CIE Mathematics Paper-1: Specimen Questions 42 - 44 of 175

Question number: 42

Essay Question▾

Describe in Detail

Solve:

Co-ordinate plane with a line

Co-ordinate plane with a line

Showing graphical lines drawn on a coordinate plane

(a) Mark the point with co-ordinates (2, 5) and label it A.

(b) Find co-ordinates of the point B.

(c) Find the gradient of the line.

Explanation

Here,

(a) Here, point is (x, y) = (2, 5) which can place in diagram as below:

Co-ordinate plane showing required point A

Co-ordinate plane showing required point A

Showing the required point drawn on a coordinate plane

(b) Here, point B is in second quarter of coordinate plane which gives minus value of x and plus value of y means (-x, y).

  • So, point B (-x, y) = B (-1, 2).

(c) Here, gradient of a line is also known as slope of the line.

  • For finding slop we need two point of line. So take A (2, 5) and B (-1, 2)
  • Now, formula for slope = AyByAxBx=522(1)=33=1 .

Question number: 43

Short Answer Question▾

Write in Short

Mike spends 30 on grocery. This amount is 23 of his allowance. Calculate his allowance.

Explanation

Here, 30 is 23 of his allowance, so take allowance is x.

So, 23×x=30x=30×32=45.

Question number: 44

Short Answer Question▾

Write in Short

The points P and Q are marked on the diagram.

Two points P and Q are on x-y axis

Two points P and Q are on x-y axis

Showing two points P and Q on x-y axis having co-ordinates

(a) Write QP as a column vector.

(b) PR=(43) . Write down the co-ordinates of R.

Explanation

Here, as shown in figure,

(a) Here, Point P = (-4, 3) and point Q = (-5, -4)

So, for vector of QP: (4(5),3(4))=(4+5, 3+4)=(1,7) , so as a column vector: (17)

(b) Here, PR=(43) , so as a row vector we can write as (4, -3), let take co-ordinate of R = (x, y) and

P (-4, 3) is given.

So, (x(4),y3)=(4,3)

x+4=4,y3=3

x=0,y=0

So, R (x, y) = R (0, 0)

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