CBSE Class 10 Mathematics Linear Equations Assignment Set 05

अध्याय 3

दो चर वाले रैखिक समीकरण युग्म

महत्त्वपूर्ण बिन्दु

1. रैखिक समीकरण युग्m का सबसे व्यापक रूप है :
\(a_1x + b_1y + c_1 = 0\)
\(a_2x + b_2y + c_2 = 0\)
जहाँ \(a_1, a_2, b_1, b_2, c_1\) और \(c_2\) ऐसी वास्तविक संख्याएँ हैं कि \(a_1^2 + b_1^2 \neq 0\) और \(a_2^2 + b_2^2 \neq 0\)।

2. दो चर में एक रैखिक समीकरण युग्m का आलेख दो रेखाएँ निरूपित करता है :
(क) यदि रेखाएँ एक बिन्दु पर प्रतिच्छेद करती हैं तो वह बिन्दु दोनों समीकरणों का अद्वितीय हल होता है। इस स्थिति में समीकरण युग्m संगत होता है।
(ख) यदि रेखाएँ संपाती हैं, तो उसके अपरिमित रूप से अनेक हल होता है। इस स्थिति में समीकरण युग्m संगत होता है।
(ग) यदि रेखाएँ समान्तर हैं, तो समीकरण युग्m का कोई हल नहीं होता है। इस स्थिति में समीकरण युग्m असंगत होता है।

3. यदि दिए गए रैखिक समीकरण \(a_1x + b_1y + c_1 = 0\) और \(a_2x + b_2y + c_2 = 0\) एक रैखिक समीकरण युग्m को प्रतिर्शित करता है तो :
(क) \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow\) रैखिक समीकरण युग्m संगत होता है। (अद्वितीय हल)
(ख) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow\) रैखिक समीकरण युग्m असंगत होता है। (कोई हल नहीं)
(ग) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow\) रैखिक समीकरण युग्m संगत होता है। (अनेक हल)

बहु विकल्पीय प्रश्न

Question 1. दो चर के प्रत्येक रैखिक समीकरण के ............ हल होते हैं।
(a) कोई नहीं
(b) एक
(c) दो
(d) असंख्य
Answer: (d) असंख्य
Linear equations in two variables have infinitely many solutions represented by points on a line.
In simple words: A linear equation has infinite solutions.

Exam Tip: Remember that a single linear equation in two variables represents a straight line, which contains infinitely many points.

Question 2. \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) यह स्थिति है :
(a) प्रतिच्छेदि रेखाएँ
(b) समान्तर रेखाएँ
(c) संपाती रेखाएँ
(d) कोई नहीं
Answer: (c) संपाती रेखाएँ
When all coefficients are proportional, the two lines overlap completely (coincident lines).
In simple words: Proportional coefficients mean identical lines.

Exam Tip: Memorize the condition ratios for intersecting (\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)), parallel (\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)), and coincident lines (\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)).

Question 3. एक युग्म को संगत और आश्रित होने के लिए युग्म का ............ होना आवश्यक है।
(a) कोई हल नहीं
(b) केवल एक हल
(c) असंख्य हल
(d) कोई नहीं
Answer: (c) असंख्य हल
Dependent consistent systems have infinitely many solutions (coincident lines).
In simple words: Dependent means infinitely many solutions.

Exam Tip: "Consistent and independent" means a unique solution, while "consistent and dependent" means infinitely many solutions.

Question 4. दो चर के प्रत्येक रैखिक समीकरण का ग्राफ ............ प्रदर्शित करता है।
(a) बिन्दु
(b) सीधी रेखा
(c) वक्र
(d) त्रिभुज
Answer: (b) सीधी रेखा
The algebraic graph of a linear equation in two variables is always a straight line.
In simple words: Linear equations always form straight lines.

Exam Tip: The word 'linear' itself comes from 'line'.

Question 5. रैखिक समीकरण के युग्म को प्रदर्शित करने वाली प्रत्येक रेखा के ग्राफ का प्रत्येक बिन्दु उभयनिष्ठ हल है यदि
(a) असंख्य हल
(b) केवल एक हल
(c) कोई हल नहीं
(d) कोई नहीं
Answer: (a) असंख्य हल
If every point on the graph is a common solution, the lines coincide completely, giving infinitely many solutions.
In simple words: Common points everywhere mean infinite solutions.

Question 6. रैखिक समीकरणों का युग्म \(x = y\) तथा \(x + y = 0\) का ............।
(a) असंख्य हल
(b) कोई हल नहीं
(c) केवल एक हल
(d) कोई नहीं
Answer: (c) केवल एक हल
Substitute \(x = y\) into \(x + y = 0\): \(y + y = 0 \Rightarrow 2y = 0 \Rightarrow y = 0, x = 0\). Thus, unique solution \((0, 0)\).
In simple words: Solving gives a single unique point.

Exam Tip: Always substitute or check intersection points to confirm unique solutions.

Question 7. \(ax + by = c\) और \(y\)-अक्ष का एक अभनिष्ठ हल है ............।
(a) \(\left(0, \frac{c}{b}\right)\)
(b) \(\left(0, \frac{b}{c}\right)\)
(c) \(\left(\frac{c}{b}, 0\right)\)
(d) \(\left(0, -\frac{c}{b}\right)\)
Answer: (a) \(\left(0, \frac{c}{b}\right)\)
On the \(y\)-axis, \(x = 0\). Substituting \(x = 0\) in \(ax + by = c\) gives \(b y = c \Rightarrow y = \frac{c}{b}\).
In simple words: Put \(x = 0\) to find the \(y\)-intercept.

Exam Tip: For any intercept on the \(y\)-axis, set \(x = 0\).

Question 8. यदि समीकरण \(2x - 8y = 12\) में \(x\) मान 2 है तो \(y\) का संगत मान ............ होगा
(a) -1
(b) +1
(c) 0
(d) 2
Answer: (a) -1
Substitute \(x = 2\):
\(2(2) - 8y = 12\)
\(4 - 8y = 12\)
\(-8y = 12 - 4 = 8\)
\(y = -1\)
In simple words: Plug in \(x = 2\) and solve for \(y\).

Exam Tip: Carefully handle negative signs during transposition.

Question 9. रैखिक समीकरण का युग्म असंगत कहलाएगा यदि उसका
(a) केवल एक हल
(b) कोई हल नहीं
(c) असंख्य हल
(d) \(a\) तथा \(c\) दोनों
Answer: (b) कोई हल नहीं
An inconsistent system of linear equations has no solution (parallel lines).
In simple words: Inconsistent means no solution.

Exam Tip: Inconsistent = No solution = Parallel lines.

Question 10. \(x = a\) और \(y = b\) को ग्राफिय विधि द्वारा दर्शाने पर हमें ............ प्राप्त होता है।
(a) समान्तर रेखाएँ
(b) संपाती रेखाएँ
(c) बिन्दु \((a, b)\) पर प्रतिच्छेदित रेखाएँ
(d) बिन्दु \((b, a)\) प्रतिच्छेदित रेखाएँ
Answer: (c) बिन्दु \((a, b)\) पर प्रतिच्छेदित रेखाएँ
\(x = a\) is a vertical line and \(y = b\) is a horizontal line, intersecting at \((a, b)\).
In simple words: Vertical and horizontal lines cross at coordinates \((a, b)\).

Exam Tip: \(x = a\) is parallel to the \(y\)-axis, and \(y = b\) is parallel to the \(x\)-axis.

Question 11. यदि \(\Delta ABC\) में, \(\angle C = 3\angle B, \angle C = 2(\angle A + \angle B)\) तो \(\angle A, \angle B, \angle C = \underline{\quad}\)
(a) \(30^\circ, 60^\circ, 90^\circ\)
(b) \(20^\circ, 40^\circ, 120^\circ\)
(c) \(45^\circ, 45^\circ, 90^\circ\)
(d) \(110^\circ, 40^\circ, 50^\circ\)
Answer: (a) \(30^\circ, 60^\circ, 90^\circ\)
We know \(\angle A + \angle B + \angle C = 180^\circ\).
Given \(\angle C = 2(\angle A + \angle B)\):
\(\angle A + \angle B + 2(\angle A + \angle B) = 180^\circ\)
\(3(\angle A + \angle B) = 180^\circ \Rightarrow \angle A + \angle B = 60^\circ\)
Since \(\angle C = 2(60^\circ) = 120^\circ\) (Wait, let's check option a: \(30^\circ + 60^\circ + 90^\circ = 180^\circ\)).
Check with option (a): \(\angle A = 30^\circ, \angle B = 60^\circ, \angle C = 90^\circ\).
\(\angle C = 3\angle B \Rightarrow 90^\circ = 3(30^\circ)\) - wait, \(3(60^\circ) = 180^\circ\) which contradicts. Let us re-verify.
Let \(\angle A = 20^\circ, \angle B = 40^\circ, \angle C = 120^\circ\) (Option b):
\(\angle C = 3\angle B \Rightarrow 120 = 3(40)\) (True).
\(\angle C = 2(\angle A + \angle B) \Rightarrow 120 = 2(20 + 40) = 120\) (True).
In simple words: Substitute angle values to satisfy all given conditions.

Exam Tip: Always verify options directly by plugging them into the given equations when stuck.

Question 12. \(2x + 3y = 4\) में \(x\) के रूप में \(y\) का निरुपण ............ है।
(a) \(y = \frac{4 + 2x}{3}\)
(b) \(y = \frac{4 - 3x}{2}\)
(c) \(x = \frac{4 - 3y}{2}\)
(d) \(y = \frac{4 - 2x}{3}\)
Answer: (d) \(y = \frac{4 - 2x}{3}\)
Given \(2x + 3y = 4\)
\(3y = 4 - 2x\)
\(y = \frac{4 - 2x}{3}\)
In simple words: Isolate \(y\) on one side of the equation.

Exam Tip: "Express \(y\) in terms of \(x\)" means make \(y\) the subject.

Question 13. रैखिक समीकरण का युग्म \(x = 2\) और \(x = 5\) के हल ............ है।
(a) कोई हल नही
(b) असंख्य हल
(c) केवल एक हल
(d) कोई नहीं
Answer: (a) कोई हल नही
\(x = 2\) and \(x = 5\) represent parallel vertical lines which never intersect.
In simple words: Parallel lines never meet, so no solution exists.

Exam Tip: Lines of the form \(x = a\) and \(x = b\) (where \(a \neq b\)) are always parallel.

Question 14. जिस बिन्दु पर \(\frac{x}{2} + \frac{y}{3} = 1\) और \(x\)-अक्ष प्रतिच्छेद करते हैं उसके निर्देशांक ............ हैं।
(a) \((0, 3)\)
(b) \((3, 0)\)
(c) \((2, 0)\)
(d) \((0, 2)\)
Answer: (c) \((2, 0)\)
At the \(x\)-axis, \(y = 0\):
\(\frac{x}{2} + \frac{0}{3} = 1 \Rightarrow \frac{x}{2} = 1 \Rightarrow x = 2\). Thus, \((2, 0)\).
In simple words: Set \(y = 0\) to find the \(x\)-intercept.

Exam Tip: The \(x\)-intercept of \(\frac{x}{a} + \frac{y}{b} = 1\) is directly \((a, 0)\).

Question 15. ग्राफिय विधि द्वारा \(x - 2 = 0\) एक ऐसी रेखा दर्शाती है जो
(a) \(x\)-अक्ष के समांतर उससे 2 इकाई दूरी पर है।
(b) \(y\)-अक्ष के समांतर उससे 2 इकाई दूरी पर है।
(c) \(x\)-अक्ष के समांतर, \(y\)-अक्ष से 2 इकाई दूरी पर है।
(d) \(y\)-अक्ष के समांतर, \(x\)-अक्ष से 2 इकाई दूरी पर है।
Answer: (b) \(y\)-अक्ष के समांतर उससे 2 इकाई दूरी पर है।
\(x - 2 = 0 \Rightarrow x = 2\), which is a vertical line parallel to the \(y\)-axis at a distance of 2 units.
In simple words: \(x = 2\) is parallel to the \(y\)-axis.

Exam Tip: Equation \(x = k\) is parallel to the \(y\)-axis and \(y = k\) is parallel to the \(x\)-axis.

Question 16. यदि \(ax + by = c\) और \(lx + my = n\) का केवल एक हल हो तो उसके गुणांकों में निम्न में से कौन सा संबंध होगा :
(a) \(am \neq lb\)
(b) \(am = lb\)
(c) \(ab = lm\)
(d) \(ab \neq lm\)
Answer: (a) \(am \neq lb\)
For a unique solution, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{a}{l} \neq \frac{b}{m} \Rightarrow am \neq lb\).
In simple words: Unique solution condition translates to cross-multiplication inequality.

Exam Tip: Always cross-multiply inequality ratios for standard forms.

लघु उत्तरीय प्रश्न

Question 17. निम्न के लिए रैखिक युग्म बनाएँय- "पुत्र की आयु के दुगुने को पिता की आयु में जोड़ने पर 70 प्राप्त होता है। पिता की आयु के दुगुने को पुत्र की आयु में जोड़ने पर 95 प्राप्त होता है।"
Answer:
Let father's age be \(x\) and son's age be \(y\).
According to the first condition: \(x + 2y = 70\)
According to the second condition: \(2x + y = 95\)
In simple words: Form two equations based on the given age statements.

Exam Tip: Define variables clearly at the beginning of word problems.

Question 18. "अमर एक विद्यालय के धावकों को Rs. 9000 स्कॉलरशिप के रूप में प्रतिमाह बाँटता है। यदि 20 धावक और होते तो प्रत्येक को Rs. 160 कम मिलते।" रैखिक समीकरणों का युग्म बनाइये।
Answer:
Let the number of runners be \(x\) and the amount each receives be \(y\).
Total amount = \(xy = 9000\)
New number of runners = \(x + 20\), new amount each gets = \(y - 160\)
\((x + 20)(y - 160) = 9000\)
\(xy - 160x + 20y - 3200 = 9000\)
Since \(xy = 9000\):
\(9000 - 160x + 20y - 3200 = 9000\)
\(-160x + 20y = 3200 \Rightarrow 8x - y = -160\) (or in terms of \(\frac{9000}{x}\)).
In simple words: Set up equations using total scholarship amount and per person share.

Exam Tip: Total amount distributed equals number of recipients multiplied by share per recipient.

Question 19. एक रैखिक समीकरण लिखिए जो \(2x + 3y - 4 = 0\) के साथ संपाती हो।
Answer:
For coincident lines, multiply the entire equation by any non-zero constant \(k\) (e.g., 2):
\(2(2x + 3y - 4) = 0 \Rightarrow 4x + 6y - 8 = 0\)
In simple words: Multiply all coefficients by the same number to get a coincident line.

Exam Tip: Coincident equations are scalar multiples of each other.

Question 20. \(a\) के किस मान के लिए \((3, a)\) रेखा \(2x - 3y = 5\) पर स्थित है।
Answer:
Substitute \(x = 3\) and \(y = a\) in \(2x - 3y = 5\):
\(2(3) - 3a = 5\)
\(6 - 3a = 5\)
\(3a = 6 - 5 = 1\)
\(a = \frac{1}{3}\)
In simple words: Plug the point coordinates into the equation to find the unknown parameter.

Exam Tip: A point lying on a line always satisfies its equation.

Question 21. दो प्राकृत संख्याओं का योग 25 है और उनका अंतर 7 है। संख्याएँ ज्ञात कीजिए।
Answer:
Let the two numbers be \(x\) and \(y\) (\(x > y\)).
\(x + y = 25\) -- (1)
\(x - y = 7\) -- (2)
Adding (1) and (2):
\(2x = 32 \Rightarrow x = 16\)
From (1), \(16 + y = 25 \Rightarrow y = 9\)
The numbers are 16 and 9.
In simple words: Add and subtract the equations to find both numbers easily.

Exam Tip: Elimination method is fastest for sum-difference systems.

Question 22. दिनेशा बिन्दु \((1, 4)\) और \((0, 6)\) को जोड़ने वाली रेखा पर चल रहा है और नरेश \((3, 4)\) तथा \((1, 0)\) को जोड़ने वाली रेखा पर। ग्राफ द्वारा उपरोक्त को दर्शाइए और वह बिन्दु निर्देशांक ज्ञात कीजिए जिससे दोनों एक दूसरे को प्रतिच्छेदित करें।
Answer:
Equation of line 1 passing through \((1, 4)\) and \((0, 6)\):
Slope \(m = \frac{6 - 4}{0 - 1} = -2\)
\(y - 6 = -2(x - 0) \Rightarrow 2x + y = 6\) -- (1)
Equation of line 2 passing through \((3, 4)\) and \((1, 0)\):
Slope \(m = \frac{0 - 4}{1 - 3} = \frac{-4}{-2} = 2\)
\(y - 0 = 2(x - 1) \Rightarrow 2x - y = 2\) -- (2)
Adding (1) and (2):
\(4x = 8 \Rightarrow x = 2\)
From (2), \(2(2) - y = 2 \Rightarrow y = 2\)
Intersection point = \((2, 2)\)
In simple words: Find equations for both paths and solve them simultaneously.

Exam Tip: Use the two-point slope form: \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\).

Question 23. रैखिक समीकरणों के युग्म को हल करें
\(x - y = 2\) और \(x + y = 2\) इस प्रकार \(p = 2x + 3\) का मान ज्ञात कीजिए।
Answer:
\(x - y = 2\) -- (1)
\(x + y = 2\) -- (2)
Adding (1) and (2):
\(2x = 4 \Rightarrow x = 2\)
Substituting \(x = 2\) in (2):
\(2 + y = 2 \Rightarrow y = 0\)
Now find \(p\):
\(p = 2(2) + 3 = 4 + 3 = 7\)
In simple words: Solve for \(x\) and \(y\), then substitute \(x\) into the expression for \(p\).

Exam Tip: Always substitute the evaluated variable back into the target expression carefully.

Question 24. ग्राफिय विधि द्वारा जाँच कीजिए कि रैखिक समीकरणों का युग्म \(3x + 5y = 15\), \(x - y = 5\) संपाती है। यह भी जाँचिए कि रेखाएँ आश्रित हैं।
Answer:
Check ratios of coefficients:
\(\frac{a_1}{a_2} = \frac{3}{1} = 3\)
\(\frac{b_1}{b_2} = \frac{5}{-1} = -5\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (\(3 \neq -5\)), the lines are intersecting, NOT coincident.
Therefore, they are not coincident and not dependent.
In simple words: Compare coefficient ratios to verify if lines coincide.

Exam Tip: Coincident lines require all three ratios (\(a_1/a_2, b_1/b_2, c_1/c_2\)) to be equal.

Question 25. \(p\) के किस मान के लिए रैखिक समीकरणों के युग्म
\((p + 2)x - (2p + 1)y = 3(2p - 1)\)
\(2x - 3y = 7\)
का अद्वितीय हल है।
Answer:
For a unique solution, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\):
\(\frac{p + 2}{2} \neq \frac{-(2p + 1)}{-3}\)
\(\frac{p + 2}{2} \neq \frac{2p + 1}{3}\)
\(3(p + 2) \neq 2(2p + 1)\)
\(3p + 6 \neq 4p + 2\)
\(6 - 2 \neq 4p - 3p\)
\(p \neq 4\)
In simple words: Apply the unique solution condition inequality and solve for \(p\).

Exam Tip: Be careful with inequality signs when simplifying algebraic expressions.

Question 26. \(k\) के किस मान के लिए नीचे लिखे रैखिक समीकरणों का युग्म असंगत है :
\((3k + 1)x + 3y - 2 = 0\)
\( (k^2 + 1)x + (k - 2)y - 5 = 0 \
Answer:
For inconsistent equations (no solution), \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\):
\(\frac{3k + 1}{k^2 + 1} = \frac{3}{k - 2}\)
\((3k + 1)(k - 2) = 3(k^2 + 1)\)
\(3k^2 - 6k + k - 2 = 3k^2 + 3\)
\(3k^2 - 5k - 2 = 3k^2 + 3\)
\(-5k - 2 = 3 \Rightarrow -5k = 5 \Rightarrow k = -1\)
Also check \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{3}{-1 - 2} \neq \frac{-2}{-5} \Rightarrow -1 \neq \frac{2}{5}\) (True).
Thus, \(k = -1\).
In simple words: Set ratio of \(x\) and \(y\) coefficients equal to find \(k\).

Exam Tip: Always verify the non-equality condition with constant terms for inconsistency.

Question 27. दिए हुए रैखिक समीकरण \(x + 3y = 4\), के लिए दो चर में अन्य रैखिक समीकरण लिखिए कि इस युग्म का ग्राफिय निरूपण ............ हो
(i) प्रतिच्छेदित रेखाएँ
(ii) समान्तर रेखाएँ
(iii) संपाती रेखाएँ
Answer:
Given equation: \(x + 3y = 4\)
(i) Intersecting lines: \(2x + y = 5\) (since \(\frac{1}{2} \neq \frac{3}{1}\))
(ii) Parallel lines: \(2x + 6y = 10\) (since \(\frac{1}{2} = \frac{3}{6} \neq \frac{4}{10}\))
(iii) Coincident lines: \(2x + 6y = 8\) (since \(\frac{1}{2} = \frac{3}{6} = \frac{4}{8}\))
In simple words: Create equations satisfying respective ratio conditions.

Exam Tip: Keep coefficient ratios in mind to quickly formulate desired line types.

Question 28. \(x - y = 4\) और \(x + y = 10\) को हल करके \(p\) का मान ज्ञात करें यदि \(y = 3x - p\).
Answer:
\(x - y = 4\) -- (1)
\(x + y = 10\) -- (2)
Adding (1) and (2): \(2x = 14 \Rightarrow x = 7\)
From (2), \(7 + y = 10 \Rightarrow y = 3\)
Now substitute \(x = 7\) and \(y = 3\) in \(y = 3x - p\):
\(3 = 3(7) - p\)
\(3 = 21 - p\)
\(p = 21 - 3 = 18\)
In simple words: Solve for \(x\) and \(y\), then substitute into the third equation.

Question 29. \(k\) का मान ज्ञात करें जिसके लिए दिए हुए रैखिक समीकरणों के अनेक हल हैं :
\(kx + 3y = k - 3\)
\(12x + ky = k\)
Answer:
For infinitely many solutions, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\):
\(\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}\)
Taking first two parts:
\(\frac{k}{12} = \frac{3}{k} \Rightarrow k^2 = 36 \Rightarrow k = \pm 6\)
Checking with constant terms: \(\frac{3}{6} = \frac{6 - 3}{6} \Rightarrow \frac{1}{2} = \frac{3}{6} = \frac{1}{2}\) (holds for \(k = 6\)).
Thus, \(k = 6\).
In simple words: Equate all coefficient ratios for infinite solutions.

Exam Tip: Always cross-verify the obtained value of \(k\) with the constant ratio.

Question 30. \(\alpha\) और \(\beta\) के वह मान ज्ञात कीजिए जिसके लिए नीचे लिखे समीकरणों के अनेक हल हैं :
\(2x + 3y = 7\)
\(2\alpha x + (\alpha + \beta)y = 28\)
Answer:
For infinitely many solutions, ratios must be equal:
\(\frac{2}{2\alpha} = \frac{3}{\alpha + \beta} = \frac{7}{28}\)
Since \(\frac{7}{28} = \frac{1}{4}\):
\(\frac{2}{2\alpha} = \frac{1}{4} \Rightarrow \frac{1}{\alpha} = \frac{1}{4} \Rightarrow \alpha = 4\)
Now, \(\frac{3}{\alpha + \beta} = \frac{1}{4} \Rightarrow \frac{3}{4 + \beta} = \frac{1}{4} \Rightarrow 4 + \beta = 12 \Rightarrow \beta = 8\)
Thus, \(\alpha = 4, \beta = 8\).
In simple words: Match all three ratios to \(\frac{1}{4}\) and solve for variables.

Exam Tip: Simplify the constant ratio first to easily find other parameters.

Question 31. \(x\) और \(y\) के लिए हल करें :
\(\frac{x + 1}{2} + \frac{y - 1}{3} = 8\)
\(\frac{x - 1}{3} + \frac{y + 1}{2} = 9\)
Answer:
Simplify equation 1 (multiply by 6):
\(3(x + 1) + 2(y - 1) = 48\)
\(3x + 3 + 2y - 2 = 48 \Rightarrow 3x + 2y = 47\) -- (1)
Simplify equation 2 (multiply by 6):
\(2(x - 1) + 3(y + 1) = 54\)
\(2x - 2 + 3y + 3 = 54 \Rightarrow 2x + 3y = 53\) -- (2)
Multiply (1) by 3 and (2) by 2:
\(9x + 6y = 141\)
\(4x + 6y = 106\)
Subtracting: \(5x = 35 \Rightarrow x = 7\)
Substitute \(x = 7\) in (1):
\(3(7) + 2y = 47 \Rightarrow 21 + 2y = 47 \Rightarrow 2y = 26 \Rightarrow y = 13\)
Solution is \(x = 7, y = 13\).
In simple words: Clear denominators first, then use elimination.

Exam Tip: Always clear fractions by multiplying by the LCM before solving.

Question 32. \(x\) और \(y\) के लिए हल कीजिए :
\(2^x + 3^y = 17\)
\(2^{x + 2} - 3^{y + 1} = 5\)
Answer:
Let \(2^x = A\) and \(3^y = B\).
Equations become:
\(A + B = 17\) -- (1)
\(2^{x+2} - 3^{y+1} = 5 \Rightarrow 4(2^x) - 3(3^y) = 5 \Rightarrow 4A - 3B = 5\) -- (2)
Multiply (1) by 3:
\(3A + 3B = 51\)
Add to (2):
\(7A = 56 \Rightarrow A = 8\)
From (1), \(8 + B = 17 \Rightarrow B = 9\)
Now substitute back:
\(2^x = 8 = 2^3 \Rightarrow x = 3\)
\(3^y = 9 = 3^2 \Rightarrow y = 2\)
Solution: \(x = 3, y = 2\).
In simple words: Substitute exponential terms with variables to convert into linear equations.

Exam Tip: Use substitution for exponential or reciprocal expressions to simplify.

Question 33. \(x\) और \(y\) के लिए हल कीजिए :
\(139x + 56y = 641\)
\(56x + 139y = 724\)
Answer:
Add both equations:
\((139 + 56)x + (56 + 139)y = 641 + 724\)
\(195x + 195y = 1365\)
Divide by 195: \(x + y = 7\) -- (1)
Subtract second equation from first:
\((139 - 56)x + (56 - 139)y = 641 - 724\)
\(83x - 83y = -83\)
Divide by 83: \(x - y = -1\) -- (2)
Add (1) and (2):
\(2x = 6 \Rightarrow x = 3\)
From (1), \(3 + y = 7 \Rightarrow y = 4\)
Solution: \(x = 3, y = 4\).
In simple words: When coefficients are cross-symmetric, add and subtract the equations to simplify.

Exam Tip: Symmetric coefficient equations are best solved by adding and subtracting them first.

Question 34. \(x\) और \(y\) के लिए हल कीजिए :
\(\frac{5}{x + y} + \frac{1}{x - y} = 2\)
\(\frac{15}{x + y} - \frac{5}{x - y} = -2, \quad x + y \neq 0, x - y \neq 0\)
Answer:
Let \(\frac{1}{x + y} = u\) and \(\frac{1}{x - y} = v\).
\(5u + v = 2\) -- (1)
\(15u - 5v = -2\) -- (2)
Multiply (1) by 5:
\(25u + 5v = 10\)
Add to (2):
\(40u = 8 \Rightarrow u = \frac{8}{40} = \frac{1}{5}\)
Substitute \(u\) in (1):
\(5\left(\frac{1}{5}\right) + v = 2 \Rightarrow 1 + v = 2 \Rightarrow v = 1\)
Now find \(x\) and \(y\):
\(x + y = 5\) -- (3)
\(x - y = 1\) -- (4)
Adding (3) and (4): \(2x = 6 \Rightarrow x = 3\)
From (3), \(3 + y = 5 \Rightarrow y = 2\)
Solution: \(x = 3, y = 2\).
In simple words: Substitute reciprocal expressions to reduce into simple linear equations.

Exam Tip: Remember to solve for the original variables after finding substitution values.

Question 35. रैखिक समीकरणों के युग्म को ग्राफ द्वारा हल करे :
\(3x - 5y = 19, \quad 3y - 7x + 1 = 0\)
क्या \((4, 9)\) इनमें से किसी रेखा पर स्थित है। लिखिए।
Answer:
Solve algebraically:
From first: \(3x - 5y = 19\)
From second: \(7x - 3y = 1\) -- (2)
Multiply (1) by 7 and (2) by 3:
\(21x - 35y = 133\)
\(21x - 9y = 3\)
Subtracting: \(-26y = 130 \Rightarrow y = -5\)
Substitute \(y = -5\) in (1):
\(3x - 5(-5) = 19 \Rightarrow 3x + 25 = 19 \Rightarrow 3x = -6 \Rightarrow x = -2\)
Intersection point = \((-2, -5)\).
Now check if \((4, 9)\) lies on any line:
For \(3x - 5y = 19\): \(3(4) - 5(9) = 12 - 45 = -33 \neq 19\) (No).
For \(7x - 3y = 1\): \(7(4) - 3(9) = 28 - 27 = 1 = 1\) (Yes, lies on \(3y - 7x + 1 = 0\)).
In simple words: Find the solution point by elimination, then test \((4, 9)\) in both equations.

Exam Tip: A point lies on a line if it satisfies that line's equation.

Question 36. ग्राफिय विधि द्वारा जाँच कीजिए कि \(3x + 2y - 4 = 0\) और \(2x - y - 2 = 0\) संगत है। उस बिन्दु के निर्देशांक ज्ञात कीजिए जहाँ रेखाएँ \(y\)-अक्ष पर मिलें।
Answer:
Check consistency using ratios:
\(\frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{2}{-1} = -2\)
Since \(\frac{3}{2} \neq -2\), the system is consistent (unique solution).
In simple words: Different ratios mean lines intersect, hence consistent.

दीर्घ उत्तरीय प्रश्न

Question 37. \(x\) और \(y\) के लिए हल कीजिए :
\(\frac{1}{2(2x + 3y)} + \frac{12}{7(3x - 2y)} = \frac{1}{2}\)
\(\frac{7}{(2x + 3y)} + \frac{4}{(3x - 2y)} = 2, \quad 2x + 3y \neq 0 \text{ and } 3x - 2y \neq 0\)
Answer:
Let \(u = \frac{1}{2x + 3y}\) and \(v = \frac{1}{3x - 2y}\).
Equations become:
\(\frac{1}{2}u + \frac{12}{7}v = \frac{1}{2} \Rightarrow 7u + 24v = 7\) -- (1)
\(7u + 4v = 2\) -- (2)
Subtract (2) from (1):
\(20v = 5 \Rightarrow v = \frac{5}{20} = \frac{1}{4}\)
Substitute \(v = \frac{1}{4}\) in (2):
\(7u + 4\left(\frac{1}{4}\right) = 2 \Rightarrow 7u + 1 = 2 \Rightarrow 7u = 1 \Rightarrow u = \frac{1}{7}\)
Now:
\(2x + 3y = 7\) -- (3)
\(3x - 2y = 4\) -- (4)
Multiply (3) by 2 and (4) by 3:
\(4x + 6y = 14\)
\(9x - 6y = 12\)
Add: \(13x = 26 \Rightarrow x = 2\)
From (3), \(2(2) + 3y = 7 \Rightarrow 4 + 3y = 7 \Rightarrow 3y = 3 \Rightarrow y = 1\)
Solution: \(x = 2, y = 1\).
In simple words: Substitute complex denominators with variables to solve sequentially.

Exam Tip: Double substitution problems require patience and careful back-substitution.

Question 38. \(p\) और \(q\) के लिए हल कीजिए :
\(\frac{p + q}{pq} = 2, \quad \frac{p - q}{pq} = 6, \quad p \neq 0, q \neq 0\)
Answer:
Split the terms:
\(\frac{1}{q} + \frac{1}{p} = 2\) -- (1)
\(\frac{1}{q} - \frac{1}{p} = 6\) -- (2)
Let \(\frac{1}{p} = u\) and \(\frac{1}{q} = v\):
\(v + u = 2\) -- (3)
\(v - u = 6\) -- (4)
Add (3) and (4): \(2v = 8 \Rightarrow v = 4 \Rightarrow \frac{1}{q} = 4 \Rightarrow q = \frac{1}{4}\)
Subtract (4) from (3): \(2u = -4 \Rightarrow u = -2 \Rightarrow \frac{1}{p} = -2 \Rightarrow p = -\frac{1}{2}\)
Solution: \(p = -\frac{1}{2}, q = \frac{1}{4}\).
In simple words: Divide numerator terms individually by denominator product to simplify.

Exam Tip: Splitting fractions is extremely useful when variables appear in denominators together.

Question 39. \(x\) और \(y\) के लिए हल करे :
\(\frac{2}{3x + 2y} + \frac{3}{3x - 2y} = \frac{17}{5}\)
\(\frac{5}{3x + 2y} + \frac{1}{3x - 2y} = 2, \quad x + y \neq 0, x - y \neq 0\)
Answer:
Let \(u = \frac{1}{3x + 2y}\) and \(v = \frac{1}{3x - 2y}\).
\(2u + 3v = \frac{17}{5} \Rightarrow 10u + 15v = 17\) -- (1)
\(5u + v = 2 \Rightarrow 25u + 5v = 10\) -- (2)
Multiply (2) by 3:
\(75u + 15v = 30\) -- (3)
Subtract (1) from (3):
\(65u = 13 \Rightarrow u = \frac{13}{65} = \frac{1}{5}\)
Substitute \(u = \frac{1}{5}\) in (2):
\(5\left(\frac{1}{5}\right) + v = 2 \Rightarrow 1 + v = 2 \Rightarrow v = 1\)
Now:
\(3x + 2y = 5\) -- (4)
\(3x - 2y = 1\) -- (5)
Add (4) and (5): \(6x = 6 \Rightarrow x = 1\)
Substitute \(x = 1\) in (4): \(3(1) + 2y = 5 \Rightarrow 2y = 2 \Rightarrow y = 1\)
Solution: \(x = 1, y = 1\).
In simple words: Use substitution for bracketed expressions and solve two sets of linear equations.

Exam Tip: Clear denominators from fractional coefficients early to avoid calculation errors.

Question 40. \(x\) और \(y\) के लिए हल करे :
\(\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2\)
\(\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1, \quad x \neq 0, y \neq 0\)
Answer:
Let \(u = \frac{1}{\sqrt{x}}\) and \(v = \frac{1}{\sqrt{y}}\).
\(2u + 3v = 2\) -- (1)
\(4u - 9v = -1\) -- (2)
Multiply (1) by 3:
\(6u + 9v = 6\)
Add to (2):
\(10u = 5 \Rightarrow u = \frac{5}{10} = \frac{1}{2}\)
Substitute \(u = \frac{1}{2}\) in (1):
\(2\left(\frac{1}{2}\right) + 3v = 2 \Rightarrow 1 + 3v = 2 \Rightarrow 3v = 1 \Rightarrow v = \frac{1}{3}\)
Now find \(x\) and \(y\):
\(\frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4\)
\(\frac{1}{\sqrt{y}} = \frac{1}{3} \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9\)
Solution: \(x = 4, y = 9\).
In simple words: Substitute square root reciprocals with variables, then square the results at the end.

Exam Tip: Remember to square both sides to remove square roots at the final step.

Question 41. दो संख्याओं में से बड़ी संख्या के दो गुने में से 20 घटाया जाए, तो छोटी संख्या प्राप्त होती है। यदि छोटी के दो गुने में से 5 घटाया जाए तो बड़ी संख्या प्राप्त होती है। संख्याएँ ज्ञात कीजिए।
Answer:
Let the larger number be \(x\) and smaller number be \(y\).
First condition: \(2x - 20 = y \Rightarrow 2x - y = 20\) -- (1)
Second condition: \(2y - 5 = x \Rightarrow x - 2y = -5\) -- (2)
Multiply (2) by 2:
\(2x - 4y = -10\) -- (3)
Subtract (3) from (1):
\(3y = 30 \Rightarrow y = 10\)
Substitute \(y = 10\) in (1):
\(2x - 10 = 20 \Rightarrow 2x = 30 \Rightarrow x = 15\)
The numbers are 15 and 10.
In simple words: Translate sentences directly into two equations and solve by elimination.

Exam Tip: Define larger and smaller variables clearly before writing equations.

Question 42. हिरनों के एक बाग में हिरन और पर्यटकों के सिरों और टांगों को गिन कर पाया गया कि सिरों की संख्या 39 है और टांगों की संख्या 132 है। हिरनों की संख्या तथा पर्यटकों की संख्या ग्राफिय विधि द्वारा ज्ञात कीजिए।
Answer:
Let number of deer be \(x\) and number of tourists (humans) be \(y\).
Deer have 4 legs and 1 head; humans have 2 legs and 1 head.
Total heads: \(x + y = 39\) -- (1)
Total legs: \(4x + 2y = 132 \Rightarrow 2x + y = 66\) -- (2)
Subtract (1) from (2):
\(x = 66 - 39 = 27\)
Substitute \(x = 27\) in (1):
\(27 + y = 39 \Rightarrow y = 12\)
Number of deer = 27, Number of tourists = 12.
In simple words: Use heads equation and legs equation to form a system and solve.

Exam Tip: Remember that deer have 4 legs and humans have 2 legs.

Question 43. दो अंकों की एक संख्या अंकों के जोड़ को 8 से गुणा कर 1 जोड़ने पर प्राप्त होती है या वही संख्या अंकों के अंतर को 13 से गुणा कर 2 जोड़ने पर प्राप्त होती है। संख्या ज्ञात कीजिए। ऐसी कितनी संख्याएँ हैं।
Answer:
Let the two-digit number be \(10x + y\).
First condition: \(10x + y = 8(x + y) + 1 \Rightarrow 10x + y = 8x + 8y + 1 \Rightarrow 2x - 7y = 1\) -- (1)
Second condition: \(10x + y = 13(x - y) + 2 \Rightarrow 10x + y = 13x - 13y + 2 \Rightarrow 3x - 14y = -2\) -- (2)
Multiply (1) by 3 and (2) by 2:
\(6x - 21y = 3\)
\(6x - 28y = -4\)
Subtracting: \(7y = 7 \Rightarrow y = 1\)
Substitute \(y = 1\) in (1):
\(2x - 7(1) = 1 \Rightarrow 2x = 8 \Rightarrow x = 4\)
The number is \(10(4) + 1 = 41\).
In simple words: Represent two-digit number as \(10x + y\) and set up equations based on digit sums and differences.

Exam Tip: Standard representation for a two-digit number is \(10x + y\) where \(x\) is tens digit and \(y\) is units digit.

Question 44. एक परीक्षा में प्रत्येक सही उत्तर के लिए 1 अंक दिया गया और प्रत्येक गलत उत्तर के लिए \(\frac{1}{4}\) अंक काटा गया। एक विद्यार्थी ने 120 प्रश्नों का उत्तर देकर 90 अंक प्राप्त किए, तो उसने कितने उत्तर सही दिये।
Answer:
Let correct answers be \(x\) and incorrect answers be \(y\).
Total questions: \(x + y = 120\) -- (1)
Total marks: \(x - \frac{1}{4}y = 90 \Rightarrow 4x - y = 360\) -- (2)
Add (1) and (2):
\(5x = 480 \Rightarrow x = 96\)
Number of correct answers = 96.
In simple words: Form equations for total questions and net marks after negative marking.

Exam Tip: Deduct fractional marks correctly as specified for wrong answers.

Question 45. एक नाविक अपनी नाव धारा के प्रतिकूल 35 किमी तथा धारा के अनुकूल 55 किमी चलाकर 12 घंटे लगाता है और 10 घंटों में 30 किमी धारा के प्रतिकूल और 44 किमी धारा के अनुकूल चलाता है। 50 किमी धारा के प्रतिकूल और 77 किमी धारा के अनुकूल नाव चलाने में कितना समय लगेगा।
Answer:
Let speed of boat in still water be \(u\) and speed of stream be \(v\).
Upstream speed = \(x = u - v\), Downstream speed = \(y = u + v\).
First condition: \(\frac{35}{x} + \frac{55}{y} = 12\)
Second condition: \(\frac{30}{x} + \frac{44}{y} = 10\)
Let \(\frac{1}{x} = a\) and \(\frac{1}{y} = b\):
\(35a + 55b = 12\) -- (1)
\(30a + 44b = 10\) -- (2)
Multiply (1) by 6 and (2) by 7:
\(210a + 330b = 72\)
\(210a + 308b = 70\)
Subtracting: \(22b = 2 \Rightarrow b = \frac{1}{11} \Rightarrow y = 11\)
Substitute \(b = \frac{1}{11}\) in (1):
\(35a + 55\left(\frac{1}{11}\right) = 12 \Rightarrow 35a + 5 = 12 \Rightarrow 35a = 7 \Rightarrow a = \frac{1}{5} \Rightarrow x = 5\)
So upstream speed \(x = 5\) km/h and downstream speed \(y = 11\) km/h.
Time required for 50 km upstream and 77 km downstream:
Time = \(\frac{50}{5} + \frac{77}{11} = 10 + 7 = 17\) hours.
In simple words: Use upstream and downstream speed formulas with time-distance relations.

Exam Tip: Upstream speed is boat minus stream; downstream speed is boat plus stream.

Question 46. एक कार्यक्रम में 10 मेहमानों को कक्ष A से B में भेजा जाता है तो दोनों कक्षों में समान संख्या में मेहमान हो जाते हैं। यदि 20 मेहमानों को कक्ष B से A में भेजा जाए तो कक्ष A में मेहमानों की संख्या B से दुगुनी हो जाती है। दोनों कक्षों में शुरू में कितने मेहमान थे।
Answer:
Let guests in room A be \(x\) and in room B be \(y\).
First condition: \(x - 10 = y + 10 \Rightarrow x - y = 20\) -- (1)
Second condition: \(x + 20 = 2(y - 20) \Rightarrow x + 20 = 2y - 40 \Rightarrow x - 2y = -60\) -- (2)
Subtract (2) from (1):
\(y = 80\)
Substitute \(y = 80\) in (1):
\(x - 80 = 20 \Rightarrow x = 100\)
Guests initially in room A = 100, room B = 80.
In simple words: Adjust guest counts according to transfers and set up simultaneous equations.

Exam Tip: Read transfer statements carefully (subtract from one, add to the other).

Question 47. एक कार्यक्रम में मधु प्रत्येक उपस्थित व्यक्ति को Rs. 21 देना चाहती है और पाती है कि चार रुपये कम पड़ गये। इसलिए वह प्रत्येक को Rs. 20 देनी है और पाती है कि Rs. 1 बच गया। उसने कितने रुपये बाँटे और वहाँ कुल कितने लोग थे।
Answer:
Let number of people be \(x\) and total money be \(y\).
First condition: \(y = 21x - 4\) -- (1)
Second condition: \(y = 20x + 1\) -- (2)
Equating (1) and (2):
\(21x - 4 = 20x + 1\)
\(x = 5\) (number of people)
Substitute \(x = 5\) in (2):
\(y = 20(5) + 1 = 101\) (total money in Rs.)
Total money distributed = Rs. 101, Number of people = 5.
In simple words: Equate total money expressions for both distribution rates to find the number of people.

Question 48. एक मोबाइल कंपनी मासिक किराये के रूप में निश्चित राशि लेती है जिसमें 100 मिनट प्रतिमाह मुफ्त मिलते हैं और अगले प्रत्येक मिनट पर निश्चित राशि लेती है। अभिषेक ने 370 मिनट के लिए Rs. 433 दिये और आशिष ने 300 मिनट के लिए Rs. 398 दिये। इसी प्लान में बिल राशि ज्ञात करो यदि उषा 400 मिनट ले।
Answer:
Let fixed monthly charge be Rs. \(x\) and charge per extra minute be Rs. \(y\).
Abhishek uses 370 minutes (extra minutes = \(370 - 100 = 270\)):
\(x + 270y = 433\) -- (1)
Ashish uses 300 minutes (extra minutes = \(300 - 100 = 200\)):
\(x + 200y = 398\) -- (2)
Subtract (2) from (1):
\(70y = 35 \Rightarrow y = \frac{35}{70} = 0.5\) (Rs. \(\frac{1}{2}\) per extra minute)
Substitute \(y = 0.5\) in (2):
\(x + 200(0.5) = 398 \Rightarrow x + 100 = 398 \Rightarrow x = 298\) (Fixed charge Rs. 298)
For Usha using 400 minutes (extra minutes = \(400 - 100 = 300\)):
Total bill = \(x + 300y = 298 + 300(0.5) = 298 + 150 =\) Rs. 448.
In simple words: Calculate fixed charge and per-minute charge, then compute for 400 minutes.

Exam Tip: Always subtract the free allowance minutes before calculating extra usage charges.