100 results found for "even integers" in Class 10.
A. (22)
यदि बीच वाला (x) है, तो पहले और तीसरे (x-2) तथा (x+2) होंगे। ((x-2)(x+2)=480) से (x=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).
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A. (18)
यदि बीच वाला (x) है, तो पहले और तीसरे (x-2) तथा (x+2) होंगे। ((x-2)(x+2)=320) से (x=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).
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The product of two consecutive positive integers is (552). Which are the integers?
B. (23) और (24)(23) and (24)
मान लें संख्याएँ (x) और (x+1) हैं। (x(x+1)=552) से (x=23) मिलता है। / Let the numbers be (x) and (x+1). From (x(x+1)=552), we get (x=23).
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The product of two consecutive positive integers is (240). Which integers are they?
B. (15) और (16)(15) and (16)
मान लें पूर्णांक (x) और (x+1) हैं। (x(x+1)=240) से (x=15) मिलता है। / Let the integers be (x) and (x+1). From (x(x+1)=240), we get (x=15).
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The product of two consecutive positive even integers is (360). What is the smaller integer?
B. (18)
संख्याएँ (x) और (x+2) मानें। (x(x+2)=360) से \(x^2+2x-360=0\) और (x=18) मिलता है। / Let the integers be (x) and (x+2). From (x(x+2)=360), \(x^2+2x-360=0\), so (x=18).
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The product of two consecutive positive even integers is (288). What is the larger integer?
A. (18)
यदि छोटा सम पूर्णांक (x) है, तो (x(x+2)=288)। (x=16) होने से बड़ा पूर्णांक (18) है। / If the smaller even integer is (x), then (x(x+2)=288). Since (x=16), the larger integer is (18).
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The product of two consecutive positive even integers is (168). What is the larger integer?
A. (14)
यदि छोटा सम पूर्णांक (x) है, तो (x(x+2)=168)। (x=12) होने से बड़ा पूर्णांक (14) है। / If the smaller even integer is (x), then (x(x+2)=168). Since (x=12), the larger integer is (14).
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Why is (p) considered even when \(p^2\) is even in the proof for \(\sqrt{2}\)?
A. क्योंकि विषम संख्या का वर्ग विषम होता हैBecause the square of an odd number is odd
यदि (p) विषम होता, तो \(p^2\) भी विषम होता। / If (p) were odd, then \(p^2\) would also be odd.
पर \(p^2\) सम मिला है, इसलिए (p) विषम नहीं हो सकता और (p) सम होगा। / Since \(p^2\) is even, (p) cannot be odd, so (p) is even.
यह छोटी सी सम-विषम बात प्रमाण में बहुत महत्त्वपूर्ण है। / This parity fact is very important in the proof.
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In the proof of \(\sqrt{2}\), if \(a^2\) is even, why must (a) be even?
A. क्योंकि यदि (a) विषम होता तो \(a^2\) भी विषम होताBecause if (a) were odd, then \(a^2\) would also be odd
विषम संख्या का वर्ग विषम होता है। / The square of an odd number is odd.
यहां \(a^2\) सम मिला है, इसलिए (a) विषम नहीं हो सकता। / Here \(a^2\) is even, so (a) cannot be odd.
अतः (a) सम होना निश्चित है। / Hence (a) must be even.
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Which statement justifies that (a) is even when \(a^2\) is even in the proof of \(\sqrt{2}\)?
A. यदि (a) विषम होता, तो \(a^2\) भी विषम होताIf (a) were odd, then \(a^2\) would also be odd
विषम संख्या का वर्ग हमेशा विषम होता है। / The square of an odd number is always odd.
यहां \(a^2\) सम मिला है, इसलिए (a) विषम नहीं हो सकता। / Here \(a^2\) is even, so (a) cannot be odd.
अतः (a) सम होना चाहिए। / Therefore (a) must be even.
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A. क्योंकि विषम संख्या का वर्ग विषम होता हैBecause the square of an odd number is odd
यदि (p) विषम होता, तो \(p^2\) भी विषम होता। / If (p) were odd, then \(p^2\) would also be odd.
लेकिन \(p^2\) सम मिला है, इसलिए (p) विषम नहीं हो सकता। / But \(p^2\) is even, so (p) cannot be odd.
इस तरह (p) सम होना तय है। / Thus (p) must be even.
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Between which two integers will \(-\frac{9}{4}\) lie on the number line?
A. (-3) और (-2)(-3) and (-2)
\(-\frac{9}{4}=-2.25\), इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक भिन्न में छोटी दिशा को ध्यान से समझें। / \(-\frac{9}{4}=-2.25\), so it lies between (-3) and (-2). Be careful with direction for negative fractions.
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Between which two integers does (2.75) lie on the number line?
A. (2) और (3)(2) and (3)
(2.75) का पूर्णांक भाग (2) है, इसलिए यह (2) और (3) के बीच है। दशमलव का पूर्णांक भाग पहला संकेत देता है। / The integer part of (2.75) is (2), so it lies between (2) and (3). The integer part gives the first clue.
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Between which two integers will \(-\frac{7}{3}\) lie on the number line?
A. (-3) और (-2)(-3) and (-2)
\(-\frac{7}{3}\approx -2.33\) है, इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक दशमलव की स्थिति ध्यान से देखें। / \(-\frac{7}{3}\approx -2.33\), so it lies between (-3) and (-2). Check the position of negative decimals carefully.
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The product of two consecutive positive odd integers is (483). What is the larger integer?
B. (23)
संख्याएँ (x) और (x+2) हैं। (x(x+2)=483) से (x=21) मिलता है इसलिए बड़ी संख्या (23) है। / The integers are (x) and (x+2). From (x(x+2)=483), (x=21), so the larger integer is (23).
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B. (13)
छोटा विषम पूर्णांक (x) हो, तो (x-2+(x+2)2=394)। इससे \(x^2+2x-195=0\), इसलिए (x=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).
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A. (14)
यदि बीच वाला पूर्णांक (x) है, तो पहले और तीसरे (x-1) तथा (x+1) होंगे। ((x-1)(x+1)=195) से (x=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).
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The product of two consecutive positive odd integers is (399). What is the smaller integer?
A. (19)
यदि छोटा विषम पूर्णांक (x) है, तो (x(x+2)=399)। \(19\times21=399\), इसलिए छोटा पूर्णांक (19) है। / If the smaller odd integer is (x), then (x(x+2)=399). Since \(19\times21=399\), the smaller integer is (19).
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The product of two consecutive positive integers is (306). What is the larger integer?
A. (18)
यदि छोटा पूर्णांक (x) है, तो (x(x+1)=306)। (x=17) होने पर बड़ा पूर्णांक (18) है। / If the smaller integer is (x), then (x(x+1)=306). When (x=17), the larger integer is (18).
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A. (12)
यदि बीच वाला पूर्णांक (x) है, तो पहले और तीसरे (x-1) तथा (x+1) होंगे। ((x-1)(x+1)=143) से (x=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).
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The product of two consecutive positive odd integers is (195). What is the smaller integer?
A. (13)
यदि छोटा विषम पूर्णांक (x) है, तो (x(x+2)=195)। (x=13) सही है क्योंकि \(13\times15=195\)। / If the smaller odd integer is (x), then (x(x+2)=195). (x=13) is correct because \(13\times15=195\).
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The product of two consecutive positive integers is (156). What is the larger integer?
A. (13)
यदि छोटा पूर्णांक (x) है, तो (x(x+1)=156)। (x=12) मिलने पर बड़ा पूर्णांक (13) है। / If the smaller integer is (x), then (x(x+1)=156). When (x=12), the larger integer is (13).
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The product of two consecutive positive integers is (380). What is the larger integer?
C. (20)
संख्याएँ (x) और (x+1) होंगी। (x(x+1)=380) से (x=19), इसलिए बड़ा पूर्णांक (20) है। / The numbers are (x) and (x+1). From (x(x+1)=380), (x=19), so the larger integer is (20).
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The product of two consecutive positive integers is (306). What is the smaller integer?
B. (17)
छोटा पूर्णांक (x) हो तो (x(x+1)=306), जिससे (x=17) मिलता है। लगातार पूर्णांकों को (x) और (x+1) लिखें। / If the smaller integer is (x), then (x(x+1)=306), giving (x=17). Write consecutive integers as (x) and (x+1).
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The product of two consecutive positive integers is (210). What is the larger integer?
C. (15)
संख्याएँ (x) और (x+1) मानें, तब (x(x+1)=210) से (x=14) है। इसलिए बड़ा पूर्णांक (15) है। / Let the numbers be (x) and (x+1), then (x(x+1)=210) gives (x=14). So the larger integer is (15).
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The product of two consecutive positive integers is (182). What is the smaller integer?
B. (13)
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। / 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.
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The product of two consecutive positive integers is (132). What is the larger integer?
C. (12)
मान लें संख्याएँ (x) और (x+1) हैं, तब (x(x+1)=132) से (x=11) है। इसलिए बड़ा पूर्णांक (12) है। / Let the numbers be (x) and (x+1), then (x(x+1)=132) gives (x=11). So the larger integer is (12).
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The product of two consecutive positive integers is (72). What is the smaller integer?
B. (8)
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। / Let the smaller integer be (x), then (x(x+1)=72) gives (x=8). In exams, write consecutive numbers as (x) and (x+1).
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A. \(2x^2+2x-144=0\)
पूर्णांक (x) और (x+1) होंगे, इसलिए (x-2+(x+1)2=145)। सरल करने पर \(2x^2+2x-144=0\) मिलता है। / The integers are (x) and (x+1), so (x-2+(x+1)2=145). Simplifying gives \(2x^2+2x-144=0\).
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A. \(2x^2+2x-84=0\)
पूर्णांक (x) और (x+1) होंगे, इसलिए (x-2+(x+1)2=85)। सरल करने पर \(2x^2+2x-84=0\) मिलता है। / The integers are (x) and (x+1), so (x-2+(x+1)2=85). Simplifying gives \(2x^2+2x-84=0\).
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The product of two consecutive positive integers is (56). Which quadratic equation is correct?
A. \(x^2+x-56=0\)
संख्याएँ (x) और (x+1) होंगी, इसलिए (x(x+1)=56)। इससे \(x^2+x-56=0\) मिलता है। / The numbers are (x) and (x+1), so (x(x+1)=56). This gives \(x^2+x-56=0\).
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A. (x(x+1)=30)
लगातार अगला पूर्णांक (x+1) होगा। गुणनफल (30) है इसलिए (x(x+1)=30) बनेगा। / The next consecutive integer is (x+1). Since the product is (30), the equation is (x(x+1)=30).
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Why is at least one number among twelve consecutive integers divisible by 12?
A. क्योंकि 12 से भाग देने पर शेषफल 0 से 11 तक चक्र में आते हैंBecause division by 12 gives remainders from 0 to 11 in a cycle
12 से भाग देने पर संभावित शेषफल 0 से 11 तक हैं। / On division by 12, possible remainders are from 0 to 11.
बारह लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Twelve consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वही 12 से विभाज्य होगी। / The number with remainder 0 is divisible by 12.
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Why is at least one number among eleven consecutive integers divisible by 11?
A. क्योंकि 11 से भाग देने पर शेषफल 0 से 10 तक चक्र में आते हैंBecause division by 11 gives remainders from 0 to 10 in a cycle
11 से भाग देने पर संभावित शेषफल 0 से 10 तक होते हैं। / On division by 11, possible remainders are from 0 to 10.
ग्यारह लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Eleven consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वही 11 से विभाज्य होगी। / The number with remainder 0 is divisible by 11.
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Why is at least one number among ten consecutive integers divisible by 10?
A. क्योंकि 10 से भाग देने पर शेषफल 0 से 9 तक चक्र में आते हैंBecause division by 10 gives remainders from 0 to 9 in a cycle
10 से भाग देने पर संभावित शेषफल 0 से 9 तक हैं। / On division by 10, possible remainders are from 0 to 9.
दस लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Ten consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वही 10 से विभाज्य होगी। / The number with remainder 0 is divisible by 10.
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Why is at least one number among nine consecutive integers divisible by 9?
A. क्योंकि 9 से भाग देने पर शेषफल 0 से 8 तक चक्र में आते हैंBecause division by 9 gives remainders from 0 to 8 in a cycle
9 से भाग देने पर संभावित शेषफल 0 से 8 तक हैं। / On division by 9, possible remainders are from 0 to 8.
नौ लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Nine consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वही 9 से विभाज्य होगी। / The number with remainder 0 is divisible by 9.
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Why is at least one number among eight consecutive integers divisible by 8?
A. क्योंकि 8 से भाग देने पर शेषफल 0 से 7 तक चक्र में आते हैंBecause division by 8 gives remainders from 0 to 7 in a cycle
8 से भाग देने पर संभावित शेषफल 0 से 7 तक होते हैं। / On division by 8, possible remainders are from 0 to 7.
आठ लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Eight consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वह 8 से विभाज्य होती है। / The number with remainder 0 is divisible by 8.
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Why is at least one number among seven consecutive integers divisible by 7?
A. क्योंकि 7 से भाग देने पर शेषफल 0 से 6 तक चक्र में आते हैंBecause division by 7 gives remainders from 0 to 6 in a cycle
7 से भाग देने पर संभावित शेषफल 0, 1, 2, 3, 4, 5, 6 हैं। / The possible remainders on division by 7 are 0, 1, 2, 3, 4, 5, and 6.
सात लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Seven consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वही 7 से विभाज्य होगी। / The number with remainder 0 is divisible by 7.
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Why is at least one number among six consecutive integers divisible by 6?
A. क्योंकि 6 से भाग देने पर शेषफल 0, 1, 2, 3, 4, 5 का चक्र आता हैBecause division by 6 gives the cycle of remainders 0, 1, 2, 3, 4, 5
6 से भाग देने पर शेषफल 0 से 5 तक हो सकते हैं। / On division by 6, possible remainders are from 0 to 5.
छह लगातार पूर्णांकों में ये सभी शेषफल एक बार आते हैं। / Six consecutive integers cover all these remainders once.
जिस संख्या का शेषफल 0 है, वह 6 से विभाज्य होगी। / The number with remainder 0 is divisible by 6.
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Why is at least one number among five consecutive integers divisible by 5?
A. क्योंकि 5 से भाग देने पर शेषफल 0, 1, 2, 3, 4 का चक्र आता हैBecause division by 5 gives the cycle of remainders 0, 1, 2, 3, 4
कोई भी पूर्णांक 5 से भाग देने पर (5q), (5q+1), (5q+2), (5q+3), या (5q+4) रूप में होता है। / Any integer divided by 5 has one of the forms (5q), (5q+1), (5q+2), (5q+3), or (5q+4).
पांच लगातार पूर्णांकों में ये पांचों शेषफल आ जाते हैं। / Five consecutive integers cover all five remainders.
जिस संख्या का शेषफल 0 है, वही 5 से विभाज्य होगी। / The number with remainder 0 is divisible by 5.
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Why is at least one number among four consecutive integers divisible by 4?
A. क्योंकि 4 से भाग देने पर शेषफल 0, 1, 2, 3 का चक्र आता हैBecause division by 4 gives the cycle of remainders 0, 1, 2, 3
कोई भी पूर्णांक 4 से भाग देने पर (4q), (4q+1), (4q+2), या (4q+3) रूप में होता है। / Any integer divided by 4 is of the form (4q), (4q+1), (4q+2), or (4q+3).
चार लगातार पूर्णांकों में ये चारों शेषफल आ जाते हैं। / Four consecutive integers cover all these four remainders.
जिस संख्या का शेषफल 0 है, वही 4 से विभाज्य होगी। / The one with remainder 0 is divisible by 4.
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Why is the product of three consecutive integers divisible by 6?
A. क्योंकि उनमें एक 2 से और एक 3 से विभाज्य होता हैBecause one of them is divisible by 2 and one is divisible by 3
दो लगातार पूर्णांकों में एक सम होता है, इसलिए 2 का गुणनखंड मिलता है। / Among two consecutive integers, one is even, so a factor 2 is present.
तीन लगातार पूर्णांकों में एक 3 से विभाज्य होता है, इसलिए 3 का गुणनखंड मिलता है। / Among three consecutive integers, one is divisible by 3, so a factor 3 is present.
2 और 3 मिलकर 6 बनाते हैं, इसलिए गुणनफल 6 से विभाज्य है। / Since 2 and 3 together make 6, the product is divisible by 6.
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Why can one of three consecutive integers be taken as of the form (3q)?
A. क्योंकि 3 से भाग देने पर शेषफल 0, 1, 2 में से एक होता हैBecause division by 3 gives one of the remainders 0, 1, 2
3 से भाग देने पर हर संख्या (3q), (3q+1), या (3q+2) रूप में होगी। / On division by 3, every integer is of the form (3q), (3q+1), or (3q+2).
तीन लगातार संख्याओं में ये तीनों शेषफल आते हैं, इसलिए एक संख्या 3 से पूर्णतः विभाजित होगी। / Three consecutive integers cover these three remainders, so one is exactly divisible by 3.
लगातार संख्याओं में शेषफल चक्र का उपयोग करें। / Use the cycle of remainders for consecutive-number problems.
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The sum of squares of two consecutive even positive numbers is (3044). What is the larger number?
C. (40)
लगातार सम संख्याएँ (x) और (x+2) हैं। (x-2+(x+2)2=3044) से (x=38), इसलिए बड़ी संख्या (40) है। / Consecutive even numbers are (x) and (x+2). From (x-2+(x+2)2=3044), (x=38), so the larger number is (40).
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The sum of squares of two consecutive even positive numbers is (2452). What is the larger number?
C. (36)
लगातार सम संख्याएँ (x) और (x+2) हैं। (x-2+(x+2)2=2452) से (x=34), इसलिए बड़ी संख्या (36) है। / Consecutive even numbers are (x) and (x+2). From (x-2+(x+2)2=2452), (x=34), so the larger number is (36).
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The sum of squares of two consecutive even positive numbers is (1924). What is the larger number?
C. (32)
लगातार सम संख्याएँ (x) और (x+2) मानें। (x-2+(x+2)2=1924) से (x=30), इसलिए बड़ी संख्या (32) है। / 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).
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The sum of squares of two consecutive even positive numbers is (1060). What is the larger number?
C. (24)
लगातार सम संख्याएँ (x) और (x+2) मानें। (x-2+(x+2)2=1060) से (x=22), इसलिए बड़ी संख्या (24) है। / 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).
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The product of two consecutive even positive numbers is (728). What is the larger number?
C. (28)
लगातार सम संख्याएँ (x) और (x+2) होंगी। (x(x+2)=728) से (x=26), इसलिए बड़ी संख्या (28) है। / Consecutive even numbers are (x) and (x+2). From (x(x+2)=728), (x=26), so the larger number is (28).
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The product of two consecutive even numbers is (528). What is the smaller number?
B. (22)
(x(x+2)=528) से (x=22) मिलता है। सम लगातार संख्याओं का अंतर (2) याद रखें। / (x(x+2)=528) gives (x=22). Remember that consecutive even numbers differ by (2).
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The product of two consecutive even positive numbers is (360). What is the larger number?
B. (20)
लगातार सम संख्याएँ (x) और (x+2) होंगी, तब (x(x+2)=360) से (x=18) है। इसलिए बड़ी संख्या (20) है। / Consecutive even numbers are (x) and (x+2), then (x(x+2)=360) gives (x=18). So the larger number is (20).
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The product of two consecutive even numbers is (224). What is the smaller number?
B. (14)
(x(x+2)=224) से (x=14) मिलता है। सम लगातार संख्याओं में (2) का अंतर रखें। / (x(x+2)=224) gives (x=14). Keep a difference of (2) for consecutive even numbers.
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The product of two consecutive even positive numbers is (168). What is the larger number?
C. (14)
लगातार सम संख्याएँ (x) और (x+2) लें, तब (x(x+2)=168) से (x=12) है। इसलिए बड़ी संख्या (14) है। / Take consecutive even numbers as (x) and (x+2), then (x(x+2)=168) gives (x=12). Hence the larger number is (14).
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B. (680)
शेष योग \(28\times29-11\times12=680\) है। सम संख्याओं के योग के लिए (n(n+1)) लगाएँ। / The remaining sum is \(28\times29-11\times12=680\). Use (n(n+1)) for the sum of even numbers.
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B. (660)
शेष योग \(24\times25-9\times10=510\) नहीं बल्कि (600-90=510) है, इसलिए दिए विकल्प गलत हैं। प्रश्न आयात से पहले सुधारें। / The remaining sum is \(24\times25-9\times10=510\), not any listed option. This question should be corrected before import.
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A. दो बारTwice
पहले \(p^2\) सम होने से (p) सम सिद्ध होता है। / First \(p^2\) even proves (p) even.
फिर \(q^2\) सम होने से (q) सम सिद्ध होता है। / Then \(q^2\) even proves (q) even.
इसलिए यह नियम प्रमाण में दो बार महत्वपूर्ण रूप से आता है। / Therefore this fact is used twice in an important way.
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A. क्योंकि (b) सम सिद्ध करने के लिए (a=2k) को समीकरण में रखना होगाBecause to prove (b) even, (a=2k) must be substituted in the equation
(a) सम होने से (b) अपने आप सम नहीं होता। / (a) being even does not automatically make (b) even.
(a=2k) रखने पर \(b^2=2k^2\) मिलता है। / After substituting (a=2k), we get \(b^2=2k^2\).
तभी (b) सम सिद्ध किया जा सकता है। / Only then can (b) be proved even.
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What is the sum of the integers divisible by (14) between (100) and (350)?
C. (4158)
पद \(112,126,\ldots,350\) हैं और कुल (18) पद बनते हैं। परीक्षा में पहला और अंतिम मान सावधानी से चुनें। / The terms are \(112,126,\ldots,350\), making (18) terms. Exam tip: choose the first and last valid terms carefully.
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What is the sum of the integers divisible by (9) between (50) and (200)?
B. (2142)
पद \(54,63,\ldots,198\) हैं और कुल (17) पद बनते हैं। परीक्षा में पहला और अंतिम मान सावधानी से चुनें। / The terms are \(54,63,\ldots,198\), making (17) terms. Exam tip: choose the first and last valid terms carefully.
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Between which two consecutive integers is \( \sqrt{131}+2 \) on the number line?
B. (13) और (14)(13) and (14)
\(11<\sqrt{131}<12\), इसलिए \(13<\sqrt{131}+2<14\)। पहले वर्गमूल की सीमा निकालें। / Since \(11<\sqrt{131}<12\), \(13<\sqrt{131}+2<14\). First find the bounds of the square root.
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Between which two integers is \( -\frac{43}{11} \) located on the number line?
C. ( -4 ) और ( -3 )( -4 ) and ( -3 )
\( -\frac{43}{11}\approx-3.909 \), इसलिए यह (-4) और (-3) के बीच है। ऋणात्मक भिन्नों को दशमलव में बदलें। / \( -\frac{43}{11}\approx-3.909 \), so it lies between (-4) and (-3). Convert negative fractions to decimals.
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Between which two consecutive integers does \( \sqrt{126}-\sqrt{80} \) lie on the number line?
B. (2) और (3)(2) and (3)
\( \sqrt{126}\approx11.22 \) और \( \sqrt{80}\approx8.94 \), इसलिए अंतर लगभग (2.28) है। पहले दोनों वर्गमूलों का अनुमान करें। / \( \sqrt{126}\approx11.22 \) and \( \sqrt{80}\approx8.94 \), so the difference is about (2.28). Estimate both square roots first.
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Between which two consecutive integers is \( \sqrt{89}+1 \) on the number line?
B. (10) और (11)(10) and (11)
\(9<\sqrt{89}<10\), इसलिए \(10<\sqrt{89}+1<11\)। पहले मूल की सीमा निकालें। / Since \(9<\sqrt{89}<10\), \(10<\sqrt{89}+1<11\). First find the root bounds.
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Between which two integers is \( -\frac{31}{9} \) located on the number line?
C. ( -4 ) और ( -3 )( -4 ) and ( -3 )
\( -\frac{31}{9}\approx-3.444 \), इसलिए यह (-4) और (-3) के बीच है। ऋणात्मक भिन्नों को दशमलव में बदलें। / \( -\frac{31}{9}\approx-3.444 \), so it lies between (-4) and (-3). Convert negative fractions to decimals.
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Between which two consecutive integers does \( \sqrt{91}-\sqrt{55} \) lie on the number line?
B. (2) और (3)(2) and (3)
\( \sqrt{91}\approx9.54 \) और \( \sqrt{55}\approx7.42 \), इसलिए अंतर लगभग (2.12) है। पहले दोनों मूलों का अनुमान करें। / \( \sqrt{91}\approx9.54 \) and \( \sqrt{55}\approx7.42 \), so the difference is about (2.12). Estimate both roots first.
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Between which two consecutive integers is \( \sqrt{6}+\frac{1}{3} \) located on the number line?
B. (2) और (3)(2) and (3)
\( \sqrt{6}\approx2.449 \) और \( \frac{1}{3}\approx0.333 \), इसलिए योग लगभग (2.782) है। मिश्रित मानों में पहले अनुमान लगाएँ। / \( \sqrt{6}\approx2.449 \) and \( \frac{1}{3}\approx0.333 \), so the sum is about (2.782). For mixed values, estimate first.
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If \(P=-\sqrt{80}\), between which two consecutive integers is (P) located?
A. ( -9 ) और ( -8 )( -9 ) and ( -8 )
क्योंकि \(8<\sqrt{80}<9\), इसलिए \(-9<-\sqrt{80}<-8\)। ऋणात्मक मूलों में अंतराल सावधानी से लिखें। / Since \(8<\sqrt{80}<9\), \(-9<-\sqrt{80}<-8\). Write intervals carefully for negative roots.
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Between which two integers is \( \sqrt{40}-2 \) on the number line?
B. (4) और (5)(4) and (5)
\(6<\sqrt{40}<7\), इसलिए \(4<\sqrt{40}-2<5\)। पहले मूल की सीमा निकालें फिर घटाएँ। / Since \(6<\sqrt{40}<7\), \(4<\sqrt{40}-2<5\). First find the root bounds and then subtract.
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Between which two consecutive integers is \( \sqrt{52} \) located on the number line?
B. (7) और (8)(7) and (8)
क्योंकि (49<52<64), इसलिए \(7<\sqrt{52}<8\)। पहले निकटतम पूर्ण वर्गों को पहचानें। / Since (49<52<64), \(7<\sqrt{52}<8\). First identify the nearest perfect squares.
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Between which two consecutive integers is \( \sqrt{37}+1 \) on the number line?
A. (7) और (8)(7) and (8)
\(6<\sqrt{37}<7\) इसलिए \(7<\sqrt{37}+1<8\)। पहले वर्गमूल की सीमा निकालें। / Since \(6<\sqrt{37}<7\), \(7<\sqrt{37}+1<8\). First find the bounds of the square root.
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Between which two integers is \( -\frac{17}{5} \) located on the number line?
A. ( -4 ) और ( -3 )( -4 ) and ( -3 )
\( -\frac{17}{5}=-3.4 \) है। यह (-4) और (-3) के बीच आता है। / \( -\frac{17}{5}=-3.4 \). It lies between (-4) and (-3).
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Between which two consecutive integers does \( \sqrt{31}-\sqrt{12} \) lie on the number line?
B. (1) और (2)(1) and (2)
\( \sqrt{31}\approx5.57 \) और \( \sqrt{12}\approx3.46 \) इसलिए अंतर लगभग (2.11) है। पहले दोनों मूलों का अनुमान करें। / \( \sqrt{31}\approx5.57 \) and \( \sqrt{12}\approx3.46 \) so the difference is about (2.11). Estimate both roots first.
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Between which two consecutive integers is \( \sqrt{5}+\frac{1}{2} \) located on the number line?
A. (2) और (3)(2) and (3)
क्योंकि \( \sqrt{5}\approx2.236\), इसलिए \( \sqrt{5}+\frac{1}{2}\approx2.736\)। अनुमान लगाकर अंतराल जल्दी पहचानें। / Since \( \sqrt{5}\approx2.236\), \( \sqrt{5}+\frac{1}{2}\approx2.736\). Use estimation to identify the interval quickly.
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Between which two integers will \( \sqrt{12}-1 \) lie on the number line?
A. (2) और (3)(2) and (3)
\(3<\sqrt{12}<4\), इसलिए \(2<\sqrt{12}-1<3\)। पहले मूल का अंतराल निकालें, फिर घटाएँ। / Since \(3<\sqrt{12}<4\), \(2<\sqrt{12}-1<3\). First find the root interval, then subtract.
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Between which two consecutive integers is \( -\sqrt{7} \) located on the number line?
A. ( -3) और (-2)( -3) and (-2)
क्योंकि \(2<\sqrt{7}<3\), इसलिए \(-3<-\sqrt{7}<-2\)। ऋणात्मक संख्या में क्रम उलटने पर ध्यान रखें। / Since \(2<\sqrt{7}<3\), we have \(-3<-\sqrt{7}<-2\). For negative numbers, remember the order reverses.
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A. (4) और (5)(4) and (5)
क्योंकि (16<18<25), इसलिए \(4<\sqrt{18}<5\)। परीक्षा में वर्गों की तुलना पहले करें। / Since (16<18<25), we get \(4<\sqrt{18}<5\). In exams, compare perfect squares first.
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On the number line, \(3+\sqrt{2}\) lies between which two integers?
A. (4) और (5)(4) and (5)
क्योंकि \(1<\sqrt{2}<2\), इसलिए \(4<3+\sqrt{2}<5\)। जोड़ करने पर अंतराल भी उतना ही खिसकता है। / Since \(1<\sqrt{2}<2\), \(4<3+\sqrt{2}<5\). Adding a number shifts the interval by that amount.
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On the number line, between which two consecutive integers will \(\frac{-11}{5}\) lie?
A. (-3) और (-2)(-3) and (-2)
\(\frac{-11}{5}=-2.2\), इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक दशमलव में स्थान का ध्यान रखें। / Since \(\frac{-11}{5}=-2.2\), it lies between (-3) and (-2). Be careful with position of negative decimals.
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If \(\sqrt{10}\) is placed on the number line, between which two integers will it lie?
A. (3) और (4)(3) and (4)
क्योंकि \(3^2=9\) और \(4^2=16\), इसलिए \(3<\sqrt{10}<4\)। पहले निकटतम पूर्ण वर्ग खोजें। / Because \(3^2=9\) and \(4^2=16\), \(3<\sqrt{10}<4\). First find the nearest perfect squares.
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Between which two integers will \(\sqrt{37}-6\) lie on the number line?
A. (0) और (1)(0) and (1)
क्योंकि \(6^2<37<7^2\), इसलिए \(6<\sqrt{37}<7\) और \(0<\sqrt{37}-6<1\) है। वर्गमूल वाले अंतराल में समान संख्या घटाकर स्थिति पाएं। / Since \(6^2<37<7^2\), \(6<\sqrt{37}<7\) and \(0<\sqrt{37}-6<1\). Subtract the same number from a root interval to locate the value.
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Between which two integers will \(1-\sqrt{3}\) lie on the number line?
B. (-1) और (0)(-1) and (0)
\(\sqrt{3}\approx1.73\), इसलिए \(1-\sqrt{3}\approx-0.73\) है। यह (-1) और (0) के बीच है। / \(\sqrt{3}\approx1.73\), so \(1-\sqrt{3}\approx-0.73\). It lies between (-1) and (0).
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Between which two integers will \(-\sqrt{11}\) lie on the number line?
C. (-4) और (-3)(-4) and (-3)
\(\sqrt{11}\) (3) और (4) के बीच है इसलिए \(-\sqrt{11}\) (-4) और (-3) के बीच होगा। ऋणात्मक चिन्ह दिशा बदल देता है। / \(\sqrt{11}\) lies between (3) and (4), so \(-\sqrt{11}\) lies between (-4) and (-3). The negative sign changes the side.
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Between which two integers will \(2+\sqrt{5}\) lie on the number line?
C. (4) और (5)(4) and (5)
\(\sqrt{5}\approx2.24\), इसलिए \(2+\sqrt{5}\approx4.24\) है। यह (4) और (5) के बीच आता है। / \(\sqrt{5}\approx2.24\), so \(2+\sqrt{5}\approx4.24\). It lies between (4) and (5).
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Between which two integers will \(\sqrt{50}\) be located on the number line?
C. (7) और (8)(7) and (8)
क्योंकि \(7^2<50<8^2\), इसलिए \(\sqrt{50}\) (7) और (8) के बीच है। पूर्ण वर्गों से अंतराल तय करें। / Since \(7^2<50<8^2\), \(\sqrt{50}\) lies between (7) and (8). Use perfect squares to decide the interval.
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Between which two consecutive integers will \(\sqrt{26}\) lie on the number line?
B. (5) और (6)(5) and (6)
क्योंकि \(5^2<26<6^2\), इसलिए \(\sqrt{26}\) (5) और (6) के बीच होगा। पास के पूर्ण वर्ग देखें। / Since \(5^2<26<6^2\), \(\sqrt{26}\) lies between (5) and (6). Check nearby perfect squares.
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Between which two integers will \(-\frac{22}{7}\) lie on the number line?
B. (-4) और (-3)(-4) and (-3)
\(-\frac{22}{7}\) लगभग (-3.14) है, इसलिए यह (-4) और (-3) के बीच है। परीक्षा में ऋणात्मक दशमलव का क्रम ध्यान रखें। / \(-\frac{22}{7}\) is about (-3.14), so it lies between (-4) and (-3). In exams, note the order of negative decimals.
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Between which two integers will \(\frac{19}{6}\) lie on the number line?
B. (3) और (4)(3) and (4)
\(\frac{19}{6}=3+\frac{1}{6}\), इसलिए यह (3) और (4) के बीच है। परीक्षा में भाग देकर पूर्ण भाग पहचानें। / \(\frac{19}{6}=3+\frac{1}{6}\), so it lies between (3) and (4). In exams, divide to identify the whole part.
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Between which two integers will \(3\sqrt{2}\) lie on the number line?
B. (4) और (5)(4) and (5)
\(3\sqrt{2}\) लगभग (4.24) है, इसलिए यह (4) और (5) के बीच है। परीक्षा में \(\sqrt{2}\) को लगभग (1.414) मानकर अनुमान लगा सकते हैं। / \(3\sqrt{2}\) is about (4.24), so it lies between (4) and (5). In exams, you may estimate \(\sqrt{2}\) as about (1.414).
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Between which two integers will \(-\sqrt{30}\) lie on the number line?
B. (-6) और (-5)(-6) and (-5)
क्योंकि \(\sqrt{30}\) (5) और (6) के बीच है, इसलिए \(-\sqrt{30}\) (-6) और (-5) के बीच होगा। परीक्षा में ऋणात्मक दिशा को ध्यान रखें। / Since \(\sqrt{30}\) lies between (5) and (6), \(-\sqrt{30}\) lies between (-6) and (-5). In exams, keep the negative direction in mind.
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Between which two integers will \(\sqrt{27}\) lie on the number line?
C. (5) और (6)(5) and (6)
क्योंकि \(5^2=25\) और \(6^2=36\), इसलिए \(\sqrt{27}\) (5) और (6) के बीच है। परीक्षा में निकट पूर्ण वर्गों को याद रखें। / Since \(5^2=25\) and \(6^2=36\), \(\sqrt{27}\) lies between (5) and (6). In exams, remember nearby perfect squares.
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Between which two integers will the zero of (q(x)=3x-7) lie on the number line?
B. (2) और (3)(2) and (3)
(3x-7=0) से \(x=\frac{7}{3}\), जो (2) और (3) के बीच है। परीक्षा में पहले शून्यक निकालें फिर स्थान तय करें। / From (3x-7=0), \(x=\frac{7}{3}\), which lies between (2) and (3). In exams, first find the zero and then locate it.
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Between which two integers will \(5-\sqrt{2}\) come on the number line?
B. (3) और (4)(3) and (4)
\(\sqrt{2}\) (1) और (2) के बीच है, इसलिए \(5-\sqrt{2}\) (3) और (4) के बीच होगा। परीक्षा में घटाने पर सीमा सावधानी से बदलें। / \(\sqrt{2}\) lies between (1) and (2), so \(5-\sqrt{2}\) lies between (3) and (4). In exams, change limits carefully while subtracting.
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Between which two integers will \(2+\sqrt{3}\) lie on the number line?
B. (3) और (4)(3) and (4)
\(\sqrt{3}\) (1) और (2) के बीच है, इसलिए \(2+\sqrt{3}\) (3) और (4) के बीच होगा। परीक्षा में जोड़ने पर पूरा अंतराल आगे खिसकता है। / \(\sqrt{3}\) lies between (1) and (2), so \(2+\sqrt{3}\) lies between (3) and (4). In exams, adding a number shifts the whole interval forward.
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Between which two integers will \(-\sqrt{18}\) lie on the number line?
C. (-5) और (-4)(-5) and (-4)
क्योंकि \(\sqrt{18}\) (4) और (5) के बीच है, इसलिए \(-\sqrt{18}\) (-5) और (-4) के बीच होगा। परीक्षा में ऋणात्मक वर्गमूल में दिशा उलटी हो जाती है। / Since \(\sqrt{18}\) lies between (4) and (5), \(-\sqrt{18}\) lies between (-5) and (-4). In exams, the direction reverses for negative square roots.
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Between which two integers will \(\sqrt{10}\) be correctly represented on the number line?
B. (3) और (4)(3) and (4)
क्योंकि \(3^2=9\) और \(4^2=16\), इसलिए \(\sqrt{10}\) (3) और (4) के बीच है। परीक्षा में वर्गमूल को पूर्ण वर्गों से घेरें। / Since \(3^2=9\) and \(4^2=16\), \(\sqrt{10}\) lies between (3) and (4). In exams, bracket square roots between perfect squares.
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Approximately between which two integers will \(2-\sqrt{2}\) lie on the number line?
B. (0) और (1)(0) and (1)
\(\sqrt{2}\) लगभग (1.41) है इसलिए \(2-\sqrt{2}\) लगभग (0.59) होगा। यह (0) और (1) के बीच है। / \(\sqrt{2}\) is about (1.41), so \(2-\sqrt{2}\) is about (0.59). It lies between (0) and (1).
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Between which two integers will \(1+\sqrt{2}\) lie on the number line?
B. (2) और (3)(2) and (3)
\(\sqrt{2}\) लगभग (1.41) है इसलिए \(1+\sqrt{2}\) लगभग (2.41) होगा। यह (2) और (3) के बीच है। / \(\sqrt{2}\) is about (1.41), so \(1+\sqrt{2}\) is about (2.41). It lies between (2) and (3).
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To place \(\frac{11}{5}\) on the number line, between which two integers will it lie?
B. (2) और (3)(2) and (3)
\(\frac{11}{5}=2.2\), इसलिए यह (2) और (3) के बीच है। अशुद्ध भिन्न को दशमलव या मिश्रित रूप में बदलें। / \(\frac{11}{5}=2.2\), so it lies between (2) and (3). Convert an improper fraction to decimal or mixed form.
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Between which two consecutive integers will \(\sqrt{10}\) be located on the number line?
B. (3) और (4)(3) and (4)
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। / Since \(3^2<10<4^2\), \(\sqrt{10}\) lies between (3) and (4). Use nearby perfect squares to locate roots.
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Between which two integers will \(-\frac{7}{4}\) lie on the number line?
A. (-2) और (-1)(-2) and (-1)
\(-\frac{7}{4}=-1.75\), इसलिए यह (-2) और (-1) के बीच होगा। ऋणात्मक भिन्न में दिशा को ध्यान से देखें। / \(-\frac{7}{4}=-1.75\), so it lies between (-2) and (-1). Be careful with direction for negative fractions.
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Between which two integers will \(\sqrt{15}\) lie on the number line?
B. (3) और (4)(3) and (4)
क्योंकि \(3^2=9\) और \(4^2=16\), इसलिए \(\sqrt{15}\) (3) और (4) के बीच है। परीक्षा में वर्गमूल की स्थिति के लिए नजदीकी पूर्ण वर्ग देखें। / Since \(3^2=9\) and \(4^2=16\), \(\sqrt{15}\) lies between (3) and (4). In exams, use nearby perfect squares to locate square roots.
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