Class 10 Pair of Linear Equations — HOTS Worksheet
Higher Order Thinking Skills (HOTS) questions covering equations reducible to linear form, literal coefficients, parameter $k$ conditions, and multi-step speed & work word problems.
Question 1: Solve the following system of equations for $x$ and $y$:
$$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2, \quad \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1 \quad (x > 0, y > 0)$$
Question 2: Solve the following pair of equations by reducing them to a pair of linear equations:
$$\frac{5}{x-1} + \frac{1}{y-2} = 2, \quad \frac{6}{x-1} - \frac{3}{y-2} = 1 \quad (x \neq 1, y \neq 2)$$
Solution: Step 1: Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$. We get:
(1) $5u + v = 2$
(2) $6u - 3v = 1$ Step 2: Multiply (1) by 3: $15u + 3v = 6$. Step 3: Add equation (2): $21u = 7 \Rightarrow u = \frac{7}{21} = \frac{1}{3}$. Step 4: From (1): $5(\frac{1}{3}) + v = 2 \Rightarrow v = 2 - \frac{5}{3} = \frac{1}{3}$. Step 5: Return to $x$ and $y$:
$\frac{1}{x-1} = \frac{1}{3} \Rightarrow x - 1 = 3 \Rightarrow x = 4$.
$\frac{1}{y-2} = \frac{1}{3} \Rightarrow y - 2 = 3 \Rightarrow y = 5$. Answer: $x = 4,\, y = 5$.
Question 3: Solve for $x$ and $y$ using elimination:
$$ax + by = a - b, \quad bx - ay = a + b$$
Solution: Step 1: Multiply the first equation by $a$ and the second equation by $b$:
(1) $a^2x + aby = a(a-b) = a^2 - ab$
(2) $b^2x - aby = b(a+b) = ab + b^2$ Step 2: Add both equations to eliminate $y$:
$(a^2 + b^2)x = (a^2 - ab) + (ab + b^2) = a^2 + b^2$
$\Rightarrow x = \frac{a^2 + b^2}{a^2 + b^2} = 1$. Step 3: Substitute $x = 1$ into $ax + by = a - b$:
$a(1) + by = a - b \Rightarrow by = -b \Rightarrow y = -1$. Answer: $x = 1,\, y = -1$.
Question 4: Solve the large-coefficient symmetric system using addition and subtraction:
$$152x - 378y = -74, \quad -378x + 152y = -604$$
Solution: Step 1 (Add the two equations):
$(152 - 378)x + (-378 + 152)y = -74 - 604$
$-226x - 226y = -678$
Divide by $-226$: $x + y = 3$ \quad (Equation A). Step 2 (Subtract second equation from first):
$[152 - (-378)]x + [-378 - 152]y = -74 - (-604)$
$530x - 530y = 530$
Divide by $530$: $x - y = 1$ \quad (Equation B). Step 3 (Solve A and B):
Adding A and B: $2x = 4 \Rightarrow x = 2$.
Subtracting B from A: $2y = 2 \Rightarrow y = 1$. Answer: $x = 2,\, y = 1$.
Question 5: Find the value of $k$ for which the following pair of linear equations has infinitely many solutions:
$$(k - 3)x + 3y = k, \quad kx + ky = 12$$
Solution: Step 1: For infinitely many solutions, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ where standard form is $a_1x + b_1y - c_1 = 0$.
Here: $\frac{k-3}{k} = \frac{3}{k} = \frac{k}{12}$. Step 2: From $\frac{3}{k} = \frac{k}{12} \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$. Step 3: Test $k = 6$:
$\frac{6-3}{6} = \frac{3}{6} = \frac{1}{2}$; \quad $\frac{3}{6} = \frac{1}{2}$; \quad $\frac{6}{12} = \frac{1}{2}$. All three ratios match $\frac{1}{2}$. Step 4: Test $k = -6$:
$\frac{-6-3}{-6} = \frac{-9}{-6} = \frac{3}{2}$, but $\frac{3}{-6} = -\frac{1}{2}$ (does not match). Answer: $k = 6$.
Question 6: A boat covers 32 km upstream and 36 km downstream in 7 hours. In 9 hours, it covers 40 km upstream and 48 km downstream. Determine the speed of the boat in still water and the speed of the stream.
Solution: Step 1: Let the speed of the boat in still water be $x$ km/h and speed of stream be $y$ km/h.
Speed upstream $= (x - y)$ km/h; Speed downstream $= (x + y)$ km/h. Step 2: Formulation:
(1) $\frac{32}{x-y} + \frac{36}{x+y} = 7$
(2) $\frac{40}{x-y} + \frac{48}{x+y} = 9$ Step 3: Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.
$32u + 36v = 7$ \quad (multiply by 4: $128u + 144v = 28$)
$40u + 48v = 9$ \quad (multiply by 3: $120u + 144v = 27$) Step 4: Subtract: $8u = 1 \Rightarrow u = \frac{1}{8}$.
Substitute $u = \frac{1}{8}$ into (1): $32(\frac{1}{8}) + 36v = 7 \Rightarrow 4 + 36v = 7 \Rightarrow 36v = 3 \Rightarrow v = \frac{1}{12}$. Step 5: $x - y = 8$ and $x + y = 12$.
Adding: $2x = 20 \Rightarrow x = 10$.
Subtracting: $2y = 4 \Rightarrow y = 2$. Answer: Speed of boat in still water $= 10\text{ km/h}$, speed of stream $= 2\text{ km/h}$.
Question 7: A train covered a certain distance at a uniform speed. If the train had been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution: Step 1: Let the original speed be $v$ km/h and scheduled time be $t$ hours. Distance $d = v \times t$. Step 2 (Case 1): Speed $= v + 10$, time $= t - 2$.
$(v + 10)(t - 2) = vt \Rightarrow vt - 2v + 10t - 20 = vt \Rightarrow -2v + 10t = 20 \Rightarrow -v + 5t = 10$ \quad (1). Step 3 (Case 2): Speed $= v - 10$, time $= t + 3$.
$(v - 10)(t + 3) = vt \Rightarrow vt + 3v - 10t - 30 = vt \Rightarrow 3v - 10t = 30$ \quad (2). Step 4: Multiply (1) by 2: $-2v + 10t = 20$.
Add to (2): $(-2v + 10t) + (3v - 10t) = 20 + 30 \Rightarrow v = 50\text{ km/h}$. Step 5: From (1): $-50 + 5t = 10 \Rightarrow 5t = 60 \Rightarrow t = 12\text{ hours}$.
Distance $d = v \times t = 50 \times 12 = 600\text{ km}$. Answer: Distance $= 600\text{ km}$ (Speed $= 50\text{ km/h}$, Time $= 12\text{ hours}$).
Question 8: The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution: Step 1: Let the tens digit be $x$ and the units digit be $y$. Original number $= 10x + y$. Reversely, number $= 10y + x$. Step 2: Given sum of digits is 9:
$x + y = 9$ \quad (1). Step 3: Nine times this number is twice the reversed number:
$9(10x + y) = 2(10y + x) \Rightarrow 90x + 9y = 20y + 2x$
$88x - 11y = 0 \Rightarrow 8x - y = 0 \Rightarrow y = 8x$ \quad (2). Step 4: Substitute (2) into (1):
$x + 8x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1$.
Then $y = 8(1) = 8$.
Original number $= 10(1) + 8 = 18$. Answer: The number is $18$.
Question 9: The angles of a cyclic quadrilateral $ABCD$ are given by $\angle A = (4y + 20)^\circ$, $\angle B = (3y - 5)^\circ$, $\angle C = (-4x)^\circ$, and $\angle D = (-7x + 5)^\circ$. Find the measures of all four angles.
Solution: Step 1: In a cyclic quadrilateral, opposite angles are supplementary (sum $= 180^\circ$):
$\angle A + \angle C = 180^\circ \Rightarrow (4y + 20) + (-4x) = 180 \Rightarrow -4x + 4y = 160 \Rightarrow -x + y = 40$ \quad (1).
$\angle B + \angle D = 180^\circ \Rightarrow (3y - 5) + (-7x + 5) = 180 \Rightarrow -7x + 3y = 180$ \quad (2). Step 2: From (1), $y = x + 40$. Substitute into (2):
$-7x + 3(x + 40) = 180 \Rightarrow -7x + 3x + 120 = 180 \Rightarrow -4x = 60 \Rightarrow x = -15$. Step 3: Find $y$: $y = -15 + 40 = 25$. Step 4: Calculate each angle:
$\angle A = 4(25) + 20 = 120^\circ$
$\angle B = 3(25) - 5 = 70^\circ$
$\angle C = -4(-15) = 60^\circ$
$\angle D = -7(-15) + 5 = 105 + 5 = 110^\circ$
(Verification: $120^\circ + 60^\circ = 180^\circ$ and $70^\circ + 110^\circ = 180^\circ$). Answer: $\angle A = 120^\circ,\, \angle B = 70^\circ,\, \angle C = 60^\circ,\, \angle D = 110^\circ$.
Question 10: 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Solution: Step 1: Let 1 woman alone take $x$ days and 1 man alone take $y$ days. 1 woman's 1 day's work $= 1/x$, 1 man's 1 day's work $= 1/y$. Step 2: From the given conditions:
(1) $\frac{2}{x} + \frac{5}{y} = \frac{1}{4}$
(2) $\frac{3}{x} + \frac{6}{y} = \frac{1}{3}$ Step 3: Put $u = 1/x$ and $v = 1/y$:
$2u + 5v = 1/4 \Rightarrow 8u + 20v = 1$ \quad (multiply by 9: $72u + 180v = 9$)
$3u + 6v = 1/3 \Rightarrow 9u + 18v = 1$ \quad (multiply by 8: $72u + 144v = 8$) Step 4: Subtract: $36v = 1 \Rightarrow v = \frac{1}{36}$.
Substitute $v = \frac{1}{36}$ into $9u + 18(\frac{1}{36}) = 1 \Rightarrow 9u + \frac{1}{2} = 1 \Rightarrow 9u = \frac{1}{2} \Rightarrow u = \frac{1}{18}$. Step 5: $x = 1/u = 18$ days, $y = 1/v = 36$ days. Answer: 1 woman takes $18$ days, 1 man takes $36$ days.