HALP PLS 1000 students are arranged in several rows (more than 16). The numbers of students in each row are consecutive positive integers. How many students are there in the first row? (A) 20 (B) 25 (C) 28 (D) 44 (E) 80

The correct option is;

(C) 28

Step-by-step explanation:

The information given are that;

The total number of students = 1000

The number of students in each row are consecutive positive numbers

Let the number of students in the first row be x,

Let the number of rows be 17

We have;

The number of students in  the last row = x + 16

The number of students in  the penultimate row = x + 15, we continue  to get;

n×x + n/2×(n-1) = 1000

(The term n/2×(n-1) is from the sum of an arithmetic progression formula)

Which gives;

17×x + 17/2×16 = 1000

17×x = 1000 - 17/2×16 = 1000 - 136 = 864

x = 864/17 = Which is not an integer,

We continue checking till we get to 25 rows which gives;

25×x + 25/2×24 = 1000

25×x + 300 = 1000

25×x  = 1000 - 300 = 700

x = 700/25 = 28

The number of student in the first row = 28 students.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.