Skip to main content

10th maths PAIR OF LINEAR EQUATION IN TWO VARIABLE MCQ

Image
By
Picture
Trend Setter MCQ's
1. Every linear equation in two variables has ___ solution(s).
(a) no (b) one (c) two (d) infinitely many

2. a1/a2 = b1/b2=c1/c2 is the condition for
(a) intersecting lines (b) parallel lines (c) coincident lines (d) none

3. For a pair to be consistent and dependent the pair must have
(a) no solution (b) unique solution (c) infinitely many solutions (d) none of these

4. Graph of every linear equation in two variables represent a ___
(a) point (b) straight line (c) curve (d) triangle

5. Each point on the graph of pair of two lines is a common solution of he lines in case of ___
(a) Infinitely many solutions (b) only one solution (c) no solution (d) none of these

6. Which of he following is the solution of the pair of linear equations 3x – 2y = 0, 5y – x = 0
(a) (5, 1) (b) (2, 3) (c) (1, 5) (d) (0, 0)

7. One of the common solution of ax + by = c and y-axis is _____
(a) (0, c/b) (b) (0,b/c ) (c) , 0 , (c/ b ) (d) (0, c/ b)

8. If the value of x in the equation 2x – 8y = 12 is 2 then the corresponding value of y will be
(a) –1 (b) +1 (c) 0 (d) 2

9. The pair of linear equations is said to be inconsistent if they have
(a) only one solution (b) no solution (c) infinitely many solutions. (d) both a and c

10. On representing x = a and y = b graphically we get ____
(a) parallel lines (b) coincident lines (c) intersecting lines at (a, b) (d) intersecting lines at (b, a)

11. How many real solutions of 2x + 3y = 5 are possible
(a) no (b) one (c) two (d) infinitely many

12. The value of k for which the system of equation 3x + 2y = – 5, x – ky = 2 has a unique solutions.
(a) K = 2/ 3 (b) K ¹ 2/3 (c) K = -2 /3 (d) K ¹ - 2/3

13. If the lines represented by the pair of linear equations 2x + 5y = 3, 2(k + 2) y + (k + 1) x = 2k are coincident then the value of k is ____
(a) –3 (b) 3 (c) 1 (d) –2

14. The coordinates of the point where x-axis and the line represented by x/2 + 4/3 = 1 intersect, are
(a) (0, 3) (b) (3, 0) (c) (2, 0) (d) (0, 2)

15. Graphically x – 2 = 0 represents a line
(a) parallel to x-axis at a distance 2 units from x-axis.    (b) parallel to y-axis at a distance 2 units from it.
(c) parallel to x-axis at a distance 2 units from y-axis.   (d) parallel to y-axis at a distance 2 units from x-axis.
Labels:

Comments

  1. rajputSeptember 22, 2012

    Very knowledgeable and descriptive blog.The linear equation can be solve by three methods.My favorite is graphical method.But there is some complex situations come when the solutions are imaginary.Can you please help me about it.
    Period

    ReplyDelete

Post a Comment

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

By
NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it
Post a Comment
Read more

Living science ratna sagar class 6 solutions

By
Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II
Post a Comment
Read more

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

By
X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page
Post a Comment
Read more

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

By
Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor
2 comments
Read more

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

By
1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca
Post a Comment
Read more