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.
| 38251. |
Polynomials means |
|
Answer» An equation which have the degree of 4 or more than 4 is known as polynomial. POLY :- means "MANY", and NOMIAL :- means "TERMS". An algebraic expressions which have non-negative power is called a polynomial |
|
| 38252. |
2x + 3y = 0 |
|
Answer» b 8 |
|
| 38253. |
Prove that number |
| Answer» | |
| 38254. |
Can two numbers have 15 as their HCF and 175 as their LCM. give reasons |
| Answer» Answer is no because hcf not divide lcm completely | |
| 38255. |
ax+by=a+b |
| Answer» Write complete question | |
| 38256. |
(y^2)÷(x+√x^2+√y^2) |
| Answer» | |
| 38257. |
Lcm of 1 and 1+root3 |
| Answer» | |
| 38258. |
Find five numbers in ap whose sum is 25 and the sum of whose square is 135 |
| Answer» | |
| 38259. |
In exercise 2.3 que 1(1) how x(square) comes below -4x(square)? |
| Answer» | |
| 38260. |
Show that any positive odd integer is in form of 4m+1 or 4m+3 |
| Answer» Let a be the positive integer and b = 4.Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.{tex}(4q)^3\\;=\\;64q^3\\;=\\;4(16q^3){/tex}= 4m, where m is some integer.{tex}(4q+1)^3\\;=\\;64q^3+48q^2+12q+1=4(\\;16q^3+12q^2+3q)+1{/tex}= 4m + 1, where m is some integer.{tex}(4q+2)^3\\;=\\;64q^3+96q^2+48q+8=4(\\;16q^3+24q^2+12q+2){/tex}= 4m, where m is some integer.{tex}(4q+3)^3\\;=\\;64q^3+144q^2+108q+27{/tex}=4×(16q3+36q2+27q+6)+3= 4m + 3, where m is some integer.Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. | |
| 38261. |
In ∆ABC,angle C=3angleB,angleC=2(angleA+angleB),then find all the angles |
|
Answer» It has been taken from practice papers from cbse guide only < A+ |
|
| 38262. |
I want maths subject in Hindi |
|
Answer» It is possible. ???????????????????? But its not possible here dear Hi abhishek |
|
| 38263. |
Method of cross multiplication method |
| Answer» Given in NCeRT | |
| 38264. |
3x-y=49x-3y=9Use elemenation methods |
| Answer» Multiply by 3 both side in equation 1 then substract equation 1&2 Ok dear | |
| 38265. |
2x-2=3 |
|
Answer» 2x-2=32x=3+2x=5/2x=2.5 2x=3+22x=5X=5÷2X=2.5 Then,, |
|
| 38266. |
How can we find coinsistent |
| Answer» Unique solution that is a1/a2is not equal to b1/b2AndInfinitly many solution that is a1/a2 = b1/b2 = c1/c2 From this we can find consistent And to find in consistent its No solution that is a1/a2 =b1/b2 is not equal to c1/c2 | |
| 38267. |
What is the requirement of the chapter introduction to trigonometry |
|
Answer» Firstly you want to learn the value of trignometric value. Solutions? |
|
| 38268. |
√2 is irrtional |
| Answer» | |
| 38269. |
Triangle ODC congruent to triangle OBA , angle BOC |
| Answer» | |
| 38270. |
Prove that any consitive integers is divisible by 3 |
| Answer» Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 3q + r ...(i)Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.Putting r = 0 in (i), we get{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ n + 2 is not divisible by 3.Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3. | |
| 38271. |
Find x and y (10y+x) + (10x+y)=132 Give the answer plz fast |
|
Answer» x+y=12 10y+x+10x+y=13210y+y+10x+x=13211y+11x=132Divide by 11x+y=12 x+y=12 Plz give me a answer for these equation |
|
| 38272. |
Find HCF of 65 and 117 |
| Answer» 13is the hcf of this qu. | |
| 38273. |
Cos45°/sec30°+cosec30° |
| Answer» 3√2-√6/8 | |
| 38274. |
3x+7y=37,5x+6y=39 |
| Answer» x=3 , y=4 | |
| 38275. |
Using Euclid division algorithm find whether the pair of numbers 847,2160 are co prime or not? |
| Answer» Here we have to find out HCF of 2160 and 847 by Using Euclid’s division Lemma, we get2160 = 847{tex}\\times{/tex}2 + 466Also 847 = 466{tex}\\times{/tex}1 + 381466 = 381{tex}\\times{/tex}1 + 85381 = 85{tex}\\times{/tex}4 + 4185 = 41{tex}\\times{/tex}2 + 341=3{tex}\\times{/tex}13 + 23 = 2{tex}\\times{/tex}1 + 12 = 1{tex}\\times{/tex}2 + 0{tex}\\therefore{/tex}HCF = 1.Hence the numbers are co-prime. | |
| 38276. |
bx+ay=a+bax (1/a -b -1/a+b) +by (1/b-a -1/b+a)=2 |
| Answer» The system of equation is given by :bx + cy = a + b ......(i){tex}ax\\left( {\\frac{1}{{a - b}} - \\frac{1}{{a + b}}} \\right) + cy\\left( {\\frac{1}{{b - a}} + \\frac{1}{{b + a}}} \\right){/tex}{tex}= \\frac{{2a}}{{a + b}}{/tex}\xa0...(ii)From equation (i)bx + cy - (a + b) = 0 ............ (iii)From equation (ii){tex} ax\\left( {\\frac{1}{{a - b}} - \\frac{1}{{a + b}}} \\right) + cy\\left( {\\frac{1}{{b - a}} + \\frac{1}{{b + a}}} \\right){/tex}\xa0{tex} - \\frac{{2a}}{{a + b}} = 0{/tex}{tex}⇒ x\\left( {\\frac{{2ab}}{{(a - b)(a + b)}}} \\right) + y\\left( {\\frac{{2ac}}{{(b - a)(b + a)}}} \\right){/tex}\xa0{tex}- \\frac{{2a}}{{a + b}} = 0{/tex}{tex} ⇒ \\frac{1}{{a + b}}\\left( {\\frac{{2abx}}{{a - b}} - \\frac{{2acy}}{{a - b}} - 2a} \\right) = 0{/tex}\xa0{tex}⇒ \\frac{{2abx}}{{a - b}} - \\frac{{2acy}}{{a - b}} - 2a = 0{/tex} 2abx - 2acy - 2a(a - b) = 0 ....(iv)From equation (iii) and (iv), we geta1 = b, b1 = c and c1 = - (a + b)and a2 = 2ab , b2 = -2ac and c3 = -2a(a - b)by cross-multiplication, we get{tex}\\frac{x}{{ - 4{a^2}c}} = \\frac{{ - y}}{{4a{b^2}}} = \\frac{{ - 1}}{{4abc}}{/tex}Now, {tex}\\frac{x}{{ - 4{a^2}c}} = \\frac{{ - 1}}{{4abc}} {/tex}{tex}⇒ x = \\frac{a}{b}{/tex}And, {tex}\\frac{{ - y}}{{4a{b^2}}} = \\frac{{ - 1}}{{4abc}} {/tex}{tex}⇒ y = \\frac{b}{c}{/tex}The solution of the system of equation are {tex}\\frac{a}{b}{/tex}\xa0and {tex}\\frac{b}{c}{/tex}. | |
| 38277. |
What is prime factor given an example |
| Answer» Any of the prime numbers that can be multiplied to give the original number.Example: The prime factors of 15 are 3 and 5 (because 3×5=15, and 3 and 5 are prime numbers). | |
| 38278. |
(1+1/tan^2A)(1-1/cos^2A)=1/sin^2A |
| Answer» | |
| 38279. |
What are co prime numbers |
|
Answer» The no which had only one common factor The numbers which have only 1 as a common factor or 1 as their HCF are called co-prime numbers. If two integer have only one common factor i.e. 1 is said to be co prime . Eg-( 7 , 9) |
|
| 38280. |
7*11*13*15+15 are composite no. |
| Answer» =7*11*13*15+15=15(7*11*13+1)=15(1002)It clearly shows that it has two factors. Hence, it is a composite numbers | |
| 38281. |
x+x=8+ +x-x=6= =13 8 |
| Answer» | |
| 38282. |
By what least number should 1029 divided to get a perfect cube. |
| Answer» | |
| 38283. |
Show that one and only of n,n+2.n+4 is divisible by 3. |
| Answer» Let the number be (3q + r){tex}n = 3 q + r \\quad 0 \\leq r < 3{/tex}{tex}\\text { or } 3 q , 3 q + 1,3 q + 2{/tex}{tex}\\text { If } n = 3 q \\text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}{tex}3 q \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 1 \\text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}{tex}( 3 q + 3 ) \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 2 \\text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}{tex}( 3 q + 6 ) \\text { is divisible by } 3{/tex}.{tex}\\therefore \\text { out of } n , ( n + 2 ) \\text { and } ( n + 4 ) \\text { only one is divisible by } 3{/tex}. | |
| 38284. |
What if hcf of 52 and 130 |
| Answer» HCF=26 | |
| 38285. |
What is hcf of 52 and 130 |
| Answer» 26 | |
| 38286. |
Value of root 2 |
| Answer» 1.41........ | |
| 38287. |
Find hcf of 65 and 117 and express in the form 65n+117m |
| Answer» HCF=13,m=2,n=-1 | |
| 38288. |
If x=7 then find the value of x+y |
|
Answer» 0 -7 7 |
|
| 38289. |
(2)*(5x+6x) |
| Answer» | |
| 38290. |
What is the pair of linear equation |
| Answer» | |
| 38291. |
55555+66666 |
| Answer» 122221 | |
| 38292. |
9th class |
| Answer» | |
| 38293. |
75/y-60/y |
| Answer» | |
| 38294. |
Show that 5-root 2 is an irrational |
| Answer» Let 5root2 be a rational number 5root2=p\\qRoot2=p\\5qNow p\\q is also a rational number But we know that root2 is irrational It contradicts our assumption Hence 5root2 is an irrational number | |
| 38295. |
Divide p(x)=x-3+4x+5g(x=x+1-x |
| Answer» | |
| 38296. |
Two number differ by 4 and their product is 192. Find the number |
|
Answer» Give answer 45,4 |
|
| 38297. |
Two number differ by 4 and their product is 192.find the number |
| Answer» 12,6 | |
| 38298. |
Take 5 odd Number and sum that is equal to the 20 |
| Answer» 1+3+5+5+3! | |
| 38299. |
Solve x and y by subtitution,elimination and cross multiplication method.1)4x+6y=15 6x-8y=14 |
| Answer» | |
| 38300. |
If a and b are to positive integers such that a |
| Answer» | |