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A math problem in the sixth grade of primary school! Wait online!
1, the known conditions of the topic are unclear. How to allocate the peaks and valleys of the third gear? The peak power of this topic is 360 degrees, and the valley power: 360/( 1+50%) = 240 (degrees) -* *: 360+240 = 600 (degrees) absolutely exceeds the third gear.

The distribution of peaks and valleys must be adjusted. If each gear is in this transmission ratio, then:

The first peak electricity fee: 0.55× 204× (1+50%)/(1+kloc-0/+50%) = 67.32 yuan.

Off-peak electricity consumption cost in the first stage: 0.35× 204×1(1+1+50%) = 28.56 yuan.

Second peak electricity fee: (0.55+0.05) × (348-204 )× (1+50%)/(1+50%) = 51.84 (Yuan).

The second off-peak electricity fee: (0.35+0.05) × (348-204 )×1(1+1+50%) = 23.04 yuan.

The third peak electricity fee: (0.55+0.3) × (600-348 )× (1+50%)/(1+kloc-0/+50%) =128.52 (RMB).

Off-peak electricity charge for the third gear: (0.35+0.3) × (600-348 )×1(1+1+50%) = 65.52 (yuan).

* * *: 67.32+28.56+51.84+23.04+128.52+65.52 = 364.80 (yuan)

2. 82.8÷0.05= 1656 (degrees) February 1 day * * * electricity consumption 1656 degrees.

1656×1(1+4) = 331.2 (degrees) February valley power.

1656× 4/(1+4) =1324.8 (degrees) February electricity consumption peak.

3, 4.8÷0.05+204=300 (degrees) April 1 day * * * Power consumption of 300 degrees.

0.486×300= 145.8 (yuan) * * * April electricity bill.

145.8-4.8 =141(yuan). If the ladder price is not implemented, the electricity fee is 14 1 yuan.

0.55×300= 165 (yuan). It is assumed that all electricity is consumed at peak hours, and the total electricity fee is 165 yuan.

165-141= 24 (yuan) is more than the actual electricity bill by 24 yuan.

0.55-0.35=0.2 (Yuan) Replacing primary peak electricity with primary peak electricity to reduce electricity bill 0.2 yuan.

24÷0.2= 120 (degrees) and the valley power is 120 degrees.

300- 120= 180 (degrees) peak electricity is 180 degrees.