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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.
| 5301. |
Find the value of a and b if 11+root 7÷11- root 7 =a+b root 7 |
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| 5302. |
(8)to the power-25-(8)to the power -26 |
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| 5303. |
Ex 2.5 solution |
| Answer» Ex 2.5 solution | |
| 5304. |
If the diagonals of parallelogram are equal then show that it is a rectangle |
| Answer» Each angles are equal | |
| 5305. |
Prove that angles in the same segment of the circle are equal |
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| 5306. |
If x=1+underroot2,then the value of (x-1/x)whole square is |
| Answer» ಹಂದಿನಂತೊನೆ | |
| 5307. |
Show how √5can be represented on the line |
| Answer» Step 1: Draw a number line. Mark O as the zero on the number line.Step 2: Mark a point A as -5 on the number line.Step 3: Mark a point C as 1 on the number line.Step 4: Mark a point B as the mid-point of AC.Step 5: With point B as the centre and radius as AB draw a semicircle.Step 6: From O draw a perpendicular line to the number line that intersects the previous drawn semi-circle at D. Here OD = square root of 5.Step 7: With O as centre and radius as OD, draw an arc that intersects the number line at point E. Here E is the point square root of 5 on the number line . | |
| 5308. |
(2a-3b)3 |
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Answer» (2a-3b)³ =using identity 7solutions= (2a)³- (3b)³-3×2a×3b(2a-3b) = 8a³ - 27b³ -30a²b + 54ab² 8a^3 +36a^2b+64ab^2+27b^3(8acube + 36a square b + 64ab square + 27 b cube) 6a-9b |
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| 5309. |
If x=5-√21/2 ,find value of *x+1/x (x plus 1by x)*x2 +1/x2 (xsquare plus one by xsquare) |
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| 5310. |
X^3+2x^2-x-2 |
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| 5311. |
3*13 |
| Answer» 36 | |
| 5312. |
Chapter 6 exercise |
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| 5313. |
identitiea formulas |
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| 5314. |
Draw An angle of 135 |
| Answer» Given: A ray OA.Required: To construct an angle of 1350\xa0at O.Steps of construction :\tProduce AO to A\'\xa0to form OA\'.\tTaking O as centre and some radius, draw an arc of a circle, which intersects OA at a point B and OA\'\xa0at a point B\'.\tTaking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.\tTaking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.\t\tDraw the ray OE passing through C.\tThen {tex}\\angle{/tex}EOA = 60o.\tDraw the ray OF passing through D.\tThen {tex}\\angle{/tex}FOE = 60o.\tTaking C and D as centres and with the radius more than {tex}1 \\over2{/tex}CD, draw arcs to intersect each other, say at G.\tDraw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e. {tex}\\angle{/tex}FOG = {tex}\\angle{/tex}EOG = {tex}1 \\over2{/tex}{tex}\\angle{/tex}FOE = {tex}1 \\over2{/tex}(60o) = 30o.\tThus {tex}\\angle{/tex}GOA = {tex}\\angle{/tex}GOE + {tex}\\angle{/tex}EOA = 30o + 60o = 90o\t{tex}\\angle{/tex}B\'OH = 90o\tTaking B\'\xa0and H as centres and with the radius more than {tex}1 \\over2{/tex}B\'H, draw arcs to intersect each other, say at I.\tDraw the ray OI. This ray OI is the bisector of the angle\xa0B\'OG i.e. {tex}\\angle{/tex}B\'OI = {tex}\\angle{/tex}GOI = {tex}1 \\over2{/tex}{tex}\\angle{/tex}B\'OG = {tex}1 \\over2{/tex}(90o) = 45o\t{tex}\\angle{/tex}IOA = {tex}\\angle{/tex}IOG + {tex}\\angle{/tex}GOA\t= 45o + 90o = 135o\tOn measuring the {tex}\\angle{/tex}IOA by protractor, find that {tex}\\angle{/tex}IOA = 135o. | |
| 5315. |
If x is equal to under root 5 minus 2 then find the value of x square + 1 by x square |
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| 5316. |
3+√5/√2 |
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| 5317. |
Factorise : 3X2 +3x2 = 22x |
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| 5318. |
P(-1)3=(-1)2 + 13 (-1) +32 (-1)+20 |
| Answer» -27 | |
| 5319. |
2x+1 |
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Answer» 2x+1=02x=-1X=-1/2 1 |
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| 5320. |
Ex 1.5 |
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| 5321. |
Is 0 a ratinal number |
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Answer» Yes 0/1 is correct 0/2 ,0/3,are wrong Yes 0 is a rational number for Ex:- 0/2,0/3...(can be written in the form of p/q) No |
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| 5322. |
Exponent of Real number |
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| 5323. |
Please give me 1st ut syllabus maths |
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Answer» I want RD Sharma 2, 3, 4 chapter answer I want RD Sharma 2,3,4chapter answer |
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| 5324. |
in theadjoining figure AoB is a straight line find the value of x |
| Answer» Tari gan | |
| 5325. |
Exercise 1.4 in 1st chapter number system that theorm figure in that which line we have to start |
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| 5326. |
3- root 5 |
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| 5327. |
How many rules are there to prove a triangle is congurance |
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Answer» SAS congruenceASA congruence AAS congruence 3 |
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| 5328. |
Cube of 6859 |
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Answer» Cube root=19 19 |
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| 5329. |
9 9 X square + 12 x y |
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| 5330. |
What is CPCT and where is it used in ch 7 TRIANGLES |
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Answer» We write in short \'CPCT\' for corresponding parts of congruent triangles.It is used in questions where you have already proven something but you need to prove it again . Ex-if you already proved congruency and you need to prove that 2 sides are congruent. You can easily use CPCT. CPCT means Congruent Parts Of Congruent Triangle . It is used when you proved two triangle in case one . If you need re prove something ( like lines are equal and parallel , angle prove and so on ) in that two triangle then you write IT IS ALREADY PROVE BY CPCT |
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| 5331. |
What is the cube root of 5 |
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Answer» 2.236 is the correct answer 2.236 2.236 2.236 is absolutely correct answer |
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| 5332. |
2x+3y=9•35 |
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Answer» 2x+3y-9.35=0a=2,b=3,c=-9.35 a=2,b=3,c= -9.35 |
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| 5333. |
Prove that Root 2 is an irrational number |
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| 5334. |
Simply: (16-1/5)5/2 |
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Answer» We have,{tex}\\left(16^{-\\frac{1}{5}}\\right)^{\\frac{5}{2}}{/tex}\xa0=\xa0{tex}16^{-\\frac{1}{5} \\times \\frac{5}{2}}{/tex}\xa0[as (am\u200b\u200b\u200b\u200b\u200b\u200b)n= am×n\xa0\u200b\u200b\u200b\u200b\u200b\u200b]\xa0=\xa0{tex}16^{-\\frac{1}{2}}{/tex}=\xa0{tex}\\frac{1}{16^{\\frac{1}{2}}}{/tex}=\xa0{tex}\\frac{1}{\\left(4^{2}\\right)^{\\frac{1}{2}}}{/tex}=\xa0{tex}\\frac{1}{4}{/tex}.\xa0 (16-1/5)5/2(80/5-1/5)5/279/5×5/2=79/2 |
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| 5335. |
Find the area of triangle two sides of which are 18cm and 10cm and the perimeter is 42cm |
| Answer» P=42 cmAlso,P=18+10+x42=28+xX=14P/2=s=21By using heron\'s formula =21root11 | |
| 5336. |
What is the latest pattern of question paper of maths for session 2019-20 |
| Answer» as per cbse guidelinesquestion paper consists of\xa01m M.C.Qs 20 questions 1 x 20 = 20\xa02m 6 questions 2 x 6 = 123m 8 questions 3 x 8 = 244m 6 questions 4 x 6 = 24 total - 20m + 12m + 24m + 24m = 80m | |
| 5337. |
Point X,Y,Z are collinear such that d(XY)17, d(YZ), 8 find d(XZ). |
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Answer» 25 X-1 |
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| 5338. |
Write the divisibility test of 2 to 11 with examples |
| Answer» 4 | |
| 5339. |
(5-√5)(5+√5) |
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Answer» 5^2-√5^2=10-5=5 20 (a - b) (a + b) = a2 - b2(5 - √5)(5 + √5) = 52 - √5 2= 25 - 5= 20 5(5+*5) - *5(5+*5) = 25+5*5 - 5*5+ 5 = 30 A |
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| 5340. |
In figure if x/y = 5 and z/x = 4, then find value of x. |
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Answer» Thank you for giving me a thanks I think in figure x/y = 5 the values are x= 5 , y= 1 and in other z/x = 4 the values are z = 20,x = 5 |
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| 5341. |
2x5ab √3ab |
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Answer» 10ab√3ab What we have to do in this Question ???? Will you explain it just a little Bit, Huh!!! |
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| 5342. |
6x^2 +5x+6 factorise |
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Answer» =6x^2+9x-4x+6=3x(2x+3)-(2x+3)=(3x-2)(2x+3) =6x^2+9x-4x+6=3x (2x+3)- 2(2x+3)=(3x-2) (2x+3) |
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| 5343. |
What is zero of a polynomial |
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Answer» A number in which the value of a polynomial \'0\' is calledEx: X+1 is-1 X-4 is 42X+1 is -1/2 0 Zeroes of a polynomial p(x) is real number ‘a’ for which polynomial p(x) if p(a) = 0.E.g.: For p(x) = x-2 , p(2) = 2-2 =0. Thus 2 is zeroes for polynomial p(x)= x-2Note: Every real number is a zero of the zero polynomial p(x)=0. |
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| 5344. |
If 10^2y = 25, then 10^-y ? |
| Answer» 1/5 | |
| 5345. |
a7+ab6 factorise it |
| Answer» Consider, a7 + ab6 = a(a6 + b6) = a[(a2)3 + (b2)3] = a[a2 + b2][a4 + a2b2 + b4] | |
| 5346. |
How to do middle term split up |
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| 5347. |
Exercise 6.3 full questions with solutions |
| Answer» Check solution here : https://mycbseguide.com/ncert-solutions.html | |
| 5348. |
√5-2√2 /√20+√32 |
| Answer» √5-2√2 /√20+√32=7.260 | |
| 5349. |
7 root 3 by 10 root 3 |
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Answer» 7/10 7/3 and 10/3 7/10 |
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| 5350. |
SIMPLIFY : 5√ x^4 4√x^3 3√x^2 √x. |
| Answer» X119/120 | |