100 results found for "algebraic-parameter" in Class 10.
यदि (3x+2y=28) और (mx-2y=12) का हल (x=5) है, तो (m) का मान क्या है?
If (3x+2y=28) and (mx-2y=12) have solution (x=5), what is (m)?
#pair-linear-equations-parameter-expert
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Putting (x=5) in the first equation gives \(y=\frac{13}{2}\). Then (5m-13=12), so (m=5).
Step 2
Why this answer is correct
The correct answer is C. (5). Putting (x=5) in the first equation gives \(y=\frac{13}{2}\). Then (5m-13=12), so (m=5).
Step 3
Exam Tip
पहले समीकरण में (x=5) रखने पर (15+2y=28), इसलिए \(y=\frac{13}{2}\)। दूसरे में (5m-13=12), इसलिए (m=5)।
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समीकरणों (px+y=17) और (3x-y=7) का हल (y=2) है। (p) का मान क्या है?
The equations (px+y=17) and (3x-y=7) have solution (y=2). What is (p)?
#pair-linear-equations-parameter-substitution
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Putting (y=2) in the second equation gives (x=3). Then (3p+2=17), so (p=5).
Step 2
Why this answer is correct
The correct answer is C. (5). Putting (y=2) in the second equation gives (x=3). Then (3p+2=17), so (p=5).
Step 3
Exam Tip
दूसरे में (y=2) रखने पर (3x-2=7), इसलिए (x=3)। पहले में (3p+2=17), इसलिए (p=5)।
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यदि (4x+ky=34) और (4x-2y=10) का हल (y=3) है, तो (k) का मान क्या होगा?
If (4x+ky=34) and (4x-2y=10) have solution (y=3), what is (k)?
#pair-linear-equations-parameter-check
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Putting (y=3) in the second equation gives (x=4). Then (16+3k=34), so verify the parameter carefully.
Step 2
Why this answer is correct
The correct answer is C. (4). Putting (y=3) in the second equation gives (x=4). Then (16+3k=34), so verify the parameter carefully.
Step 3
Exam Tip
दूसरे में (y=3) रखने पर (4x-6=10), इसलिए (x=4)। पहले में (16+3k=34), इसलिए (k=6), विकल्प जांचें।
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यदि (ax+3y=25) और (2x-3y=5) का हल (x=5) है, तो (a) का मान क्या है?
If (ax+3y=25) and (2x-3y=5) have solution (x=5), what is the value of (a)?
#pair-linear-equations-parameter
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Putting (x=5) in the second equation gives \(y=\frac{5}{3}\). Then (5a+5=25), so (a=4).
Step 2
Why this answer is correct
The correct answer is B. (4). Putting (x=5) in the second equation gives \(y=\frac{5}{3}\). Then (5a+5=25), so (a=4).
Step 3
Exam Tip
दूसरे समीकरण में (x=5) रखने पर (10-3y=5), इसलिए \(y=\frac{5}{3}\)। पहले में (5a+5=25), इसलिए (a=4)।
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यदि (2x+3y=13) और (mx-3y=17) का हल (x=5) है, तो (m) का मान क्या है?
If (2x+3y=13) and (mx-3y=17) have solution (x=5), what is (m)?
#pair-linear-equations
#parameter
#substitution
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Putting (x=5) in the first equation gives (y=1). Then (5m-3=17), so (m=4).
Step 2
Why this answer is correct
The correct answer is B. (4). Putting (x=5) in the first equation gives (y=1). Then (5m-3=17), so (m=4).
Step 3
Exam Tip
पहले समीकरण में (x=5) रखने पर (10+3y=13), इसलिए (y=1)। दूसरे में (5m-3=17), इसलिए (m=4)।
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समीकरणों (px+y=14) और (2x-y=1) का हल (y=5) है। (p) का मान क्या होगा?
The equations (px+y=14) and (2x-y=1) have solution (y=5). What is (p)?
#pair-linear-equations
#parameter
#check
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (y=5) in the second equation gives (x=3). Then (3p+5=14), so (p=3); match the option carefully.
Step 2
Why this answer is correct
The correct answer is B. (2). Putting (y=5) in the second equation gives (x=3). Then (3p+5=14), so (p=3); match the option carefully.
Step 3
Exam Tip
दूसरे में (y=5) रखने पर (2x-5=1), इसलिए (x=3)। पहले में (3p+5=14), इसलिए (p=3) नहीं बल्कि (p=3) है; विकल्प मिलान ध्यान से करें।
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यदि (4x+ky=26) और (4x-3y=2) का हल (y=4) है, तो (k) का मान क्या है?
If (4x+ky=26) and (4x-3y=2) have solution (y=4), what is (k)?
#pair-linear-equations
#parameter
#substitution
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Putting (y=4) in the second equation gives \(x=\frac{7}{2}\). Then (14+4k=26), so (k=3).
Step 2
Why this answer is correct
The correct answer is B. (3). Putting (y=4) in the second equation gives \(x=\frac{7}{2}\). Then (14+4k=26), so (k=3).
Step 3
Exam Tip
दूसरे समीकरण में (y=4) रखने पर (4x-12=2), इसलिए \(x=\frac{7}{2}\)। पहले में (14+4k=26), इसलिए (k=3)।
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यदि (ax+2y=17) और (3x-2y=7) का हल (x=4) है, तो (a) का मान क्या है?
If the solution of (ax+2y=17) and (3x-2y=7) has (x=4), what is the value of (a)?
#pair-linear-equations
#parameter
#substitution
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=4) in the second equation gives \(y=\frac{5}{2}\). Then (4a+5=17), so (a=3).
Step 2
Why this answer is correct
The correct answer is C. (3). Putting (x=4) in the second equation gives \(y=\frac{5}{2}\). Then (4a+5=17), so (a=3).
Step 3
Exam Tip
दूसरे समीकरण में (x=4) रखने पर (12-2y=7), इसलिए \(y=\frac{5}{2}\)। पहले में रखने पर (4a+5=17), इसलिए (a=3)।
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समीकरणों (9x+15y=45) और (kx+5y=18) का कोई हल न हो, इसके लिए (k) का मान क्या है?
For (9x+15y=45) and (kx+5y=18) to have no solution, what is the value of (k)?
#linear equations
#no solution
#parameter
#expert
#class 10
A (k=2)
B (k=3)
C (k=4)
D (k=5)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.
Step 2
Why this answer is correct
The correct answer is B. (k=3). The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.
Step 3
Exam Tip
पहला समीकरण (3x+5y=15) बनता है। (k=3) पर दूसरा (3x+5y=18) होगा, इसलिए कोई हल नहीं।
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यदि (px+5y=43) और (3x-y=17) का हल (x=6,\ y=1) है, तो (p) का मान क्या है?
If (px+5y=43) and (3x-y=17) have solution (x=6,\ y=1), what is the value of (p)?
#linear equations
#parameter
#substitution
#expert
#class 10
A \(p=\frac{17}{3}\)
B (p=6)
C \(p=\frac{19}{3}\)
D \(p=\frac{20}{3}\)
Explanation opens after your attempt
Correct Answer
C. \(p=\frac{19}{3}\)
Step 1
Concept
Put (x=6,\ y=1) in (px+5y=43). Then (6p+5=43), so \(p=\frac{19}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(p=\frac{19}{3}\). Put (x=6,\ y=1) in (px+5y=43). Then (6p+5=43), so \(p=\frac{19}{3}\).
Step 3
Exam Tip
(x=6,\ y=1) को (px+5y=43) में रखें। (6p+5=43), इसलिए \(p=\frac{19}{3}\)।
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समीकरणों (4x+ay=16) और (8x+10y=45) का कोई हल न हो, इसके लिए (a) का मान क्या है?
For (4x+ay=16) and (8x+10y=45) to have no solution, what is the value of (a)?
#linear equations
#no solution
#parameter
#expert
#class 10
A (a=4)
B (a=5)
C (a=6)
D (a=7)
Explanation opens after your attempt
Step 1
Concept
To make coefficients proportional, (4:8=a:10) must hold. This gives (a=5), while constants are not in the same ratio.
Step 2
Why this answer is correct
The correct answer is B. (a=5). To make coefficients proportional, (4:8=a:10) must hold. This gives (a=5), while constants are not in the same ratio.
Step 3
Exam Tip
गुणांक समानुपाती करने के लिए (4:8=a:10) होना चाहिए। इससे (a=5), जबकि स्थिरांक समान अनुपात में नहीं हैं।
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यदि (4x+ky=55) का हल (x=9,\ y=5) है, तो (k) का मान क्या है?
If (x=9,\ y=5) is a solution of (4x+ky=55), what is the value of (k)?
#linear equations
#parameter
#substitution
#expert
#class 10
A \(k=\frac{17}{5}\)
B \(k=\frac{18}{5}\)
C \(k=\frac{19}{5}\)
D (k=4)
Explanation opens after your attempt
Correct Answer
C. \(k=\frac{19}{5}\)
Step 1
Concept
Substituting (x=9,\ y=5) gives (36+5k=55). Therefore \(k=\frac{19}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(k=\frac{19}{5}\). Substituting (x=9,\ y=5) gives (36+5k=55). Therefore \(k=\frac{19}{5}\).
Step 3
Exam Tip
(x=9,\ y=5) रखने पर (36+5k=55) मिलता है। इसलिए \(k=\frac{19}{5}\)।
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समीकरणों (ax+9y=27) और (2x+3y=9) के अनंत हल होने के लिए (a) का मान क्या है?
What is the value of (a) for (ax+9y=27) and (2x+3y=9) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#expert
#class 10
A (a=4)
B (a=5)
C (a=6)
D (a=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (3) times the second. Therefore (a=6).
Step 2
Why this answer is correct
The correct answer is C. (a=6). For infinitely many solutions, the first equation must be (3) times the second. Therefore (a=6).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (3) गुना होना चाहिए। इसलिए (a=6)।
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समीकरणों (12x+18y=54) और (2x+3y=c) का कोई हल न हो, इसके लिए (c) का कौन-सा मान सही है?
For (12x+18y=54) and (2x+3y=c) to have no solution, which value of (c) is correct?
#linear equations
#no solution
#parameter
#expert
#class 10
A (c=8)
B (c=9)
C (c=10)
D (c=11)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (2x+3y=9). When (c=10), the left side is the same but the right side is different.
Step 2
Why this answer is correct
The correct answer is C. (c=10). The first equation becomes (2x+3y=9). When (c=10), the left side is the same but the right side is different.
Step 3
Exam Tip
पहला समीकरण (2x+3y=9) बनता है। (c=10) होने पर समान बायां पक्ष और अलग दायां पक्ष होगा।
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यदि (3x+my=29) का हल (x=5,\ y=2) है, तो (m) का मान क्या होगा?
If (x=5,\ y=2) is a solution of (3x+my=29), what will be the value of (m)?
#linear equations
#parameter
#substitution
#expert
#class 10
A (m=5)
B (m=6)
C (m=7)
D (m=8)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=5,\ y=2) gives (15+2m=29). Therefore (m=7).
Step 2
Why this answer is correct
The correct answer is C. (m=7). Substituting (x=5,\ y=2) gives (15+2m=29). Therefore (m=7).
Step 3
Exam Tip
(x=5,\ y=2) रखने पर (15+2m=29) मिलता है। इसलिए (m=7)।
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समीकरणों (px-10y=30) और (3x-5y=15) के अनंत हल होने के लिए (p) का मान क्या है?
What is the value of (p) for (px-10y=30) and (3x-5y=15) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#expert
#class 10
A (p=4)
B (p=5)
C (p=6)
D (p=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (2) times the second. Hence (p=6).
Step 2
Why this answer is correct
The correct answer is C. (p=6). For infinitely many solutions, the first equation must be (2) times the second. Hence (p=6).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (2) गुना होना चाहिए। इसलिए (p=6)।
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समीकरणों (6x+ay=24) और (2x+3y=11) का कोई हल न हो, इसके लिए (a) का मान क्या है?
For (6x+ay=24) and (2x+3y=11) to have no solution, what is the value of (a)?
#linear equations
#no solution
#parameter
#expert
#class 10
A (a=6)
B (a=7)
C (a=8)
D (a=9)
Explanation opens after your attempt
Step 1
Concept
For no solution, variable coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9).
Step 2
Why this answer is correct
The correct answer is D. (a=9). For no solution, variable coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9).
Step 3
Exam Tip
कोई हल न होने के लिए चर गुणांक समानुपाती और स्थिरांक असमानुपाती होने चाहिए। (6:2=3), इसलिए (a=9)।
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यदि (kx+4y=38) और (x-y=3) का हल (x=7,\ y=4) है, तो (k) का मान क्या होगा?
If (kx+4y=38) and (x-y=3) have solution (x=7,\ y=4), what will be the value of (k)?
#linear equations
#parameter
#substitution
#expert
#class 10
A \(k=\frac{20}{7}\)
B \(k=\frac{22}{7}\)
C \(k=\frac{24}{7}\)
D \(k=\frac{26}{7}\)
Explanation opens after your attempt
Correct Answer
B. \(k=\frac{22}{7}\)
Step 1
Concept
Put the given solution in (kx+4y=38). (7k+16=38), so \(k=\frac{22}{7}\).
Step 2
Why this answer is correct
The correct answer is B. \(k=\frac{22}{7}\). Put the given solution in (kx+4y=38). (7k+16=38), so \(k=\frac{22}{7}\).
Step 3
Exam Tip
दिए हल को (kx+4y=38) में रखें। (7k+16=38), इसलिए \(k=\frac{22}{7}\)।
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समीकरणों (6x+12y=30) और (kx+2y=8) का कोई हल न हो, इसके लिए (k) का मान क्या है?
For (6x+12y=30) and (kx+2y=8) to have no solution, what is the value of (k)?
#linear equations
#no solution
#parameter
#hard
#class 10
A (k=1)
B (k=2)
C (k=3)
D (k=4)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (x+2y=5). At (k=1), the second becomes (x+2y=8), so there is no solution.
Step 2
Why this answer is correct
The correct answer is A. (k=1). The first equation becomes (x+2y=5). At (k=1), the second becomes (x+2y=8), so there is no solution.
Step 3
Exam Tip
पहला समीकरण (x+2y=5) बनता है। (k=1) पर दूसरा (x+2y=8) होगा, इसलिए कोई हल नहीं।
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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 (5x+ay=11) और (10x+6y=30) का कोई हल न हो, इसके लिए (a) का मान क्या होगा?
For (5x+ay=11) and (10x+6y=30) to have no solution, what should be the value of (a)?
#linear equations
#no solution
#parameter
#hard
#class 10
A (a=2)
B (a=3)
C (a=4)
D (a=5)
Explanation opens after your attempt
Step 1
Concept
To make coefficients proportional, (5:10=a:6) must hold. This gives (a=3), while (11:30) is not the same ratio.
Step 2
Why this answer is correct
The correct answer is B. (a=3). To make coefficients proportional, (5:10=a:6) must hold. This gives (a=3), while (11:30) is not the same ratio.
Step 3
Exam Tip
गुणांक समानुपाती करने के लिए (5:10=a:6) होना चाहिए। इससे (a=3), जबकि (11:30) समान अनुपात में नहीं 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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समीकरणों (ax+6y=14) और (2x+3y=7) के अनंत हल होने के लिए (a) का मान क्या है?
What is the value of (a) for (ax+6y=14) and (2x+3y=7) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#hard
#class 10
A (a=2)
B (a=3)
C (a=4)
D (a=6)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (2) times the second. Therefore (a=4).
Step 2
Why this answer is correct
The correct answer is C. (a=4). For infinitely many solutions, the first equation must be (2) times the second. Therefore (a=4).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (2) गुना होना चाहिए। इसलिए (a=4)।
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समीकरणों (8x+12y=40) और (2x+3y=c) का कोई हल न हो, इसके लिए (c) का कौन-सा मान सही है?
For (8x+12y=40) and (2x+3y=c) to have no solution, which value of (c) is correct?
#linear equations
#no solution
#parameter
#hard
#class 10
A (c=8)
B (c=10)
C (c=11)
D (c=12)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (2x+3y=10). At (c=11), the left side is the same but the right side is different, so there is no solution.
Step 2
Why this answer is correct
The correct answer is C. (c=11). The first equation becomes (2x+3y=10). At (c=11), the left side is the same but the right side is different, so there is no solution.
Step 3
Exam Tip
पहला समीकरण (2x+3y=10) बनता है। (c=11) पर समान बायां पक्ष और अलग दायां पक्ष होगा, इसलिए कोई हल नहीं।
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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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समीकरणों (px-8y=24) और (3x-4y=12) के अनंत हल होने के लिए (p) का मान क्या है?
What is the value of (p) for (px-8y=24) and (3x-4y=12) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#hard
#class 10
A (p=4)
B (p=5)
C (p=6)
D (p=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (2) times the second. Therefore (p=6).
Step 2
Why this answer is correct
The correct answer is C. (p=6). For infinitely many solutions, the first equation must be (2) times the second. Therefore (p=6).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (2) गुना होना चाहिए। इसलिए (p=6)।
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समीकरणों (6x+ay=18) और (2x+3y=11) का कोई हल न हो, इसके लिए (a) का मान क्या होगा?
For (6x+ay=18) and (2x+3y=11) to have no solution, what should be the value of (a)?
#linear equations
#no solution
#parameter
#hard
#class 10
A (a=6)
B (a=8)
C (a=9)
D (a=12)
Explanation opens after your attempt
Step 1
Concept
For no solution, coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9), and (18:11) is not the same ratio.
Step 2
Why this answer is correct
The correct answer is C. (a=9). For no solution, coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9), and (18:11) is not the same ratio.
Step 3
Exam Tip
कोई हल न होने के लिए गुणांक समानुपाती और स्थिरांक असमानुपाती होने चाहिए। (6:2=3), इसलिए (a=9) होगा और (18:11) समान अनुपात में नहीं है।
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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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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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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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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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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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समीकरणों (5x+10y=20) और (kx+2y=7) का कोई हल न हो, इसके लिए (k) का मान क्या है?
For (5x+10y=20) and (kx+2y=7) to have no solution, what is the value of (k)?
#linear equations
#no solution
#parameter
#hard
#class 10
A (k=0)
B (k=1)
C (k=2)
D (k=5)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (x+2y=4). At (k=1), the second becomes (x+2y=7), so there is no solution.
Step 2
Why this answer is correct
The correct answer is B. (k=1). The first equation becomes (x+2y=4). At (k=1), the second becomes (x+2y=7), so there is no solution.
Step 3
Exam Tip
पहला समीकरण (x+2y=4) बनता है। (k=1) पर दूसरा (x+2y=7) होगा, इसलिए कोई हल नहीं।
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यदि (px+2y=18) और (3x-y=5) का हल (x=4,\ y=7) है, तो (p) का मान क्या है?
If (px+2y=18) and (3x-y=5) have solution (x=4,\ y=7), what is the value of (p)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (p=1)
B (p=2)
C (p=3)
D (p=4)
Explanation opens after your attempt
Step 1
Concept
Put (x=4,\ y=7) in (px+2y=18). (4p+14=18), so (p=1).
Step 2
Why this answer is correct
The correct answer is A. (p=1). Put (x=4,\ y=7) in (px+2y=18). (4p+14=18), so (p=1).
Step 3
Exam Tip
(x=4,\ y=7) को (px+2y=18) में रखें। (4p+14=18), इसलिए (p=1)।
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समीकरणों (3x+ay=7) और (6x+8y=20) का कोई हल न हो, इसके लिए (a) का मान क्या होगा?
For (3x+ay=7) and (6x+8y=20) to have no solution, what should be the value of (a)?
#linear equations
#no solution
#parameter
#hard
#class 10
A (a=2)
B (a=3)
C (a=5)
D (a=4)
Explanation opens after your attempt
Step 1
Concept
To make coefficients proportional, (3:6=a:8) must hold. This gives (a=4), and constants (7:20) are not in the same ratio.
Step 2
Why this answer is correct
The correct answer is D. (a=4). To make coefficients proportional, (3:6=a:8) must hold. This gives (a=4), and constants (7:20) are not in the same ratio.
Step 3
Exam Tip
गुणांक समानुपाती करने के लिए (3:6=a:8) होना चाहिए। इससे (a=4), और स्थिरांक (7:20) समान अनुपात में नहीं हैं।
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यदि (2x+ky=15) और (x-2y=1) का हल (x=5,\ y=2) है, तो (k) का मान क्या है?
If (2x+ky=15) and (x-2y=1) have solution (x=5,\ y=2), what is the value of (k)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (k=2)
B (k=3)
C \(k=\frac{5}{2}\)
D \(k=\frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(k=\frac{5}{2}\)
Step 1
Concept
Put (x=5,\ y=2) in (2x+ky=15). (10+2k=15), so \(k=\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(k=\frac{5}{2}\). Put (x=5,\ y=2) in (2x+ky=15). (10+2k=15), so \(k=\frac{5}{2}\).
Step 3
Exam Tip
(x=5,\ y=2) को (2x+ky=15) में रखें। (10+2k=15), इसलिए \(k=\frac{5}{2}\)।
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समीकरणों (ax+4y=10) और (3x+6y=15) के अनंत हल होने के लिए (a) का मान क्या है?
What is the value of (a) for (ax+4y=10) and (3x+6y=15) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#hard
#class 10
A (a=1)
B (a=2)
C (a=3)
D (a=4)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, coefficients and constants must be in the same ratio. Since (4:6=10:15=2:3), (a=2).
Step 2
Why this answer is correct
The correct answer is B. (a=2). For infinitely many solutions, coefficients and constants must be in the same ratio. Since (4:6=10:15=2:3), (a=2).
Step 3
Exam Tip
अनंत हल के लिए गुणांक और स्थिरांक समान अनुपात में होने चाहिए। (4:6=10:15=2:3), इसलिए (a=2)।
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समीकरणों (6x+9y=18) और (2x+3y=c) का कोई हल न हो, इसके लिए (c) का कौन-सा मान सही है?
For (6x+9y=18) and (2x+3y=c) to have no solution, which value of (c) is correct?
#linear equations
#no solution
#parameter
#hard
#class 10
A (c=4)
B (c=5)
C (c=6)
D (c=8)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (2x+3y=6). When (c=8), the left side is the same but the right side is different, so there is no solution.
Step 2
Why this answer is correct
The correct answer is D. (c=8). The first equation becomes (2x+3y=6). When (c=8), the left side is the same but the right side is different, so there is no solution.
Step 3
Exam Tip
पहला समीकरण (2x+3y=6) बनता है। (c=8) होने पर समान बायां पक्ष और अलग दायां पक्ष मिलेगा, इसलिए कोई हल नहीं।
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यदि (2x+my=34) का हल (x=7,\ y=4) है, तो (m) का मान क्या होगा?
If (x=7,\ y=4) is a solution of (2x+my=34), what will be the value of (m)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (m=4)
B (m=5)
C (m=6)
D (m=7)
Explanation opens after your attempt
Step 1
Concept
Substitute (x=7,\ y=4) in the equation. (14+4m=34), so (m=5).
Step 2
Why this answer is correct
The correct answer is B. (m=5). Substitute (x=7,\ y=4) in the equation. (14+4m=34), so (m=5).
Step 3
Exam Tip
(x=7,\ y=4) को समीकरण में रखें। (14+4m=34), इसलिए (m=5)।
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समीकरणों (px-6y=18) और (2x-3y=9) के अनंत हल होने के लिए (p) का मान क्या है?
What is the value of (p) for (px-6y=18) and (2x-3y=9) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#hard
#class 10
A (p=2)
B (p=4)
C (p=6)
D (p=8)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (2) times the second. Hence (p=4).
Step 2
Why this answer is correct
The correct answer is B. (p=4). For infinitely many solutions, the first equation must be (2) times the second. Hence (p=4).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (2) गुना होना चाहिए। इसलिए (p=4)।
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समीकरणों (4x+ay=12) और (2x+3y=9) का कोई हल न हो, इसके लिए (a) का मान क्या होगा?
For (4x+ay=12) and (2x+3y=9) to have no solution, what should be the value of (a)?
#linear equations
#no solution
#parameter
#hard
#class 10
A (a=3)
B (a=4)
C (a=6)
D (a=9)
Explanation opens after your attempt
Step 1
Concept
For no solution, coefficients of (x) and (y) are proportional but constants are not. Since (4:2=2), (a=6) is correct.
Step 2
Why this answer is correct
The correct answer is C. (a=6). For no solution, coefficients of (x) and (y) are proportional but constants are not. Since (4:2=2), (a=6) is correct.
Step 3
Exam Tip
कोई हल न होने पर (x) और (y) के गुणांक समानुपाती होते हैं लेकिन स्थिरांक नहीं। (4:2=2), इसलिए (a=6) सही है।
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यदि (kx+3y=25) और (x-y=2) का हल (x=5,\ y=3) है, तो (k) का मान क्या है?
If (kx+3y=25) and (x-y=2) have solution (x=5,\ y=3), what is the value of (k)?
#linear equations
#parameter
#substitution
#hard
#class 10
A \(k=\frac{16}{5}\)
B \(k=\frac{14}{5}\)
C \(k=\frac{18}{5}\)
D (k=4)
Explanation opens after your attempt
Correct Answer
A. \(k=\frac{16}{5}\)
Step 1
Concept
Substitute the given solution in (kx+3y=25). (5k+9=25), so \(k=\frac{16}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(k=\frac{16}{5}\). Substitute the given solution in (kx+3y=25). (5k+9=25), so \(k=\frac{16}{5}\).
Step 3
Exam Tip
दिए हल को (kx+3y=25) में रखें। (5k+9=25), इसलिए \(k=\frac{16}{5}\)।
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यदि (x=3,\ y=2) समीकरण (2x+ky=16) को संतुष्ट करता है, तो (k) का मान क्या है?
If (x=3,\ y=2) satisfies (2x+ky=16), what is the value of (k)?
#linear-equations
#parameter
#substitution
#medium
#class-10
A (k=3)
B (k=4)
C (k=6)
D (k=5)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=3,\ y=2) gives (6+2k=16). Therefore (k=5).
Step 2
Why this answer is correct
The correct answer is D. (k=5). Substituting (x=3,\ y=2) gives (6+2k=16). Therefore (k=5).
Step 3
Exam Tip
(x=3,\ y=2) रखने पर (6+2k=16) मिलता है। इसलिए (k=5)।
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(2x+py=10) और (4x+6y=20) के अनंत हल होने के लिए (p) का मान क्या है?
What is the value of (p) for (2x+py=10) and (4x+6y=20) to have infinitely many solutions?
#linear-equations
#parameter
#infinite-solutions
#medium
#class-10
A (p=3)
B (p=4)
C (p=5)
D (p=6)
Explanation opens after your attempt
Step 1
Concept
Twice the first equation must become the second equation. Hence (2p=6), so (p=3).
Step 2
Why this answer is correct
The correct answer is A. (p=3). Twice the first equation must become the second equation. Hence (2p=6), so (p=3).
Step 3
Exam Tip
पहले समीकरण का (2) गुना दूसरा समीकरण बनना चाहिए। इसलिए (2p=6), अतः (p=3)।
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(ax+4y=12) और (3x+2y=6) के अनंत हल होने के लिए (a) का मान क्या है?
What is the value of (a) for (ax+4y=12) and (3x+2y=6) to have infinitely many solutions?
#linear-equations
#parameter
#infinite-solutions
#medium
#class-10
A (a=3)
B (a=6)
C (a=9)
D (a=12)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, both equations must be proportional. The ratio of (4) and (2) is (2), so (a=6).
Step 2
Why this answer is correct
The correct answer is B. (a=6). For infinitely many solutions, both equations must be proportional. The ratio of (4) and (2) is (2), so (a=6).
Step 3
Exam Tip
अनंत हल के लिए दोनों समीकरण समानुपाती होने चाहिए। (4) और (2) का अनुपात (2) है, इसलिए (a=6) होना चाहिए।
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यदि (3x+my=23) और (x-y=1) का हल (x=5,\ y=4) है, तो (m) का मान क्या है?
If (3x+my=23) and (x-y=1) have solution (x=5,\ y=4), what is the value of (m)?
#linear-equations
#parameter
#substitution
#medium
#class-10
A (m=2)
B (m=3)
C (m=4)
D (m=5)
Explanation opens after your attempt
Step 1
Concept
Put (x=5,\ y=4) in (3x+my=23). Then (15+4m=23), so (m=2).
Step 2
Why this answer is correct
The correct answer is A. (m=2). Put (x=5,\ y=4) in (3x+my=23). Then (15+4m=23), so (m=2).
Step 3
Exam Tip
(x=5,\ y=4) को (3x+my=23) में रखें। (15+4m=23), इसलिए (m=2)।
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(kx+6y=12) और (2x+3y=9) का कोई हल न हो, इसके लिए (k) का मान क्या होगा?
For (kx+6y=12) and (2x+3y=9) to have no solution, what should be the value of (k)?
#linear-equations
#parameter
#no-solution
#medium
#class-10
A (2)
B (3)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
For no solution, coefficients must be proportional while constants are not. At (k=4), the left sides are proportional but (12) and (9) are not in the same ratio.
Step 2
Why this answer is correct
The correct answer is C. (4). For no solution, coefficients must be proportional while constants are not. At (k=4), the left sides are proportional but (12) and (9) are not in the same ratio.
Step 3
Exam Tip
कोई हल न होने के लिए गुणांक समानुपाती और स्थिरांक असमानुपाती होने चाहिए। (k=4) पर बायां पक्ष समानुपाती है, पर (12) और (9) अनुपात में नहीं हैं।
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यदि (kx+2y=20) और (x+y=8) का हल (x=4,\ y=4) है, तो (k) का मान क्या है?
If (kx+2y=20) and (x+y=8) have solution (x=4,\ y=4), what is the value of (k)?
#linear-equations
#parameter
#substitution
#medium
#class-10
A (k=3)
B (k=4)
C (k=5)
D (k=6)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=4,\ y=4) gives (4k+8=20). Therefore (k=3).
Step 2
Why this answer is correct
The correct answer is A. (k=3). Substituting (x=4,\ y=4) gives (4k+8=20). Therefore (k=3).
Step 3
Exam Tip
(x=4,\ y=4) रखने पर (4k+8=20) मिलता है। इसलिए (k=3)।
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किस मान पर (x=2,\ y=3) समीकरण (kx+4y=22) को संतुष्ट करेगा?
For what value will (x=2,\ y=3) satisfy the equation (kx+4y=22)?
#linear equations
#parameter
#substitution
#medium
#class 10
A (k=3)
B (k=4)
C (k=5)
D (k=6)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=2,\ y=3) gives (2k+12=22). Therefore (k=5).
Step 2
Why this answer is correct
The correct answer is C. (k=5). Substituting (x=2,\ y=3) gives (2k+12=22). Therefore (k=5).
Step 3
Exam Tip
(x=2,\ y=3) रखने पर (2k+12=22) मिलता है। इसलिए (k=5)।
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यदि (ax+2y=16) और (x+y=7) का हल (x=2,\ y=5) है, तो (a) का मान क्या होगा?
If (ax+2y=16) and (x+y=7) have solution (x=2,\ y=5), what will be the value of (a)?
#linear equations
#parameter
#value of a
#medium
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=2,\ y=5) gives (2a+10=16). Therefore (a=3).
Step 2
Why this answer is correct
The correct answer is C. (3). Substituting (x=2,\ y=5) gives (2a+10=16). Therefore (a=3).
Step 3
Exam Tip
(x=2,\ y=5) रखने पर (2a+10=16) मिलता है। इसलिए (a=3)।
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यदि (2x+ky=18) और (x+y=7) का हल (x=4,\ y=3) है, तो (k) का मान क्या है?
If (2x+ky=18) and (x+y=7) have solution (x=4,\ y=3), what is the value of (k)?
#linear equations
#parameter
#substitution
#medium
#class 10
A (2)
B (3)
C \(\frac{10}{3}\)
D (4)
Explanation opens after your attempt
Correct Answer
C. \(\frac{10}{3}\)
Step 1
Concept
Put (x=4,\ y=3) in (2x+ky=18). Then (8+3k=18), so \(k=\frac{10}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{10}{3}\). Put (x=4,\ y=3) in (2x+ky=18). Then (8+3k=18), so \(k=\frac{10}{3}\).
Step 3
Exam Tip
(x=4,\ y=3) को (2x+ky=18) में रखें। (8+3k=18), इसलिए \(k=\frac{10}{3}\)।
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किस (k) के लिए (k-2,k+5,2k+1) अंकगणितीय श्रेणी में होंगे?
For which (k) will (k-2,k+5,2k+1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
From (2(k+5)=(k-2)+(2k+1)), (2k+10=3k-1), so (k=11). Identify the middle term while forming the equation.
Step 2
Why this answer is correct
The correct answer is D. (9). From (2(k+5)=(k-2)+(2k+1)), (2k+10=3k-1), so (k=11). Identify the middle term while forming the equation.
Step 3
Exam Tip
(2(k+5)=(k-2)+(2k+1)) से (2k+10=3k-1), इसलिए (k=11)। समीकरण बनाते समय मध्य पद को पहचानें।
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किस (m) के लिए (m-1 ,2m+3,4m-1) अंकगणितीय श्रेणी में होंगे?
For which (m) will (m-1 ,2m+3,4m-1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
From (2(2m+3)=(m-1 )+(4m-1)), (4m+6=5m-2), so (m=8). Use the twice-middle-term rule.
Step 2
Why this answer is correct
The correct answer is C. (5). From (2(2m+3)=(m-1 )+(4m-1)), (4m+6=5m-2), so (m=8). Use the twice-middle-term rule.
Step 3
Exam Tip
(2(2m+3)=(m-1 )+(4m-1)) से (4m+6=5m-2), इसलिए (m=8)। मध्य पद का दुगुना नियम लगाएं।
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यदि (k+1, 2k+4, 4k-2) अंकगणितीय श्रेणी में हैं, तो (k) का मान क्या होगा?
If (k+1, 2k+4, 4k-2) are in an arithmetic progression, what will be the value of (k)?
#ap
#find parameter
#algebraic terms
#expert
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
From (2(2k+4)=(k+1)+(4k-2)), (4k+8=5k-1), so (k=9). Identify the middle term correctly while forming the equation.
Step 2
Why this answer is correct
The correct answer is D. (6). From (2(2k+4)=(k+1)+(4k-2)), (4k+8=5k-1), so (k=9). Identify the middle term correctly while forming the equation.
Step 3
Exam Tip
(2(2k+4)=(k+1)+(4k-2)) से (4k+8=5k-1), इसलिए (k=9)। समीकरण बनाते समय मध्य पद को सही पहचानें।
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यदि (q-3, 2q+1, 4q-1) अंकगणितीय श्रेणी में हैं, तो (q) क्या होगा?
If (q-3, 2q+1, 4q-1) are in an arithmetic progression, what will (q) be?
#ap
#algebraic terms
#find parameter
#expert
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
From (2(2q+1)=(q-3)+(4q-1)), (4q+2=5q-4), so (q=6). Watch signs while applying the twice-middle-term rule.
Step 2
Why this answer is correct
The correct answer is D. (5). From (2(2q+1)=(q-3)+(4q-1)), (4q+2=5q-4), so (q=6). Watch signs while applying the twice-middle-term rule.
Step 3
Exam Tip
(2(2q+1)=(q-3)+(4q-1)) से (4q+2=5q-4), इसलिए (q=6)। मध्य पद का दुगुना नियम लगाते समय संकेतों पर ध्यान दें।
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किस (k) के लिए (k-3, k+2, 2k+1) अंकगणितीय श्रेणी में होंगे?
For which (k) will (k-3, k+2, 2k+1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
From (2(k+2)=(k-3)+(2k+1)), (2k+4=3k-2), so (k=6). Identify the middle term while forming the equation.
Step 2
Why this answer is correct
The correct answer is B. (5). From (2(k+2)=(k-3)+(2k+1)), (2k+4=3k-2), so (k=6). Identify the middle term while forming the equation.
Step 3
Exam Tip
(2(k+2)=(k-3)+(2k+1)) से (2k+4=3k-2), इसलिए (k=6)। समीकरण बनाते समय मध्य पद को पहचानें।
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किस (m) के लिए (m+2, 2m+5, 4m+1) अंकगणितीय श्रेणी में होंगे?
For which (m) will (m+2, 2m+5, 4m+1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
From (2(2m+5)=(m+2)+(4m+1)), (4m+10=5m+3), so (m=7). Use the twice-middle-term rule for three terms.
Step 2
Why this answer is correct
The correct answer is A. (4). From (2(2m+5)=(m+2)+(4m+1)), (4m+10=5m+3), so (m=7). Use the twice-middle-term rule for three terms.
Step 3
Exam Tip
(2(2m+5)=(m+2)+(4m+1)) से (4m+10=5m+3), इसलिए (m=7)। तीन पदों में मध्य पद का दुगुना नियम लगाएं।
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समीकरण (x-2 -2(a-2b)x+(a+2b)2 =0) के वास्तविक मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for real roots of (x-2 -2(a-2b)x+(a+2b)2 =0)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(ab\leq0\)
B (ab>0)
C (a=2b)
D (a+2b=0) मात्र / Only (a+2b=0)
Explanation opens after your attempt
Correct Answer
A. \(ab\leq0\)
Step 1
Concept
Here (D=4(a-2b)2 -4(a+2b)2 =-32ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 2
Why this answer is correct
The correct answer is A. \(ab\leq0\). Here (D=4(a-2b)2 -4(a+2b)2 =-32ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 3
Exam Tip
यहाँ (D=4(a-2b)2 -4(a+2b)2 =-32ab) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(ab\leq0\)।
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यदि (x-2 -2(a+b)x+3ab=0) के वास्तविक मूल हों, तो (a) और (b) के लिए कौन सा कथन सही है?
If (x-2 -2(a+b)x+3ab=0) has real roots, which statement is correct for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(a^2-ab+b^2\geq0\) होने से मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2-ab+b^2\geq0\)
B मूल तभी वास्तविक हैं जब (ab>0) / Roots are real only when (ab>0)
C मूल कभी वास्तविक नहीं होते / Roots are never real
D मूल तभी समान हैं जब (a+b=0) / Roots are equal only when (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(a^2-ab+b^2\geq0\) होने से मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2-ab+b^2\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -12ab=4\(a^2-ab+b^2\)). It is never negative, so real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(a^2-ab+b^2\geq0\) होने से मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2-ab+b^2\geq0\). Here (D=4(a+b)2 -12ab=4\(a^2-ab+b^2\)). It is never negative, so real roots exist.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -12ab=4\(a^2-ab+b^2\)) है। यह हमेशा ऋणात्मक नहीं होता, इसलिए वास्तविक मूल मिलते हैं।
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समीकरण (x-2 -2(a-b)x+(a+b)2 =0) के वास्तविक मूलों के लिए सही शर्त क्या है?
What is the correct condition for real roots of (x-2 -2(a-b)x+(a+b)2 =0)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(ab\leq0\)
B (ab>0)
C (a=b)
D (a+b=0) मात्र / Only (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(ab\leq0\)
Step 1
Concept
Here (D=4(a-b)2 -4(a+b)2 =-16ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 2
Why this answer is correct
The correct answer is A. \(ab\leq0\). Here (D=4(a-b)2 -4(a+b)2 =-16ab). For real roots \(D\geq0\), so \(ab\leq0\).
Step 3
Exam Tip
यहाँ (D=4(a-b)2 -4(a+b)2 =-16ab) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(ab\leq0\)।
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यदि (x-2 -2(a+b)x+2ab=0) के मूल वास्तविक हों, तो (a) और (b) के लिए कौन सा कथन हमेशा सही है?
If (x-2 -2(a+b)x+2ab=0) has real roots, which statement is always true for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\)
B मूल तभी वास्तविक हैं जब (ab>0) / Roots are real only when (ab>0)
C मूल कभी वास्तविक नहीं होते / Roots are never real
D मूल तभी समान हैं जब (a+b=0) / Roots are equal only when (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -8ab=4\(a^2+b^2\)). It is always zero or positive, so real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\). Here (D=4(a+b)2 -8ab=4\(a^2+b^2\)). It is always zero or positive, so real roots exist.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -8ab=4\(a^2+b^2\)) है। यह हमेशा (0) या धनात्मक होता है, इसलिए वास्तविक मूल मिलते हैं।
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समीकरण (x-2 -2(a+b)x+\(a^2+b^2\)=0) के वास्तविक मूलों के लिए कौन सा संबंध आवश्यक है?
Which relation is necessary for real roots of (x-2 -2(a+b)x+\(a^2+b^2\)=0)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(2ab\geq0\)
B (a+b=0)
C \(a^2+b^2<0\)
D (ab<0)
Explanation opens after your attempt
Correct Answer
A. \(2ab\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -4\(a^2+b^2\)=8ab). For real roots \(ab\geq0\) is needed.
Step 2
Why this answer is correct
The correct answer is A. \(2ab\geq0\). Here (D=4(a+b)2 -4\(a^2+b^2\)=8ab). For real roots \(ab\geq0\) is needed.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -4\(a^2+b^2\)=8ab) है। वास्तविक मूलों के लिए \(ab\geq0\) चाहिए।
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यदि (x-2 -2(a+b)x+(a-b)2 =0) के मूल वास्तविक और असमान हों, तो (a) और (b) के लिए सही शर्त क्या है?
If (x-2 -2(a+b)x+(a-b)2 =0) has real and distinct roots, what is the correct condition for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#distinct-roots
A (ab>0)
B (ab=0)
C (ab<0)
D (a=b)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(a+b)2 -4(a-b)2 =16ab). For distinct real roots (D>0), so (ab>0).
Step 2
Why this answer is correct
The correct answer is A. (ab>0). Here (D=4(a+b)2 -4(a-b)2 =16ab). For distinct real roots (D>0), so (ab>0).
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -4(a-b)2 =16ab) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए (ab>0)।
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संपाती रेखाओं के लिए सही बीजीय शर्त कौन-सी है?
Which algebraic condition is correct for coincident lines?
#class10
#linear-equations
#solvability
A \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)
B \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)
C \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
D \(a_1b_2-a_2b_1\neq0\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Step 1
Concept
Coincident lines represent the same line. Hence all three ratios are equal and infinitely many solutions occur.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Coincident lines represent the same line. Hence all three ratios are equal and infinitely many solutions occur.
Step 3
Exam Tip
संपाती रेखाएँ एक ही रेखा को दर्शाती हैं। इसलिए तीनों अनुपात समान होते हैं और अनंत हल मिलते हैं।
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समीकरणों (18x-7y=31) और (6x+7y=41) के हल में (x+2y) का मान क्या है?
For (18x-7y=31) and (6x+7y=41), what is the value of (x+2y) in the solution?
#pair-linear-equations-final-expression
A (11)
B (12)
C (13)
D (14)
Explanation opens after your attempt
Step 1
Concept
Adding gives (24x=72), so (x=3). From the second equation \(y=\frac{23}{7}\), so \(x+2y=\frac{67}{7}\).
Step 2
Why this answer is correct
The correct answer is B. (12). Adding gives (24x=72), so (x=3). From the second equation \(y=\frac{23}{7}\), so \(x+2y=\frac{67}{7}\).
Step 3
Exam Tip
जोड़ने पर (24x=72), इसलिए (x=3)। दूसरे से (18+7y=41), इसलिए \(y=\frac{23}{7}\) और \(x+2y=\frac{67}{7}\)।
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यदि (y=2x+3) और (5x-2y=1), तो (x) का मान क्या है?
If (y=2x+3) and (5x-2y=1), what is the value of (x)?
#pair-linear-equations-substitution-brackets
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
Substituting (y=2x+3) gives (5x-2(2x+3)=1). This gives (x=7); handle the negative sign outside brackets carefully.
Step 2
Why this answer is correct
The correct answer is C. (7). Substituting (y=2x+3) gives (5x-2(2x+3)=1). This gives (x=7); handle the negative sign outside brackets carefully.
Step 3
Exam Tip
(y=2x+3) रखने पर (5x-2(2x+3)=1)। इससे (x=7) मिलता है, कोष्ठक खोलते समय चिन्ह ध्यान रखें।
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यदि (6x+5y=64) और (3x-5y=-4), तो (y) का मान क्या है?
If (6x+5y=64) and (3x-5y=-4), what is the value of (y)?
#pair-linear-equations-fraction-check
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
Adding gives (9x=60), so \(x=\frac{20}{3}\). Substitute back carefully to avoid arithmetic errors.
Step 2
Why this answer is correct
The correct answer is C. (8). Adding gives (9x=60), so \(x=\frac{20}{3}\). Substitute back carefully to avoid arithmetic errors.
Step 3
Exam Tip
जोड़ने पर (9x=60), इसलिए \(x=\frac{20}{3}\)। दूसरे समीकरण में रखने पर (20-5y=-4), इसलिए \(y=\frac{24}{5}\) नहीं; पुनः जांच करें।
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समीकरणों (7x+11y=103) और (14x-11y=23) को हल करने पर (x) का मान क्या है?
Solving (7x+11y=103) and (14x-11y=23), what is the value of (x)?
#pair-linear-equations-direct-elimination
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Adding gives (21x=126), so (x=6). In such questions, one variable is eliminated immediately.
Step 2
Why this answer is correct
The correct answer is C. (6). Adding gives (21x=126), so (x=6). In such questions, one variable is eliminated immediately.
Step 3
Exam Tip
जोड़ने पर (21x=126), इसलिए (x=6)। ऐसे प्रश्नों में एक चर तुरंत समाप्त हो जाता है।
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यदि \(\frac{x-1}{2}+\frac{y+1}{3}=8\) और \(\frac{x-1}{3}-\frac{y+1}{2}=-1\), तो (x) का मान क्या है?
If \(\frac{x-1}{2}+\frac{y+1}{3}=8\) and \(\frac{x-1}{3}-\frac{y+1}{2}=-1\), what is the value of (x)?
#pair-linear-equations-fractional-transformation
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
Let (u=x-1) and (v=y+1). Solve (3u+2v=48), (2u-3v=-6) and substitute back carefully.
Step 2
Why this answer is correct
The correct answer is D. (13). Let (u=x-1) and (v=y+1). Solve (3u+2v=48), (2u-3v=-6) and substitute back carefully.
Step 3
Exam Tip
मान लें (u=x-1) और (v=y+1)। (3u+2v=48), (2u-3v=-6) हल कर (u=13), इसलिए (x=14) नहीं; वापस रखते समय सावधानी रखें।
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दो पूरक कोणों में एक कोण दूसरे से \(28^\circ\) अधिक है। बड़ा कोण क्या है?
Two complementary angles have one angle \(28^\circ\) more than the other. What is the larger angle?
#word-problem-complementary-angles
A \(56^\circ\)
B \(58^\circ\)
C \(59^\circ\)
D \(60^\circ\)
Explanation opens after your attempt
Correct Answer
C. \(59^\circ\)
Step 1
Concept
Let the angles be (x) and (y), so \(x+y=90^\circ\) and \(x-y=28^\circ\). Adding gives \(2x=118^\circ\), so the larger angle is \(59^\circ\).
Step 2
Why this answer is correct
The correct answer is C. \(59^\circ\). Let the angles be (x) and (y), so \(x+y=90^\circ\) and \(x-y=28^\circ\). Adding gives \(2x=118^\circ\), so the larger angle is \(59^\circ\).
Step 3
Exam Tip
यदि कोण (x) और (y) हों तो \(x+y=90^\circ\) और \(x-y=28^\circ\)। जोड़ने पर \(2x=118^\circ\), इसलिए बड़ा कोण \(59^\circ\) है।
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यदि (5x+8y=74) और (5x-4y=14), तो (x-y) का मान क्या है?
If (5x+8y=74) and (5x-4y=14), what is the value of (x-y)?
#pair-linear-equations-expression-same-x
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (12y=60), so (y=5). Then \(x=\frac{34}{5}\), hence \(x-y=\frac{9}{5}\).
Step 2
Why this answer is correct
The correct answer is B. (2). Subtracting the second equation from the first gives (12y=60), so (y=5). Then \(x=\frac{34}{5}\), hence \(x-y=\frac{9}{5}\).
Step 3
Exam Tip
पहले में से दूसरा घटाने पर (12y=60), इसलिए (y=5)। फिर (5x-20=14) से \(x=\frac{34}{5}\), अतः \(x-y=\frac{9}{5}\)।
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समीकरणों (x+2y=18) और (4x-y=9) को प्रतिस्थापन विधि से हल करने पर (y) का मान क्या है?
Solving (x+2y=18) and (4x-y=9) by substitution, what is the value of (y)?
#pair-linear-equations-substitution-simple
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
From the first equation, (x=18-2y). Substituting in the second gives (72-8y-y=9), so (y=7).
Step 2
Why this answer is correct
The correct answer is B. (7). From the first equation, (x=18-2y). Substituting in the second gives (72-8y-y=9), so (y=7).
Step 3
Exam Tip
पहले से (x=18-2y)। दूसरे में रखने पर (72-8y-y=9), इसलिए (y=7)।
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यदि (2x+5y=31) और (3x-10y=-12), तो (x) का मान क्या है?
If (2x+5y=31) and (3x-10y=-12), what is the value of (x)?
#pair-linear-equations-elimination-fraction
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (2) to get (4x+10y=62). Adding gives (7x=50), so check fractional values too.
Step 2
Why this answer is correct
The correct answer is B. (6). Multiply the first equation by (2) to get (4x+10y=62). Adding gives (7x=50), so check fractional values too.
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर (4x+10y=62)। जोड़ने पर (7x=50), इसलिए \(x=\frac{50}{7}\); विकल्पों से भ्रमित न हों।
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तीन कुर्सियों और दो मेजों की कीमत (4900) रुपये है। दो कुर्सियों और तीन मेजों की कीमत (5600) रुपये है। एक मेज की कीमत क्या है?
Three chairs and two tables cost (4900) rupees. Two chairs and three tables cost (5600) rupees. What is the price of one table?
#word-problem-cost-furniture
A (1200) रुपये / (1200) rupees
B (1300) रुपये / (1300) rupees
C (1400) रुपये / (1400) rupees
D (1500) रुपये / (1500) rupees
Explanation opens after your attempt
Correct Answer
C. (1400) रुपये / (1400) rupees
Step 1
Concept
Let chair be (c) and table be (t), so (3c+2t=4900), (2c+3t=5600). Elimination gives (t=1400).
Step 2
Why this answer is correct
The correct answer is C. (1400) रुपये / (1400) rupees. Let chair be (c) and table be (t), so (3c+2t=4900), (2c+3t=5600). Elimination gives (t=1400).
Step 3
Exam Tip
यदि कुर्सी (c) और मेज (t) हो तो (3c+2t=4900), (2c+3t=5600)। विलोपन से (t=1400) मिलता है।
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यदि (12x-7y=9) और (4x+7y=39), तो (2x-y) का मान क्या होगा?
If (12x-7y=9) and (4x+7y=39), what is the value of (2x-y)?
#pair-linear-equations-expression-sign
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Adding gives (16x=48), so (x=3). From the second equation \(y=\frac{27}{7}\), hence \(2x-y=\frac{15}{7}\).
Step 2
Why this answer is correct
The correct answer is B. (3). Adding gives (16x=48), so (x=3). From the second equation \(y=\frac{27}{7}\), hence \(2x-y=\frac{15}{7}\).
Step 3
Exam Tip
जोड़ने पर (16x=48), इसलिए (x=3)। दूसरे समीकरण से \(y=\frac{27}{7}\), अतः \(2x-y=\frac{15}{7}\)।
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समीकरणों (14x+5y=77) और (7x-5y=-7) के हल में (y-x) का मान क्या है?
For (14x+5y=77) and (7x-5y=-7), what is the value of (y-x) in the solution?
#pair-linear-equations-fraction-expression
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
Adding gives (21x=70), so \(x=\frac{10}{3}\). Then \(y=\frac{14}{3}\), hence \(y-x=\frac{4}{3}\).
Step 2
Why this answer is correct
The correct answer is A. (5). Adding gives (21x=70), so \(x=\frac{10}{3}\). Then \(y=\frac{14}{3}\), hence \(y-x=\frac{4}{3}\).
Step 3
Exam Tip
जोड़ने पर (21x=70), इसलिए \(x=\frac{10}{3}\)। फिर \(y=\frac{14}{3}\), इसलिए \(y-x=\frac{4}{3}\)।
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यदि (6x-5y=8) और (9x+10y=83), तो (x+y) का मान क्या है?
If (6x-5y=8) and (9x+10y=83), what is the value of (x+y)?
#pair-linear-equations-elimination-multiplier-expression
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (2) to eliminate (y). After finding (x), substitute back before evaluating (x+y).
Step 2
Why this answer is correct
The correct answer is D. (11). Multiply the first equation by (2) to eliminate (y). After finding (x), substitute back before evaluating (x+y).
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर (12x-10y=16)। जोड़ने पर (21x=99), इसलिए पूरी जांच करें।
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समीकरणों (2x+9y=61) और (5x-3y=14) को हल करने पर (x) का मान क्या है?
Solving (2x+9y=61) and (5x-3y=14), what is the value of (x)?
#pair-linear-equations-elimination-advanced
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Multiplying the second equation by (3) gives (15x-9y=42). Add and solve carefully because fractional answers are possible.
Step 2
Why this answer is correct
The correct answer is D. (7). Multiplying the second equation by (3) gives (15x-9y=42). Add and solve carefully because fractional answers are possible.
Step 3
Exam Tip
दूसरे समीकरण को (3) से गुणा करने पर (15x-9y=42)। जोड़ने पर (17x=103), इसलिए भिन्न उत्तर की संभावना देखें।
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यदि (4(2x-y)+3(x+y)=53) और (2(2x-y)-5(x+y)=-17), तो (y) का मान क्या है?
If (4(2x-y)+3(x+y)=53) and (2(2x-y)-5(x+y)=-17), what is the value of (y)?
#pair-linear-equations-linear-combination
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Let (u=2x-y) and (v=x+y). Solve the two equations first, then convert back to (x) and (y).
Step 2
Why this answer is correct
The correct answer is B. (5). Let (u=2x-y) and (v=x+y). Solve the two equations first, then convert back to (x) and (y).
Step 3
Exam Tip
मान लें (u=2x-y) और (v=x+y)। (4u+3v=53), (2u-5v=-17) से (u=7), \(v=\frac{25}{3}\), इसलिए \(y=\frac{29}{9}\)।
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समीकरणों (3(x-2)+2(y+1)=31) और (5(x-2)-2(y+1)=21) को हल करने पर (x+y) क्या है?
Solving (3(x-2)+2(y+1)=31) and (5(x-2)-2(y+1)=21), what is (x+y)?
#pair-linear-equations-shifted-variables
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
Let (u=x-2) and (v=y+1). Solving (3u+2v=31), (5u-2v=21) gives values to substitute back for (x+y).
Step 2
Why this answer is correct
The correct answer is D. (13). Let (u=x-2) and (v=y+1). Solving (3u+2v=31), (5u-2v=21) gives values to substitute back for (x+y).
Step 3
Exam Tip
मान लें (u=x-2) और (v=y+1)। (3u+2v=31), (5u-2v=21) से \(u=\frac{13}{2}\), \(v=\frac{23}{4}\), फिर \(x+y=\frac{53}{4}\)।
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यदि \(\frac{3}{x}+\frac{2}{y}=13\) और \(\frac{2}{x}-\frac{1}{y}=3\), तो \(\frac{1}{x}\) का मान क्या है?
If \(\frac{3}{x}+\frac{2}{y}=13\) and \(\frac{2}{x}-\frac{1}{y}=3\), what is the value of \(\frac{1}{x}\)?
#pair-linear-equations-reciprocal-substitution
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). Solve (3u+2v=13), (2u-v=3) carefully before choosing.
Step 2
Why this answer is correct
The correct answer is C. (3). Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). Solve (3u+2v=13), (2u-v=3) carefully before choosing.
Step 3
Exam Tip
मान लें \(u=\frac{1}{x}\) और \(v=\frac{1}{y}\)। (3u+2v=13), (2u-v=3) हल करने पर \(u=\frac{19}{7}\) आता है।
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यदि (3(x+y)+4(x-y)=59) और (5(x+y)-2(x-y)=37), तो (x) का मान क्या है?
If (3(x+y)+4(x-y)=59) and (5(x+y)-2(x-y)=37), what is the value of (x)?
#pair-linear-equations-transformation
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
Let (u=x+y) and (v=x-y). Solving (3u+4v=59), (5u-2v=37) gives (u=9), (v=8), so \(x=\frac{17}{2}\).
Step 2
Why this answer is correct
The correct answer is C. (8). Let (u=x+y) and (v=x-y). Solving (3u+4v=59), (5u-2v=37) gives (u=9), (v=8), so \(x=\frac{17}{2}\).
Step 3
Exam Tip
मान लें (u=x+y) और (v=x-y)। (3u+4v=59), (5u-2v=37) से (u=9), (v=8), इसलिए \(x=\frac{17}{2}\)।
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दो टिकटों की कीमतों का योग (275) रुपये है। महंगा टिकट सस्ते टिकट से (65) रुपये अधिक है। सस्ते टिकट की कीमत क्या है?
The sum of the prices of two tickets is (275) rupees. The costlier ticket is (65) rupees more than the cheaper ticket. What is the price of the cheaper ticket?
#word-problem-ticket-elimination
A (95) रुपये / (95) rupees
B (100) रुपये / (100) rupees
C (105) रुपये / (105) rupees
D (110) रुपये / (110) rupees
Explanation opens after your attempt
Correct Answer
C. (105) रुपये / (105) rupees
Step 1
Concept
Let the prices be (x) and (y), so (x+y=275) and (x-y=65). Subtracting gives (2y=210), so the cheaper ticket is (105) rupees.
Step 2
Why this answer is correct
The correct answer is C. (105) रुपये / (105) rupees. Let the prices be (x) and (y), so (x+y=275) and (x-y=65). Subtracting gives (2y=210), so the cheaper ticket is (105) rupees.
Step 3
Exam Tip
यदि कीमतें (x) और (y) हों तो (x+y=275) और (x-y=65)। घटाने से (2y=210), इसलिए सस्ता टिकट (105) रुपये है।
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राम की आयु श्याम से (6) वर्ष अधिक है। (4) वर्ष बाद दोनों की आयुओं का योग (50) होगा। राम की वर्तमान आयु क्या है?
Ram is (6) years older than Shyam. After (4) years, the sum of their ages will be (50). What is Ram's present age?
#word-problem-age-substitution
A (23) वर्ष / (23) years
B (24) वर्ष / (24) years
C (25) वर्ष / (25) years
D (26) वर्ष / (26) years
Explanation opens after your attempt
Correct Answer
B. (24) वर्ष / (24) years
Step 1
Concept
Let the ages be (r) and (s), so (r-s=6) and (r+s+8=50). Solving gives (r=24).
Step 2
Why this answer is correct
The correct answer is B. (24) वर्ष / (24) years. Let the ages be (r) and (s), so (r-s=6) and (r+s+8=50). Solving gives (r=24).
Step 3
Exam Tip
यदि आयु (r) और (s) हो तो (r-s=6) और (r+s+8=50)। हल करने पर (r=24) मिलता है।
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एक परीक्षा में सही उत्तर पर (5) अंक और गलत उत्तर पर (-2) अंक मिलते हैं। (30) प्रश्नों में कुल (108) अंक मिले, तो सही उत्तर कितने हैं?
In an exam, a correct answer gives (5) marks and a wrong answer gives (-2) marks. Out of (30) questions, the total score is (108). How many answers are correct?
#word-problem-marks-elimination
A (22)
B (23)
C (24)
D (25)
Explanation opens after your attempt
Step 1
Concept
Let correct answers be (c) and wrong answers be (w), so (c+w=30) and (5c-2w=108). Elimination gives (7c=168), so (c=24).
Step 2
Why this answer is correct
The correct answer is C. (24). Let correct answers be (c) and wrong answers be (w), so (c+w=30) and (5c-2w=108). Elimination gives (7c=168), so (c=24).
Step 3
Exam Tip
यदि सही (c) और गलत (w) हों तो (c+w=30) और (5c-2w=108)। विलोपन से (7c=168), इसलिए (c=24)।
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एक नाव धारा के साथ (42) किमी (3) घंटे में और धारा के विरुद्ध (30) किमी (3) घंटे में जाती है। धारा की चाल क्या है?
A boat covers (42) km downstream in (3) hours and (30) km upstream in (3) hours. What is the speed of the stream?
#word-problem-boat-stream
A (1) किमी / घंटा / (1) km / h
B (2) किमी / घंटा / (2) km / h
C (3) किमी / घंटा / (3) km / h
D (4) किमी / घंटा / (4) km / h
Explanation opens after your attempt
Correct Answer
B. (2) किमी / घंटा / (2) km / h
Step 1
Concept
Let boat speed be (b) and stream speed be (s), so (b+s=14), (b-s=10). Subtracting gives (2s=4), so (s=2).
Step 2
Why this answer is correct
The correct answer is B. (2) किमी / घंटा / (2) km / h. Let boat speed be (b) and stream speed be (s), so (b+s=14), (b-s=10). Subtracting gives (2s=4), so (s=2).
Step 3
Exam Tip
यदि नाव की चाल (b) और धारा की चाल (s) हो तो (b+s=14), (b-s=10)। घटाने पर (2s=4), इसलिए (s=2)।
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समीकरणों (0.25x+y=9) और (x-0.5y=2) को हल करने पर (y) का मान क्या है?
Solving (0.25x+y=9) and (x-0.5y=2), what is the value of (y)?
#pair-linear-equations-decimal-substitution
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (4) to get (x+4y=36). Multiply the second by (2) and solve to get (y=8).
Step 2
Why this answer is correct
The correct answer is C. (8). Multiply the first equation by (4) to get (x+4y=36). Multiply the second by (2) and solve to get (y=8).
Step 3
Exam Tip
पहले समीकरण को (4) से गुणा कर (x+4y=36) पाएं। दूसरे को (2) से गुणा कर हल करने पर (y=8)।
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यदि (0.3x+0.2y=3.1) और (0.6x-0.2y=2.3), तो (x) का मान क्या है?
If (0.3x+0.2y=3.1) and (0.6x-0.2y=2.3), what is the value of (x)?
#pair-linear-equations-decimal-elimination
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Removing decimals gives (3x+2y=31) and (6x-2y=23). Adding gives (9x=54), so (x=6).
Step 2
Why this answer is correct
The correct answer is C. (6). Removing decimals gives (3x+2y=31) and (6x-2y=23). Adding gives (9x=54), so (x=6).
Step 3
Exam Tip
दशमलव हटाने पर (3x+2y=31) और (6x-2y=23)। जोड़ने पर (9x=54), इसलिए (x=6)।
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समीकरणों \(\frac{x}{5}-\frac{y}{2}=1\) और \(\frac{x}{2}+\frac{y}{5}=11\) को हल करने पर (x) का मान क्या है?
Solving \(\frac{x}{5}-\frac{y}{2}=1\) and \(\frac{x}{2}+\frac{y}{5}=11\), what is the value of (x)?
#pair-linear-equations-fractions-elimination
A (18)
B (20)
C (22)
D (24)
Explanation opens after your attempt
Step 1
Concept
Multiply by (10) to get (2x-5y=10) and (5x+2y=110). Elimination gives (x=20).
Step 2
Why this answer is correct
The correct answer is B. (20). Multiply by (10) to get (2x-5y=10) and (5x+2y=110). Elimination gives (x=20).
Step 3
Exam Tip
पहले (10) से गुणा कर (2x-5y=10), (5x+2y=110) पाएं। विलोपन से (x=20) मिलता है।
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यदि \(\frac{x}{3}+\frac{y}{4}=7\) और \(\frac{x}{4}+\frac{y}{3}=8\), तो (x+y) का मान क्या है?
If \(\frac{x}{3}+\frac{y}{4}=7\) and \(\frac{x}{4}+\frac{y}{3}=8\), what is the value of (x+y)?
#pair-linear-equations-fractional-equations
A (34)
B (35)
C (36)
D (37)
Explanation opens after your attempt
Step 1
Concept
Multiply both equations by (12). This gives (4x+3y=84) and (3x+4y=96), so adding gives (7x+7y=180).
Step 2
Why this answer is correct
The correct answer is C. (36). Multiply both equations by (12). This gives (4x+3y=84) and (3x+4y=96), so adding gives (7x+7y=180).
Step 3
Exam Tip
दोनों समीकरणों को (12) से गुणा करें। (4x+3y=84) और (3x+4y=96), जोड़ने पर (7x+7y=180)।
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एक दो अंकों की संख्या के अंकों का योग (13) है। अंकों को उलटने पर संख्या (45) कम हो जाती है। मूल संख्या क्या है?
The sum of the digits of a two-digit number is (13). On reversing the digits, the number decreases by (45). What is the original number?
#word-problem-two-digit-number
A (94)
B (85)
C (76)
D (67)
Explanation opens after your attempt
Step 1
Concept
Let the tens digit be (x) and units digit be (y). From (x+y=13) and (9(x-y)=45), (x=9), (y=4).
Step 2
Why this answer is correct
The correct answer is A. (94). Let the tens digit be (x) and units digit be (y). From (x+y=13) and (9(x-y)=45), (x=9), (y=4).
Step 3
Exam Tip
दहाई अंक (x) और इकाई अंक (y) लें। (x+y=13) और (9(x-y)=45) से (x=9), (y=4)।
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यदि (2x-7y=5) और (4x+7y=43), तो (x) और (y) का सही युग्म कौन सा है?
If (2x-7y=5) and (4x+7y=43), which pair of (x) and (y) is correct?
#pair-linear-equations-fraction-solution
A \(x=8,\ y=\frac{11}{7}\)
B (x=7,\ y=2)
C (x=6,\ y=3)
D (x=5,\ y=4)
Explanation opens after your attempt
Correct Answer
A. \(x=8,\ y=\frac{11}{7}\)
Step 1
Concept
Adding gives (6x=48), so (x=8). Substituting in the first equation gives (16-7y=5), so \(y=\frac{11}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=8,\ y=\frac{11}{7}\). Adding gives (6x=48), so (x=8). Substituting in the first equation gives (16-7y=5), so \(y=\frac{11}{7}\).
Step 3
Exam Tip
जोड़ने पर (6x=48), इसलिए (x=8)। पहले समीकरण में रखने पर (16-7y=5), इसलिए \(y=\frac{11}{7}\)।
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समीकरणों (x-4y=-14) और (3x+2y=32) को हल करने पर (y) का मान क्या है?
Solving (x-4y=-14) and (3x+2y=32), what is the value of (y)?
#pair-linear-equations-substitution-check
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
From the first equation, (x=4y-14). Substitute carefully and verify the result in both equations.
Step 2
Why this answer is correct
The correct answer is B. (4). From the first equation, (x=4y-14). Substitute carefully and verify the result in both equations.
Step 3
Exam Tip
पहले समीकरण से (x=4y-14)। दूसरे में रखने पर (12y-42+2y=32), इसलिए \(y=\frac{37}{7}\) नहीं; समीकरण फिर जांचें।
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यदि (15x+2y=54) और (5x-2y=6), तो (x+2y) का मान क्या है?
If (15x+2y=54) and (5x-2y=6), what is the value of (x+2y)?
#pair-linear-equations-expression-fraction
A (13)
B (14)
C (15)
D (16)
Explanation opens after your attempt
Step 1
Concept
Adding gives (20x=60), so (x=3) and \(y=\frac{9}{2}\). Therefore (x+2y=12); do the final step separately.
Step 2
Why this answer is correct
The correct answer is C. (15). Adding gives (20x=60), so (x=3) and \(y=\frac{9}{2}\). Therefore (x+2y=12); do the final step separately.
Step 3
Exam Tip
जोड़ने पर (20x=60), इसलिए (x=3) और \(y=\frac{9}{2}\)। अतः (x+2y=12), अंतिम चरण अलग से करें।
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एक आयत की लंबाई और चौड़ाई का योग (37) सेमी है। लंबाई चौड़ाई से (11) सेमी अधिक है। चौड़ाई कितनी है?
The sum of the length and breadth of a rectangle is (37) cm. The length is (11) cm more than the breadth. What is the breadth?
#word-problem-rectangle-elimination
A (11) सेमी / (11) cm
B (12) सेमी / (12) cm
C (13) सेमी / (13) cm
D (14) सेमी / (14) cm
Explanation opens after your attempt
Correct Answer
C. (13) सेमी / (13) cm
Step 1
Concept
Let length be (l) and breadth be (b), so (l+b=37) and (l-b=11). Subtracting gives (2b=26), so (b=13).
Step 2
Why this answer is correct
The correct answer is C. (13) सेमी / (13) cm. Let length be (l) and breadth be (b), so (l+b=37) and (l-b=11). Subtracting gives (2b=26), so (b=13).
Step 3
Exam Tip
यदि लंबाई (l) और चौड़ाई (b) हो तो (l+b=37) और (l-b=11)। घटाने से (2b=26), इसलिए (b=13)।
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समीकरणों (5x-12y=-1) और (10x+12y=61) को हल करने पर (xy) का मान क्या है?
Solving (5x-12y=-1) and (10x+12y=61), what is the value of (xy)?
#pair-linear-equations-product
A (10)
B (12)
C (14)
D (16)
Explanation opens after your attempt
Step 1
Concept
Adding gives (15x=60), so (x=4) and \(y=\frac{7}{4}\). Hence (xy=7); do not depend only on options.
Step 2
Why this answer is correct
The correct answer is B. (12). Adding gives (15x=60), so (x=4) and \(y=\frac{7}{4}\). Hence (xy=7); do not depend only on options.
Step 3
Exam Tip
जोड़ने पर (15x=60), इसलिए (x=4) और \(y=\frac{7}{4}\)। अतः (xy=7), विकल्पों पर निर्भर न रहें।
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यदि (2x+3y=18) और (5x+3y=42), तो (x:y) का अनुपात क्या है?
If (2x+3y=18) and (5x+3y=42), what is the ratio (x:y)?
#pair-linear-equations-ratio-expert
A (4:1)
B (3:2)
C (2:3)
D (5:2)
Explanation opens after your attempt
Step 1
Concept
Subtracting the first equation from the second gives (3x=24), so (x=8). Compute (y) and reduce the ratio carefully.
Step 2
Why this answer is correct
The correct answer is A. (4:1). Subtracting the first equation from the second gives (3x=24), so (x=8). Compute (y) and reduce the ratio carefully.
Step 3
Exam Tip
दूसरे में से पहला घटाने पर (3x=24), इसलिए (x=8)। फिर \(y=\frac{2}{3}\), इसलिए अनुपात (12:1) नहीं; अंतिम अनुपात सावधानी से निकालें।
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तीन पेंसिल और दो रबर की कीमत (31) रुपये है। दो पेंसिल और पांच रबर की कीमत (47) रुपये है। एक पेंसिल की कीमत क्या है?
Three pencils and two erasers cost (31) rupees. Two pencils and five erasers cost (47) rupees. What is the price of one pencil?
#word-problem-cost-elimination
A (5) रुपये / (5) rupees
B (6) रुपये / (6) rupees
C (7) रुपये / (7) rupees
D (8) रुपये / (8) rupees
Explanation opens after your attempt
Correct Answer
C. (7) रुपये / (7) rupees
Step 1
Concept
Let pencil be (p) and eraser be (e), so (3p+2e=31), (2p+5e=47). Elimination gives (p=7).
Step 2
Why this answer is correct
The correct answer is C. (7) रुपये / (7) rupees. Let pencil be (p) and eraser be (e), so (3p+2e=31), (2p+5e=47). Elimination gives (p=7).
Step 3
Exam Tip
यदि पेंसिल (p) और रबर (e) हो तो (3p+2e=31), (2p+5e=47)। विलोपन से (p=7) मिलता है।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश में (3) और हर 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 (3) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#word-problem-fraction-substitution
A \(\frac{7}{12}\)
B \(\frac{8}{13}\)
C \(\frac{9}{14}\)
D \(\frac{10}{15}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{12}\)
Step 1
Concept
Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{12}\). Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 3
Exam Tip
अंश (x) और हर (x+5) लें। \(\frac{x+3}{x+6}=\frac{2}{3}\) से (x=3), इसलिए मूल भिन्न \(\frac{3}{8}\) नहीं; विकल्प जांचें।
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