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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 36951. |
If HCF of 65 and 117 is expressible in the form 65n_117, then find the value of n |
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Answer» 2 22 |
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| 36952. |
48 ÷47×54+67 |
| Answer» 122.14? | |
| 36953. |
What is formula of surface area of cube |
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Answer» 6 a 2 6a² 6a² 6 a square h... 6a |
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| 36954. |
what is the property of an issocelles triangle |
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Answer» \tIt has two equal sides.\tIt has two equal angles, that is, the base angles.\tWhen the third angle is 90 degree, it is called a right isosceles triangle. Ye toh vaise aapko pta hi hona chahiye tha ...kyoki ye basics toh 7th se hi kra dete hai.. An isosceles triangle is a triangle that has two equal side lengths. hurry up |
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| 36955. |
What is NTCP? |
| Answer» National Tobacco Control Programme (NTCP) | |
| 36956. |
1/a-1/b |
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| 36957. |
Find greatest 6 digit no. Exactly divisible by 24 15 and 36 |
| Answer» First of all,we will find the LCM of 24,15,36 =360The greatest 6 digit number is 999999On dividing this by 360,we get 279 as remainderHence, 999999-279=999720 is divisible by 360therefore, the required number is 999720.(Answer) | |
| 36958. |
Cbsencert retranslation for 10 class en |
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| 36959. |
if cosA+sinA=√2sinA prove that cosA-sinA=√2cosA |
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| 36960. |
What is suyllbus of class X basic maths |
| Answer» Same | |
| 36961. |
How to prove ✓7 irrational |
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Answer» let us assume that √7 be rational.then it must in the form of p / q [q ≠ 0] [p and q are co-prime]√7 = p / q=> √7 x q = psquaring on both sides=> 7q2= p2\xa0 ------> (1)p2 is divisible by 7p is divisible by 7p = 7c [c is a positive integer] [squaring on both sides ]p2 = 49 c2 --------- > (2)substitute p2 in equ (1) we get7q2 = 49 c2q2 = 7c2=> q is divisible by 7thus q and p have a common factor 7.there is a contradiction as our assumption p & q are co prime but it has a common factor.so that √7 is an irrational. Ok last mai then root 7p be rational then irrational And last write hence prove Root 7 =p/q where is not equal to 0 Let root 7 be a rational no in the form p/q Ok To Pura bata do adha answer kyon bata rage ho Let root 7 be irrational |
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| 36962. |
(x+2)(2x-1) +3what is a dividend verify |
| Answer» 2x square +3x+1 | |
| 36963. |
Find the quadratic polynomial whose zeroes are 2+root 3 and 2-root3 |
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| 36964. |
State where( underoot 6+underoot 9) is rational or irrational |
| Answer» Let root 6 + root 9 V rational root 9 equal to 3 root 6 + 3 equal to P by Q suppose P and Q have a common factor other than one root 6 + 3 equal to a upon B where A and B are co primes root 6 equal to a upon B + 3 root 6 equal to a + 3 B upon B since A and B are integers we get a + 3 B upon B is rational root 6 is rational but this contradict fact that root 6 is irrational this contradiction arise because of incorrect assumption that root 6 + root 3 is irrational hence root 6 + 3 is irrational | |
| 36965. |
If the HCF of 152 and 272 is expressible in the form 272x8+152x find x |
| Answer» On applying the Euclid’s division lemma to find HCF of 152, 272, we get{tex}272 = 152\\times1 + 120{/tex}Here the remainder =\xa00.Using Euclid’s division lemma to find the HCF of 152 and 120, we get{tex}152 = 120\\times1 + 32{/tex}Again the remainder =\xa00.Using division lemma to find the HCF of 120 and 32, we get{tex}120 = 32\\times3 + 24{/tex}Similarly,{tex}32 = 24\\times1 + 8{/tex}{tex}24 = 8\\times3 + 0{/tex}HCF of 272 and 152 is 8.272{tex}\\times{/tex}8 + 152x = H.C.F. of the numbers{tex}\\Rightarrow {/tex}{tex}8 = 272\\times8 + 152x{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}8 - 272\\times8 = 152x{/tex}{tex}\\Rightarrow 8(1- 272) = 152x{/tex}{tex}\\Rightarrow x = \\frac { - 2168 } { 152 } = \\frac { - 271 } { 19 }{/tex}\xa0 | |
| 36966. |
[email\xa0protected]+8 |
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| 36967. |
What is the 1/2 (x1(y2-y3)+x2(y3-y1)+x3(y1-y2) |
| Answer» this is the formula for area of triangle in coordinate geometry | |
| 36968. |
Find the quadratic polynomial sum of whose zeroes is √3 and their product is (2/3). |
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| 36969. |
Define Real number, Primes number, Co-prime, Composite number. Explain with the help of examples. |
| Answer» A natural number which has exactly two factors, i.e. 1 and the number itself, is a prime number. For example: 2, 3, 5, 7, 11, 19, 37, 41, 313, 241 etc.Every non-prime number is a composite number. Composite numbers are those natural numbers which have more than two factors. Such numbers are divisible by other numbers as well. For example: 4, 6, 8, 10, 12, 14, 500, 6000 etc.Numbers, which do not have any common factor between them other than one, are called co-prime numbers. For example 16 and 25 do not have any common factor other than one. Similarly 84 and 65 do not have any common factor and are hence co-prime.All rational and all irrational number makes the collection of real numbers. It is denoted by the letter R . For example: 1, -2/3, 3/4, √2, √2 + 5\xa0 | |
| 36970. |
In a rectangular ABCD, AB=13,BC=7,AC=x+y,DA=3x+y, find x&y |
| Answer» Here AC = x + y or CD = x + y | |
| 36971. |
If there are two positive integers x and y express them in terms of primes as x=pq^3 and y=p^3q |
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| 36972. |
If one zero of a polynomial (a^2 +9)x^2 +13x+ 6a is the reciprocal of the other, find the value of a |
| Answer» 3 | |
| 36973. |
If -1 and 2 are two zeroes of the polynomial 2/ |
| Answer» Given polynomial is p(x) = 2x3\xa0- x2- 5x - 2\xa0and -1 and 2 are zeroes of polynomial.{tex}\\therefore{/tex}\xa0{x - (-1)} (x - 2)= ( x + 1) (x - 2) = x2 - 2x + x - 2 = x2- x - 2 is a factor of p(x)For other zeroes, 2x\xa0+ 1 = 0{tex}\\Rightarrow x = \\frac { - 1 } { 2 }{/tex}{tex}\\therefore{/tex}\xa0Other zero = {tex}\\frac { - 1 } { 2 }{/tex} | |
| 36974. |
Chapter 2 2011 question |
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| 36975. |
Basic of polnomial |
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| 36976. |
Solve the pair of linear equation by the substitution method. 3x-y=39x-3y=9 |
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Answer» Please note given set of equations have infinte solutions ok but it is the NCERT Q Elimination by substitution :-9x-3y=9 -------> [ 1 ]\xa0Equ.3x - y = 3 ------> [ 2 ]Equ.3x = 3 + y\xa0,x = [3 + y]/ 3Put this value in equation [1 ],9\xa0× {\xa0[3 + y]/ 3 } - 3y = 9,[27+9 y ] / 3 - 3y = 9,9 [ 3 + 1 y ] / 3 = 9,3 [ 3 + 1 y ] = 9,9 + 3y = 9,3y = 9 - 9,3y = 0,y = 0,x = y + 3 / 3 = 0 + 3 / 3 = 1Hence ,x = 1 , y = 0 Naman g to solve with substitution method we must have at least 2 equation, but here both equation are equivalent to each other . |
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| 36977. |
Polinomial samajho |
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| 36978. |
Express 429 as a product of it prime factors |
| Answer» 429=3×13×11 | |
| 36979. |
Explain why (1×2×3×4×5×6)+5 is a composite number |
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Answer» Because it can be factorize Ha |
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| 36980. |
What is an real number?? |
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Answer» Ok thanks You may also say that the combination of rational and irrational are real nber The no. Which is represented on number line are real number |
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| 36981. |
How to prove that √5 is irrational? |
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Answer» let root 5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime]root 5=p/q=> root 5 × q = psquaring on both sides=> 5×q×q = p×p ------> 1p×p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p×p = 25c×c --------- > 2sub p×p in 15×q×q = 25×c×cq×q = 5×c×c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso\xa0√5 is an irrational How to prove underroot 5 is irrational |
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| 36982. |
A number when divided by 61 gives 27as quotient and 32as remainder find the number |
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Answer» স । ৬,যনকসজা ক ৃ 1679 |
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| 36983. |
Find the value of k when x³+5x²+x+2 is ➗ by5x²+2x+K |
| Answer» ५\' , र म. क* কললচ। ম ন= গ$ ুুপল^ | |
| 36984. |
√3-√2√2 |
| Answer» √3-2 | |
| 36985. |
Hcf and Lcm of two numbers are same.find the difference between the two numbers |
| Answer» Zero | |
| 36986. |
What is the lcm of smallest prime number and smallest composite number. |
| Answer» \tThe smallest composite number is 4.\tThe smallest prime number is 2.\u200b\u200b\u200b\u200bThe LCM of 4 and 2 is 4. | |
| 36987. |
Prove that cot2a(secA-1)/1+sinA=sec2A(1-sinA/1+sinA) |
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| 36988. |
135or255 which HCF |
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| 36989. |
Prove that √2 is an irrational number. |
| Answer» Given √2 is irrational number.Let √2 = a / b wher a,b are integers b ≠ 0we also suppose that a / b is written in the simplest formNow √2 = a / b ⇒ 2 = a2 / b2\xa0⇒ 2b2 = a2∴ 2b2 is divisible by 2⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2ca2 = 4c2 ⇒ 2b2 = 4c2 ⇒\xa0b2 = 2c2∴ 2c2 is divisible by 2∴ b2 is divisible by 2∴ b is divisible by 2∴a are b are divisible by 2 .this contradicts our supposition that a/b is written in the simplest formHence our supposition is wrong∴ √2 is irrational number. | |
| 36990. |
If(x+1) is a factor of 2x3+ax2+2bx+1,then find the values of a &b given that 2a-3b=4 |
| Answer» a=5 and b=2. | |
| 36991. |
If a and b are two positive integers such that a = 14b, find the HCF of a and b |
| Answer» By Euclid\'s division lemma a = bq + r where 0<=r < b Here a = 14b Therefore, 14b = 14*b + 0 So, HCF = b | |
| 36992. |
5x-4y+8=0 can you represent in graph |
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| 36993. |
P(x)=x3-3x2-x+3 |
| Answer» (x – 1), (x – 3) | |
| 36994. |
4 root 3 ka whole square square |
| Answer» 48 | |
| 36995. |
2*_3*ansqe |
| Answer» I do not understand what you are asking? | |
| 36996. |
Show that square of an odd positive integer is 8 m + 1 |
| Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q. | |
| 36997. |
What is the HCF of 1032 &408 and express it in the form 1032m-408*5 |
| Answer» Given integers are 408 and 1032 where 408 < 1032By applying Euclid’s division lemma, we get 1032 = 408 {tex}\\times{/tex}\xa02 + 216.Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as408 = 216 {tex}\\times{/tex}\xa01 + 192.Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192216 = 192 {tex}\\times{/tex}\xa01 + 24.Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24\xa0192 = 24 × 8 + 0.Now the remainder has become 0. Therefore, the H.C.F of 408 and 1032 = 24.Therefore,24 = 1032m - 408 {tex}\\times{/tex}\xa051032m = 24 + 408 {tex}\\times{/tex}\xa051032m = 24 + 20401032m = 2064 {tex}m = \\frac{{2064}}{{1032}}{/tex}Therefore, m = 2. | |
| 36998. |
2x+3y =8. ;4x+6y =7. Solve by elimination method |
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| 36999. |
If a= 2n + 13, b= n+ 7 where n is a natural number then hcf of a and b isA) 2. B) 1. C) 4. D) 3 |
| Answer» Taking different values of n, we find that A and B are co-prime.When {tex}n = 1, A =1 5, B = 8{/tex}When {tex}n = 2, A = 17, B = 9{/tex}When {tex}n = 3, A = 19, B = 10{/tex} and so on....Therefore, {tex}HCF = 1{/tex} | |
| 37000. |
If x=y then find value of a and b in polynomial ax+by=10 |
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