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Answers to questions 14, 13 and 14 in the new observation of mathematics in the eighth grade.
A store sold several pieces of A-type electrical appliances at the original price (cost+profit) in the first quarter, and each piece earned an average profit of 25%. In the second quarter, due to a slight increase in profit, the number of pieces sold by A Electric Appliance was only 5/6 of that in the first quarter, but the total profit was the same as that in the first quarter.

(1) What is the average profit per piece of Class A electrical appliances sold by this counter in the second quarter?

(2) In the third quarter, the counter sold 90% of the price in the first quarter. Therefore, the number of pieces sold increased by 65,438+0.5 times compared with the first quarter. What percentage is the profit of Class A household appliances sold in the third quarter higher than that of Class A household appliances sold in the first quarter?

Solution: (1) Let the cost be A, the number of pieces sold be B, and the profit rate in the second quarter be C.

Then profit =a×25%= 1/4a.

In the second quarter, 5/6b electrical appliances were sold.

Total profit in the first quarter = 1/4ab

Profit in the second quarter =ac×5/6b=5/6abc

According to the meaning of the question

1/4ab=5/6abc

c= 1/4×6/5

c=3/ 10=30%

(2) Pricing in the first quarter =a( 1+25%)=5/4a.

Pricing in the third quarter =5/4a×90%=9/8a

Sold (1.5+ 1)b=2.5b pieces in the third quarter.

Total profit in the third quarter =9/8a×2.5b-2.5ab=5/ 16ab.

Total profit growth in the third quarter (5/16ab-1/4ab)/(kloc-0//4ab) = (116)/(1/4) = 0.25.

Party A and Party B set out from two places, A and B respectively, and face to face at the same time. They met for the first time 50 kilometers away from Party A's midpoint, and turned back immediately after arriving at the two places. So, Party A and Party B meet for the second time at the distance of A 100 meters, and find the distance between the two places.

Two people, A and B, set out from A to B, A can't, B rides a bike. If a walks 6 kilometers, they will arrive at b at the same time after 45 minutes from b; If A takes the poem of 1 first, and B catches up with A half an hour after departure, find the distance between A and B. ..

Let the speed of a be a km/h and the speed of b be b km/h.

45 minutes =3/4 hours

6+3/4a=3/4b

a=(b-a)x 1/2

simplify

b-a=8( 1)

3a=b(2)

( 1)+(2)

2a=8

A=4 km/h

B = 3x4 =12km/h

AB distance = 12x 3/4 = 9 km.

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