Wan duo xing cheng zi Wei Bao underground parking

I don't know what grade the question is. Suppose you haven't learned to set the unknown, you can do this:

( 1)

From the ground 1 parking space and underground 1 parking space, it needs 6000 yuan, that is, 2 above ground+2 underground = 0.6× 2 =1.20 thousand.

The cost of adding three above-ground parking spaces and two underground parking spaces is 6.5438+0.3 million yuan, which means that the above-ground parking space is 654.38+0 = 654.38+0.3-654.38+0.2 = 0.654.38+0.0 million yuan.

So underground 1 = 0.6-0. 1 = 0.5 million.

(2) Give it a try. Assuming that repairing 25 floors above ground and 25 floors below ground requires 1.5 million, which is 20,000 more than the budgeted limit of 1.3 million, and repairing 1 floor above ground saves 04,000 than repairing 1 floor below ground, then at least 20,000/.4 million =5 needs to be increased above ground. tabulation/listing

Total amount above ground and underground

30 20 13

3 1 19 12.6

32 18 12.2

When the ground is increased to 33, it is 65438+200,000 less than the budget.

Therefore, the above three schemes are feasible.