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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.
| 4301. |
If 1+4=5 2+5=113+6=21 thenFind 21+24=???? Find then tell by comment |
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Answer» The correct ans. Is 528 because 24×21+24=528 Wrong 45 |
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| 4302. |
23+25 |
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Answer» 48 It is 48 |
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| 4303. |
GIVENTanA=m TanBSinA=n SinBTO PROVECos^2A=[m^2-1]\\ [n^2-1] |
| Answer» Given,\xa0tan A = n tan B{tex} \\Rightarrow{/tex} tanB = {tex}\\frac{1}{n}{/tex}tan A{tex}\\Rightarrow{/tex}\xa0cotB =\xa0{tex}\\frac { n } { \\tan A }{/tex}..........(1)Also given,\xa0sin A = m sin B{tex}\\Rightarrow{/tex}\xa0sin B =\xa0{tex}\\frac{1}{m}{/tex}sin A{tex}\\Rightarrow{/tex}\xa0cosec B =\xa0{tex}\\frac { m } { \\sin A }{/tex}.....(2)We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-{tex} \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } } { \\tan ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = sin2A{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = 1 - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = n2cos2A - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = (n2 - 1) cos2A{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex}\xa0cos2A | |
| 4304. |
(2+2)-(2+2)×1 |
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Answer» How is your pre board going on ? Hey frnds from where you all are from ? 4 - 4*1=0 0 0 |
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| 4305. |
In board exam, steps of construction in geometry carry any marks? |
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Answer» Yes Yes Yes Noooo Ya 1 or 2 marks Yes |
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| 4306. |
How to get new sample pappers |
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Answer» Yes www.cbseacademics.nic.in |
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| 4307. |
marking scheme for class 10 |
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Answer» number system.......6algebra......20coordinate geometry......6geometry........15trignometry.......12mensuration.....10probablity and statestics.....11 What scheme |
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| 4308. |
Find the LCM and HCF of 2, 5, 50. |
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Answer» LCM 50 and hcf 1 Lcm 50 and hcf 1 LCM:150HCF:1 |
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| 4309. |
Sum of a number and it\'s square out squareroot is 240 find number |
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| 4310. |
Geometry |
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| 4311. |
Cosec2 63°+tan2 24°/cot2 66° +Sec2 63°+cos63°sin27°+sin 27° sec63°/2 (cosec2 65°_tan2 25°) |
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Answer» 1 + tan2\xa00\xa0= sec2\xa00\xa0{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}\u200b\u200b\u200b\u200b\u200b Sec2 (90-24) +tan2 24/tan2 (90-24)+Sec2 27Sin2 63+cos 63.cos (90-63)+sin 27. Cosec 27/2 (Cosec2 65-cot2 65)Sec2 24+ tan2 24/tan2 24+Sec2 27+Sin2 63+cos2 63 + SIN 27. COSEC 27/2×11+1+1/2=2/2=11+1=2 Ans. |
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| 4312. |
show that exactly one of the number n, n+2, n+4 is divisible by 3 |
| Answer» Do by algoritum ch1 | |
| 4313. |
Find the value of k so that the equations 3x-2y-7=0, kx+5y+8=0 have a unique solution |
| Answer» Given equations are3x - 2y = 0 ...........\xa0(1),kx + 5y = 0 ..........\xa0(2)From (1){tex}\\Rightarrow{/tex}\xa03x = 2yx=\xa0{tex}\\frac{2y}{{3}}{/tex}Now we put the value of x in equation (2).\xa0k({tex}\\frac{2y}{{3}}{/tex}) + 5y = 0k({tex}\\frac{2y}{{3}}{/tex}) = - 5y2ky = -3{tex}\\times{/tex}5y2k = -15k = {tex}\\frac{-15}{{2}}{/tex}So the value of k = -7.5 | |
| 4314. |
How to find middle term of an arithmetic progression |
| Answer» first of all , calculate the no .of terms if n is odd then n+1 divided by 2 , if n is even then n/2 and n/2+1 | |
| 4315. |
Value of tan15° |
| Answer» tan(15°) = tan(45°-30°)We know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B)⇒tan(45°-30°) = (tan45°- tan30°)/(1+tan45°tan30°)= {1- (1/√3)} / {1+(1/√3)}∴ tan15° = (√3 - 1) / (√3 + 1)May help you | |
| 4316. |
Find the area of triangle with vertices (0,4), (0,0) and (3,0). |
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Answer» 0 i think.. 6 |
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| 4317. |
If nth term of an AP is 78 fund 4th term |
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Answer» Mujhe bhi Aida hi lagta h I think question pura nhi h... |
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| 4318. |
3x^2+5^2+7 |
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| 4319. |
prove tangent recant theorem |
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Answer» Not possible here No I will nat prove it what you can do |
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| 4320. |
Evaluate 3tan²30+tan 60th+cosec30-tan45/////////cot²45 |
| Answer» {tex}=\\frac { 3 \\tan ^ { 2 } 30 ^ { \\circ } + \\tan ^ { 2 } 60 ^ { \\circ } + cosec 30 ^ { \\circ } - \\tan 45 ^ { \\circ } } { \\cot ^ { 2 } 45 ^ { \\circ } }{/tex}{tex}= \\frac { 3 \\times \\left( \\frac { 1 } { \\sqrt { 3 } } \\right) ^ { 2 } + ( \\sqrt { 3 } ) ^ { 2 } + 2 - 1 } { ( 1 ) ^ { 2 } }{/tex}{tex} = \\frac { 3 \\times \\frac { 1 } { 3 } + 3 + 2 - 1 } { 1 }{/tex}{tex}=1+3+2-1{/tex}{tex}=6-1{/tex}{tex}=5{/tex} | |
| 4321. |
If cosec theta- sin theta=a cube, sec theta - cos theta =b cube , prove that asq.bsq.(asq.+bsq.) =1 |
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| 4322. |
Find hcf if 525&3000,is composite 3,5,15,25,75 |
| Answer» It is given that 3, 5, 15, 25 and 75 are the common factors of 525 and 3000.\xa0So 75 is the highest common factor.Hence HCF of 525 and 3000 is 75. | |
| 4323. |
cosx[1/1-sinx-1/1+sinx]can be written in form ktanx and find k |
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| 4324. |
RQ and TP are perpendiculars to PQ and TS is perpendicular to PR. prove that ST. AC = PS.PQ |
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| 4325. |
25/5(2+3) |
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Answer» 25 1 |
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| 4326. |
If Cos A =2/5,find the value of 4+4 tan 2A |
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Answer» 4+4tan2a=4 (1+tan2a)=sec2a=1/cos2a=(1/2/5)2=(5/2)2=25/4 25 Find sin2A by identity sin2A=1-cos2A they find tan2A Consider angle A of ∆ABC, U WILL GET D ANSWER |
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| 4327. |
HCF of 2 and 3 |
| Answer» 1 | |
| 4328. |
If Cos A=2/5, find the value of 4+4 tan2A |
| Answer» 25 | |
| 4329. |
Hcf of first prime and composite number |
| Answer» The smallest prime number is 2 and the smallest\xa0composite number is\xa0{tex}4=2^2{/tex}Hence the\xa0required HCF (4\xa0, 2) = 2 | |
| 4330. |
ncert exemplar guide for maths is good to refer or not |
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Answer» it\'s very good guide it can be Nice** It is nicr |
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| 4331. |
Find the eleventh term from last term of an ap 27,23,19.... |
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Answer» An matlab last term jo ki eleven haii shayad First find its an term that will be its last term; then just find it Question is wrong bro.... Because the last term is not given..... BRO.... Question is wrong bro.... Pehle a-27,d(-4), then an-a plus n-1 *d |
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| 4332. |
Books name for class10 maths by cbse? |
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Answer» *maths Mathis textbook for class 10 ncert |
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| 4333. |
Use Euclid s division lemma to find the HCF of 135&225 |
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Answer» 45 135\t(225)1 - 135 95(135)1 - 95 45(95)2 - 90 5(45)9 - 45 XThen, will be 5 225=135×1+90135=90×1+4590=45×2+0So, H.C.F. of 225 &135 is 45 |
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| 4334. |
Ncert maths ex- 6.4 question no. 5 please.... |
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| 4335. |
If secA+tanA =2 then find sinA+cosA |
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| 4336. |
0/0 |
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Answer» Not 0/0 1/0 is not defined Not defined 0 |
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| 4337. |
What is the altitude of equaliterl triangle of each side 6 cm |
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| 4338. |
Centroid of a triagle is at originA( a,b) B (b,C)C(C,a) than find the value of a |
| Answer» Centroid is a point where all the three medians of the triangle intersect. So,the centroid\xa0of triangle can be found by finding the average of the x-coordinate\'s value and the average of the y-coordinate\'s value of all the vertices of the triangleThe vertices of\xa0{tex}\\Delta ABC{/tex}\xa0are (a, b), (b, c) and (c, a)\xa0Therefore Centroid is{tex}\\left( \\frac { x _ { 1 } + x _ { 2 } + x _ { 3 } } { 3 } , \\frac { y _ { 1 } + y _ { 2 } + y _ { 3 } } { 3 } \\right){/tex}or\xa0{tex}\\left( \\frac { a + b + c } { 3 } , \\frac { b + c + a } { 3 } \\right){/tex}But centroid is (0, 0).{tex}\\Rightarrow{/tex}\xa0a + b + c = 0 | |
| 4339. |
✓x+y=11 and x+✓y=7 find the value of x and y |
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| 4340. |
If tan theta + 8 equal to 2 prove that tan theta + 2 theta equal to zero |
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| 4341. |
How to make perfect in math |
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Answer» Do daily practice of 2hr...minimum Only do and do practice |
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| 4342. |
If p(x)=ax^3+bx^2+cxc+d is polynomial & a+b+c+d=0,then what is the factor of p(x)? |
| Answer» Sorry i dont understand that | |
| 4343. |
If the point (x,y)is equidistant from the the point (a+b,b-a) and (a-b,a+b) prove that bx=ay |
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Answer» Take (x,y)A (a+b,b-a)B and (a-b,a+b)C then distance AB=AC Use distance formuale |
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| 4344. |
Can you please release more sample papers book |
| Answer» Only cbse we can\'t | |
| 4345. |
A hemispherical tank full of water is tempted |
| Answer» So | |
| 4346. |
Solve the equation: (a-b)x+(a+b)=a2-2ab-b2 |
| Answer» x = a + b{tex}\\therefore{/tex} (a - b)(a + b) + (a + b)y = a2 - b2 - 2aba2 - b2 + (a + b)y = a2 - b2 - 2ab{tex}y = \\frac{{ - 2ab}}{{a + b}}{/tex} | |
| 4347. |
2×3etff |
| Answer» etff ???? | |
| 4348. |
The COMic two no. is 192 and their product is 3072 . Find their HCF |
| Answer» What is 192? | |
| 4349. |
Guys maths mein konsi book se practice karte ho |
| Answer» RD Sharma | |
| 4350. |
Cos^2+cos^2 *cot^2=to prove cot^2 |
| Answer» Cos^2+cos^2 *cot^2=to prove cot^2 | |