Question:medium
A store sells four items (A, B, C, D) over three months (Jan, Feb, Mar).
Total sales of A over the three months = 300 units.
Feb sales of C are 50 more than Jan sales of C.
Mar sales of B are half of Feb sales of A.
Total sales across all months for all items = 1320 units.
What are the sales in Feb for item A?
A store sells four items (A, B, C, D) over three months (Jan, Feb, Mar).
Total sales of A over the three months = 300 units.
Feb sales of C are 50 more than Jan sales of C.
Mar sales of B are half of Feb sales of A.
Total sales across all months for all items = 1320 units.
What are the sales in Feb for item A?
Total sales of A over the three months = 300 units.
Feb sales of C are 50 more than Jan sales of C.
Mar sales of B are half of Feb sales of A.
Total sales across all months for all items = 1320 units.
What are the sales in Feb for item A?
Show Hint
When one variable influences several others (like A_F), isolate it in the total equation. Feasibility and non-negativity often force a unique solution.
Updated On: Jul 4, 2026
Show Solution
Correct Answer: 200
Solution and Explanation
This one doesn't need step-by-step arithmetic so much as a quick check on how much the clues actually pin down. There are 12 unknown monthly figures (4 items times 3 months), but only 4 facts are given: the yearly total for A, a link between C's Jan and Feb, a link between B's March and A's Feb, and the grand total. Four equations can never fix twelve unknowns to a single value for any one of them, including Feb sales of A specifically, whatever value you assume for it, the rest of the table can always be filled in to keep every stated condition true. So no unique number for "Feb sales of item A" exists from this data; the correct response is that the question is data-insufficient, not a specific figure.
Was this answer helpful?
2
Top Questions on Data Analysis
- The schematic diagram below shows 12 rectangular houses in a housing complex. House numbers are mentioned in the rectangles representing the houses. The houses are located in six columns - Column-A through Column-F, and two rows - Row-1 and Row- 2 . The houses are divided into two blocks - Block XX and Block YY. The diagram also shows two roads, one passing in front of the houses in Row-2 and another between the two blocks.
Some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The road adjacency value of a house is the number of its sides adjacent to a road. For example, the road adjacency values of C2, F2, and B1 are 2, 1, and 0, respectively. The neighbour count of a house is the number of sides of that house adjacent to occupied houses in the same block. For example, E1 and C1 can have the maximum possible neighbour counts of 3 and 2, respectively.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbour count).
The following information is also known.
1. The maximum quoted price of a house in Block XX is Rs. 24 lakhs. The minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column-E.
2. Row-1 has two occupied houses, one in each block.
3. Both houses in Column-E are vacant. Each of Column-D and Column-F has at least one occupied house.
4. There is only one house with parking space in Block YY. - A visa processing office (VPO) accepts visa applications in four categories - US, UK, Schengen, and Others. The applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.
There are ten counters in the office, four dedicated to US applications, and two each for UK applications, Schengen applications and Others applications. Applicants are called in for processing sequentially on a first-come-first-served basis whenever a counter gets freed for their category. The processing time for an application is the same within each category. But it may vary across the categories. Each US and UK application requires 10 minutes of processing time. Depending on the number of applications in a category and time required to process an application for that category, it is possible that an applicant for a slot may be processed later.
On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the office, exactly six out of the ten counters were either processing applications, or had finished processing one and ready to start processing the next.
Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category.
The following additional information is known about that day.
1. All slots were full.
2. The number of US applications was the same in all the slots. The same was true for the other three categories.
3. 50% of the applications were US applications.
4. All applicants except Ira, Vijay and Nandini arrived on time.
5. Vijay was called to a counter at 9:25 am. - There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of
coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied.
i) The minimum among the numbers of coins in the three sacks in the box is 1.
ii) The median of the numbers of coins in the three sacks is 1.
iii) The maximum among the numbers of coins in the three sacks in the box is 9. - There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2.
Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same.
The total number of coins in each column is also the same. Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a * if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by ** if two or more of these conditions are satisfied.
i) The minimum among the numbers of coins in the three sacks in the box is 1.
ii) The median of the numbers of coins in the three sacks is 1.
iii) The maximum among the numbers of coins in the three sacks in the box is 9. - Want to practice more? Try solving extra ecology questions todayView All Questions