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InterviewSolution
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.
| 1051. |
What do you mean by composite numbers? |
| Answer» The number which has 2 or more than 2 factors | |
| 1052. |
If. 6 times the 6thterm of an AP equal to9th term show that it\'s 15term is zero |
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Answer» Sorry,but i think ur question is incorrect. Please check it 6(a+5d)=a+8d => 6a+30d-a-8d=0 => 5a+22d=0 => 5a=-22d |
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| 1053. |
3.5 ka 3 no all soltion |
| Answer» | |
| 1054. |
There are 5terms in ap.find the last term |
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Answer» So please send the list of numbers U not give any list of number Give a list of number |
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| 1055. |
Prove under root 5 is irrational. |
| Answer» Let us assume that √5 is a rational number.Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0⇒√5=p/qOn squaring both the sides we get,⇒5=p²/q²⇒5q²=p² —————–(i)p²/5= q²So 5 divides pp is a multiple of 5⇒p=5m⇒p²=25m² ————-(ii)From equations (i) and (ii), we get,5q²=25m²⇒q²=5m²⇒q² is a multiple of 5⇒q is a multiple of 5Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number√5 is an irrational number | |
| 1056. |
Solve by cross multiplication method , x-y+1=0 ,3x+2y-12=0.? |
| Answer» Koi to bta do answer plz friends help me | |
| 1057. |
Zero polynomials |
| Answer» A polynomial having all coefficients equal to zero , is called a zero polynomial. For eg.1 f (x)=0The degree of a zero polynomial is not defined. | |
| 1058. |
Kesa ho |
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Answer» ?? Hyeee Aap konsi class ma hu Thik h aap kese ho |
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| 1059. |
Express tan A in terms of cos A |
| Answer» √1-cos^2A/cosA | |
| 1060. |
Draw a right triangle and name it. Write all the t-ratios with respect to boththe acute angles. |
| Answer» | |
| 1061. |
the exponent of 2 in the prime factorisation of 144 is |
| Answer» 2^4*3^2 | |
| 1062. |
Me sharma |
| Answer» What is national number | |
| 1063. |
How many distinct solution if at all does the following system of equations possess ?9x-7y+1=0 |
| Answer» Infinite solutions because9x-7y+1=0→9x+7y =-1 we can put any value of x and then we can find any solution for y. so there are infinite values that we can put to X and thus there are infinite solutions for y | |
| 1064. |
999999999999999×99999999999999999 |
| Answer» 1E32answer | |
| 1065. |
If alfa nd bita are zeros of x²-5x+6 evalute alfa⁴ - bitta⁴ |
| Answer» \xa0The polynomial is,Step-by-step explanation:If\xa0\xa0and\xa0\xa0are the roots of a quadratic equation,Then the quadratic equation is,Here, the given quadratic equation is,Which having the zeroes,\xa0\xa0and\xa0.For finding the zeroes,Thus, the zeroes of the given quadratic equation are 2 and 3,(2)4 - (3)416 - 81= -65 | |
| 1066. |
√2+√5 is irrational prove that |
| Answer» Let us assume that √2+√5 is a rational number.A rational number can be written in the form of p/q where p,q are integers and q≠0√2+√5 = p/qOn squaring both sides we get,(√2+√5)² = (p/q)²√2²+√5²+2(√5)(√2) = p²/q²2+5+2√10 = p²/q²7+2√10 = p²/q²2√10 = p²/q² – 7√10 = (p²-7q²)/2qp,q are integers then (p²-7q²)/2q is a rational number.Then √10 is also a rational number.But this contradicts the fact that √10 is an irrational number.Our assumption is incorrect√2+√5 is an irrational number.Hence proved. | |
| 1067. |
What is important points in ch.5 before starting 5.1 ex |
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Answer» Understanding AP Bhai tu question paper hi mang le |
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| 1068. |
88888888+8888888 |
| Answer» 97777776 | |
| 1069. |
How many terms of AP 17 15 13 11 must be added to get some 72 ?explain that answer |
| Answer» We are given a = 17, d = -2, Sn\xa0= 72 and we have to find n.Using Sn\xa0= n[2a +(n -1)d]/2, we get72 = n[2.17 +(n -1)(-2)]/2 = n(36 -2n)/2 = n(18 -n)=> n² -18n +72 = 0 => (n -6)(n -12) = 0=> n = 6, 12Both values of n, being positive integers are valid.\xa0We get double answer because sum of 7th to 12th terms is zero, as some terms are positive and some are negative. | |
| 1070. |
Which term of AP:121,117,113....... is its first negative term? |
| Answer» a = 121d = -4let an= 0thenan = a + (n-1)d=> 0 = 121 + (n-1)(-4)=> 0 = 121 - 4n + 4=> 4n = 125=> n = 125/4=> n = 31.25as n must be natural number then n= 3232nd term is First negative number.a32 = 121 + (31)-4= 121 - 124 = -3 | |
| 1071. |
Find the value of cos145 |
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Answer» which one of the following does not belong to the territory sector Which one of the following is not an activity of primary sector On the trig unit circle, arc 145 is in Quadrant IIcos 145 = - cos (35)Calculator gives -> cos 35 = 0.82Therefor, cos 145 = - 0.82. |
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| 1072. |
Exercise 3.3 quition no. 3 |
| Answer» | |
| 1073. |
200x +400y =1How is this possible |
| Answer» It can be possible when x=-399/200 and y=1 and many more solutions... | |
| 1074. |
If 13 sin A=5 find 325 sinA - 130 cosA - 12 tan A |
| Answer» If sin A=5/13P=5cm H=13cm B=12cmCos A=12/13Tan A= 5/12325 sinA-130 cosA- 12 tanA Now you can put value and find the exact answer | |
| 1075. |
2x+3y-2x-2y=+4+5 |
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Answer» Since x=0 as 2x and -2x subtracted y=9 |
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| 1076. |
How do solve quadratic polynomial |
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Answer» There are 3 ways.1) middle term split2) completing the squares 3) quadratic formula By factorisation of the middle term |
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| 1077. |
2+45 |
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Answer» But ye class 10 ka question nahi 47 47 47 |
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| 1078. |
Find the value of a so that point (3,a) lies on the line represented by 2x-3y+5=0 |
| Answer» 2x-3y+5=0Put x = 3 and y= a2(3) -3a+5=06 - 3a + 5 = 011 - 3a =03a = 11a = 11/3 | |
| 1079. |
How will we get to know if we need to subtract or add the equation in elimination method? |
| Answer» The\xa0elimination method\xa0of solving systems of equations is also called the addition\xa0method. To solve a system of equations by\xa0elimination\xa0we transform the system such that one variable "cancels out". ... In order to solve for y, take the value for x and substitute it back into either one of the original equations. In the\xa0elimination method you\xa0either\xa0add\xa0or\xa0subtract\xa0the\xa0equations\xa0to get an\xa0equation\xa0in one variable.\xa0When\xa0the coefficients of one variable are opposites\xa0you add the equations\xa0to\xa0eliminate\xa0a variable and\xa0when\xa0the coefficients of one variable are equal\xa0you subtract\xa0the\xa0equations\xa0to\xa0eliminate\xa0a variable. | |
| 1080. |
Zeroes of p(x)=7x+3 is |
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Answer» Its a linear equation..So there will be only one zero that is -3/7 -3/7 |
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| 1081. |
√1\\6√1\\6√1\\6… prove that it is a quadratic equation |
| Answer» Can you rewrite the question? | |
| 1082. |
proove that root 2 is a irrational number |
| Answer» Let\xa0√2 be a rational number\xa0Therefore,\xa0√2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q\xa0≠ 0On squaring both sides, we get p²= 2q² ...(1)Clearly, 2 is a factor of 2q²⇒ 2 is a factor of p² [since, 2q²=p²]⇒ 2 is a factor of p\xa0Let p =2 m for all m ( where m is a positive integer)Squaring both sides, we get p²= 4 m² ...(2)From (1) and (2), we get 2q² = 4m² ⇒ q²= 2m²Clearly, 2 is a factor of 2m²⇒ 2 is a factor of q² [since, q² = 2m²]⇒ 2 is a factor of q\xa0Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1 Therefore, Our supposition is wrongHence\xa0√2 is not a rational number i.e., irrational number.Read more on Brainly.in - https://brainly.in/question/2367037#readmore | |
| 1083. |
Chapter 3all |
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Answer» NCERT solution What u want notes or ncert solution |
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| 1084. |
Aanuvasikta kisses kahte hai |
| Answer» Apni gf ka nam likhe ho kya bhai? | |
| 1085. |
2x+3y=4. 2x+2y=-10 |
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Answer» X=-19Y=14 X= -19Y= 14 Sorry x=-19 Y=14 and z=-17 |
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| 1086. |
24-5 |
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Answer» 18 Waah ? 19 ?? 19 |
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| 1087. |
How to split middle teams |
| Answer» In Quadratic Factorization using Splitting of Middle Term which is x term is the sum of two factors and product equal to last term. | |
| 1088. |
The n term of an AP is 7-4n. Find its common differences |
| Answer» Value of a is given? | |
| 1089. |
cosec A by cosecA-1 +cosecA by cosecA +1 =2+2 tan sq. A =2sec A |
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Answer» CosecA/CosecA-1 + CosecA/CosecA+1 = 2+2tan²A = 2Sec²ATry solving now..If you still don\'t get then ping me I\'ll send you the solution Its very easy.But at last you\'ve written (=2SecA) it should be (=2Sec²A) |
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| 1090. |
Find the value of k if the pair of linear equations 2x-3y=8 and 2(k-4)x-ky=k+B are inconsistend |
| Answer» The value of K will be,6 | |
| 1091. |
y= x²-2x - 8 |
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Answer» X=4 X=4 X²-4x+2x-8 => x(x-4)+2(x-4) => x=-2 or 4 Y=x2-(4-2)x-8 = X2-4x+2x-8 = X(x-4)+2(x-4) =(X+2)(x-4)Ignore (x+2) Then x=4 |
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| 1092. |
Use prime factorisation method to find hcf of 26 and 91 |
| Answer» 182 | |
| 1093. |
7cm,24Cm,25, |
| Answer» Yes,this can make a right triangle | |
| 1094. |
The radius of a circle with centre (-2,3) is 5 units the point (2,5) lies |
| Answer» The point (2,5) lies inside the circle.To check whether a point lies inside or outside the circle, we write the equation of circle and substitute the point in the equation.If the value of the equation is equal to r² then the point lies on the circle.If the value of equation is less than r² then it lies inside the circle.And if the value of the equation is greater than r² then it lies outside the circle.Writing the equation of circle first,(x -h)2 +(x-k)2 = r2 , where (h,k) is the center of circle and r is the radius of the circle.Substituting the values we get,(x -(-2))2 +(x-3)2 = 52Above equation is the equation of the circle.Now we have to find whether the point lies inside or outside the circle, we substitute the point in the equation of circle.(2 -(-2))2 +(5-3)2 = 5216+4 = 20 , which is less than r² = 5² =25.Therefore the point (2,5) lies inside the circle.\xa0 | |
| 1095. |
Real number me se ex.1.2 |
| Answer» | |
| 1096. |
If Sn=3n+5n find ap and a20 |
| Answer» | |
| 1097. |
If 10th term is 21 and sum of its first 10 terms is 120. Find nth term. |
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Answer» Thanks Let’s consider a to be the first term and d be the common difference.And we know that, sum of first n terms is:Sn\xa0= n/2(2a + (n − 1)d)\xa0and nth\xa0term is given by: an\xa0= a + (n – 1)dNow, from the question we have S10\xa0= 120⟹\xa0120 = 10/2(2a + (10 − 1)d)⟹\xa0120 = 5(2a + 9d)⟹ 24 = 2a + 9d\xa0…. (1)Also given that, a10\xa0= 21⟹\xa021 = a + (10 – 1)d⟹\xa021 = a + 9d\xa0…. (2)Subtracting (2) from (1), we get24 – 21 = 2a + 9d – a – 9d⟹a = 3Now, on putting a = 3 in equation (2), we get3 + 9d = 219d = 18d = 2Thus, we have the first term(a) = 3 and the common difference(d) = 2Therefore, the nth\xa0term is given by an = a + (n – 1)d = 3 + (n – 1)2= 3 + 2n -2= 2n + 1Hence, the nth\xa0term of the A.P is (an) = 2n + 1. |
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| 1098. |
15cotA=8,find sinA and secA |
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Answer» 15cotA=8Cot A =8/15 = adjacent side/opposite sideTherefore, adjacent side =8 And, opposite side = 15Hypotenuse = ?By phythagoras theorem, (Hypotenuse )^2 = (side)^2 + (side)^2(Hypotenuse)^2 = (8)^2 + (15)^2 (Hypotenuse)^2 = 64 +225(Hypotenuse)^2 = 289(Hypotenuse) = root 289Hypotenuse = 17Therefore, sin A = opp side/ Hypotenuse Sin A = 15/17Sec A = Hypotenuse/ adjacent side Sec A= 17/8Hope it helps you ? SinA=15/17 and secA=17/8 |
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| 1099. |
Explain chemical equation |
| Answer» It presents a symbolic way of chemical reactions in simple r from | |
| 1100. |
tan a + tan 2a + tan 3a = 3 tan a. tan 2a. tan 3a, is it correct, if yes how? |
| Answer» | |