InterviewSolution
Saved Bookmarks
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.
| 101. |
Which of the following rational numbers have terminating decimal a)7/250b)16/225 c)5/18 d)2/21 |
| Answer» (i) We have,Theorem states:\xa0Let\xa0be a rational number, such that the prime factorization of\xa0q\xa0is not of the form, where\xa0m\xa0and\xa0n\xa0are non-negative integers.Then,\xa0x\xa0has a decimal expression which does not have terminating decimal.(ii) We have,Theorem states:\xa0Let\xa0be a rational number, such that the prime factorization of\xa0q\xa0is not of the form, where\xa0m\xa0and\xa0n\xa0are non-negative integers.Then,\xa0x\xa0has a decimal expression which does not have terminating decimal.(iii) We have,Theorem states:\xa0Let\xa0be a rational number, such that the prime factorization of\xa0q\xa0is not of the form, where\xa0m\xa0and\xa0n\xa0are non-negative integers.Then,\xa0x\xa0has a decimal expression which does not have terminating decimal.(iv) We have,Theorem states:\xa0Let\xa0be a rational number, such that the prime factorization of\xa0q\xa0is of the form, where\xa0m\xa0and\xa0n\xa0are non-negative integers.Then,\xa0x\xa0has a decimal expression which terminates after\xa0k\xa0places of decimals, where\xa0k\xa0is the larger of\xa0m\xa0and\xa0n.Then,\xa0x\xa0has a decimal expression which will have terminating decimal after 3 places of decimal.\xa0 | |
| 102. |
1(0) 453 |
| Answer» 0 | |
| 103. |
In triangle ABC right angle |
| Answer» Your question is incomplete . I think so..... | |
| 104. |
Real no. Theorem 1.4 in brief |
| Answer» | |
| 105. |
The HCF (a, b) = 2 and LCM (a, b) = 27. What is the value a × b |
|
Answer» The value of a*b is 54...hope it would be helpful✌ PLEASE ANSWER IT FAST AS TOMORROW IS MY PAPER |
|
| 106. |
In a triangle ABC, write cos (B+C/2) in terms of Angel a |
| Answer» We know that,A+ B+ C = 180°B+C = 180° - ADividing both sides by 2 B+C /2 = 90° - A/2Taking cos on both sides,cos(B+C/2) = cos(90° - A/2)This gives,cos(B+C/2)= sin(A/2) | |
| 107. |
In fig. PQ and PR are tangents to circle with centre O such that angleQPR =50 |
| Answer» In quad. PQOR,∠R+\xa0∠O+\xa0∠Q+\xa0∠P = 360∠P +\xa0∠O = 90∠O = 130Now,In ∆ROQ,∠RQO +\xa0∠QOR +\xa0∠ORQ = 180°Now as the sides OQ = RO because of radius of same circle... Hence.... OQR = 25° | |
| 108. |
A dice is tossed find the probability of getting a number is even or a multiple of 3 |
|
Answer» 1/6 Reason:- A number which is both an even and a multiple of 3 is 6Therefore, P(E) = number of possible outcomes/ total number of outcomes =》 P(E) 1/6 2 by 6 kaise aayega 2 by 3 4 by6 = 2 by 3 2 by 6 |
|
| 109. |
Zeroes of quadratic polynomial and verify relationship between zeroes and coefficients 4u²+8u |
|
Answer» Here, 4u^2+8u=0 =4u(u+2)=0 4u=0 or u+2=0 U=0 or u=-2 :. Zeroes of the given quadratic equation is 0 and -2 Then, alpha=0 and beta=-2 Where, a=4, b=8, c=0 The relationship between zeroes and coefficient of given quadratic equation: Alpha+beta=-b/a :.0+(-2)=-8/4 -2=-2 Alpha×beta=c/a :.0×-2=0/4 0=0 Here, 4u^2+8u=0=4u(u+2)=04u=0 or u+2=0U=0 or u=-2:. Zeroes of the given quadratic equation is 0 and -2Then, alpha=0 and beta=-2Where, a=4, b=8, c=0The relationship between zeroes and coefficient of given quadratic equation:Alpha+beta=-b/a:.0+(-2)=-8/4-2=-2Alpha×beta=c/a:.0×-2=0/40=0 |
|
| 110. |
For the following APs, write the first term and the common difference: (i) 3, 1, – 1, – 3,... |
|
Answer» a=3, d=-2 First term (a) :- 3D :- a²-a¹ = 1-3 = -2 a=3 ;d= a2- a1=1-3=-2 |
|
| 111. |
Use Euclid division algorithm to find the HCF of 135and225 |
| Answer» By Euclid division algorithm, a=bq+r, 225>135 225=135×1+90 135=90×1+45 90=45×2+0Since, r=0:. HCF of 135 and 225 is 45 | |
| 112. |
For what value k, will the system of equations X+2y=53x+ky+15=0Have a unique solution |
|
Answer» K is not equal to 6For all the values of k other than 6 The equation has a unique solution Here, a1=1,b1=2,c1=-5a2=3, b2=k, c2=15For a pair of linear equation to have a unique solution:a1/a2 not equal to b1/b2=1/3 not equal to 2/k ( solve this equation) ,we getK=6:.The value of k=6 |
|
| 113. |
Chapter 3,4,5 ncert solutions in very easy way |
| Answer» | |
| 114. |
Volume of cylinder |
|
Answer» A cylinder can be seen as a collection of multiple congruent disks, stacked one above the other. In order to calculate the space occupied by a cylinder, we calculate the space occupied by each disk and then add them up. Thus,\xa0the volume of the cylinder can be given by the product of the area of base and height.For any cylinder with base radius ‘r’, and height ‘h’, the volume will be base times the height.Therefore, the cylinder’s volume of base radius ‘r’, and height ‘h’ = (area of base) × height of the cylinderSince the base is the circle, it can be written asVolume = πr2\xa0× hTherefore,\xa0the volume of a cylinder =\xa0πr2h cubic units. I mean πr^2h πr^2 |
|
| 115. |
What is f of x |
|
Answer» Yes, please write complete question. ?✌ Please write complete question. ?✌ |
|
| 116. |
For any positive integer n,prove that n^3-n is divisible by 6 |
|
Answer» |
|
| 117. |
solve graphically and find the coordinates of the vertixes of the triangle.a)y=x y=2x x+y=6 |
| Answer» | |
| 118. |
2/√x |
| Answer» | |
| 119. |
AOBC is a rectangle whose three vertical are A(0,3)O(0,0)B(5,0).Find the length of a its diagonal |
| Answer» If AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0), then the length of its diagonal isSolution:Now, length of the diagonal AB = Distance between the points A(0, 3) and B(5, 0).\xa0∴ Distance between the points (x,, y,) and (x , y ),Hence, the required length of its diagonal is √34. | |
| 120. |
Prove ³√6 is a irrational number |
| Answer» | |
| 121. |
The circumference of a circle exceed the diameter by 16.8cm.Find the circumference of the circle. |
|
Answer» Circumference of the circle exceed the diameterby 16.8. 2pir |
|
| 122. |
Koee answer bata do |
| Answer» ??? | |
| 123. |
Write 10 term of ap 2,6,10,14 |
|
Answer» a=2d=4n=10a^n=a+(n-1)d a=2d=6-2d=4an=a+(n-1)*da10=2+(10-1)*4a10=2+9*4a10=2+36a10=38 A.P:- 2,6,10,14,18,22,26,30,34,38. |
|
| 124. |
Write formula to find a slant height of cone |
|
Answer» l^2 =h^2 +r^2 L=√r2+h2 Answer - under root h square minus r square L ka square is equal to under root r ka square plus h ka square |
|
| 125. |
what is the formula of standard quadratic equation |
|
Answer» ax2+bx+c=0, a is not=to 0 ax^2+ bx + c = 0 The standard form of a quadratic equation is\xa0ax2+bx+c=0, where a,b and c are real numbers and\xa0a≠0.‘a’ is the coefficient of\xa0x2. It is called the quadratic coefficient. ‘b’ is the coefficient of x. It is called the linear coefficient. ‘c’ is the constant term. |
|
| 126. |
Prime factor of 156 |
|
Answer» 2 × 2 × 3 × 13 are the factor of 156. 2^2*3*13 |
|
| 127. |
3x+2k+5 find the value of k |
| Answer» Write complete question ?? plz?? | |
| 128. |
If 3 power (x+y) =9 and 31 (x-y) =3, then (x,y) = |
| Answer» X=3,Y=0 | |
| 129. |
Find M if root 3 minus M is a factor zero of a polynomial equals to 2 x square minus 2 root 3 MX |
| Answer» | |
| 130. |
Please suggest a reference book for mathematics |
|
Answer» NCERT Exemplar Together with U can download ncert class 10 book offine all books are which u want S Chand.. Pearson etc... Rd Sharma ,Rs Aggrawal, |
|
| 131. |
Chapter 5,, e.x 5:3 question number 3 |
| Answer» Solution :- (i) given that a=5 , d=3, an=50 we know nth term of an AP, an= a+(n-1)d =50=5+(n-1)3 =50-5=(n-1)3 =45=(n-1)3 =45/3=n-1 =15=n-1 =n=15+1=n=16Now,sum of n terms, Sn=n/2 [a+an] Sn=16/2 [5+50] sn=16/2 ×55 Sn=8×55=440 . (ii) given that,a=7,a13 =35 we know nth term of an AP, an =a+n-1 =a13=7+(13-1)d=35=7=12d=35-7=12d=28=12d=d=28/12=7/3Now,sum of n terms Sn =n/2 [a+an] S13=13/2 [7+25] S13=13/2×42S13=273.(iv) given that a3=15, S10 =125 we know, nth term of an AP, An=a+(n-1)= a3=a+(3-1)d--------------> (1)Sum of n terms Sn=n/2 [2a+(n-1)d]=S10=10/2 [2a+(10-1)d]=S10=5(2a+9d)-------------> (2)multiplying equ.(1) by (2) we get 30=2a+4d ----------> (3)Subtracting equ.(2) from equ (3) we get 2a+4d=30 2a+9d=25 (-) (-) _____________ -5d =5 d=___5__= -1 -5subtracting the value of d in the equ (1)we get 15=a+2(-1) =15=a+2=a=15+2=a=a=17Therefore a10 =a+(10-1)da10=17+(10-1)(-1)a10=17+(10-1)(-1)a10=17+9×(-1)a10=17-9a10=8 thus , d= -1 and a10=8 (v) given that, d=5, Sq=75 we know sum of n terms, Sn=n/2 [2a+(n-1)d] = Sq=9/2 [2a+(9-1)]=75=9/2 [2a+40] =75×2=9[2a+40]=150×18a+360=18a=150-360=18a= -210=a= -210/18=-35/3we know, an =a+(9-1)da9=-35/3 +(9-1)5a9= -35/3 +40a9= -35+120/3 =85/3 thus a=-35/3 and a9=85/3(vi) given a=2, d=8, Sn=90we know,sum of n terms,Sn =n/2 [2a+(n-1)] =90=n/2 [2 (2)+(n-1)4]=90= n/2×2 [2+(n-1)4]=90=n [2+(4n-4]=90= n[4n-2]=90=4n^ - 2n=4n^-2n-90=0=2n^-n-45=0=2n^-10n+9n-45 (splitting the middle term)=2n(n-5)+9(n-5)=0=(n-5)(Zn+9)=0=n-5=0 we know nth term of an AP = an =an=2+(5-1)8=an=2+2 4×8=an= 2+32=an=34 thus, n=5 and an=34(viii) given that an=4,d=2, Sn=-14we know nth term of an AP,an =an=a+(n-1)2=4=a+2n-2=6=a+2n=a=6-2n ------------>(1)we know sum of n terms Sn=n/2[2a+(n-1)]= -14=n/2×2[a+(n-1)]= -14=n[a+(n-1)] -------------->subtracting equ (1) in equ.(2) we get -14= [6-2n+n-1]= -14=n [5-n]= -14=5n-n2=n2-5n-14=0=n2-7n +2n-14=0=n(n-7)+2 (n-7)=0=(n-7)(n+2)=0=n-7=0=n=7 subtracting the value of\'n\' in the equ (1)we get =a=6-2(7)=a=6-14=a= -8 thus n=7 and a= -8(ix) given that a=3, n=8 and s=192 we know the sum of n terms of an AP Sn= n/2[a+an]=192=8/2 [3+an]=192=8/2[3+an]=192/4=[3+an]=192/4=3+an=48=3+an=an=48-3=an=45we know nth term of an APan=a+(n-1)d=45=3+(8-1)d=45=7d=d=6 thus d=6 | |
| 132. |
If sin 0=cos 0than the value of tan 0+cot 0 is |
| Answer» Giorgi | |
| 133. |
If p times pth is same as q times qth term and first term is Pq, then find the sum of (p+q)terms. |
| Answer» p[a+(p−1)a]=q[a+(q−1)d]ora(p−q)+(p2−q2)d+(q−p)d=0ora(p−a)+(p+a)(p−q)−(p−q)d=0ora+(p+q−1)d=0This is the\xa0(p+q)\xa0th term. | |
| 134. |
For an AP if a=4/3 and d=2/3....then what is an?? |
| Answer» | |
| 135. |
Find the HCF and LCM of 510 and 92 and also very it |
|
Answer» 510 = 92 x 5 + 5092 = 50 x 1 + 4250 = 42 x 1 + 842 = 8 x 5 + 28 = 2 x 4 + 0∴ HCF of 510 and 92 = 2Product of two numbers = Product of their LCM and HCF510 x 92 = 2 x LCMLCM = (510 x 92) / 2 = 23460∴ LCM of 510 and 92 = 23460. 510 = 92 x 5 + 5092 = 50 x 1 + 4250 = 42 x 1 + 842 = 8 x 5 + 28 = 2 x 4 + 0∴ HCF of 510 and 92 = 2Product of two numbers = Product of their LCM and HCF510 x 92 = 2 x LCMLCM = (510 x 92) / 2 = 23460∴ LCM of 510 and 92 = 23460. |
|
| 136. |
HCF of (65,117) |
|
Answer» Ans=13?? By Euclid\'s division algorithm\xa0117 = 65x1 + 52.65 = 52x1 + 1352 = 13x4 + 0Therefore 13 is the HCF (65, 117). 13?? |
|
| 137. |
Find the HCF and LCM 519 AND 92 And also verify it |
| Answer» 510 = 92 x 5 + 5092 = 50 x 1 + 4250 = 42 x 1 + 842 = 8 x 5 + 28 = 2 x 4 + 0∴ HCF of 510 and 92 = 2Product of two numbers = Product of their LCM and HCF510 x 92 = 2 x LCMLCM = (510 x 92) / 2 = 23460∴ LCM of 510 and 92 = 23460. | |
| 138. |
If sin(A+B)=1 & cos(A-B)=1, 0° |
|
Answer» G A=B=45 |
|
| 139. |
Explain why 7*11*13+13 and 7*6*5*4*3*2*1+5 are composite numbers? |
|
Answer» Find the number of zeroes x2 -2x - 8 Since 7*11*13+13=13(7*11+1)=13(77+1)=13(78)=13*78We no products of prime no. And co preme is composite no.Similarly7*6*5*4*3*2*1+5=5(7*6*4*3*2*1+1)=5(1008+1)=5(1009)=5*1009Product of prime numbers are composite |
|
| 140. |
Find area of quadrilateral 2|x| + 3|y|=6 ? |
| Answer» | |
| 141. |
Prove sec ∅√(1-sin2∅)= 1 |
|
Answer» LHS= secA√(1-sin²A) [using identity cos²A=1-sin²A] = secA√cos²A = secA.cosA = 1/cosA.cosA =1 = RHSHence proved. Here LHS=secA√(1-sin²A)Since 1-sin²A=cos²A and secA=1/cosA=1/cosA√cos²A=1/cosA *cosA=1=RHSHERE I TAKE A FOR THITHA |
|
| 142. |
what is the syallabus for term 1 |
| Answer» Yes I | |
| 143. |
X²-x+3 |
| Answer» x²-x+3a=1, b= -1 &c=3.D=b²-4ac =(-1)²-4(1)(3) =1-12 =-11Since discriminate is smaller than 0.Therefore no real roots can exists for this equation. | |
| 144. |
ncert book ch 12 example 4 |
| Answer» Open page 232 of tge ncert | |
| 145. |
If I have taken basic maths this year, then I will be eligible for taking PCM or not in 11th? |
|
Answer» No ways No NO No |
|
| 146. |
If p,q,r on AP then find value of p-q/q-r |
| Answer» Given: p, q & r in AP.To Find: p-q/q-rSolution:Since p,q and r are in AP.Therefore difference must be equal.(p-q )=(q-r) ;(p-q)/(q-r)=1.Thus, answer is 1. | |
| 147. |
. The quadratic equation whose roots are 3 and -3 is |
| Answer» Sum of roots = 3-3 = 0product of roots = 3*-3=-9quadratic equation is x²-(sum of roots)x+product of roots=x²-0x-9=x²-9 | |
| 148. |
289÷17 |
|
Answer» 17 is the correct answer ?? 17 17 Ans=17, and 17² =289 Ans. =17 |
|
| 149. |
Who is the author of geometry |
|
Answer» Euclid was the author of Geometry. Euclid hiii |
|
| 150. |
What is the ramanujan special number |
|
Answer» Mr. Hardy quipped that he came in a taxi with the number \'1729\' which seemed a fairly ordinary number. Ramanujan said that it was not. 1729, the Hardy-Ramanujan Number, is the smallest number which can be expressed as the sum of two different cubes in two different ways.1729 is the sum of the cubes of 10 and 9 - cube of 10 is 1000 and cube of 9 is 729; adding the two numbers results in 1729.1729 is also the sum of the cubes of 12 and 1- cube of 12 is 1728 and cube of 1 is 1; adding the two results in 1729. 1729 Which is the cube root of 10 cube plus 9 cube and also 1 cube plus 12 cube |
|