JEE Mathematics Linear Equations and Graphs MCQs Set 01

Answer: (a) \( \left(\tfrac23,0\right) \)

Explanation:
1. At the x-axis \( y=0 \), so \( 3x-2=0 \).
2. Hence \( x=\tfrac23 \).
3. The graph cuts the x-axis at \( \left(\tfrac23,0\right) \); it cuts the y-axis at \( (0,-2) \).

Teacher's Note:
1. Put y = 0 for the x-intercept and x = 0 for the y-intercept.
2. The point (0, -2) is the y-intercept, not the x-intercept.

Answer: (c) \( y=3x-2 \) and \( y=3x+2 \)

Explanation:
1. The slopes are \( 3,\ -3 \) and \( 3 \) respectively.
2. Lines with equal slopes are parallel.
3. So \( y=3x-2 \) and \( y=3x+2 \) are parallel.

Teacher's Note:
1. In \( y=mx+c \) the coefficient m is the slope.
2. Parallel lines have the same slope and different intercepts.

Answer: (d) \( (2,10) \)

Explanation:
1. Substitute each point.
2. \( x=1:\ y=2 \); \( x=-1:\ y=-16 \); \( x=0:\ y=-7 \), so these three lie on the graph.
3. For \( x=2 \), \( y=11\ne10 \), so \( (2,10) \) does not lie on it.

Teacher's Note:
1. A point is on a graph only if its coordinates satisfy the equation.
2. Test every option, since the three correct points are a check on each other.

Answer: (d) \( (-1,10) \)

Explanation:
1. For \( x=0 \), \( y=11 \), so \( (0,0) \) is not on the graph.
2. For \( x=-1 \), \( y=10 \), so \( (-1,2) \) is not on it but \( (-1,10) \) is.
3. For \( x=1 \), \( y=12\ne10 \). Hence the point is \( (-1,10) \).

Teacher's Note:
1. Substitute x and compare the result with the given y.
2. Points that fail the test are not on the graph.

Answer: (c) \( a=7,\ b=5 \)

Explanation:
1. From \( (0,5) \), \( b=5 \).
2. From \( (1,12) \), \( a+5=12 \), so \( a=7 \).
3. Check: \( 7\times2+5=19 \) and \( 7\times(-3)+5=-16 \), so \( a=7,\ b=5 \).

Teacher's Note:
1. The point with x = 0 gives the constant b directly.
2. Always verify the law with the remaining data points.

Answer: (c) 40

Explanation:
1. The law is \( y=7x+5 \).
2. At \( x=5 \), \( y=7\times5+5=40 \).
3. So \( y=40 \).

Teacher's Note:
1. First find a and b from the data, then substitute.
2. The slope is 7, so each unit step in x raises y by 7.

Answer: (c) 50

Explanation:
1. The vertices are the intersections: \( (5,-3) \) from \( x+y=2 \) and \( y=-3 \); \( (-5,-3) \) from \( x=-5 \) and \( y=-3 \); \( (-5,7) \) from \( x=-5 \) and \( x+y=2 \).
2. The base along \( y=-3 \) has length 10 and the height along \( x=-5 \) is 10.
3. Area \( =\tfrac12\times10\times10=50 \).

Teacher's Note:
1. Lines of the form x = c and y = c are parallel to the axes.
2. The triangle is right-angled at (-5, -3), so use half the product of the legs.

Answer: (b) 1

Explanation:
1. \( f(x)=x \) means \( y=x \), which has the form \( y=mx \) with \( m=1 \).
2. It passes through \( (0,0) \) and \( (1,1) \).
3. So its slope is 1 and it makes \( 45^\circ \) with the x-axis.

Teacher's Note:
1. The graph of y = mx is a line through the origin.
2. Equal x and y values mean the line bisects the first and third quadrants.

Answer: (b) \( (0.9,-0.7) \)

Explanation:
1. Equate the two expressions: \( 2-3x=\dfrac x3-1 \).
2. So \( 3=\dfrac{10x}{3} \) and \( x=0.9 \).
3. Then \( y=2-2.7=-0.7 \), and the point is \( (0.9,-0.7) \).

Teacher's Note:
1. At the point of intersection both equations give the same y.
2. A careful graph reads this point to one decimal place.

Answer: (c) decreases, so the function is decreasing

Explanation:
1. The slope is \( -2<0 \).
2. Each increase of 1 in \( x \) reduces \( y \) by 2.
3. So \( y=-2x \) is a decreasing function over the whole number line, while \( y=2x \) is increasing.

Teacher's Note:
1. The sign of the slope decides whether a straight-line function increases or decreases.
2. Compare values at x = 0, 1, 2 to check.

Answer: (c) no solution, since the lines are parallel

Explanation:
1. Both lines have slope 4 but different intercepts 3 and 0.
2. Parallel distinct lines never meet.
3. So the system has no solution.

Teacher's Note:
1. Equal slopes with unequal intercepts mean parallel lines.
2. Subtracting the equations gives 3 = 0, which is impossible.

Answer: (a) \( (4,-1) \)

Explanation:
1. From the first equation \( y=2x-9 \).
2. Substitute into the second: \( x+4x-18=2 \), so \( x=4 \) and \( y=-1 \).
3. Check: \( 8+1=9 \) and \( 4-2=2 \). The solution is \( (4,-1) \).

Teacher's Note:
1. The solution of two linear equations is the intersection point of their lines.
2. Always substitute the point into both equations.

Answer: (c) \( (5,2) \)

Explanation:
1. Multiply the first by 2 and the second by 3: \( 4x-6y=8 \) and \( 9x+6y=57 \).
2. Adding, \( 13x=65 \), so \( x=5 \) and \( y=2 \).
3. The solution is \( (5,2) \).

Teacher's Note:
1. Eliminate one variable to check the intersection found from the graph.
2. Verify in both equations: 10 - 6 = 4 and 15 + 4 = 19.

Answer: (d) \( (1,2) \)

Explanation:
1. From the first equation \( y=4-2x \).
2. Substitute: \( 8-4x-3x=1 \), so \( 7x=7 \) and \( x=1 \), \( y=2 \).
3. The solution is \( (1,2) \).

Teacher's Note:
1. Substitution works well when one equation is easily solved for y.
2. Check: 2 + 2 = 4 and 4 - 3 = 1.

Answer: (b) Rs. 5,000

Explanation:
1. The amount rises by \( 9200-7400=1800 \) in 3 years, i.e. Rs. 600 per year.
2. The interest for 4 years is Rs. 2400.
3. Principal \( =7400-2400=Rs.\ 5000 \).

Teacher's Note:
1. On the graph of amount against time, the line cuts the amount axis at the principal.
2. The yearly increase equals the yearly simple interest.

Answer: (d) 12%

Explanation:
1. The principal is Rs. 5,000 and the yearly interest is Rs. 600.
2. Rate \( =\dfrac{600}{5000}\times100 \).
3. So the rate is 12% per annum.

Teacher's Note:
1. Rate is yearly interest divided by principal.
2. The slope of the amount-time graph equals the yearly interest.

Answer: (d) 10 years

Explanation:
1. The amount is \( 5000+600t \).
2. Setting \( 5000+600t=11000 \) gives \( t=10 \).
3. So it amounts to Rs. 11,000 in 10 years.

Teacher's Note:
1. Form the linear relation amount = P + (yearly interest) x t.
2. Read the time at which the line reaches the required amount.

Answer: (d) Rs. 8,000

Explanation:
1. The amount is \( 5000+600t \).
2. At \( t=5 \), the amount is \( 5000+3000 \).
3. So the amount is Rs. 8,000.

Teacher's Note:
1. Interpolate on the same straight line.
2. The amount after 5 years lies between those for 4 and 7 years.

Answer: (b) 6 PM, 48 km from Delhi

Explanation:
1. Let \( t \) be hours after noon. A has gone \( 8t \) km and B has gone \( 12(t-2) \) km.
2. Equating, \( 8t=12t-24 \), so \( t=6 \).
3. They meet at 6 PM, at a distance \( 8\times6=48 \) km from Delhi.

Teacher's Note:
1. B overtakes A when both have covered the same distance.
2. On a distance-time graph this is where the two lines cross.

Answer: (d) 4 PM and 8 PM

Explanation:
1. The gap is \( |8t-12(t-2)|=|24-4t| \).
2. Setting it equal to 8 gives \( 24-4t=8 \) or \( 24-4t=-8 \), so \( t=4 \) or \( t=8 \).
3. They are 8 km apart at 4 PM and 8 PM.

Teacher's Note:
1. The distance between them can be 8 km before B overtakes A and again after.
2. Use the modulus of the difference.

Answer: (d) B is 4 km behind A and C is 6 km behind B

Explanation:
1. At 5 PM, A has gone \( 8\times5=40 \) km, B has gone \( 12\times3=36 \) km and C has gone \( 15\times2=30 \) km.
2. So A is ahead of B by 4 km.
3. B is ahead of C by 6 km, which gives B 4 km behind A and C 6 km behind B.

Teacher's Note:
1. Find each rider's time of travel, then multiply by the speed.
2. Compare the three distances from Delhi.

Answer: (b) 4 PM

Explanation:
1. B has gone \( 12(t-2) \) km and C has gone \( 15(t-3) \) km, where \( t \) is hours after noon.
2. The gap is \( 12t-24-15t+45=21-3t \).
3. Setting \( 21-3t=9 \) gives \( t=4 \), so the time is 4 PM.

Teacher's Note:
1. C starts at 3 PM, so its travel time is t - 3 hours.
2. Check at 4 PM: B has gone 24 km and C has gone 15 km, a gap of 9 km.

Answer: (c) 150 km

Explanation:
1. For constant speed, \( y=vt \), so the faster car has \( y=60t \).
2. At \( t=2.5 \), \( y=60\times2.5 \).
3. The distance is 150 km.

Teacher's Note:
1. Distance equals speed times time.
2. On the graph, the faster car has the steeper line through the origin.

Answer: (d) \( (0,0) \)

Explanation:
1. At \( (0,0) \): \( 6>0 \), true.
2. At \( (0,3) \): \( -9 \); at \( (-5,0) \): \( -9 \); at \( (1,4) \): \( -11 \), all false.
3. So the point is \( (0,0) \); the solution region lies on the origin side, below the line.

Teacher's Note:
1. Use a test point, usually the origin, to decide which side of the line is the solution.
2. The inequality can also be written as \( y<\tfrac{3x+6}{5} \).

Answer: (b) \( (-4,0) \)

Explanation:
1. The conditions reduce to \( x<\tfrac65 \) and \( x<-3 \), so together \( x<-3 \).
2. Only \( (-4,0) \) has \( x<-3 \).
3. At \( (0,0) \) and \( (-2,1) \) the second condition fails, and at \( (2,1) \) both fail. The point is \( (-4,0) \).

Teacher's Note:
1. Solve each inequality separately and take the common region.
2. Here the second condition is stronger.

Answer: (d) \( (3,0) \)

Explanation:
1. At \( (3,0) \): \( -4<0 \) and \( 9-6=3>0 \), both true.
2. At \( (0,0) \): the second gives \( -6 \), false. At \( (0,2) \): the first gives \( 1 \), false. At \( (1,1) \): the second gives \( -2 \), false.
3. So the point is \( (3,0) \).

Teacher's Note:
1. Test each point in both inequalities.
2. The common region lies below y = x + 1 and above y = 6 - 3x.