100 results found for "consecutive-integers" in Class 10.
दो क्रमागत धनात्मक पूर्णांकों का गुणनफल (552) है। वे पूर्णांक कौन-से हैं?
The product of two consecutive positive integers is (552). Which are the integers?
#quadratic equations
#word problems
#consecutive integers
A (22) और (23) / (22) and (23)
B (23) और (24) / (23) and (24)
C (24) और (25) / (24) and (25)
D (21) और (22) / (21) and (22)
Explanation opens after your attempt
Correct Answer
B. (23) और (24) / (23) and (24)
Step 1
Concept
Let the numbers be (x) and (x+1). From (x(x+1)=552), we get (x=23).
Step 2
Why this answer is correct
The correct answer is B. (23) और (24) / (23) and (24). Let the numbers be (x) and (x+1). From (x(x+1)=552), we get (x=23).
Step 3
Exam Tip
मान लें संख्याएँ (x) और (x+1) हैं। (x(x+1)=552) से (x=23) मिलता है।
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दो क्रमागत धनात्मक पूर्णांकों का गुणनफल (240) है। वे पूर्णांक कौन से हैं?
The product of two consecutive positive integers is (240). Which integers are they?
#quadratic equations
#consecutive integers
#application
A (14) और (15) / (14) and (15)
B (15) और (16) / (15) and (16)
C (16) और (17) / (16) and (17)
D (12) और (13) / (12) and (13)
Explanation opens after your attempt
Correct Answer
B. (15) और (16) / (15) and (16)
Step 1
Concept
Let the integers be (x) and (x+1). From (x(x+1)=240), we get (x=15).
Step 2
Why this answer is correct
The correct answer is B. (15) और (16) / (15) and (16). Let the integers be (x) and (x+1). From (x(x+1)=240), we get (x=15).
Step 3
Exam Tip
मान लें पूर्णांक (x) और (x+1) हैं। (x(x+1)=240) से (x=15) मिलता है।
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दो क्रमागत विषम धनात्मक पूर्णांकों का गुणनफल (483) है। बड़ी संख्या क्या है?
The product of two consecutive positive odd integers is (483). What is the larger integer?
#quadratic equations
#consecutive odd integers
#product
A (21)
B (23)
C (25)
D (27)
Explanation opens after your attempt
Step 1
Concept
The integers are (x) and (x+2). From (x(x+2)=483), (x=21), so the larger integer is (23).
Step 2
Why this answer is correct
The correct answer is B. (23). The integers are (x) and (x+2). From (x(x+2)=483), (x=21), so the larger integer is (23).
Step 3
Exam Tip
संख्याएँ (x) और (x+2) हैं। (x(x+2)=483) से (x=21) मिलता है इसलिए बड़ी संख्या (23) है।
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दो क्रमागत सम धनात्मक पूर्णांकों का गुणनफल (360) है। छोटी संख्या क्या है?
The product of two consecutive positive even integers is (360). What is the smaller integer?
#quadratic equations
#consecutive even integers
#product
A (16)
B (18)
C (20)
D (22)
Explanation opens after your attempt
Step 1
Concept
Let the integers be (x) and (x+2). From (x(x+2)=360), \(x^2+2x-360=0\), so (x=18).
Step 2
Why this answer is correct
The correct answer is B. (18). Let the integers be (x) and (x+2). From (x(x+2)=360), \(x^2+2x-360=0\), so (x=18).
Step 3
Exam Tip
संख्याएँ (x) और (x+2) मानें। (x(x+2)=360) से \(x^2+2x-360=0\) और (x=18) मिलता है।
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तीन लगातार धनात्मक पूर्णांकों में पहले और तीसरे का गुणनफल (195) है। बीच वाला पूर्णांक क्या है?
In three consecutive positive integers, the product of the first and the third is (195). What is the middle integer?
#quadratic-equations
#word-problems
#three-consecutive-integers
A (14)
B (13)
C (15)
D (12)
Explanation opens after your attempt
Step 1
Concept
If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=195), (x=14).
Step 2
Why this answer is correct
The correct answer is A. (14). If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=195), (x=14).
Step 3
Exam Tip
यदि बीच वाला पूर्णांक (x) है, तो पहले और तीसरे (x-1) तथा (x+1) होंगे। ((x-1)(x+1)=195) से (x=14)।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (306) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive integers is (306). What is the larger integer?
#quadratic-equations
#word-problems
#consecutive-integers
A (18)
B (17)
C (19)
D (16)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x(x+1)=306). When (x=17), the larger integer is (18).
Step 2
Why this answer is correct
The correct answer is A. (18). If the smaller integer is (x), then (x(x+1)=306). When (x=17), the larger integer is (18).
Step 3
Exam Tip
यदि छोटा पूर्णांक (x) है, तो (x(x+1)=306)। (x=17) होने पर बड़ा पूर्णांक (18) है।
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तीन लगातार धनात्मक पूर्णांकों में पहले और तीसरे का गुणनफल (143) है। बीच वाला पूर्णांक क्या है?
In three consecutive positive integers, the product of the first and the third is (143). What is the middle integer?
#quadratic-equations
#word-problems
#three-consecutive-integers
A (12)
B (11)
C (13)
D (10)
Explanation opens after your attempt
Step 1
Concept
If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=143), (x=12).
Step 2
Why this answer is correct
The correct answer is A. (12). If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=143), (x=12).
Step 3
Exam Tip
यदि बीच वाला पूर्णांक (x) है, तो पहले और तीसरे (x-1) तथा (x+1) होंगे। ((x-1)(x+1)=143) से (x=12)।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (156) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive integers is (156). What is the larger integer?
#quadratic-equations
#word-problems
#consecutive-integers
A (13)
B (12)
C (14)
D (11)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x(x+1)=156). When (x=12), the larger integer is (13).
Step 2
Why this answer is correct
The correct answer is A. (13). If the smaller integer is (x), then (x(x+1)=156). When (x=12), the larger integer is (13).
Step 3
Exam Tip
यदि छोटा पूर्णांक (x) है, तो (x(x+1)=156)। (x=12) मिलने पर बड़ा पूर्णांक (13) है।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (380) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive integers is (380). What is the larger integer?
#quadratic equations
#consecutive integers
#application
A (18)
B (19)
C (20)
D (21)
Explanation opens after your attempt
Step 1
Concept
The numbers are (x) and (x+1). From (x(x+1)=380), (x=19), so the larger integer is (20).
Step 2
Why this answer is correct
The correct answer is C. (20). The numbers are (x) and (x+1). From (x(x+1)=380), (x=19), so the larger integer is (20).
Step 3
Exam Tip
संख्याएँ (x) और (x+1) होंगी। (x(x+1)=380) से (x=19), इसलिए बड़ा पूर्णांक (20) है।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (306) है। छोटा पूर्णांक क्या है?
The product of two consecutive positive integers is (306). What is the smaller integer?
#quadratic equations
#word problems
#consecutive integers
A (16)
B (17)
C (18)
D (19)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x(x+1)=306), giving (x=17). Write consecutive integers as (x) and (x+1).
Step 2
Why this answer is correct
The correct answer is B. (17). If the smaller integer is (x), then (x(x+1)=306), giving (x=17). Write consecutive integers as (x) and (x+1).
Step 3
Exam Tip
छोटा पूर्णांक (x) हो तो (x(x+1)=306), जिससे (x=17) मिलता है। लगातार पूर्णांकों को (x) और (x+1) लिखें।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (210) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive integers is (210). What is the larger integer?
#quadratic equations
#consecutive integers
#application
A (13)
B (14)
C (15)
D (16)
Explanation opens after your attempt
Step 1
Concept
Let the numbers be (x) and (x+1), then (x(x+1)=210) gives (x=14). So the larger integer is (15).
Step 2
Why this answer is correct
The correct answer is C. (15). Let the numbers be (x) and (x+1), then (x(x+1)=210) gives (x=14). So the larger integer is (15).
Step 3
Exam Tip
संख्याएँ (x) और (x+1) मानें, तब (x(x+1)=210) से (x=14) है। इसलिए बड़ा पूर्णांक (15) है।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (182) है। छोटा पूर्णांक क्या है?
The product of two consecutive positive integers is (182). What is the smaller integer?
#quadratic equations
#word problems
#consecutive integers
A (12)
B (13)
C (14)
D (15)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x(x+1)=182), which gives (x=13). Writing consecutive integers as (x) and (x+1) is the correct method.
Step 2
Why this answer is correct
The correct answer is B. (13). If the smaller integer is (x), then (x(x+1)=182), which gives (x=13). Writing consecutive integers as (x) and (x+1) is the correct method.
Step 3
Exam Tip
छोटा पूर्णांक (x) हो तो (x(x+1)=182), जिससे (x=13) मिलता है। लगातार पूर्णांकों को (x) और (x+1) लिखना सही तरीका है।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (132) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive integers is (132). What is the larger integer?
#quadratic equations
#word problems
#consecutive integers
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
Let the numbers be (x) and (x+1), then (x(x+1)=132) gives (x=11). So the larger integer is (12).
Step 2
Why this answer is correct
The correct answer is C. (12). Let the numbers be (x) and (x+1), then (x(x+1)=132) gives (x=11). So the larger integer is (12).
Step 3
Exam Tip
मान लें संख्याएँ (x) और (x+1) हैं, तब (x(x+1)=132) से (x=11) है। इसलिए बड़ा पूर्णांक (12) है।
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दो लगातार धनात्मक पूर्णांकों का गुणनफल (72) है। छोटा पूर्णांक क्या है?
The product of two consecutive positive integers is (72). What is the smaller integer?
#quadratic equations
#word problems
#consecutive integers
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
Let the smaller integer be (x), then (x(x+1)=72) gives (x=8). In exams, write consecutive numbers as (x) and (x+1).
Step 2
Why this answer is correct
The correct answer is B. (8). Let the smaller integer be (x), then (x(x+1)=72) gives (x=8). In exams, write consecutive numbers as (x) and (x+1).
Step 3
Exam Tip
मान लें छोटा पूर्णांक (x) है, तब (x(x+1)=72) से (x=8) मिलता है। परीक्षा में लगातार संख्याओं को (x) और (x+1) लिखें।
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दो क्रमागत धनात्मक पूर्णांकों के वर्गों का योग (145) है। यदि छोटा पूर्णांक (x) है, तो समीकरण कौन-सा है?
The sum of squares of two consecutive positive integers is (145). If the smaller integer is (x), which equation is correct?
#quadratic-equations
#consecutive-integers
#word-problem
#medium
A \(2x^2+2x-144=0\)
B \(2x^2+2x-145=0\)
C \(x^2+x-145=0\)
D \(2x^2-x-145=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+2x-144=0\)
Step 1
Concept
The integers are (x) and (x+1), so (x-2 +(x+1)2 =145). Simplifying gives \(2x^2+2x-144=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+2x-144=0\). The integers are (x) and (x+1), so (x-2 +(x+1)2 =145). Simplifying gives \(2x^2+2x-144=0\).
Step 3
Exam Tip
पूर्णांक (x) और (x+1) होंगे, इसलिए (x-2 +(x+1)2 =145)। सरल करने पर \(2x^2+2x-144=0\) मिलता है।
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दो क्रमागत पूर्णांकों के वर्गों का योग (85) है। यदि छोटा पूर्णांक (x) है, तो समीकरण कौन-सा है?
The sum of squares of two consecutive integers is (85). If the smaller integer is (x), which equation is correct?
#quadratic-equations
#consecutive-integers
#word-problem
#medium
A \(2x^2+2x-84=0\)
B \(2x^2+2x-85=0\)
C \(x^2+x-85=0\)
D \(2x^2-x-85=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+2x-84=0\)
Step 1
Concept
The integers are (x) and (x+1), so (x-2 +(x+1)2 =85). Simplifying gives \(2x^2+2x-84=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+2x-84=0\). The integers are (x) and (x+1), so (x-2 +(x+1)2 =85). Simplifying gives \(2x^2+2x-84=0\).
Step 3
Exam Tip
पूर्णांक (x) और (x+1) होंगे, इसलिए (x-2 +(x+1)2 =85)। सरल करने पर \(2x^2+2x-84=0\) मिलता है।
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दो क्रमागत धनात्मक पूर्णांकों का गुणनफल (56) है। सही द्विघात समीकरण कौन-सा है?
The product of two consecutive positive integers is (56). Which quadratic equation is correct?
#quadratic-equations
#consecutive-integers
#word-problem
#medium
A \(x^2+x-56=0\)
B \(x^2-x-56=0\)
C (2x+1=56)
D \(x^2+2x-56=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+x-56=0\)
Step 1
Concept
The numbers are (x) and (x+1), so (x(x+1)=56). This gives \(x^2+x-56=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x-56=0\). The numbers are (x) and (x+1), so (x(x+1)=56). This gives \(x^2+x-56=0\).
Step 3
Exam Tip
संख्याएँ (x) और (x+1) होंगी, इसलिए (x(x+1)=56)। इससे \(x^2+x-56=0\) मिलता है।
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दो लगातार धन पूर्णांकों का गुणनफल (30) है। यदि छोटा पूर्णांक (x) है तो समीकरण क्या होगा?
The product of two consecutive positive integers is (30). If the smaller integer is (x), what is the equation?
#quadratic equations
#word problem
#consecutive integers
A (x(x+1)=30)
B (x+x+1=30)
C (x(x-1)=30)
D (2x=30)
Explanation opens after your attempt
Correct Answer
A. (x(x+1)=30)
Step 1
Concept
The next consecutive integer is (x+1). Since the product is (30), the equation is (x(x+1)=30).
Step 2
Why this answer is correct
The correct answer is A. (x(x+1)=30). The next consecutive integer is (x+1). Since the product is (30), the equation is (x(x+1)=30).
Step 3
Exam Tip
लगातार अगला पूर्णांक (x+1) होगा। गुणनफल (30) है इसलिए (x(x+1)=30) बनेगा।
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बारह लगातार पूर्णांकों में से कम से कम एक संख्या 12 से विभाज्य क्यों होती है?
Why is at least one number among twelve consecutive integers divisible by 12?
#euclids-division-lemma
#consecutive-integers
#proof
A क्योंकि 12 से भाग देने पर शेषफल 0 से 11 तक चक्र में आते हैं / Because division by 12 gives remainders from 0 to 11 in a cycle
B क्योंकि सभी बारह संख्याएं 12 से विभाज्य होती हैं / Because all twelve numbers are divisible by 12
C क्योंकि उनका योग हमेशा 12 होता है / Because their sum is always 12
D क्योंकि हर तीसरी संख्या 12 से विभाज्य होती है / Because every third number is divisible by 12
Explanation opens after your attempt
Correct Answer
A. क्योंकि 12 से भाग देने पर शेषफल 0 से 11 तक चक्र में आते हैं / Because division by 12 gives remainders from 0 to 11 in a cycle
Step 1
Concept
On division by 12, possible remainders are from 0 to 11.
Step 2
Why this answer is correct
Twelve consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 12. चरण 1: 12 से भाग देने पर संभावित शेषफल 0 से 11 तक हैं। चरण 2: बारह लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 12 से विभाज्य होगी।
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ग्यारह लगातार पूर्णांकों में से कम से कम एक संख्या 11 से विभाज्य क्यों होती है?
Why is at least one number among eleven consecutive integers divisible by 11?
#euclids-division-lemma
#consecutive-integers
#divisibility-proof
A क्योंकि 11 से भाग देने पर शेषफल 0 से 10 तक चक्र में आते हैं / Because division by 11 gives remainders from 0 to 10 in a cycle
B क्योंकि सभी ग्यारह संख्याएं 11 से विभाज्य होती हैं / Because all eleven numbers are divisible by 11
C क्योंकि उनका योग हमेशा 11 होता है / Because their sum is always 11
D क्योंकि हर दूसरी संख्या 11 से विभाज्य होती है / Because every second number is divisible by 11
Explanation opens after your attempt
Correct Answer
A. क्योंकि 11 से भाग देने पर शेषफल 0 से 10 तक चक्र में आते हैं / Because division by 11 gives remainders from 0 to 10 in a cycle
Step 1
Concept
On division by 11, possible remainders are from 0 to 10.
Step 2
Why this answer is correct
Eleven consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 11. चरण 1: 11 से भाग देने पर संभावित शेषफल 0 से 10 तक होते हैं। चरण 2: ग्यारह लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 11 से विभाज्य होगी।
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दस लगातार पूर्णांकों में से कम से कम एक संख्या 10 से विभाज्य क्यों होती है?
Why is at least one number among ten consecutive integers divisible by 10?
#euclids-division-lemma
#consecutive-integers
#proof
A क्योंकि 10 से भाग देने पर शेषफल 0 से 9 तक चक्र में आते हैं / Because division by 10 gives remainders from 0 to 9 in a cycle
B क्योंकि सभी दस संख्याएं 10 से विभाज्य होती हैं / Because all ten numbers are divisible by 10
C क्योंकि उनका योग हमेशा 10 होता है / Because their sum is always 10
D क्योंकि हर दूसरी संख्या 10 से विभाज्य होती है / Because every second number is divisible by 10
Explanation opens after your attempt
Correct Answer
A. क्योंकि 10 से भाग देने पर शेषफल 0 से 9 तक चक्र में आते हैं / Because division by 10 gives remainders from 0 to 9 in a cycle
Step 1
Concept
On division by 10, possible remainders are from 0 to 9.
Step 2
Why this answer is correct
Ten consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 10. चरण 1: 10 से भाग देने पर संभावित शेषफल 0 से 9 तक हैं। चरण 2: दस लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 10 से विभाज्य होगी।
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नौ लगातार पूर्णांकों में से कम से कम एक संख्या 9 से विभाज्य क्यों होती है?
Why is at least one number among nine consecutive integers divisible by 9?
#euclids-division-lemma
#consecutive-integers
#divisibility-proof
A क्योंकि 9 से भाग देने पर शेषफल 0 से 8 तक चक्र में आते हैं / Because division by 9 gives remainders from 0 to 8 in a cycle
B क्योंकि सभी नौ संख्याएं 9 से विभाज्य होती हैं / Because all nine numbers are divisible by 9
C क्योंकि उनका योग हमेशा 9 होता है / Because their sum is always 9
D क्योंकि हर तीसरी संख्या 9 से विभाज्य होती है / Because every third number is divisible by 9
Explanation opens after your attempt
Correct Answer
A. क्योंकि 9 से भाग देने पर शेषफल 0 से 8 तक चक्र में आते हैं / Because division by 9 gives remainders from 0 to 8 in a cycle
Step 1
Concept
On division by 9, possible remainders are from 0 to 8.
Step 2
Why this answer is correct
Nine consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 9. चरण 1: 9 से भाग देने पर संभावित शेषफल 0 से 8 तक हैं। चरण 2: नौ लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 9 से विभाज्य होगी।
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आठ लगातार पूर्णांकों में से कम से कम एक संख्या 8 से विभाज्य क्यों होती है?
Why is at least one number among eight consecutive integers divisible by 8?
#euclids-division-lemma
#consecutive-integers
#proof
A क्योंकि 8 से भाग देने पर शेषफल 0 से 7 तक चक्र में आते हैं / Because division by 8 gives remainders from 0 to 7 in a cycle
B क्योंकि सभी आठ संख्याएं 8 से विभाज्य होती हैं / Because all eight numbers are divisible by 8
C क्योंकि उनका गुणनफल 8 होता है / Because their product is 8
D क्योंकि हर तीसरी संख्या 8 से विभाज्य होती है / Because every third number is divisible by 8
Explanation opens after your attempt
Correct Answer
A. क्योंकि 8 से भाग देने पर शेषफल 0 से 7 तक चक्र में आते हैं / Because division by 8 gives remainders from 0 to 7 in a cycle
Step 1
Concept
On division by 8, possible remainders are from 0 to 7.
Step 2
Why this answer is correct
Eight consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 8. चरण 1: 8 से भाग देने पर संभावित शेषफल 0 से 7 तक होते हैं। चरण 2: आठ लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वह 8 से विभाज्य होती है।
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सात लगातार पूर्णांकों में से कम से कम एक संख्या 7 से विभाज्य क्यों होती है?
Why is at least one number among seven consecutive integers divisible by 7?
#euclids-division-lemma
#consecutive-integers
#divisibility-proof
A क्योंकि 7 से भाग देने पर शेषफल 0 से 6 तक चक्र में आते हैं / Because division by 7 gives remainders from 0 to 6 in a cycle
B क्योंकि सभी सात संख्याएं 7 से विभाज्य होती हैं / Because all seven numbers are divisible by 7
C क्योंकि उनका योग हमेशा 7 होता है / Because their sum is always 7
D क्योंकि हर दूसरी संख्या 7 से विभाज्य होती है / Because every second number is divisible by 7
Explanation opens after your attempt
Correct Answer
A. क्योंकि 7 से भाग देने पर शेषफल 0 से 6 तक चक्र में आते हैं / Because division by 7 gives remainders from 0 to 6 in a cycle
Step 1
Concept
The possible remainders on division by 7 are 0, 1, 2, 3, 4, 5, and 6.
Step 2
Why this answer is correct
Seven consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 7. चरण 1: 7 से भाग देने पर संभावित शेषफल 0, 1, 2, 3, 4, 5, 6 हैं। चरण 2: सात लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 7 से विभाज्य होगी।
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छह लगातार पूर्णांकों में से कम से कम एक संख्या 6 से विभाज्य क्यों होती है?
Why is at least one number among six consecutive integers divisible by 6?
#euclids-division-lemma
#consecutive-integers
#proof
A क्योंकि 6 से भाग देने पर शेषफल 0, 1, 2, 3, 4, 5 का चक्र आता है / Because division by 6 gives the cycle of remainders 0, 1, 2, 3, 4, 5
B क्योंकि सभी छह संख्याएं 6 से विभाज्य होती हैं / Because all six numbers are divisible by 6
C क्योंकि उनका योग हमेशा 6 होता है / Because their sum is always 6
D क्योंकि हर दूसरी संख्या 6 से विभाज्य होती है / Because every second number is divisible by 6
Explanation opens after your attempt
Correct Answer
A. क्योंकि 6 से भाग देने पर शेषफल 0, 1, 2, 3, 4, 5 का चक्र आता है / Because division by 6 gives the cycle of remainders 0, 1, 2, 3, 4, 5
Step 1
Concept
On division by 6, possible remainders are from 0 to 5.
Step 2
Why this answer is correct
Six consecutive integers cover all these remainders once.
Step 3
Exam Tip
The number with remainder 0 is divisible by 6. चरण 1: 6 से भाग देने पर शेषफल 0 से 5 तक हो सकते हैं। चरण 2: छह लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वह 6 से विभाज्य होगी।
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पांच लगातार पूर्णांकों में से कम से कम एक संख्या 5 से विभाज्य क्यों होती है?
Why is at least one number among five consecutive integers divisible by 5?
#euclids-division-lemma
#consecutive-integers
#divisibility-proof
A क्योंकि 5 से भाग देने पर शेषफल 0, 1, 2, 3, 4 का चक्र आता है / Because division by 5 gives the cycle of remainders 0, 1, 2, 3, 4
B क्योंकि सभी पांच संख्याएं 5 से विभाज्य होती हैं / Because all five numbers are divisible by 5
C क्योंकि उनका गुणनफल 5 होता है / Because their product is 5
D क्योंकि हर पूर्णांक 5 से विभाज्य होता है / Because every integer is divisible by 5
Explanation opens after your attempt
Correct Answer
A. क्योंकि 5 से भाग देने पर शेषफल 0, 1, 2, 3, 4 का चक्र आता है / Because division by 5 gives the cycle of remainders 0, 1, 2, 3, 4
Step 1
Concept
Any integer divided by 5 has one of the forms (5q), (5q+1), (5q+2), (5q+3), or (5q+4).
Step 2
Why this answer is correct
Five consecutive integers cover all five remainders.
Step 3
Exam Tip
The number with remainder 0 is divisible by 5. चरण 1: कोई भी पूर्णांक 5 से भाग देने पर (5q), (5q+1), (5q+2), (5q+3), या (5q+4) रूप में होता है। चरण 2: पांच लगातार पूर्णांकों में ये पांचों शेषफल आ जाते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 5 से विभाज्य होगी।
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चार लगातार पूर्णांकों में से कम से कम एक संख्या 4 से विभाज्य क्यों होती है?
Why is at least one number among four consecutive integers divisible by 4?
#euclids-division-lemma
#consecutive-integers
#proof
A क्योंकि 4 से भाग देने पर शेषफल 0, 1, 2, 3 का चक्र आता है / Because division by 4 gives the cycle of remainders 0, 1, 2, 3
B क्योंकि सभी चार संख्याएं 4 से विभाज्य होती हैं / Because all four numbers are divisible by 4
C क्योंकि उनका योग 4 होता है / Because their sum is 4
D क्योंकि हर दूसरी संख्या 4 से विभाज्य होती है / Because every second number is divisible by 4
Explanation opens after your attempt
Correct Answer
A. क्योंकि 4 से भाग देने पर शेषफल 0, 1, 2, 3 का चक्र आता है / Because division by 4 gives the cycle of remainders 0, 1, 2, 3
Step 1
Concept
Any integer divided by 4 is of the form (4q), (4q+1), (4q+2), or (4q+3).
Step 2
Why this answer is correct
Four consecutive integers cover all these four remainders.
Step 3
Exam Tip
The one with remainder 0 is divisible by 4. चरण 1: कोई भी पूर्णांक 4 से भाग देने पर (4q), (4q+1), (4q+2), या (4q+3) रूप में होता है। चरण 2: चार लगातार पूर्णांकों में ये चारों शेषफल आ जाते हैं। चरण 3: जिस संख्या का शेषफल 0 है, वही 4 से विभाज्य होगी।
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तीन लगातार पूर्णांकों का गुणनफल 6 से विभाज्य क्यों होता है?
Why is the product of three consecutive integers divisible by 6?
#euclids-division-lemma
#consecutive-integers
#divisibility-proof
A क्योंकि उनमें एक 2 से और एक 3 से विभाज्य होता है / Because one of them is divisible by 2 and one is divisible by 3
B क्योंकि तीनों 6 से विभाज्य होते हैं / Because all three are divisible by 6
C क्योंकि उनका योग 6 होता है / Because their sum is 6
D क्योंकि हर पूर्णांक 6 से विभाज्य होता है / Because every integer is divisible by 6
Explanation opens after your attempt
Correct Answer
A. क्योंकि उनमें एक 2 से और एक 3 से विभाज्य होता है / Because one of them is divisible by 2 and one is divisible by 3
Step 1
Concept
Among two consecutive integers, one is even, so a factor 2 is present.
Step 2
Why this answer is correct
Among three consecutive integers, one is divisible by 3, so a factor 3 is present.
Step 3
Exam Tip
Since 2 and 3 together make 6, the product is divisible by 6. चरण 1: दो लगातार पूर्णांकों में एक सम होता है, इसलिए 2 का गुणनखंड मिलता है। चरण 2: तीन लगातार पूर्णांकों में एक 3 से विभाज्य होता है, इसलिए 3 का गुणनखंड मिलता है। चरण 3: 2 और 3 मिलकर 6 बनाते हैं, इसलिए गुणनफल 6 से विभाज्य है।
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तीन लगातार पूर्णांकों में से किसी एक का रूप (3q) क्यों माना जा सकता है?
Why can one of three consecutive integers be taken as of the form (3q)?
#euclids-division-lemma
#consecutive-integers
#proof
A क्योंकि 3 से भाग देने पर शेषफल 0, 1, 2 में से एक होता है / Because division by 3 gives one of the remainders 0, 1, 2
B क्योंकि हर संख्या 3 से विभाजित होती है / Because every number is divisible by 3
C क्योंकि शेषफल हमेशा 3 होता है / Because the remainder is always 3
D क्योंकि भागफल हमेशा 3 होता है / Because the quotient is always 3
Explanation opens after your attempt
Correct Answer
A. क्योंकि 3 से भाग देने पर शेषफल 0, 1, 2 में से एक होता है / Because division by 3 gives one of the remainders 0, 1, 2
Step 1
Concept
On division by 3, every integer is of the form (3q), (3q+1), or (3q+2).
Step 2
Why this answer is correct
Three consecutive integers cover these three remainders, so one is exactly divisible by 3.
Step 3
Exam Tip
Use the cycle of remainders for consecutive-number problems. चरण 1: 3 से भाग देने पर हर संख्या (3q), (3q+1), या (3q+2) रूप में होगी। चरण 2: तीन लगातार संख्याओं में ये तीनों शेषफल आते हैं, इसलिए एक संख्या 3 से पूर्णतः विभाजित होगी। चरण 3: लगातार संख्याओं में शेषफल चक्र का उपयोग करें।
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यदि (x-2 -(2a+3)x+\(a^2+3a+2\)=0) की जड़ें लगातार पूर्णांक हैं, तो वे कौन-सी हैं?
If the roots of (x-2 -(2a+3)x+\(a^2+3a+2\)=0) are consecutive integers, which are they?
#quadratic-roots
#consecutive-roots
#parametric
A (a+1) और (a+2) / (a+1) and (a+2)
B (a) और (a+3) / (a) and (a+3)
C (2a) और (3) / (2a) and (3)
D (a-1) और (a+4) / (a-1) and (a+4)
Explanation opens after your attempt
Correct Answer
A. (a+1) और (a+2) / (a+1) and (a+2)
Step 1
Concept
The product is (a-2 +3a+2=(a+1)(a+2)) and the sum is (2a+3). Hence the roots are (a+1) and (a+2).
Step 2
Why this answer is correct
The correct answer is A. (a+1) और (a+2) / (a+1) and (a+2). The product is (a-2 +3a+2=(a+1)(a+2)) and the sum is (2a+3). Hence the roots are (a+1) and (a+2).
Step 3
Exam Tip
गुणनफल (a-2 +3a+2=(a+1)(a+2)) है और योग (2a+3) है। इसलिए जड़ें (a+1) और (a+2) हैं।
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दो क्रमागत विषम धनात्मक पूर्णांकों के वर्गों का योग (394) है। छोटा पूर्णांक क्या है?
The sum of the squares of two consecutive positive odd integers is (394). What is the smaller integer?
#quadratic equations
#odd integers
#application
A (11)
B (13)
C (15)
D (17)
Explanation opens after your attempt
Step 1
Concept
Let the smaller odd integer be (x), then (x-2 +(x+2)2 =394). This gives \(x^2+2x-195=0\), so (x=13).
Step 2
Why this answer is correct
The correct answer is B. (13). Let the smaller odd integer be (x), then (x-2 +(x+2)2 =394). This gives \(x^2+2x-195=0\), so (x=13).
Step 3
Exam Tip
छोटा विषम पूर्णांक (x) हो, तो (x-2 +(x+2)2 =394)। इससे \(x^2+2x-195=0\), इसलिए (x=13)।
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तीन लगातार सम पूर्णांकों में पहले और तीसरे का गुणनफल (480) है। बीच वाला सम पूर्णांक क्या है?
In three consecutive even integers, the product of the first and the third is (480). What is the middle even integer?
#quadratic-equations
#word-problems
#even-integers
A (22)
B (20)
C (24)
D (18)
Explanation opens after your attempt
Step 1
Concept
If the middle one is (x), the first and third are (x-2) and (x+2). From ((x-2)(x+2)=480), (x=22).
Step 2
Why this answer is correct
The correct answer is A. (22). If the middle one is (x), the first and third are (x-2) and (x+2). From ((x-2)(x+2)=480), (x=22).
Step 3
Exam Tip
यदि बीच वाला (x) है, तो पहले और तीसरे (x-2) तथा (x+2) होंगे। ((x-2)(x+2)=480) से (x=22)।
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दो लगातार धनात्मक सम पूर्णांकों का गुणनफल (288) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive even integers is (288). What is the larger integer?
#quadratic-equations
#word-problems
#even-integers
A (18)
B (16)
C (20)
D (14)
Explanation opens after your attempt
Step 1
Concept
If the smaller even integer is (x), then (x(x+2)=288). Since (x=16), the larger integer is (18).
Step 2
Why this answer is correct
The correct answer is A. (18). If the smaller even integer is (x), then (x(x+2)=288). Since (x=16), the larger integer is (18).
Step 3
Exam Tip
यदि छोटा सम पूर्णांक (x) है, तो (x(x+2)=288)। (x=16) होने से बड़ा पूर्णांक (18) है।
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दो लगातार धनात्मक विषम पूर्णांकों का गुणनफल (399) है। छोटा पूर्णांक क्या है?
The product of two consecutive positive odd integers is (399). What is the smaller integer?
#quadratic-equations
#word-problems
#odd-integers
A (19)
B (21)
C (17)
D (23)
Explanation opens after your attempt
Step 1
Concept
If the smaller odd integer is (x), then (x(x+2)=399). Since \(19\times21=399\), the smaller integer is (19).
Step 2
Why this answer is correct
The correct answer is A. (19). If the smaller odd integer is (x), then (x(x+2)=399). Since \(19\times21=399\), the smaller integer is (19).
Step 3
Exam Tip
यदि छोटा विषम पूर्णांक (x) है, तो (x(x+2)=399)। \(19\times21=399\), इसलिए छोटा पूर्णांक (19) है।
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तीन लगातार सम पूर्णांकों में पहले और तीसरे का गुणनफल (320) है। बीच वाला सम पूर्णांक क्या है?
In three consecutive even integers, the product of the first and the third is (320). What is the middle even integer?
#quadratic-equations
#word-problems
#even-integers
A (18)
B (16)
C (20)
D (14)
Explanation opens after your attempt
Step 1
Concept
If the middle one is (x), the first and third are (x-2) and (x+2). From ((x-2)(x+2)=320), (x=18).
Step 2
Why this answer is correct
The correct answer is A. (18). If the middle one is (x), the first and third are (x-2) and (x+2). From ((x-2)(x+2)=320), (x=18).
Step 3
Exam Tip
यदि बीच वाला (x) है, तो पहले और तीसरे (x-2) तथा (x+2) होंगे। ((x-2)(x+2)=320) से (x=18)।
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दो लगातार धनात्मक सम पूर्णांकों का गुणनफल (168) है। बड़ा पूर्णांक क्या है?
The product of two consecutive positive even integers is (168). What is the larger integer?
#quadratic-equations
#word-problems
#even-integers
A (14)
B (12)
C (16)
D (10)
Explanation opens after your attempt
Step 1
Concept
If the smaller even integer is (x), then (x(x+2)=168). Since (x=12), the larger integer is (14).
Step 2
Why this answer is correct
The correct answer is A. (14). If the smaller even integer is (x), then (x(x+2)=168). Since (x=12), the larger integer is (14).
Step 3
Exam Tip
यदि छोटा सम पूर्णांक (x) है, तो (x(x+2)=168)। (x=12) होने से बड़ा पूर्णांक (14) है।
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दो लगातार धनात्मक विषम पूर्णांकों का गुणनफल (195) है। छोटा पूर्णांक क्या है?
The product of two consecutive positive odd integers is (195). What is the smaller integer?
#quadratic-equations
#word-problems
#odd-integers
A (13)
B (15)
C (11)
D (17)
Explanation opens after your attempt
Step 1
Concept
If the smaller odd integer is (x), then (x(x+2)=195). (x=13) is correct because \(13\times15=195\).
Step 2
Why this answer is correct
The correct answer is A. (13). If the smaller odd integer is (x), then (x(x+2)=195). (x=13) is correct because \(13\times15=195\).
Step 3
Exam Tip
यदि छोटा विषम पूर्णांक (x) है, तो (x(x+2)=195)। (x=13) सही है क्योंकि \(13\times15=195\)।
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संख्या रेखा पर \( \sqrt{131}+2 \) किस दो लगातार पूर्णांकों के बीच है?
Between which two consecutive integers is \( \sqrt{131}+2 \) on the number line?
#number-line
#root-expression
#integer-bounds
A (12) और (13) / (12) and (13)
B (13) और (14) / (13) and (14)
C (14) और (15) / (14) and (15)
D (11) और (12) / (11) and (12)
Explanation opens after your attempt
Correct Answer
B. (13) और (14) / (13) and (14)
Step 1
Concept
Since \(11<\sqrt{131}<12\), \(13<\sqrt{131}+2<14\). First find the bounds of the square root.
Step 2
Why this answer is correct
The correct answer is B. (13) और (14) / (13) and (14). Since \(11<\sqrt{131}<12\), \(13<\sqrt{131}+2<14\). First find the bounds of the square root.
Step 3
Exam Tip
\(11<\sqrt{131}<12\), इसलिए \(13<\sqrt{131}+2<14\)। पहले वर्गमूल की सीमा निकालें।
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संख्या रेखा पर \( \sqrt{126}-\sqrt{80} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers does \( \sqrt{126}-\sqrt{80} \) lie on the number line?
#number-line
#irrational-difference
#estimation
A (1) और (2) / (1) and (2)
B (2) और (3) / (2) and (3)
C (3) और (4) / (3) and (4)
D (0) और (1) / (0) and (1)
Explanation opens after your attempt
Correct Answer
B. (2) और (3) / (2) and (3)
Step 1
Concept
\( \sqrt{126}\approx11.22 \) and \( \sqrt{80}\approx8.94 \), so the difference is about (2.28). Estimate both square roots first.
Step 2
Why this answer is correct
The correct answer is B. (2) और (3) / (2) and (3). \( \sqrt{126}\approx11.22 \) and \( \sqrt{80}\approx8.94 \), so the difference is about (2.28). Estimate both square roots first.
Step 3
Exam Tip
\( \sqrt{126}\approx11.22 \) और \( \sqrt{80}\approx8.94 \), इसलिए अंतर लगभग (2.28) है। पहले दोनों वर्गमूलों का अनुमान करें।
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संख्या रेखा पर \( \sqrt{89}+1 \) किस दो लगातार पूर्णांकों के बीच है?
Between which two consecutive integers is \( \sqrt{89}+1 \) on the number line?
#number-line
#root-expression
#integer-bounds
A (9) और (10) / (9) and (10)
B (10) और (11) / (10) and (11)
C (11) और (12) / (11) and (12)
D (8) और (9) / (8) and (9)
Explanation opens after your attempt
Correct Answer
B. (10) और (11) / (10) and (11)
Step 1
Concept
Since \(9<\sqrt{89}<10\), \(10<\sqrt{89}+1<11\). First find the root bounds.
Step 2
Why this answer is correct
The correct answer is B. (10) और (11) / (10) and (11). Since \(9<\sqrt{89}<10\), \(10<\sqrt{89}+1<11\). First find the root bounds.
Step 3
Exam Tip
\(9<\sqrt{89}<10\), इसलिए \(10<\sqrt{89}+1<11\)। पहले मूल की सीमा निकालें।
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संख्या रेखा पर \( \sqrt{91}-\sqrt{55} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers does \( \sqrt{91}-\sqrt{55} \) lie on the number line?
#number-line
#irrational-difference
#estimation
A (1) और (2) / (1) and (2)
B (2) और (3) / (2) and (3)
C (0) और (1) / (0) and (1)
D (3) और (4) / (3) and (4)
Explanation opens after your attempt
Correct Answer
B. (2) और (3) / (2) and (3)
Step 1
Concept
\( \sqrt{91}\approx9.54 \) and \( \sqrt{55}\approx7.42 \), so the difference is about (2.12). Estimate both roots first.
Step 2
Why this answer is correct
The correct answer is B. (2) और (3) / (2) and (3). \( \sqrt{91}\approx9.54 \) and \( \sqrt{55}\approx7.42 \), so the difference is about (2.12). Estimate both roots first.
Step 3
Exam Tip
\( \sqrt{91}\approx9.54 \) और \( \sqrt{55}\approx7.42 \), इसलिए अंतर लगभग (2.12) है। पहले दोनों मूलों का अनुमान करें।
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संख्या रेखा पर \( \sqrt{6}+\frac{1}{3} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers is \( \sqrt{6}+\frac{1}{3} \) located on the number line?
#number-line
#irrational-expression
#estimation
A (1) और (2) / (1) and (2)
B (2) और (3) / (2) and (3)
C (3) और (4) / (3) and (4)
D (4) और (5) / (4) and (5)
Explanation opens after your attempt
Correct Answer
B. (2) और (3) / (2) and (3)
Step 1
Concept
\( \sqrt{6}\approx2.449 \) and \( \frac{1}{3}\approx0.333 \), so the sum is about (2.782). For mixed values, estimate first.
Step 2
Why this answer is correct
The correct answer is B. (2) और (3) / (2) and (3). \( \sqrt{6}\approx2.449 \) and \( \frac{1}{3}\approx0.333 \), so the sum is about (2.782). For mixed values, estimate first.
Step 3
Exam Tip
\( \sqrt{6}\approx2.449 \) और \( \frac{1}{3}\approx0.333 \), इसलिए योग लगभग (2.782) है। मिश्रित मानों में पहले अनुमान लगाएँ।
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यदि \(P=-\sqrt{80}\), तो (P) किस दो लगातार पूर्णांकों के बीच स्थित है?
If \(P=-\sqrt{80}\), between which two consecutive integers is (P) located?
#number-line
#negative-square-root
#integer-bounds
A ( -9 ) और ( -8 ) / ( -9 ) and ( -8 )
B ( -8 ) और ( -7 ) / ( -8 ) and ( -7 )
C (8) और (9) / (8) and (9)
D ( -10 ) और ( -9 ) / ( -10 ) and ( -9 )
Explanation opens after your attempt
Correct Answer
A. ( -9 ) और ( -8 ) / ( -9 ) and ( -8 )
Step 1
Concept
Since \(8<\sqrt{80}<9\), \(-9<-\sqrt{80}<-8\). Write intervals carefully for negative roots.
Step 2
Why this answer is correct
The correct answer is A. ( -9 ) और ( -8 ) / ( -9 ) and ( -8 ). Since \(8<\sqrt{80}<9\), \(-9<-\sqrt{80}<-8\). Write intervals carefully for negative roots.
Step 3
Exam Tip
क्योंकि \(8<\sqrt{80}<9\), इसलिए \(-9<-\sqrt{80}<-8\)। ऋणात्मक मूलों में अंतराल सावधानी से लिखें।
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संख्या रेखा पर \( \sqrt{52} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers is \( \sqrt{52} \) located on the number line?
#number-line
#square-roots
#integer-bounds
A (6) और (7) / (6) and (7)
B (7) और (8) / (7) and (8)
C (5) और (6) / (5) and (6)
D (8) और (9) / (8) and (9)
Explanation opens after your attempt
Correct Answer
B. (7) और (8) / (7) and (8)
Step 1
Concept
Since (49<52<64), \(7<\sqrt{52}<8\). First identify the nearest perfect squares.
Step 2
Why this answer is correct
The correct answer is B. (7) और (8) / (7) and (8). Since (49<52<64), \(7<\sqrt{52}<8\). First identify the nearest perfect squares.
Step 3
Exam Tip
क्योंकि (49<52<64), इसलिए \(7<\sqrt{52}<8\)। पहले निकटतम पूर्ण वर्गों को पहचानें।
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संख्या रेखा पर \( \sqrt{37}+1 \) किस दो लगातार पूर्णांकों के बीच है?
Between which two consecutive integers is \( \sqrt{37}+1 \) on the number line?
#number-line
#root-expression
#integer-bounds
A (7) और (8) / (7) and (8)
B (6) और (7) / (6) and (7)
C (8) और (9) / (8) and (9)
D (5) और (6) / (5) and (6)
Explanation opens after your attempt
Correct Answer
A. (7) और (8) / (7) and (8)
Step 1
Concept
Since \(6<\sqrt{37}<7\), \(7<\sqrt{37}+1<8\). First find the bounds of the square root.
Step 2
Why this answer is correct
The correct answer is A. (7) और (8) / (7) and (8). Since \(6<\sqrt{37}<7\), \(7<\sqrt{37}+1<8\). First find the bounds of the square root.
Step 3
Exam Tip
\(6<\sqrt{37}<7\) इसलिए \(7<\sqrt{37}+1<8\)। पहले वर्गमूल की सीमा निकालें।
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संख्या रेखा पर \( \sqrt{31}-\sqrt{12} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers does \( \sqrt{31}-\sqrt{12} \) lie on the number line?
#number-line
#irrational-difference
#estimation
A (2) और (3) / (2) and (3)
B (1) और (2) / (1) and (2)
C (3) और (4) / (3) and (4)
D (0) और (1) / (0) and (1)
Explanation opens after your attempt
Correct Answer
B. (1) और (2) / (1) and (2)
Step 1
Concept
\( \sqrt{31}\approx5.57 \) and \( \sqrt{12}\approx3.46 \) so the difference is about (2.11). Estimate both roots first.
Step 2
Why this answer is correct
The correct answer is B. (1) और (2) / (1) and (2). \( \sqrt{31}\approx5.57 \) and \( \sqrt{12}\approx3.46 \) so the difference is about (2.11). Estimate both roots first.
Step 3
Exam Tip
\( \sqrt{31}\approx5.57 \) और \( \sqrt{12}\approx3.46 \) इसलिए अंतर लगभग (2.11) है। पहले दोनों मूलों का अनुमान करें।
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संख्या रेखा पर \( \sqrt{5}+\frac{1}{2} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers is \( \sqrt{5}+\frac{1}{2} \) located on the number line?
#number-line
#irrational-expression
#estimation
A (2) और (3) / (2) and (3)
B (1) और (2) / (1) and (2)
C (3) और (4) / (3) and (4)
D (4) और (5) / (4) and (5)
Explanation opens after your attempt
Correct Answer
A. (2) और (3) / (2) and (3)
Step 1
Concept
Since \( \sqrt{5}\approx2.236\), \( \sqrt{5}+\frac{1}{2}\approx2.736\). Use estimation to identify the interval quickly.
Step 2
Why this answer is correct
The correct answer is A. (2) और (3) / (2) and (3). Since \( \sqrt{5}\approx2.236\), \( \sqrt{5}+\frac{1}{2}\approx2.736\). Use estimation to identify the interval quickly.
Step 3
Exam Tip
क्योंकि \( \sqrt{5}\approx2.236\), इसलिए \( \sqrt{5}+\frac{1}{2}\approx2.736\)। अनुमान लगाकर अंतराल जल्दी पहचानें।
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संख्या रेखा पर \( -\sqrt{7} \) किस दो लगातार पूर्णांकों के बीच स्थित है?
Between which two consecutive integers is \( -\sqrt{7} \) located on the number line?
#number-line
#negative-roots
#comparison
A ( -3) और (-2) / ( -3) and (-2)
B ( -2) और (-1) / ( -2) and (-1)
C (2) और (3) / (2) and (3)
D ( -4) और (-3) / ( -4) and (-3)
Explanation opens after your attempt
Correct Answer
A. ( -3) और (-2) / ( -3) and (-2)
Step 1
Concept
Since \(2<\sqrt{7}<3\), we have \(-3<-\sqrt{7}<-2\). For negative numbers, remember the order reverses.
Step 2
Why this answer is correct
The correct answer is A. ( -3) और (-2) / ( -3) and (-2). Since \(2<\sqrt{7}<3\), we have \(-3<-\sqrt{7}<-2\). For negative numbers, remember the order reverses.
Step 3
Exam Tip
क्योंकि \(2<\sqrt{7}<3\), इसलिए \(-3<-\sqrt{7}<-2\)। ऋणात्मक संख्या में क्रम उलटने पर ध्यान रखें।
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संख्या रेखा पर \( \sqrt{18} \) को सबसे सही किस दो लगातार पूर्णांकों के बीच दिखाया जाएगा?
Between which two consecutive integers should \( \sqrt{18} \) be shown most accurately on the number line?
#number-line
#real-numbers
#square-roots
A (4) और (5) / (4) and (5)
B (3) और (4) / (3) and (4)
C (5) और (6) / (5) and (6)
D (2) और (3) / (2) and (3)
Explanation opens after your attempt
Correct Answer
A. (4) और (5) / (4) and (5)
Step 1
Concept
Since (16<18<25), we get \(4<\sqrt{18}<5\). In exams, compare perfect squares first.
Step 2
Why this answer is correct
The correct answer is A. (4) और (5) / (4) and (5). Since (16<18<25), we get \(4<\sqrt{18}<5\). In exams, compare perfect squares first.
Step 3
Exam Tip
क्योंकि (16<18<25), इसलिए \(4<\sqrt{18}<5\)। परीक्षा में वर्गों की तुलना पहले करें।
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संख्या रेखा पर \(\frac{-11}{5}\) किस दो क्रमागत पूर्णांकों के बीच आएगा?
On the number line, between which two consecutive integers will \(\frac{-11}{5}\) lie?
#polynomials
#number-line
#negative-rational
#fraction
A (-3) और (-2) / (-3) and (-2)
B (-2) और (-1) / (-2) and (-1)
C (2) और (3) / (2) and (3)
D (-1) और (0) / (-1) and (0)
Explanation opens after your attempt
Correct Answer
A. (-3) और (-2) / (-3) and (-2)
Step 1
Concept
Since \(\frac{-11}{5}=-2.2\), it lies between (-3) and (-2). Be careful with position of negative decimals.
Step 2
Why this answer is correct
The correct answer is A. (-3) और (-2) / (-3) and (-2). Since \(\frac{-11}{5}=-2.2\), it lies between (-3) and (-2). Be careful with position of negative decimals.
Step 3
Exam Tip
\(\frac{-11}{5}=-2.2\), इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक दशमलव में स्थान का ध्यान रखें।
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संख्या रेखा पर \(\sqrt{26}\) किस दो क्रमागत पूर्णांकों के बीच होगा?
Between which two consecutive integers will \(\sqrt{26}\) lie on the number line?
#number-line
#irrational-numbers
#square-root
#comparison
A (4) और (5) / (4) and (5)
B (5) और (6) / (5) and (6)
C (6) और (7) / (6) and (7)
D (7) और (8) / (7) and (8)
Explanation opens after your attempt
Correct Answer
B. (5) और (6) / (5) and (6)
Step 1
Concept
Since \(5^2<26<6^2\), \(\sqrt{26}\) lies between (5) and (6). Check nearby perfect squares.
Step 2
Why this answer is correct
The correct answer is B. (5) और (6) / (5) and (6). Since \(5^2<26<6^2\), \(\sqrt{26}\) lies between (5) and (6). Check nearby perfect squares.
Step 3
Exam Tip
क्योंकि \(5^2<26<6^2\), इसलिए \(\sqrt{26}\) (5) और (6) के बीच होगा। पास के पूर्ण वर्ग देखें।
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संख्या रेखा पर \(\sqrt{10}\) किस दो क्रमागत पूर्णांकों के बीच स्थित होगा?
Between which two consecutive integers will \(\sqrt{10}\) be located on the number line?
#number-line
#square-root
#irrational-numbers
#comparison
A (2) और (3) / (2) and (3)
B (3) और (4) / (3) and (4)
C (4) और (5) / (4) and (5)
D (5) और (6) / (5) and (6)
Explanation opens after your attempt
Correct Answer
B. (3) और (4) / (3) and (4)
Step 1
Concept
Since \(3^2<10<4^2\), \(\sqrt{10}\) lies between (3) and (4). Use nearby perfect squares to locate roots.
Step 2
Why this answer is correct
The correct answer is B. (3) और (4) / (3) and (4). Since \(3^2<10<4^2\), \(\sqrt{10}\) lies between (3) and (4). Use nearby perfect squares to locate roots.
Step 3
Exam Tip
क्योंकि \(3^2<10<4^2\), इसलिए \(\sqrt{10}\), (3) और (4) के बीच होगा। वर्गमूल की स्थिति के लिए पास के पूर्ण वर्ग देखें।
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संख्या रेखा पर \(\sqrt{5}\) किस दो क्रमागत पूर्णांकों के बीच होगा?
Between which two consecutive integers will \(\sqrt{5}\) lie on the number line?
#number-line
#irrational-numbers
#square-root
#easy
A (1) और (2) / (1) and (2)
B (2) और (3) / (2) and (3)
C (3) और (4) / (3) and (4)
D (4) और (5) / (4) and (5)
Explanation opens after your attempt
Correct Answer
B. (2) और (3) / (2) and (3)
Step 1
Concept
Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3). Use squares to locate square roots quickly.
Step 2
Why this answer is correct
The correct answer is B. (2) और (3) / (2) and (3). Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3). Use squares to locate square roots quickly.
Step 3
Exam Tip
क्योंकि \(2^2<5<3^2\), इसलिए \(\sqrt{5}\), (2) और (3) के बीच होगा। वर्गों से वर्गमूल की स्थिति जल्दी मिलती है।
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दो लगातार धनात्मक पूर्णांकों के वर्गों का योग (1513) है। छोटा पूर्णांक क्या है?
The sum of squares of two consecutive positive integers is (1513). What is the smaller integer?
#quadratic equations
#word problems
#sum of squares
A (25)
B (26)
C (27)
D (28)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x-2 +(x+1)2 =1513), giving (x=27). Write consecutive integers as (x) and (x+1).
Step 2
Why this answer is correct
The correct answer is C. (27). If the smaller integer is (x), then (x-2 +(x+1)2 =1513), giving (x=27). Write consecutive integers as (x) and (x+1).
Step 3
Exam Tip
छोटा पूर्णांक (x) हो तो (x-2 +(x+1)2 =1513), जिससे (x=27) मिलता है। लगातार पूर्णांकों को (x) और (x+1) लिखें।
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दो लगातार धनात्मक पूर्णांकों के वर्गों का योग (1201) है। छोटा पूर्णांक क्या है?
The sum of squares of two consecutive positive integers is (1201). What is the smaller integer?
#quadratic equations
#word problems
#sum of squares
A (22)
B (23)
C (24)
D (25)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x-2 +(x+1)2 =1201), giving (x=24). Write consecutive integers as (x) and (x+1).
Step 2
Why this answer is correct
The correct answer is C. (24). If the smaller integer is (x), then (x-2 +(x+1)2 =1201), giving (x=24). Write consecutive integers as (x) and (x+1).
Step 3
Exam Tip
छोटा पूर्णांक (x) हो तो (x-2 +(x+1)2 =1201), जिससे (x=24) मिलता है। लगातार पूर्णांकों को (x) और (x+1) लिखें।
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दो लगातार धनात्मक पूर्णांकों के वर्गों का योग (841) है। छोटा पूर्णांक क्या है?
The sum of squares of two consecutive positive integers is (841). What is the smaller integer?
#quadratic equations
#word problems
#sum of squares
A (18)
B (19)
C (20)
D (21)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x-2 +(x+1)2 =841), giving (x=20). Write consecutive numbers as (x) and (x+1).
Step 2
Why this answer is correct
The correct answer is C. (20). If the smaller integer is (x), then (x-2 +(x+1)2 =841), giving (x=20). Write consecutive numbers as (x) and (x+1).
Step 3
Exam Tip
छोटा पूर्णांक (x) हो तो (x-2 +(x+1)2 =841), जिससे (x=20) मिलता है। लगातार संख्याओं को (x) और (x+1) लिखें।
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दो लगातार धनात्मक पूर्णांकों के वर्गों का योग (365) है। छोटा पूर्णांक क्या है?
The sum of squares of two consecutive positive integers is (365). What is the smaller integer?
#quadratic equations
#word problems
#sum of squares
A (11)
B (12)
C (13)
D (14)
Explanation opens after your attempt
Step 1
Concept
If the smaller integer is (x), then (x-2 +(x+1)2 =365), giving (x=13). Write consecutive integers as (x) and (x+1).
Step 2
Why this answer is correct
The correct answer is C. (13). If the smaller integer is (x), then (x-2 +(x+1)2 =365), giving (x=13). Write consecutive integers as (x) and (x+1).
Step 3
Exam Tip
छोटा पूर्णांक (x) हो तो (x-2 +(x+1)2 =365), जिससे (x=13) मिलता है। लगातार पूर्णांकों को (x) और (x+1) लिखें।
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यदि (x-2 -(2k+5)x+\(k^2+5k+6\)=0) के मूल लगातार पूर्णांक हैं, तो वे मूल कौन-से होंगे?
If the roots of (x-2 -(2k+5)x+\(k^2+5k+6\)=0) are consecutive integers, what will those roots be?
#quadratic-equations
#roots
#parameter-pattern
#expert
A (k+2) और (k+3) / (k+2) and (k+3)
B (k+1) और (k+4) / (k+1) and (k+4)
C (k-2) और (k+3) / (k-2) and (k+3)
D (k) और (k+5) / (k) and (k+5)
Explanation opens after your attempt
Correct Answer
A. (k+2) और (k+3) / (k+2) and (k+3)
Step 1
Concept
The sum of roots is (2k+5) and the product is \(k^2+5k+6\). ((k+2)+(k+3)=2k+5) and ((k+2)(k+3)=k-2 +5k+6) are correct.
Step 2
Why this answer is correct
The correct answer is A. (k+2) और (k+3) / (k+2) and (k+3). The sum of roots is (2k+5) and the product is \(k^2+5k+6\). ((k+2)+(k+3)=2k+5) and ((k+2)(k+3)=k-2 +5k+6) are correct.
Step 3
Exam Tip
मूलों का योग (2k+5) और गुणनफल \(k^2+5k+6\) है। ((k+2)+(k+3)=2k+5) और ((k+2)(k+3)=k-2 +5k+6) सही है।
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यदि (x-2 -(2k+3)x+\(k^2+3k\)=0) के मूल लगातार धनात्मक पूर्णांक हैं, तो (k) का मान क्या होगा?
If the roots of (x-2 -(2k+3)x+\(k^2+3k\)=0) are consecutive positive integers, what is the value of (k)?
#quadratic-equations
#parameter
#roots-pattern
#expert
A (3)
B (2)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
The sum is (2k+3) and the product is \(k^2+3k\). Direct checking with the options shows (k=3) gives the required equation pattern.
Step 2
Why this answer is correct
The correct answer is A. (3). The sum is (2k+3) and the product is \(k^2+3k\). Direct checking with the options shows (k=3) gives the required equation pattern.
Step 3
Exam Tip
मूलों का योग (2k+3) और गुणनफल \(k^2+3k\) है। ये (k) और (k+3) नहीं बल्कि (k) तथा (k+3) जैसे नहीं बनते, जाँच से (k=3) पर मूल (3) और (6) नहीं इसलिए सही विकल्प नहीं है।
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यदि (5,11,n,23) अंकगणितीय श्रेणी के क्रमागत पद हैं तो (n) क्या है?
If (5,11,n,23) are consecutive terms of an arithmetic progression, what is (n)?
#ap
#missing term
#consecutive terms
#medium
A (15)
B (16)
C (17)
D (18)
Explanation opens after your attempt
Step 1
Concept
The common difference is (11-5=6), so (n=11+6=17). It is easy to find (d) from the first two terms.
Step 2
Why this answer is correct
The correct answer is C. (17). The common difference is (11-5=6), so (n=11+6=17). It is easy to find (d) from the first two terms.
Step 3
Exam Tip
सार्व अंतर (11-5=6) है इसलिए (n=11+6=17)। पहले दो पदों से (d) निकालना आसान होता है।
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यदि किसी अंकगणितीय श्रेणी में लगातार पद (27) और (35) हैं तो (d) क्या है?
If two consecutive terms in an arithmetic progression are (27) and (35), what is (d)?
#ap
#consecutive terms
#find d
#medium
A (8)
B (27)
C (35)
D (62)
Explanation opens after your attempt
Step 1
Concept
(d=35-27=8). When consecutive terms are given, subtract the earlier term from the later term.
Step 2
Why this answer is correct
The correct answer is A. (8). (d=35-27=8). When consecutive terms are given, subtract the earlier term from the later term.
Step 3
Exam Tip
(d=35-27=8) है। लगातार पद दिए हों तो बाद वाले पद में से पहले वाला पद घटाएं।
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यदि किसी समांतर श्रेढ़ी के लगातार दो पद (18) और (25) हैं, तो सार्व अंतर क्या है?
If two consecutive terms of an arithmetic progression are (18) and (25), what is the common difference?
#arithmetic-progression
#consecutive-terms
#class10
A (5)
B (6)
C (7)
D (43)
Explanation opens after your attempt
Step 1
Concept
The common difference is (25-18=7). When consecutive terms are given, subtract directly.
Step 2
Why this answer is correct
The correct answer is C. (7). The common difference is (25-18=7). When consecutive terms are given, subtract directly.
Step 3
Exam Tip
सार्व अंतर (25-18=7) है। लगातार पद दिए हों तो सीधे घटाव करें।
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यदि (6,10,n,18) अंकगणितीय श्रेणी के क्रमागत पद हैं तो (n) क्या है?
If (6,10,n,18) are consecutive terms of an arithmetic progression, what is (n)?
#ap
#missing term
#consecutive terms
#medium
A (12)
B (13)
C (15)
D (14)
Explanation opens after your attempt
Step 1
Concept
The common difference is (10-6=4), so (n=10+4=14). It is easy to find (d) from the first two terms.
Step 2
Why this answer is correct
The correct answer is D. (14). The common difference is (10-6=4), so (n=10+4=14). It is easy to find (d) from the first two terms.
Step 3
Exam Tip
सार्व अंतर (10-6=4) है इसलिए (n=10+4=14)। पहले दो पदों से (d) निकालना आसान होता है।
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यदि किसी अंकगणितीय श्रेणी में लगातार पद (12) और (17) हैं तो (d) क्या है?
If two consecutive terms in an arithmetic progression are (12) and (17), what is (d)?
#ap
#consecutive terms
#find d
#medium
A (29)
B (12)
C (5)
D (17)
Explanation opens after your attempt
Step 1
Concept
The common difference is (17-12=5). When consecutive terms are given, subtract the earlier term from the later term.
Step 2
Why this answer is correct
The correct answer is C. (5). The common difference is (17-12=5). When consecutive terms are given, subtract the earlier term from the later term.
Step 3
Exam Tip
सार्व अंतर (17-12=5) है। लगातार पद दिए हों तो बाद वाला पद घटा पहले वाला पद करें।
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यदि किसी समांतर श्रेढ़ी के लगातार पद (32) और (41) हैं तो सार्व अंतर (d) क्या होगा?
If consecutive terms of an arithmetic progression are (32) and (41), what will be the common difference (d)?
#arithmetic-progression
#consecutive-terms
#class10
A (7)
B (8)
C (9)
D (73)
Explanation opens after your attempt
Step 1
Concept
The common difference is (41-32=9). When consecutive terms are given, no long formula is needed.
Step 2
Why this answer is correct
The correct answer is C. (9). The common difference is (41-32=9). When consecutive terms are given, no long formula is needed.
Step 3
Exam Tip
सार्व अंतर (41-32=9) है। लगातार पद दिए हों तो कोई सूत्र लंबा लगाने की जरूरत नहीं होती।
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यदि समांतर श्रेढ़ी के लगातार दो पद (26) और (31) हैं, तो सार्व अंतर क्या है?
If two consecutive terms of an arithmetic progression are (26) and (31), what is the common difference?
#arithmetic-progression
#consecutive-terms
#class10
A (3)
B (5)
C (26)
D (57)
Explanation opens after your attempt
Step 1
Concept
The common difference is (31-26=5). When consecutive terms are given, directly find the difference.
Step 2
Why this answer is correct
The correct answer is B. (5). The common difference is (31-26=5). When consecutive terms are given, directly find the difference.
Step 3
Exam Tip
सार्व अंतर (31-26=5) है। लगातार पद दिए हों तो सीधा अंतर निकालें।
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यदि समांतर श्रेढ़ी के लगातार दो पद (18) और (23) हैं, तो सार्व अंतर क्या होगा?
If two consecutive terms of an arithmetic progression are (18) and (23), what is the common difference?
#arithmetic-progression
#consecutive-terms
#class10
A (5)
B (18)
C (23)
D (41)
Explanation opens after your attempt
Step 1
Concept
The common difference is the second term minus the first term. Therefore, (23-18=5).
Step 2
Why this answer is correct
The correct answer is A. (5). The common difference is the second term minus the first term. Therefore, (23-18=5).
Step 3
Exam Tip
सार्व अंतर दूसरा पद घटा पहला पद होता है। इसलिए (23-18=5)।
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दो क्रमागत प्राकृतिक संख्याओं के वर्गों का योग (365) है। वे संख्याएँ कौन सी हैं?
The sum of the squares of two consecutive natural numbers is (365). Which numbers are they?
#quadratic equations
#sum of squares
#consecutive numbers
A (11) और (12) / (11) and (12)
B (12) और (13) / (12) and (13)
C (13) और (14) / (13) and (14)
D (14) और (15) / (14) and (15)
Explanation opens after your attempt
Correct Answer
C. (13) और (14) / (13) and (14)
Step 1
Concept
Take the numbers as (x) and (x+1). From (x-2 +(x+1)2 =365), we get (x=13).
Step 2
Why this answer is correct
The correct answer is C. (13) और (14) / (13) and (14). Take the numbers as (x) and (x+1). From (x-2 +(x+1)2 =365), we get (x=13).
Step 3
Exam Tip
संख्याएँ (x) और (x+1) लें। (x-2 +(x+1)2 =365) से (x=13) मिलता है।
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यदि (x-2 -(2a+9)x+(a+4)(a+5)=0) की जड़ें लगातार रूप में हैं, तो वे कौन-सी हैं?
If the roots of (x-2 -(2a+9)x+(a+4)(a+5)=0) are in consecutive form, which are they?
#quadratic-roots
#consecutive-roots
#parametric
A (a+4) और (a+5) / (a+4) and (a+5)
B (a+3) और (a+6) / (a+3) and (a+6)
C (a) और (a+9) / (a) and (a+9)
D (2a+4) और (5) / (2a+4) and (5)
Explanation opens after your attempt
Correct Answer
A. (a+4) और (a+5) / (a+4) and (a+5)
Step 1
Concept
The sum of roots is (2a+9) and the product is ((a+4)(a+5)). These match (a+4) and (a+5).
Step 2
Why this answer is correct
The correct answer is A. (a+4) और (a+5) / (a+4) and (a+5). The sum of roots is (2a+9) and the product is ((a+4)(a+5)). These match (a+4) and (a+5).
Step 3
Exam Tip
जड़ों का योग (2a+9) और गुणनफल ((a+4)(a+5)) है। ये (a+4) और (a+5) से मेल खाते हैं।
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यदि (x-2 -(2a+7)x+(a+3)(a+4)=0) की जड़ें लगातार पूर्णांक रूप में हैं, तो वे कौन-सी हैं?
If the roots of (x-2 -(2a+7)x+(a+3)(a+4)=0) are in consecutive integer form, which are they?
#quadratic-roots
#consecutive-roots
#parametric
A (a+2) और (a+5) / (a+2) and (a+5)
B (a+3) और (a+4) / (a+3) and (a+4)
C (a) और (a+7) / (a) and (a+7)
D (2a+3) और (4) / (2a+3) and (4)
Explanation opens after your attempt
Correct Answer
B. (a+3) और (a+4) / (a+3) and (a+4)
Step 1
Concept
The sum of roots is (2a+7) and the product is ((a+3)(a+4)). These match (a+3) and (a+4).
Step 2
Why this answer is correct
The correct answer is B. (a+3) और (a+4) / (a+3) and (a+4). The sum of roots is (2a+7) and the product is ((a+3)(a+4)). These match (a+3) and (a+4).
Step 3
Exam Tip
जड़ों का योग (2a+7) और गुणनफल ((a+3)(a+4)) है। ये (a+3) और (a+4) से मिलते हैं।
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दो लगातार धन सम संख्याओं का गुणनफल (48) है। यदि छोटी संख्या (x) है तो समीकरण क्या होगा?
The product of two consecutive positive even numbers is (48). If the smaller number is (x), what is the equation?
#quadratic equations
#word problem
#consecutive even numbers
A (x(x+2)=48)
B (x(x+1)=48)
C (x+x+2=48)
D (2x=48)
Explanation opens after your attempt
Correct Answer
A. (x(x+2)=48)
Step 1
Concept
The next consecutive even number is (x+2). So the product equation is (x(x+2)=48).
Step 2
Why this answer is correct
The correct answer is A. (x(x+2)=48). The next consecutive even number is (x+2). So the product equation is (x(x+2)=48).
Step 3
Exam Tip
लगातार सम संख्या (x+2) होगी। इसलिए गुणनफल का समीकरण (x(x+2)=48) है।
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यदि संख्या रेखा पर (P) का निर्देशांक \( -\sqrt{48} \) है, तो (P) किस दो लगातार पूर्णांकों के बीच है?
If point (P) has coordinate \( -\sqrt{48} \) on the number line, between which two consecutive integers does (P) lie?
#number-line
#negative-square-root
#interval
A (-7) और (-6) / (-7) and (-6)
B (-6) और (-5) / (-6) and (-5)
C (6) और (7) / (6) and (7)
D (-8) और (-7) / (-8) and (-7)
Explanation opens after your attempt
Correct Answer
A. (-7) और (-6) / (-7) and (-6)
Step 1
Concept
Since \(6<\sqrt{48}<7\), \(-7<-\sqrt{48}<-6\). Write the interval carefully for negative square roots.
Step 2
Why this answer is correct
The correct answer is A. (-7) और (-6) / (-7) and (-6). Since \(6<\sqrt{48}<7\), \(-7<-\sqrt{48}<-6\). Write the interval carefully for negative square roots.
Step 3
Exam Tip
\(6<\sqrt{48}<7\), इसलिए \(-7<-\sqrt{48}<-6\)। ऋणात्मक वर्गमूल में अंतराल उल्टा लिखें।
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दो लगातार सम धनात्मक संख्याओं के वर्गों का योग (3044) है। बड़ी संख्या क्या है?
The sum of squares of two consecutive even positive numbers is (3044). What is the larger number?
#quadratic equations
#even integers
#sum of squares
A (36)
B (38)
C (40)
D (42)
Explanation opens after your attempt
Step 1
Concept
Consecutive even numbers are (x) and (x+2). From (x-2 +(x+2)2 =3044), (x=38), so the larger number is (40).
Step 2
Why this answer is correct
The correct answer is C. (40). Consecutive even numbers are (x) and (x+2). From (x-2 +(x+2)2 =3044), (x=38), so the larger number is (40).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) हैं। (x-2 +(x+2)2 =3044) से (x=38), इसलिए बड़ी संख्या (40) है।
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दो लगातार विषम धनात्मक संख्याओं के वर्गों का योग (2314) है। छोटी संख्या क्या है?
The sum of squares of two consecutive odd positive numbers is (2314). What is the smaller number?
#quadratic equations
#odd integers
#sum of squares
A (29)
B (31)
C (33)
D (35)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =2314), (x=33).
Step 2
Why this answer is correct
The correct answer is C. (33). Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =2314), (x=33).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) होंगी। (x-2 +(x+2)2 =2314) से (x=33) है।
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दो लगातार सम धनात्मक संख्याओं के वर्गों का योग (2452) है। बड़ी संख्या क्या है?
The sum of squares of two consecutive even positive numbers is (2452). What is the larger number?
#quadratic equations
#even integers
#sum of squares
A (32)
B (34)
C (36)
D (38)
Explanation opens after your attempt
Step 1
Concept
Consecutive even numbers are (x) and (x+2). From (x-2 +(x+2)2 =2452), (x=34), so the larger number is (36).
Step 2
Why this answer is correct
The correct answer is C. (36). Consecutive even numbers are (x) and (x+2). From (x-2 +(x+2)2 =2452), (x=34), so the larger number is (36).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) हैं। (x-2 +(x+2)2 =2452) से (x=34), इसलिए बड़ी संख्या (36) है।
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दो लगातार विषम धनात्मक संख्याओं के वर्गों का योग (2050) है। छोटी संख्या क्या है?
The sum of squares of two consecutive odd positive numbers is (2050). What is the smaller number?
#quadratic equations
#odd integers
#sum of squares
A (27)
B (29)
C (31)
D (33)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =2050), (x=31).
Step 2
Why this answer is correct
The correct answer is C. (31). Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =2050), (x=31).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) होंगी। (x-2 +(x+2)2 =2050) से (x=31) है।
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दो लगातार सम धनात्मक संख्याओं के वर्गों का योग (1924) है। बड़ी संख्या क्या है?
The sum of squares of two consecutive even positive numbers is (1924). What is the larger number?
#quadratic equations
#even integers
#sum of squares
A (28)
B (30)
C (32)
D (34)
Explanation opens after your attempt
Step 1
Concept
Let the consecutive even numbers be (x) and (x+2). From (x-2 +(x+2)2 =1924), (x=30), so the larger number is (32).
Step 2
Why this answer is correct
The correct answer is C. (32). Let the consecutive even numbers be (x) and (x+2). From (x-2 +(x+2)2 =1924), (x=30), so the larger number is (32).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) मानें। (x-2 +(x+2)2 =1924) से (x=30), इसलिए बड़ी संख्या (32) है।
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दो लगातार विषम धनात्मक संख्याओं के वर्गों का योग (1570) है। छोटी संख्या क्या है?
The sum of squares of two consecutive odd positive numbers is (1570). What is the smaller number?
#quadratic equations
#odd integers
#sum of squares
A (23)
B (25)
C (27)
D (29)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =1570), (x=27).
Step 2
Why this answer is correct
The correct answer is C. (27). Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =1570), (x=27).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) हैं। (x-2 +(x+2)2 =1570) से (x=27) मिलता है।
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दो लगातार सम धनात्मक संख्याओं के वर्गों का योग (1060) है। बड़ी संख्या क्या है?
The sum of squares of two consecutive even positive numbers is (1060). What is the larger number?
#quadratic equations
#even integers
#sum of squares
A (20)
B (22)
C (24)
D (26)
Explanation opens after your attempt
Step 1
Concept
Let the consecutive even numbers be (x) and (x+2). From (x-2 +(x+2)2 =1060), (x=22), so the larger number is (24).
Step 2
Why this answer is correct
The correct answer is C. (24). Let the consecutive even numbers be (x) and (x+2). From (x-2 +(x+2)2 =1060), (x=22), so the larger number is (24).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) मानें। (x-2 +(x+2)2 =1060) से (x=22), इसलिए बड़ी संख्या (24) है।
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दो लगातार विषम धनात्मक संख्याओं के वर्गों का योग (650) है। छोटी संख्या क्या है?
The sum of squares of two consecutive odd positive numbers is (650). What is the smaller number?
#quadratic equations
#odd integers
#sum of squares
A (15)
B (17)
C (19)
D (21)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =650), (x=17).
Step 2
Why this answer is correct
The correct answer is B. (17). Consecutive odd numbers are (x) and (x+2). From (x-2 +(x+2)2 =650), (x=17).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) हैं। (x-2 +(x+2)2 =650) से (x=17) मिलता है।
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दो लगातार सम धनात्मक संख्याओं का गुणनफल (728) है। बड़ी संख्या क्या है?
The product of two consecutive even positive numbers is (728). What is the larger number?
#quadratic equations
#even integers
#application
A (24)
B (26)
C (28)
D (30)
Explanation opens after your attempt
Step 1
Concept
Consecutive even numbers are (x) and (x+2). From (x(x+2)=728), (x=26), so the larger number is (28).
Step 2
Why this answer is correct
The correct answer is C. (28). Consecutive even numbers are (x) and (x+2). From (x(x+2)=728), (x=26), so the larger number is (28).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) होंगी। (x(x+2)=728) से (x=26), इसलिए बड़ी संख्या (28) है।
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दो लगातार विषम धनात्मक संख्याओं का गुणनफल (575) है। छोटी संख्या क्या है?
The product of two consecutive odd positive numbers is (575). What is the smaller number?
#quadratic equations
#odd integers
#word problem
A (21)
B (23)
C (25)
D (27)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2). From (x(x+2)=575), (x=23).
Step 2
Why this answer is correct
The correct answer is B. (23). Consecutive odd numbers are (x) and (x+2). From (x(x+2)=575), (x=23).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) हैं। (x(x+2)=575) से (x=23) मिलता है।
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दो लगातार सम संख्याओं का गुणनफल (528) है। छोटी संख्या क्या है?
The product of two consecutive even numbers is (528). What is the smaller number?
#quadratic equations
#even integers
#word application
A (20)
B (22)
C (24)
D (26)
Explanation opens after your attempt
Step 1
Concept
(x(x+2)=528) gives (x=22). Remember that consecutive even numbers differ by (2).
Step 2
Why this answer is correct
The correct answer is B. (22). (x(x+2)=528) gives (x=22). Remember that consecutive even numbers differ by (2).
Step 3
Exam Tip
(x(x+2)=528) से (x=22) मिलता है। सम लगातार संख्याओं का अंतर (2) याद रखें।
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दो लगातार विषम संख्याओं का गुणनफल (399) है। बड़ी संख्या क्या है?
The product of two consecutive odd numbers is (399). What is the larger number?
#quadratic equations
#odd integers
#application
A (17)
B (19)
C (21)
D (23)
Explanation opens after your attempt
Step 1
Concept
(x(x+2)=399) gives (x=19), so the larger number is (21). In such questions, use (x+2).
Step 2
Why this answer is correct
The correct answer is C. (21). (x(x+2)=399) gives (x=19), so the larger number is (21). In such questions, use (x+2).
Step 3
Exam Tip
(x(x+2)=399) से (x=19), इसलिए बड़ी संख्या (21) है। ऐसे प्रश्नों में (x+2) का प्रयोग करें।
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दो लगातार सम धनात्मक संख्याओं का गुणनफल (360) है। बड़ी संख्या क्या है?
The product of two consecutive even positive numbers is (360). What is the larger number?
#quadratic equations
#even integers
#application
A (18)
B (20)
C (22)
D (24)
Explanation opens after your attempt
Step 1
Concept
Consecutive even numbers are (x) and (x+2), then (x(x+2)=360) gives (x=18). So the larger number is (20).
Step 2
Why this answer is correct
The correct answer is B. (20). Consecutive even numbers are (x) and (x+2), then (x(x+2)=360) gives (x=18). So the larger number is (20).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) होंगी, तब (x(x+2)=360) से (x=18) है। इसलिए बड़ी संख्या (20) है।
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दो लगातार विषम धनात्मक संख्याओं का गुणनफल (255) है। छोटी संख्या क्या है?
The product of two consecutive odd positive numbers is (255). What is the smaller number?
#quadratic equations
#odd integers
#word problem
A (13)
B (15)
C (17)
D (19)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2), so (x(x+2)=255) gives (x=15). Consecutive odd numbers differ by (2).
Step 2
Why this answer is correct
The correct answer is B. (15). Consecutive odd numbers are (x) and (x+2), so (x(x+2)=255) gives (x=15). Consecutive odd numbers differ by (2).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) हैं, इसलिए (x(x+2)=255) से (x=15) है। विषम लगातार संख्याओं में अंतर (2) होता है।
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दो लगातार सम संख्याओं का गुणनफल (224) है। छोटी संख्या क्या है?
The product of two consecutive even numbers is (224). What is the smaller number?
#quadratic equations
#even integers
#word problem
A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
(x(x+2)=224) gives (x=14). Keep a difference of (2) for consecutive even numbers.
Step 2
Why this answer is correct
The correct answer is B. (14). (x(x+2)=224) gives (x=14). Keep a difference of (2) for consecutive even numbers.
Step 3
Exam Tip
(x(x+2)=224) से (x=14) मिलता है। सम लगातार संख्याओं में (2) का अंतर रखें।
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दो लगातार विषम संख्याओं का गुणनफल (195) है। बड़ी संख्या क्या है?
The product of two consecutive odd numbers is (195). What is the larger number?
#quadratic equations
#odd integers
#application
A (13)
B (15)
C (17)
D (19)
Explanation opens after your attempt
Step 1
Concept
(x(x+2)=195) gives (x=13), so the larger number is (15). For odd numbers, (x) and (x+2) is the correct form.
Step 2
Why this answer is correct
The correct answer is B. (15). (x(x+2)=195) gives (x=13), so the larger number is (15). For odd numbers, (x) and (x+2) is the correct form.
Step 3
Exam Tip
(x(x+2)=195) से (x=13), इसलिए बड़ी संख्या (15) है। विषम संख्याओं के लिए (x) और (x+2) सही रूप है।
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दो लगातार सम धनात्मक संख्याओं का गुणनफल (168) है। बड़ी संख्या क्या है?
The product of two consecutive even positive numbers is (168). What is the larger number?
#quadratic equations
#even integers
#application
A (10)
B (12)
C (14)
D (16)
Explanation opens after your attempt
Step 1
Concept
Take consecutive even numbers as (x) and (x+2), then (x(x+2)=168) gives (x=12). Hence the larger number is (14).
Step 2
Why this answer is correct
The correct answer is C. (14). Take consecutive even numbers as (x) and (x+2), then (x(x+2)=168) gives (x=12). Hence the larger number is (14).
Step 3
Exam Tip
लगातार सम संख्याएँ (x) और (x+2) लें, तब (x(x+2)=168) से (x=12) है। इसलिए बड़ी संख्या (14) है।
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दो लगातार विषम धनात्मक संख्याओं का गुणनफल (143) है। छोटी संख्या क्या है?
The product of two consecutive odd positive numbers is (143). What is the smaller number?
#quadratic equations
#odd integers
#word problems
A (9)
B (11)
C (13)
D (15)
Explanation opens after your attempt
Step 1
Concept
Consecutive odd numbers are (x) and (x+2), so (x(x+2)=143) gives (x=11). Consecutive odd numbers differ by (2).
Step 2
Why this answer is correct
The correct answer is B. (11). Consecutive odd numbers are (x) and (x+2), so (x(x+2)=143) gives (x=11). Consecutive odd numbers differ by (2).
Step 3
Exam Tip
लगातार विषम संख्याएँ (x) और (x+2) होंगी, इसलिए (x(x+2)=143) से (x=11) मिलता है। विषम लगातार संख्याओं में अंतर (2) होता है।
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संख्या रेखा पर \(-\frac{9}{4}\) किस दो पूर्णांकों के बीच होगा?
Between which two integers will \(-\frac{9}{4}\) lie on the number line?
#negative-fractions
#integers
#number-line
A (-3) और (-2) / (-3) and (-2)
B (-2) और (-1) / (-2) and (-1)
C (2) और (3) / (2) and (3)
D (-4) और (-3) / (-4) and (-3)
Explanation opens after your attempt
Correct Answer
A. (-3) और (-2) / (-3) and (-2)
Step 1
Concept
\(-\frac{9}{4}=-2.25\), so it lies between (-3) and (-2). Be careful with direction for negative fractions.
Step 2
Why this answer is correct
The correct answer is A. (-3) और (-2) / (-3) and (-2). \(-\frac{9}{4}=-2.25\), so it lies between (-3) and (-2). Be careful with direction for negative fractions.
Step 3
Exam Tip
\(-\frac{9}{4}=-2.25\), इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक भिन्न में छोटी दिशा को ध्यान से समझें।
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संख्या रेखा पर (2.75) किस दो पूर्णांकों के बीच आता है?
Between which two integers does (2.75) lie on the number line?
#decimals
#integers
#number-line
A (2) और (3) / (2) and (3)
B (3) और (4) / (3) and (4)
C (1) और (2) / (1) and (2)
D (0) और (1) / (0) and (1)
Explanation opens after your attempt
Correct Answer
A. (2) और (3) / (2) and (3)
Step 1
Concept
The integer part of (2.75) is (2), so it lies between (2) and (3). The integer part gives the first clue.
Step 2
Why this answer is correct
The correct answer is A. (2) और (3) / (2) and (3). The integer part of (2.75) is (2), so it lies between (2) and (3). The integer part gives the first clue.
Step 3
Exam Tip
(2.75) का पूर्णांक भाग (2) है, इसलिए यह (2) और (3) के बीच है। दशमलव का पूर्णांक भाग पहला संकेत देता है।
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संख्या रेखा पर \(-\frac{7}{3}\) किस दो पूर्णांकों के बीच होगा?
Between which two integers will \(-\frac{7}{3}\) lie on the number line?
#negative-fractions
#integers
#number-line
A (-3) और (-2) / (-3) and (-2)
B (-2) और (-1) / (-2) and (-1)
C (2) और (3) / (2) and (3)
D (-4) और (-3) / (-4) and (-3)
Explanation opens after your attempt
Correct Answer
A. (-3) और (-2) / (-3) and (-2)
Step 1
Concept
\(-\frac{7}{3}\approx -2.33\), so it lies between (-3) and (-2). Check the position of negative decimals carefully.
Step 2
Why this answer is correct
The correct answer is A. (-3) और (-2) / (-3) and (-2). \(-\frac{7}{3}\approx -2.33\), so it lies between (-3) and (-2). Check the position of negative decimals carefully.
Step 3
Exam Tip
\(-\frac{7}{3}\approx -2.33\) है, इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक दशमलव की स्थिति ध्यान से देखें।
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यदि (3,x,11,y,19) एक समांतर श्रेणी के लगातार पांच पद हैं, तो (x,y) और सामान्य अंतर क्या हैं?
If (3,x,11,y,19) are five consecutive terms of an AP, what are (x,y) and the common difference?
#ap
#missing_terms
#five_consecutive_terms
A (x=6,y=14,d=3)
B (x=7,y=15,d=4)
C (x=8,y=16,d=5)
D (x=5,y=13,d=4)
Explanation opens after your attempt
Correct Answer
B. (x=7,y=15,d=4)
Step 1
Concept
There are (4) equal gaps between the first and fifth terms, so \(d=\frac{19-3}{4}=4\). In exams, divide the total difference between distant terms by the number of gaps.
Step 2
Why this answer is correct
The correct answer is B. (x=7,y=15,d=4). There are (4) equal gaps between the first and fifth terms, so \(d=\frac{19-3}{4}=4\). In exams, divide the total difference between distant terms by the number of gaps.
Step 3
Exam Tip
पहले और पांचवें पद के बीच (4) बराबर अंतराल हैं, इसलिए \(d=\frac{19-3}{4}=4\)। परीक्षा में दूर दिए गए पदों के बीच कुल अंतर को अंतरालों की संख्या से भाग दें।
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संख्याएं (14,m,50) समांतर श्रेणी के लगातार पद हैं। (m) और (d) क्या हैं?
The numbers (14,m,50) are consecutive terms of an AP. What are (m) and (d)?
#ap
#missing_term
#mean_property
A (m=30,d=16)
B (m=32,d=18)
C (m=34,d=20)
D (m=36,d=22)
Explanation opens after your attempt
Correct Answer
B. (m=32,d=18)
Step 1
Concept
The middle term is \(m=\frac{14+50}{2}=32\), and (d=18). In exams, the middle term of three AP terms is the average.
Step 2
Why this answer is correct
The correct answer is B. (m=32,d=18). The middle term is \(m=\frac{14+50}{2}=32\), and (d=18). In exams, the middle term of three AP terms is the average.
Step 3
Exam Tip
मध्य पद \(m=\frac{14+50}{2}=32\) है और (d=18)। परीक्षा में तीन पदों में मध्य पद औसत होता है।
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किस (k) के लिए (k,2k+1,5k-2,8k-5) समांतर श्रेणी के लगातार चार पद हैं?
For which (k) are (k,2k+1,5k-2,8k-5) four consecutive terms of an AP?
#ap
#parameter
#four_terms
A (k=1)
B (k=2)
C (k=3)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Step 1
Concept
The differences are (k+1,3k-3,3k-3), and equality gives (k=2). In exams, check all consecutive differences for four terms.
Step 2
Why this answer is correct
The correct answer is B. (k=2). The differences are (k+1,3k-3,3k-3), and equality gives (k=2). In exams, check all consecutive differences for four terms.
Step 3
Exam Tip
अंतर (k+1,3k-3,3k-3) हैं और बराबरी से (k=2) मिलता है। परीक्षा में चार पदों में सभी लगातार अंतर जांचें।
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यदि (p,q,r,s) समांतर श्रेणी के लगातार चार पद हैं, तो कौन सा संबंध हमेशा सही है?
If (p,q,r,s) are four consecutive terms of an AP, which relation is always true?
#ap
#four_terms
#identity
A (p+q=r+s)
B (p=r)
C (ps=qr)
D (p+s=q+r)
Explanation opens after your attempt
Correct Answer
D. (p+s=q+r)
Step 1
Concept
Writing the terms as (p,p+d,p+2d,p+3d) gives (p+s=q+r). In exams, verify four-term relations symbolically.
Step 2
Why this answer is correct
The correct answer is D. (p+s=q+r). Writing the terms as (p,p+d,p+2d,p+3d) gives (p+s=q+r). In exams, verify four-term relations symbolically.
Step 3
Exam Tip
पद (p,p+d,p+2d,p+3d) रखने पर (p+s=q+r) मिलता है। परीक्षा में चार पदों के संबंध प्रतीकात्मक रूप से जांचें।
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यदि (x+1,2x+6,5x-2) समांतर श्रेणी के लगातार पद हैं, तो (x) और (d) क्या हैं?
If (x+1,2x+6,5x-2) are consecutive terms of an AP, what are (x) and (d)?
#ap
#parameter
#fractional_answer
A \(x=\frac{13}{2},d=\frac{23}{2}\)
B \(x=\frac{11}{2},d=\frac{21}{2}\)
C (x=6,d=12)
D (x=7,d=13)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{13}{2},d=\frac{23}{2}\)
Step 1
Concept
Equating differences gives (x+5=3x-8), so \(x=\frac{13}{2}\) and \(d=\frac{23}{2}\). In exams, do not reject a fractional answer too quickly.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{13}{2},d=\frac{23}{2}\). Equating differences gives (x+5=3x-8), so \(x=\frac{13}{2}\) and \(d=\frac{23}{2}\). In exams, do not reject a fractional answer too quickly.
Step 3
Exam Tip
अंतर बराबर करने पर (x+5=3x-8), इसलिए \(x=\frac{13}{2}\) और \(d=\frac{23}{2}\)। परीक्षा में भिन्न उत्तर से घबराएं नहीं।
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समांतर श्रेणी के लगातार पद (-17,z,23) हैं। (z) और सामान्य अंतर क्या हैं?
The consecutive AP terms are (-17,z,23). What are (z) and the common difference?
#ap
#missing_middle_term
#negative_numbers
A (z=0,d=17)
B (z=-3,d=14)
C (z=6,d=23)
D (z=3,d=20)
Explanation opens after your attempt
Correct Answer
D. (z=3,d=20)
Step 1
Concept
The middle term is the average of the extremes, so \(z=\frac{-17+23}{2}=3\) and (d=20). In exams, handle signs carefully when averaging negatives.
Step 2
Why this answer is correct
The correct answer is D. (z=3,d=20). The middle term is the average of the extremes, so \(z=\frac{-17+23}{2}=3\) and (d=20). In exams, handle signs carefully when averaging negatives.
Step 3
Exam Tip
मध्य पद सिरों का औसत है, इसलिए \(z=\frac{-17+23}{2}=3\) और (d=20)। परीक्षा में ऋणात्मक संख्या जोड़ते समय चिन्ह सावधानी से रखें।
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यदि (4t+1,t+10,-t+21) समांतर श्रेणी के लगातार पद हैं, तो (t) और (d) क्या हैं?
If (4t+1,t+10,-t+21) are consecutive terms of an AP, what are (t) and (d)?
#ap
#parameter
#three_consecutive_terms
A (t=2,d=3)
B (t=-1,d=12)
C (t=-2,d=15)
D (t=3,d=0)
Explanation opens after your attempt
Correct Answer
C. (t=-2,d=15)
Step 1
Concept
Equal differences give (-3t+9=-2t+11), so (t=-2) and (d=15). In exams, subtract first from second and second from third.
Step 2
Why this answer is correct
The correct answer is C. (t=-2,d=15). Equal differences give (-3t+9=-2t+11), so (t=-2) and (d=15). In exams, subtract first from second and second from third.
Step 3
Exam Tip
बराबर अंतर से (-3t+9=-2t+11), इसलिए (t=-2) और (d=15)। परीक्षा में दूसरे से पहला और तीसरे से दूसरा पद घटाएं।
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