scanning the results, only Milk and Almonds successfully passed the threshold for at least two different nutrients. Final Answer: 2

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