यदि (px+3y=27) और (2x-y=9) का हल (x=5,\ y=1) है, तो (p) का मान क्या है?
If (px+3y=27) and (2x-y=9) have solution (x=5,\ y=1), what is the value of (p)?
#linear equations
#parameter
#substitution
#hard
#class 10
A \(p=\frac{24}{5}\)
B \(p=\frac{22}{5}\)
C \(p=\frac{26}{5}\)
D \(p=\frac{28}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(p=\frac{24}{5}\)
Step 1
Concept
Put (x=5,\ y=1) in (px+3y=27). (5p+3=27), so \(p=\frac{24}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(p=\frac{24}{5}\). Put (x=5,\ y=1) in (px+3y=27). (5p+3=27), so \(p=\frac{24}{5}\).
Step 3
Exam Tip
(x=5,\ y=1) को (px+3y=27) में रखें। (5p+3=27), इसलिए \(p=\frac{24}{5}\)।
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समीकरणों \(\frac{3x-y}{4}=5\) और \(\frac{x+2y}{3}=7\) से (x) का मान क्या है?
What is the value of (x) from \(\frac{3x-y}{4}=5\) and \(\frac{x+2y}{3}=7\)?
#linear equations
#transformed equations
#substitution
#hard
#class 10
A \(x=\frac{61}{7}\)
B \(x=\frac{62}{7}\)
C \(x=\frac{63}{7}\)
D \(x=\frac{64}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{61}{7}\)
Step 1
Concept
The equations become (3x-y=20) and (x+2y=21). Substitution gives \(x=\frac{61}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{61}{7}\). The equations become (3x-y=20) and (x+2y=21). Substitution gives \(x=\frac{61}{7}\).
Step 3
Exam Tip
दिए समीकरण (3x-y=20) और (x+2y=21) बनते हैं। प्रतिस्थापन से \(x=\frac{61}{7}\) मिलता है।
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यदि (3x+ky=40) और (x+2y=13) का हल \(x=6,\ y=\frac{7}{2}\) है, तो (k) का मान क्या है?
If (3x+ky=40) and (x+2y=13) have solution \(x=6,\ y=\frac{7}{2}\), what is the value of (k)?
#linear equations
#parameter
#substitution
#hard
#class 10
A \(k=\frac{34}{7}\)
B \(k=\frac{38}{7}\)
C \(k=\frac{44}{7}\)
D \(k=\frac{48}{7}\)
Explanation opens after your attempt
Correct Answer
C. \(k=\frac{44}{7}\)
Step 1
Concept
Put the given solution in (3x+ky=40). \(18+\frac{7k}{2}=40\), so \(k=\frac{44}{7}\).
Step 2
Why this answer is correct
The correct answer is C. \(k=\frac{44}{7}\). Put the given solution in (3x+ky=40). \(18+\frac{7k}{2}=40\), so \(k=\frac{44}{7}\).
Step 3
Exam Tip
दिए हल को (3x+ky=40) में रखें। \(18+\frac{7k}{2}=40\), इसलिए \(k=\frac{44}{7}\)।
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यदि (2(x+y)+3(x-y)=41) और (3(x+y)-2(x-y)=34), तो (y) का मान क्या है?
If (2(x+y)+3(x-y)=41) and (3(x+y)-2(x-y)=34), what is the value of (y)?
#linear equations
#transformation
#substitution
#hard
#class 10
A \(y=\frac{23}{13}\)
B \(y=\frac{25}{13}\)
C \(y=\frac{27}{13}\)
D \(y=\frac{29}{13}\)
Explanation opens after your attempt
Correct Answer
B. \(y=\frac{25}{13}\)
Step 1
Concept
Let (x+y=s) and (x-y=d). Solving gives \(s=\frac{184}{13}\) and \(d=\frac{134}{13}\), so \(y=\frac{25}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(y=\frac{25}{13}\). Let (x+y=s) and (x-y=d). Solving gives \(s=\frac{184}{13}\) and \(d=\frac{134}{13}\), so \(y=\frac{25}{13}\).
Step 3
Exam Tip
(x+y=s) और (x-y=d) मानकर हल करें। \(s=\frac{184}{13}\) और \(d=\frac{134}{13}\), इसलिए \(y=\frac{25}{13}\)।
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समीकरणों \(\frac{x}{4}+\frac{y}{7}=6\) और (x-y=5) से (x) का मान क्या है?
What is the value of (x) from \(\frac{x}{4}+\frac{y}{7}=6\) and (x-y=5)?
#linear equations
#fraction equations
#substitution
#hard
#class 10
A \(x=\frac{188}{11}\)
B \(x=\frac{178}{11}\)
C \(x=\frac{168}{11}\)
D \(x=\frac{158}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{188}{11}\)
Step 1
Concept
Multiply the first equation by (28) to get (7x+4y=168). Using (x=y+5) gives \(x=\frac{188}{11}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{188}{11}\). Multiply the first equation by (28) to get (7x+4y=168). Using (x=y+5) gives \(x=\frac{188}{11}\).
Step 3
Exam Tip
पहले समीकरण को (28) से गुणा कर (7x+4y=168) बनाइए। (x=y+5) रखने पर \(x=\frac{188}{11}\) मिलता है।
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यदि (x=6,\ y=2) समीकरण (2x+my=26) को संतुष्ट करता है, तो (m) का मान क्या है?
If (x=6,\ y=2) satisfies (2x+my=26), what is the value of (m)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (m=5)
B (m=6)
C (m=7)
D (m=8)
Explanation opens after your attempt
Step 1
Concept
Substitute (x=6,\ y=2) in the equation. (12+2m=26), so (m=7).
Step 2
Why this answer is correct
The correct answer is C. (m=7). Substitute (x=6,\ y=2) in the equation. (12+2m=26), so (m=7).
Step 3
Exam Tip
(x=6,\ y=2) को समीकरण में रखें। (12+2m=26), इसलिए (m=7)।
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यदि (kx+5y=42) और (x-y=3) का हल (x=8,\ y=5) है, तो (k) का मान क्या है?
If (kx+5y=42) and (x-y=3) have solution (x=8,\ y=5), what is the value of (k)?
#linear equations
#parameter
#substitution
#hard
#class 10
A \(k=\frac{15}{8}\)
B \(k=\frac{17}{8}\)
C \(k=\frac{19}{8}\)
D \(k=\frac{21}{8}\)
Explanation opens after your attempt
Correct Answer
B. \(k=\frac{17}{8}\)
Step 1
Concept
Put the given solution in (kx+5y=42). Then (8k+25=42), so \(k=\frac{17}{8}\).
Step 2
Why this answer is correct
The correct answer is B. \(k=\frac{17}{8}\). Put the given solution in (kx+5y=42). Then (8k+25=42), so \(k=\frac{17}{8}\).
Step 3
Exam Tip
दिए हल को (kx+5y=42) में रखिए। (8k+25=42), इसलिए \(k=\frac{17}{8}\)।
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समीकरणों \(\frac{x}{2}+\frac{y}{5}=6\) और (x-y=4) का हल क्या है?
What is the solution of \(\frac{x}{2}+\frac{y}{5}=6\) and (x-y=4)?
#linear equations
#fraction equations
#substitution
#hard
#class 10
A \(x=\frac{60}{7},\ y=\frac{32}{7}\)
B \(x=\frac{64}{7},\ y=\frac{36}{7}\)
C \(x=\frac{72}{7},\ y=\frac{44}{7}\)
D \(x=\frac{68}{7},\ y=\frac{40}{7}\)
Explanation opens after your attempt
Correct Answer
D. \(x=\frac{68}{7},\ y=\frac{40}{7}\)
Step 1
Concept
Clear denominators to get (5x+2y=60) and use (x=y+4). This gives \(y=\frac{40}{7}\) and \(x=\frac{68}{7}\).
Step 2
Why this answer is correct
The correct answer is D. \(x=\frac{68}{7},\ y=\frac{40}{7}\). Clear denominators to get (5x+2y=60) and use (x=y+4). This gives \(y=\frac{40}{7}\) and \(x=\frac{68}{7}\).
Step 3
Exam Tip
हर हटाकर (5x+2y=60) बनाइए और (x=y+4) रखिए। इससे \(y=\frac{40}{7}\) और \(x=\frac{68}{7}\) मिलता है।
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यदि (7x+2y=36) और (3x-4y=2) हैं, तो (2x+y) का मान क्या है?
If (7x+2y=36) and (3x-4y=2), what is the value of (2x+y)?
#linear equations
#substitution
#elimination
#class 10
A (8)
B (10)
C (12)
D (14)
Explanation opens after your attempt
Step 1
Concept
Solving gives (x=4) and (y=4), so (2x+y=12). In exams, compute the required expression after finding the variables.
Step 2
Why this answer is correct
The correct answer is C. (12). Solving gives (x=4) and (y=4), so (2x+y=12). In exams, compute the required expression after finding the variables.
Step 3
Exam Tip
हल करने पर (x=4) और (y=4) मिलता है, इसलिए (2x+y=12)। परीक्षा में अंतिम मांगे गए मान को अलग से निकालें।
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एक पिता और पुत्र की वर्तमान आयु का योग (56) वर्ष है। (4) वर्ष पहले पिता की आयु पुत्र की आयु की (5) गुनी थी। पुत्र की वर्तमान आयु क्या है?
The sum of the present ages of a father and son is (56) years. (4) years ago, the father's age was (5) times the son's age. What is the son's present age?
#linear equations
#age problem
#substitution
#class 10
A (10) वर्ष / (10) years
B (12) वर्ष / (12) years
C (14) वर्ष / (14) years
D (16) वर्ष / (16) years
Explanation opens after your attempt
Correct Answer
B. (12) वर्ष / (12) years
Step 1
Concept
Form (f+s=56) and (f-4=5(s-4 )), then solve. In exams, apply addition or subtraction correctly for past and future ages.
Step 2
Why this answer is correct
The correct answer is B. (12) वर्ष / (12) years. Form (f+s=56) and (f-4=5(s-4 )), then solve. In exams, apply addition or subtraction correctly for past and future ages.
Step 3
Exam Tip
(f+s=56) और (f-4=5(s-4 )) बनाकर हल करें। परीक्षा में पहले और बाद की आयु में सही जोड़-घटाव करें।
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यदि (3x+2y=25) और (mx-y=10) का हल (y=5) है, तो (m) का मान क्या होगा?
If (3x+2y=25) and (mx-y=10) have solution (y=5), what will be the value of (m)?
#linear equations
#parameter
#substitution
#class 10
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Putting (y=5) in the first equation gives (x=5). Then (5m-5=10) gives (m=3).
Step 2
Why this answer is correct
The correct answer is B. (3). Putting (y=5) in the first equation gives (x=5). Then (5m-5=10) gives (m=3).
Step 3
Exam Tip
(y=5) को पहले समीकरण में रखने से (x=5) मिलता है। फिर (5m-5=10) से (m=3) मिलता है।
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समीकरणों (10x+7y=87) और (2x-y=7) का हल क्या है?
What is the solution of the equations (10x+7y=87) and (2x-y=7)?
#linear equations
#substitution
#solution
#class 10
A (x=5, y=3)
B (x=6, y=5)
C (x=7, y=7)
D (x=8, y=9)
Explanation opens after your attempt
Correct Answer
B. (x=6, y=5)
Step 1
Concept
From (2x-y=7), put (y=2x-7) and solve the first equation. In exams, substitute the isolated variable into the correct equation.
Step 2
Why this answer is correct
The correct answer is B. (x=6, y=5). From (2x-y=7), put (y=2x-7) and solve the first equation. In exams, substitute the isolated variable into the correct equation.
Step 3
Exam Tip
(2x-y=7) से (y=2x-7) रखें और पहला समीकरण हल करें। परीक्षा में अलग किए गए चर को सही समीकरण में रखें।
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यदि (4x+ay=35) और (x-y=1) का हल (x=6) है, तो (a) का मान क्या है?
If (4x+ay=35) and (x-y=1) have solution (x=6), what is the value of (a)?
#linear equations
#parameter
#substitution
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=6) gives (y=5). Then (24+5a=35) gives \(a=\frac{11}{5}\), so check option calculations carefully.
Step 2
Why this answer is correct
The correct answer is B. (2). Putting (x=6) gives (y=5). Then (24+5a=35) gives \(a=\frac{11}{5}\), so check option calculations carefully.
Step 3
Exam Tip
(x=6) रखने पर (y=5) मिलता है। फिर (24+5a=35) से \(a=\frac{11}{5}\) मिलता है, इसलिए विकल्पों की गणना सावधानी से जाँचें।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश में (2) और हर में (1) जोड़ने पर भिन्न \(\frac{3}{4}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{3}{4}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{5}{8}\)
B \(\frac{6}{9}\)
C \(\frac{7}{10}\)
D \(\frac{8}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{7}{10}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{7}{10}\). Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y-x=3) और \(\frac{x+2}{y+1}=\frac{3}{4}\) बनता है। परीक्षा में क्रॉस गुणा के बाद सरल रैखिक समीकरण हल करें।
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यदि (3x+11y=67) और (6x-y=23), तो (x:y) का अनुपात क्या है?
If (3x+11y=67) and (6x-y=23), what is the ratio (x:y)?
#linear equations
#substitution
#ratio
#class 10
A (2:1)
B (3:2)
C (4:3)
D (5:4)
Explanation opens after your attempt
Step 1
Concept
From the second equation, put (y=6x-23) to get (x=6), (y=4). In exams, write the ratio in simplest form.
Step 2
Why this answer is correct
The correct answer is B. (3:2). From the second equation, put (y=6x-23) to get (x=6), (y=4). In exams, write the ratio in simplest form.
Step 3
Exam Tip
दूसरे समीकरण से (y=6x-23) रखकर (x=6), (y=4) मिलता है। परीक्षा में अनुपात को सरल रूप में लिखें।
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दो अंकों की संख्या में दहाई अंक इकाई अंक से (4) अधिक है। अंकों को उलटने पर संख्या मूल संख्या से (36) कम हो जाती है। मूल संख्या क्या है?
In a two-digit number, the tens digit is (4) more than the units digit. On reversing the digits, the number becomes (36) less than the original number. What is the original number?
#linear equations
#digits
#substitution
#class 10
A (51)
B (62)
C (73)
D (84)
Explanation opens after your attempt
Step 1
Concept
Let the tens digit be (x) and the units digit be (y), so (x-y=4). In exams, write the original number as (10x+y) and the reversed number as (10y+x).
Step 2
Why this answer is correct
The correct answer is B. (62). Let the tens digit be (x) and the units digit be (y), so (x-y=4). In exams, write the original number as (10x+y) and the reversed number as (10y+x).
Step 3
Exam Tip
दहाई अंक (x) और इकाई अंक (y) मानकर (x-y=4) बनता है। परीक्षा में मूल संख्या (10x+y) और उलटी संख्या (10y+x) लिखें।
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समीकरणों (14x+3y=59) और (2x+y=11) को हल करने पर (x) और (y) के मान क्या होंगे?
On solving the equations (14x+3y=59) and (2x+y=11), what are the values of (x) and (y)?
#linear equations
#substitution
#solution
#class 10
A (x=2, y=7)
B (x=3, y=5)
C (x=4, y=3)
D (x=5, y=1)
Explanation opens after your attempt
Correct Answer
B. (x=3, y=5)
Step 1
Concept
From (2x+y=11), put (y=11-2x) in the first equation. In exams, combine all terms correctly after substitution.
Step 2
Why this answer is correct
The correct answer is B. (x=3, y=5). From (2x+y=11), put (y=11-2x) in the first equation. In exams, combine all terms correctly after substitution.
Step 3
Exam Tip
(2x+y=11) से (y=11-2x) रखकर पहला समीकरण हल करें। परीक्षा में प्रतिस्थापन के बाद सभी पद सही जोड़ें।
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यदि (kx+2y=16) और (3x-y=5) का हल (x=4) है, तो (k) का मान क्या है?
If (kx+2y=16) and (3x-y=5) have solution (x=4), what is the value of (k)?
#linear equations
#parameter
#substitution
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=4) in the second equation gives (y=7). Then the first equation gives (4k+14=16), so \(k=\frac{1}{2}\); check options carefully in exams.
Step 2
Why this answer is correct
The correct answer is B. (2). Putting (x=4) in the second equation gives (y=7). Then the first equation gives (4k+14=16), so \(k=\frac{1}{2}\); check options carefully in exams.
Step 3
Exam Tip
(x=4) रखने पर दूसरे समीकरण से (y=7) मिलता है। फिर पहले समीकरण से (4k+14=16), इसलिए \(k=\frac{1}{2}\) नहीं बल्कि विकल्पों में कोई नहीं दिखता; सही गणना से विकल्प जाँचें।
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एक पिता की आयु पुत्र की आयु के (4) गुने से (2) वर्ष अधिक है। (8) वर्ष बाद पिता की आयु पुत्र की आयु की (3) गुनी होगी। वर्तमान में पुत्र की आयु क्या है?
A father's age is (2) years more than (4) times his son's age. After (8) years, the father's age will be (3) times the son's age. What is the son's present age?
#linear equations
#age problem
#substitution
#class 10
A (6) वर्ष / (6) years
B (7) वर्ष / (7) years
C (8) वर्ष / (8) years
D (9) वर्ष / (9) years
Explanation opens after your attempt
Correct Answer
A. (6) वर्ष / (6) years
Step 1
Concept
Form (f=4s+2) and (f+8=3(s+8)), then solve. In exams, add the same number of years to both ages for future age.
Step 2
Why this answer is correct
The correct answer is A. (6) वर्ष / (6) years. Form (f=4s+2) and (f+8=3(s+8)), then solve. In exams, add the same number of years to both ages for future age.
Step 3
Exam Tip
(f=4s+2) और (f+8=3(s+8)) बनाकर हल करें। परीक्षा में भविष्य की आयु में दोनों की आयु में समान वर्ष जोड़ें।
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दो संख्याओं में पहली संख्या दूसरी की (2) गुनी से (3) कम है। दोनों का योग (42) है। छोटी संख्या क्या है?
The first number is (3) less than twice the second number. Their sum is (42). What is the smaller number?
#linear equations
#word problem
#substitution
#class 10
A (13)
B (14)
C (15)
D (16)
Explanation opens after your attempt
Step 1
Concept
Form (x=2y-3) and (x+y=42), then solve. In exams, first convert the relation statement into an equation.
Step 2
Why this answer is correct
The correct answer is C. (15). Form (x=2y-3) and (x+y=42), then solve. In exams, first convert the relation statement into an equation.
Step 3
Exam Tip
(x=2y-3) और (x+y=42) बनाकर हल करें। परीक्षा में संबंध वाले वाक्य को पहले समीकरण में बदलें।
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यदि (ax+3y=25) और (2x-y=5) का हल (y=3) है, तो (a) का मान ज्ञात करें।
If (ax+3y=25) and (2x-y=5) have solution (y=3), find the value of (a).
#linear equations
#parameter
#substitution
#class 10
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Putting (y=3) in (2x-y=5) gives (x=4). Then (ax+3y=25) gives (a=4).
Step 2
Why this answer is correct
The correct answer is C. (4). Putting (y=3) in (2x-y=5) gives (x=4). Then (ax+3y=25) gives (a=4).
Step 3
Exam Tip
(y=3) को (2x-y=5) में रखने से (x=4) मिलता है। फिर (ax+3y=25) से (a=4) आता है।
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यदि (2x+ky=19) और (x+y=7) का हल (x=5) है, तो (k) का मान क्या होगा?
If (2x+ky=19) and (x+y=7) have solution (x=5), what will be the value of (k)?
#linear equations
#parameter
#substitution
#class 10
A (2)
B (3)
C \(\frac{9}{2}\)
D (5)
Explanation opens after your attempt
Correct Answer
C. \(\frac{9}{2}\)
Step 1
Concept
Putting (x=5) in (x+y=7) gives (y=2). Then (2x+ky=19) gives \(k=\frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{9}{2}\). Putting (x=5) in (x+y=7) gives (y=2). Then (2x+ky=19) gives \(k=\frac{9}{2}\).
Step 3
Exam Tip
(x=5) को (x+y=7) में रखने से (y=2) मिलता है। फिर (2x+ky=19) से \(k=\frac{9}{2}\) मिलता है।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश और हर दोनों में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (1) is added to both numerator and denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{9}{14}\)
B \(\frac{8}{13}\)
C \(\frac{7}{12}\)
D \(\frac{6}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{14}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{14}\). Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y=x+5) और \(\frac{x+1}{y+1}=\frac{2}{3}\) बनता है। परीक्षा में भिन्न को समीकरण में बदलते समय क्रॉस गुणा करें।
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दो अंकों की संख्या में अंकों का योग (11) है। अंकों को उलटने पर बनी संख्या मूल संख्या से (27) कम है। मूल संख्या क्या है?
In a two-digit number, the sum of digits is (11). The number formed by reversing the digits is (27) less than the original number. What is the original number?
#linear equations
#digits
#substitution
#class 10
A (74)
B (83)
C (92)
D (65)
Explanation opens after your attempt
Step 1
Concept
Let the tens digit be (x) and the units digit be (y), giving (x+y=11) and (9x-9y=27). In exams, write a two-digit number as (10x+y).
Step 2
Why this answer is correct
The correct answer is A. (74). Let the tens digit be (x) and the units digit be (y), giving (x+y=11) and (9x-9y=27). In exams, write a two-digit number as (10x+y).
Step 3
Exam Tip
दहाई अंक (x) और इकाई अंक (y) मानकर (x+y=11) और (9x-9y=27) बनता है। परीक्षा में दो अंकों की संख्या को (10x+y) लिखें।
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समीकरणों (9x-2y=23) और (4x+y=17) को हल करने पर (x+2y) का मान क्या होगा?
On solving (9x-2y=23) and (4x+y=17), what will be the value of (x+2y)?
#linear equations
#substitution
#expression
#class 10
A (23)
B (25)
C (27)
D (29)
Explanation opens after your attempt
Step 1
Concept
Use (y=17-4x) from the second equation to get (x=3), (y=5). In exams, calculate the asked expression after finding the solution.
Step 2
Why this answer is correct
The correct answer is D. (29). Use (y=17-4x) from the second equation to get (x=3), (y=5). In exams, calculate the asked expression after finding the solution.
Step 3
Exam Tip
दूसरे समीकरण से (y=17-4x) रखें और (x=3), (y=5) पाएँ। परीक्षा में हल के बाद सीधे मांगा गया व्यंजक निकालें।
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यदि (5x+3y=44) और (2x-y=3), तो (xy) का मान ज्ञात कीजिए।
If (5x+3y=44) and (2x-y=3), find the value of (xy).
#linear equations
#substitution
#product
#class 10
A (32)
B (35)
C (36)
D (40)
Explanation opens after your attempt
Step 1
Concept
From the second equation, put (y=2x-3), giving (x=5) and (y=7). In exams, recheck both values before finding (xy).
Step 2
Why this answer is correct
The correct answer is B. (35). From the second equation, put (y=2x-3), giving (x=5) and (y=7). In exams, recheck both values before finding (xy).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-3) रखकर (x=5) और (y=7) मिलता है। परीक्षा में (xy) निकालते समय दोनों मान फिर से जाँचें।
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समीकरणों (6x+7y=39) और (2x-y=1) में (y) का मान क्या है?
In the equations (6x+7y=39) and (2x-y=1), what is the value of (y)?
#linear equations
#substitution
#y value
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
From (2x-y=1), put (y=2x-1) in the first equation. In exams, isolate one variable clearly first.
Step 2
Why this answer is correct
The correct answer is C. (3). From (2x-y=1), put (y=2x-1) in the first equation. In exams, isolate one variable clearly first.
Step 3
Exam Tip
(2x-y=1) से (y=2x-1) रखकर पहला समीकरण हल करें। परीक्षा में पहले एक चर को स्पष्ट रूप से अलग करें।
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यदि (x+y=14) और (3x-2y=7), तो (x-y) का मान क्या होगा?
If (x+y=14) and (3x-2y=7), what is the value of (x-y)?
#linear equations
#substitution
#expression
#class 10
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
Put (y=14-x) in (3x-2y=7) and solve. In exams, give the final answer in the asked form (x-y).
Step 2
Why this answer is correct
The correct answer is D. (3). Put (y=14-x) in (3x-2y=7) and solve. In exams, give the final answer in the asked form (x-y).
Step 3
Exam Tip
(y=14-x) रखकर (3x-2y=7) हल करें। परीक्षा में अंतिम उत्तर पूछे गए रूप (x-y) में ही दें।
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समीकरणों (3x+2y=16) और (5x-y=11) को हल करने पर (x) और (y) के मान क्या होंगे?
On solving the equations (3x+2y=16) and (5x-y=11), what are the values of (x) and (y)?
#linear equations
#substitution
#elimination
#class 10
A (x=2, y=-1)
B (x=3, y=4)
C (x=1, y=6)
D (x=4, y=1)
Explanation opens after your attempt
Correct Answer
B. (x=3, y=4)
Step 1
Concept
From (5x-y=11), put (y=5x-11) in the first equation and solve. In exams, combine terms carefully after substitution.
Step 2
Why this answer is correct
The correct answer is B. (x=3, y=4). From (5x-y=11), put (y=5x-11) in the first equation and solve. In exams, combine terms carefully after substitution.
Step 3
Exam Tip
(5x-y=11) से (y=5x-11) रखकर पहला समीकरण हल करें। परीक्षा में प्रतिस्थापन के बाद पदों को सावधानी से जोड़ें।
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यदि (x=3y-4) और (2x+5y=37), तो (y) का मान क्या है?
If (x=3y-4) and (2x+5y=37), what is the value of (y)?
#linear equations
#substitution
#fraction value
#hard
#class 10
A \(y=\frac{39}{11}\)
B \(y=\frac{42}{11}\)
C \(y=\frac{45}{11}\)
D \(y=\frac{48}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{45}{11}\)
Step 1
Concept
Substitute (x=3y-4) in the second equation. (6y-8+5y=37), so \(y=\frac{45}{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{45}{11}\). Substitute (x=3y-4) in the second equation. (6y-8+5y=37), so \(y=\frac{45}{11}\).
Step 3
Exam Tip
(x=3y-4) को दूसरे समीकरण में रखें। (6y-8+5y=37), इसलिए \(y=\frac{45}{11}\)।
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