13, a school-run factory 10 produced 200 sets of A products. By improving the production technology, the total output in February and March increased by the same percentage point compared with the previous month, so the total output in the first quarter reached 1400 units. What is the percentage?

Solution: let this percentage be a.

200+200( 1+a)+200( 1+a)？ = 1400

Let 1+a=t

t？ +t-6=0

(t-2)(t+3)=0

T=2 or t=-3 (truncation)

So 1+a=2.

a= 1= 100%

14, with two numbers. Their sum is 13, and their product is -48. What are these two numbers?

Solution: Let one number be a and the other number be13-a.

a( 13-a)=-48

Answer? - 13a-48=0

(a- 16)(a+3)=0

A=-3 or a= 16

When a=-3, another number is 16.

When a= 16, another number is -3.

15. In order to expand the 1 swimming pool with a length of 200m and a width of 60m into a large water amusement park with a circumference of 600m, the length of the swimming pool is increased by xm. So when x =, the area of water swimming pool is 20000 square meters. If we can find the value of x? If you can't explain why.

Solution: after the length is increased, it is 100+x meters.

At this time, the width is (600/2- 100-x)=200-x meters.

( 100+x)(200-x)=20000

20000+200x- 100x-x？ =20000

x？ - 100x=0

x(x- 100)=0

X= 100 or x=0 (omitted)

The length is increased by 100m and the width is increased by 200- 100-60 = 40m.

16, Class 1, Grade 9, Group 2 At a party, each student presented a photo to the other students. This activity sent out 90 photos. How many students are there in this group?

Solution: There are X students.

Each student received x- 1 photo.

A * * * has x students.

Then the total number = number of students × number of photos received by each student.

The formula x(x- 1)=90.

x？ -x-90=0

(x- 10)(x+9)=0

X= 10 or -9 (truncated)

So there are 10 students.

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