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Exam 001

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MATHEMATICS

CLASS IX CBSE Board

Jawahar Navodaya Vidyalaya


Linear equations in two variables and Lines and Angles

Time allowed : 2 hours
Maximum Marks : 60
General Instructions :

1. All questions are compulsory.
2. This question paper consists of 12 questions of 5 marks each
3. Use of calculator is not permitted.

SECTION – A

marks

Question numbers 1 to 12 carry 5 marks each.

  1. Write each of the following equations in the form of linear equation and indicate a,b,c in each case
    a) 4=3x4=3x
    b) 10=x+y5-10 = -x + \dfrac{y}{5}
  2. It is given XYZ=64°\angle XYZ = 64 \degree and XYXY is produced to point P. Draw a figure from the given information, if ray YQYQ bisects ZYP\angle ZYP, find XYQ\angle XYQ and reflex QYP\angle QYP
  3. Find the value of kk if x=2,y=1x=2, y=1 is a solution of the equation 2x+3y=k2x + 3y = k
  4. If x+y=w+zx + y = w + z, then prove that POQPOQ is a line.











  5. The taxi fare in a city is as follows. For the first Kilometer, the fare is ₹18 and for the subsequent distance it is ₹15 per km. Write the linear equation for this information and draw its graph.
  6. The linear equat that converts Celsius to Fahrenheit
    C=(F32)59C = \left ( F - 32 \right ) \dfrac {5}{9}
    i) If the temp is 60°60 \degree, what is the temperature in Fahrenheit?
    ii) Is there a temperature which is numerically the same in both Fahrenheit and Celsius. If yes, find it.
  7. AOBAOB is a line, Ray OCOC is \perp to line ABAB. ODOD is another ray lying between rays OAOA and OCOC. Prove that
    COD=12(BODAOD)\angle COD = \dfrac{1}{2} \left ( \angle BOD - \angle AOD \right )
  8. Solve the equation 2x+1=x32x + 1 = x - 3 and represent the solution on
    i) the number line
    ii) cartesian plane
  9. PQST,PQR=110°PQ \parallel ST, \angle PQR = 110 \degree and RST=130°\angle RST = 130 \degree. Find QRS\angle QRS
  10. The sides ABAB and ACAC of ΔABC\Delta ABC are produced to points EE and DD resp. If bisectors BOBO and COCO of CBF\angle CBF and BCD\angle BCD resp. meet at point OO, then prove that BOC=90°12BAC\angle BOC = 90 \degree - \dfrac{1}{2} \angle BAC











  11. Side QRQR of ΔPQR\Delta PQR is produced to a point SS. If the bisecture of PQR\angle PQR and PRS\angle PRS meet at point T. Then prove that 12QPR=QTR\dfrac {1}{2} \angle QPR = \angle QTR
  12. PQPQ and RSRS are two mirror placed parallel to each other. An incident ray ABAB strikes the mirror PQPQ at BB, the reflected ray moves along the path BCBC and strikes the mirror RSRS at CC and again reflects back along CDCD. Prove that ABCDAB \parallel CD

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