Class 9 Maths Ganita Manjari Chapter 13 Solutions — Two Variables, One Line

This has exactly $a=3$, $b=0$ and $c=-\frac{1}{5}$. Multiplying by 5 gives the equivalent equation $15x-1=0$, but its coefficients are different.

(i) Charlie is correct: $t=2p$. The notebook’s cost is twice the pen’s cost, so the larger cost t equals two copies of p. Meera’s equation reverses the relationship. (ii) Yes, $x+y=176$ represents the combined score. Here x and y are non-negative integers. The equation states their total; it does not identify their individual scores.

The left side equals the right side, so (4,3) is a solution.

(i) For $7x-3y=21$, setting $y=0$ gives $x=3$; setting $x=0$ gives $y=-7$. Thus (3,0) and (0,−7). (ii) For $2x+3y=5$, choose $x=1$ to get $y=1$, and $x=4$ to get $y=-1$. Thus (1,1) and (4,−1). These are sample choices among infinitely many.

Plot each pair and draw the corresponding line. The graphs below use labelled axes and show both chosen points.

C(2,3) is not on the line and does not satisfy the equation. A graph of an equation consists exactly of its solutions, so no point off this line can satisfy $3x-7y=21$.

(i) Multiplying $3x+4y=7$ by 2 gives $6x+8y=14$. Conversely divide the second equation by 2. Therefore each solution satisfies both equations. (ii) For $k\ne 0$, multiplication by k and division by k are reversible, so ax + by = c and kax + kby = kc have exactly the same solution set. If $k=0$, the second equation becomes $0=0$ and the conclusion fails.

The values are approximately 0.857, 0.545 and 0.800. From least to greatest steepness: B, C, A. Compare rise/run, not height alone.

Both measurements are in metres, so the slope is dimensionless.

The slope readings assume the same unit per grid interval on the two axes, as intended by the supplied square grids.

This is the horizontal run, not the sloping ramp length. It uses the ratio supplied in the exercise.

x is the cost of one ticket and y the cost of one snack box, in rupees. Optional check: $x=245$, $y=120$ gives both totals.

x is the fixed charge in rupees and y the charge in rupees per kilometre. Subtracting gives $2y=38$, so $y=19$ and $x=8$. This is the model specified in the question.

Here x and y count tickets, so both must be non-negative integers. Subtract 100 times the first equation from the second: $50x=5000$, giving $x=100$ and $y=100$.

Verification: $7x-15y=\frac{343-285}{29}=2$ and $x+2y=\frac{49+38}{29}=3$. If instead $y=\frac{3-x}{2}$ is substituted, multiplying by 2 gives $14x-45+15x=4$, so $29x=49$ and the same solution results.

The angles are 26°, 104°, 50°, all positive and summing to 180°.

The actual incomes are 9x = ₹18,000 and 7x = ₹14,000, not x and y. Expenditures are ₹16,000 and ₹12,000; each saving is ₹2,000.

Reflection: to eliminate x, multiply the first equation by 7 and the second by 9. Subtraction gives $y=4000$, then $x=2000$. Substitution with $x=\frac{2000+4y}{9}$ gives $7(2000+4y)-27y=18000$, again $y=4000$. Elimination avoids fractions here; substitution is convenient when a coefficient is 1.

Let the digits be x and y. The sum with its reverse gives $11\left(x+y\right)=66$, hence $x+y=6$. If $x-y=2$, adding gives $x=4$ and $y=2$, producing 42. But the original wording says only that the digits differ by 2: 24 also satisfies it. Thus there are two numbers, 24 and 42. The textbook’s uniqueness statement applies only after adding the assumption that the tens digit is larger.

Check: $-8+13=5$ and $-8-13=-21$.

The usual intended positive numbers are 39 and 13. Since the question does not explicitly require positive numbers, −13 and −39 also have difference 26 and one is three times the other. The equation $x=3y$ assumes the larger number is the triple, which excludes that second case.

One bat costs ₹1200 and one ball ₹80. Check: $8400+480=8880$ and $3600+400=4000$.

Fixed charge ₹25; rate ₹13/km; a 25 km trip costs ₹350.

The fraction is $\frac79$. Checks: $\frac{7+2}{9+2}=\frac{9}{11}$ and $\frac{7+3}{9+3}=\frac{10}{12}=\frac{5}{6}$. Denominators used in cross multiplication must be nonzero.

The fraction is $\frac35$. Then $\frac{4}{4}=1$ and $\frac{3}{6}=\frac{1}{2}$, as required.

Present ages: Nuri 50 years, Sonu 20 years. Five years ago $45=3\times 15$; ten years later $60=2\times 30$.

The number is 18; $9\times 18=162=2\times 81$.

There are 10 notes of ₹50 and 15 notes of ₹100. Their total is ₹500+₹1500 = ₹2000.

The fixed charge for the first three days is ₹15; each additional day costs ₹3. Seven days means four extra days, not seven.

Let x count boys and y count girls. Then $x+y=10$ and $y-x=4$. Plot (0,10),(10,0) for the first line and (0,4),(6,10) for the second. The intersection is (3,7), so there are 3 boys and 7 girls. Check: $3+7=10$ and $7-3=4$.

The dimensions are 20 m by 16 m. The full perimeter is 72 m.

Sample answers: (i) $x-y=0$ gives an intersection with $2x+3y=8$ at $\left(\frac{8}{5},\frac{8}{5}\right)$. (ii) $2x+3y-10=0$ is distinct and parallel. (iii) $4x+6y-16=0$ is twice the given equation and coincides with it. Many other examples are valid.

One citron costs 8 and one wood-apple costs 5 in the problem’s monetary units. Adding and subtracting takes advantage of the exchanged coefficients. Checks: $72+35=107$ and $56+45=101$.

A pencil costs ₹3 and a pen ₹5.

Let s be stool height and c the cat’s height in the same upright posture. The first picture gives $s+c=85\,\mathrm{cm}$. The second measures the difference between the stool top and the cat’s head, so $s-c=25\,\mathrm{cm}$.

The stool is 55 cm high.

(i) The slope is 3. (ii) The y-intercept is 0: the model starts with zero water depth at time 0. (iii)(a) At 5 minutes $y=15\,\mathrm{cm}$. (b) For 21 cm, $x=\frac{21}{3}=7$ minutes. These predictions assume the stated constant-rate model remains applicable. (iv) (4,10) is not on the line because $3\times 4=12$, not 10. (v) At $x=3$, $y=9$, so the intersection is (3,9).

(i) Use Celsius C on the horizontal axis and Fahrenheit F on the vertical axis. Plot (−40,−40), (0,32), (20,68), (40,104), then draw the line.

(ii) At 30°C:

(iii) At 95°F:

(iv) $0{}^{\circ}\mathrm{C}=32{}^{\circ}\mathrm{F}$. At 0°F:

(v) Put $C=F=t$:

Thus −40°C and −40°F are numerically equal.

For $2x+y=6$ use (0,6),(3,0); for $2x-y-2=0$ use (0,−2),(1,0). The graph is the same pair as Exercise 13.5 Question 4(iii), and the intersection is (2,2). Addition gives $4x=8$ and then $y=2$, verifying the graph.

Insufficient information in the supplied cover. The included cover image shows two lines but labels only one equation, $15x-10y-115=0$. The other line has no readable equation and the axes have no numbered unit scale. Therefore an exact intersection cannot be determined from this copy without guessing. Once the second equation or a calibrated graph is supplied, solve the two equations together. This item is flagged rather than assigned an invented coordinate.

Set $y=0$. If $m\ne 0$:

If $m=0$ and $c\ne 0$, the horizontal line has no x-intercept. If $m=c=0$, the line is the x-axis itself, so every point on it is an intersection; there is no single intercept value.

For non-negative monthly minutes t: Plan B is cheaper when $0\le t$ < 200; they both cost ₹90 at t=200; Plan A is cheaper for t>200. This uses the fees and continuous per-minute model stated in the question.

(i) Infinitely many lines have a given finite slope m: $y=mx+d$ for any real d. (ii) Exactly one has that slope and passes through $\left(x_{0},y_{0}\right)$:

If “given slope” means a vertical direction instead, all vertical lines have undefined slope and exactly one, $x=x_{0}$, passes through the specified point.

A unique solution requires $\Delta \ne 0$, hence $p\ne 4$. At $p=4$, the first equation gives $4x+4y=-8$ while twice the second gives $4x+4y=-4$, so there is no solution, not infinitely many.

The first equation must be k times $3x-y=14$. Compare coefficients and constants:

Check: the first equation becomes $6x-2y=28$, exactly twice the second.

The second equation must be a scalar multiple of $x+2y=3$. Its multiplier is k−1, so:

The constant also matches; $k=3$ gives $2x+4y=6$. Checking only the variable coefficients would not be sufficient.

(i) Yes; there are several shared points and segments. After n complete moves, Robot 1 is at (3n,4n), and after m complete moves Robot 2 is at (10+m,2m). But each path includes its separate horizontal and vertical movements, not only those endpoints. The first common point along the paths is (24,28). The complete intersection set is:

To verify, Robot 1’s horizontal segment is $y=4n$, $3n\le x\le 3n+3$, and its vertical segment is $x=3n+3$, $4n\le y\le 4n+4$. Robot 2’s corresponding segments are $y=2m$, $10+m\le x\le 11+m$ and $x=11+m$, $2m\le y\le 2m+2$, for non-negative integers n,m. Intersecting these interval descriptions gives the table.

Joining only the completed-move positions by straight lines gives $y=\frac{4x}{3}$ and $y=2\left(x-10\right)$, which intersect at (30,40). That point is on both actual paths, but it is not their only intersection. A common geometric point does not imply a simultaneous collision, since speeds and timings are unspecified.

(ii) Yes: the shared initial segment is $y=0$, $7\le x\le 8$. Robot 1 first travels from (3,0) to (8,0); Robot 2 from (7,0) to (12,0). Afterwards one path lies above and the other below the x-axis, so there are no further common portions. Replacing their movements by straight lines would change the question.

Use the intended meaning “a is six times b”. The elapsed-time relation is $a+15=60+b$, and $a=6b$.

He first looked at 2:54 and then at 3:09. The phrase “six times greater” is imprecise; “six times as great” clearly states the intended multiplier.

The number is 78; $87-78=9$.

Opposite angles are supplementary:

Hence $\angle A=37^\circ$, $\angle B=48^\circ$, $\angle C=143^\circ$, $\angle D=132^\circ$. The opposite pairs sum to 180° and all four sum to 360°.

Let v be the usual speed in km/h and t the usual time in hours. The same distance is vt in each case.