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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.
| 26301. |
Ho w to save a paper in our phone |
| Answer» Pdf download ker lo | |
| 26302. |
if cos(alpha+beta)=0 then find the value of sin(alpha-beta) |
| Answer» | |
| 26303. |
What is the sum of a triangle |
| Answer» Sum of angle of triangle=180 | |
| 26304. |
For what value of k will the pair of equations have no solution |
| Answer» equation???? | |
| 26305. |
2(a+b) |
| Answer» (2a+2b) | |
| 26306. |
In triangle ABC, AD perpendicular BC and BD=1upon3CD. prove that 2AC square =2AB square +BC square |
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| 26307. |
AS perpendicular on ZY and BY perpendicular on ZY such that AY=ZY and BY=XT. Prove that AB=ZX |
| Answer» | |
| 26308. |
Prove that √3 is irrational? |
| Answer» yha prove kre??? | |
| 26309. |
Show that product of three positive integer is divisible by 6 |
| Answer» Let three consecutive numbers be x, (x + 1) and (x + 2)Let x = 6q + r 0 {tex}\\leq r < 6{/tex}{tex}\\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}{tex}\\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}if x = 6q then which is divisible by 6{tex}\\text { if } x = 6 q + 1{/tex}{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}{tex}= 6 ( 3 q + 1 ) \\cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}which is divisible by 6if x = 6q + 2{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}{tex}= 6 ( 2 q + 1 ) \\cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}Which is divisible by 6{tex}\\text { if } x = 6 q + 3{/tex}{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}which is divisible by 6{tex}\\text { if } x = 6 q + 4{/tex}{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}which is divisible by 6if x = 6q + 5{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}which is divisible by 6{tex}\\therefore {/tex}\xa0the product of any three natural numbers is divisible by 6. | |
| 26310. |
Show that exactly one of the number n,n+2 orn+4 is divisble 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}. | |
| 26311. |
4u²+8u |
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Answer» Sorry bhai nahi sister See in ncert book in chapter 02 exercise 2.2merebhai |
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| 26312. |
2power 559 unit digit |
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| 26313. |
What is the total surface area of a frustum of a cone |
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Answer» πl(r+R)+πRsqr+πrsqr Book m h |
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| 26314. |
S and T trisect the side QR of a right triangle PQR, prove that : 8PT^2 = 3PR^2 + 5PS^2 |
| Answer» Given that a right angle\xa0{tex}\\triangle PQR{/tex}\xa0in which\xa0{tex}\\angle P Q R = 90 ^ { \\circ }{/tex}And S and T are the points of trisection of QR. Thus, QS = ST = TRTo Prove: 8PT2 = 3PR2 + 5PS2.Proof: Let QS = ST = TR = x unitsThen, QS = x units, QT = 2x units and QR = 3x units.\xa0From right angle triangles PQS, PQT and PQR where\xa0{tex}\\angle Q= 90^°{/tex} in all, by using Pythagoras Theorem, we havePS2 = PQ2 + QS2 ... (i)\xa0PT2 = PQ2 + QT2... (ii)\xa0and PR2 = PQ2 + QR2\xa0\u200b\u200b\u200b\u200b\u200b\u200b... (iii)Multiplying (iii) by 3, (i) by 5 and (ii) by 8Adding thus obtained (iii) and (i) and Subtracting (ii) from it, we get{tex}\\therefore{/tex}\xa03PR2 + 5PS2 - 8PT2=3(PQ2 + QR2) + 5(PQ2 + QS2) - 8(PQ2 + QT2)= 3QR2 + 5QS2 - 8QT2{tex}= 3 \\times ( 3 x ) ^ { 2 } + 5 ( x ) ^ { 2 } - 8 \\times ( 2 x ) ^ { 2 }{/tex}\xa0[{tex}\\because{/tex}\xa0QR = 3x, QS = x and QT = 2x](27x2\xa0+ 5x2\xa0- 32x2\xa0) = 0Thus, 3PR2 + 5PS2 - 8PT2 = 0Hence, 8PT2 = 3PR2 + 5PS2Hence Proved\xa0\xa0 | |
| 26315. |
Is 7×6×5×4×3×2×1+5 a composite number? Justify your answer |
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Answer» Let 7*6*5*4*3*2*1+5 = n Now taking 5 common => 5(7*6*4*3*2*1+1)=> 5 (1008+1)=> 5*1009This shows factors of n are n,1 ,5and 1009Since it has more than 2 factors therefore its a composite no. Its too easy dear.dont worry net perbhi search kiya to mil jayenga Maths ki 10th ki ncert book me hai Same here Aaj tk maths me full lata rha par kewal composite no. Hi smj nhi aaya Solve it pta chl jayga |
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| 26316. |
Solve for x and y : 4/x + 3y = 8 ; 6/x - 4y = -5 |
| Answer» the answer is | |
| 26317. |
CosA/1+sinA + 1+sinA/cosA =2secA |
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| 26318. |
Sec2€+cosec2€=sec2€×cosec2€ |
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| 26319. |
Find the least number which is divisible by all the numbers from 1 to 10 both included |
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Answer» Yes.. Ans.2520 Take LCM of 1,2,3,4,5,6,7,8,9,10 |
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| 26320. |
Which book to study for examination |
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Answer» NCERT + Last year question papers S.chand :science NCERT and fr reference S.C is best for Science and Oswaal que.bank for maths...they r vry effective and good stdy material Ncert + sample paper Ncert Ncert |
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| 26321. |
Sin theta + cos theta =2^ then sec theta + cosec theta = |
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| 26322. |
In triangle ABC, DE // AB If AD=2x, DC=x+3, BE =2x- 1 , CE=x Find x |
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Answer» X=-3/7 X=7/3 |
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| 26323. |
What is the Time Table of class 10 |
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Answer» 6 March hindi12 ..English16....science 20 ..kannda 23...social....28 ...math..? Sme time leter |
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| 26324. |
What is the syllabus of maths class 10 for 2018 boards |
| Answer» 28 march 2018 | |
| 26325. |
1/2a+b+x=1/2a+1/b+1/2x |
| Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a | |
| 26326. |
VBQ question from all the chapter? |
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Answer» Sory anamika ? of science Chapter???? |
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| 26327. |
(X,Y )(1,2)(7,0)are collinear. What is the relation between X and Y? |
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Answer» Area of ∆ formed =0 [x1=X ,x2=1 ,x3=7 ,y1=Y ,y2=2 ,y3=0]1/2{x1(y2-y3)+x2 (y3-y1)+x3 (y1-y2)} =0X(2-0)+1(0-Y)+7(Y-2)=02X-0+1-Y+7Y-14=02X-13-6Y=0Hence ,X and Y form a linear equation. Collinear ki property lags do |
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| 26328. |
LCD of 18 |
| Answer» | |
| 26329. |
If x, y, z are in AP., then find the value of(x+y-z) (y+z-x) |
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| 26330. |
Gn.... |
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Answer» GN Gn Gn |
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| 26331. |
Find the perimeter of a quadrant of a circle with radius r |
| Answer» πr/2 +2r(πr +4r)/222r + 28r /2 50r/2=25r | |
| 26332. |
Given that HCF (306,657)=a find LCM? |
| Answer» HCF*LCM=product of two numberA*LCM=306*657LCM=201042/a (a=3 because HCF of 306 and 627 is 3 and in q hcf,=a so equate )=201042/3 67014=lcm | |
| 26333. |
TanQ /secQ+1 + cot Q / cosQ+1 = cocec-sinQ tanQ |
| Answer» very easy and important question. ☺️☺️ | |
| 26334. |
My I\'d is sakshi1241 |
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Answer» ????????????????Ooooooooooooooooooooooooooo0000000000000ooooooooooooooooooooooo. ?????????????????????????????????????????????????? Of brainly app |
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| 26335. |
Perimeter of sector and perimeter of right triangle |
| Answer» Given in book !! | |
| 26336. |
Hlo....... |
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Answer» Me tooo Oh...awesone... What...?????? And you Awesome Away Hlo devil kaise ho...??? Hiii |
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| 26337. |
Hey guys , recently joined my cbse . Hello with my best greetings ,? |
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Answer» Hey...??? Hello hello Hiiiiii??? Hii Hiii?????☺☺☺☺☺?????? |
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| 26338. |
Hlo guys koi h....???? |
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Answer» Yeah! Yes |
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| 26339. |
Determine the ap whose 3rd term is 5 and 7th term is 9 |
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Answer» a+2d=5 and a+6d= 9 then after solving the equation d=1 so ap = 3456 a+2d =5......a+6d =9......solve these eq. We get a=3 d=1 so that AP is 3,4,5,6...... A ki value 1D ki value 3 d=1,a=3,the ap is 3,4,5....... |
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| 26340. |
If sinA =cosA then find the value of A |
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Answer» 1 45 |
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| 26341. |
If asinthetha is equal to bcosthetha show that a×a=b×b |
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| 26342. |
Tough |
| Answer» | |
| 26343. |
Prove that one of any three consecutive positive integers must be divided 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. | |
| 26344. |
Pass marks kitna hai |
| Answer» Passing marks out of 100 is 33 but from 80 is 27 | |
| 26345. |
Any1 give me the blue print of maths plz help I may pass from this |
| Answer» | |
| 26346. |
Kon kon maths mein fail karega |
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Answer» Aisa to koi bhi nhi chahega ???? Tume apne chances dikhte h kya Aap aisa kaise puch sakte hoo Pass marks kitna hai Bilkul bhi nahi Im not |
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| 26347. |
[email\xa0protected]_ [email\xa0protected]+1/[email\xa0protected][email\xa0protected]_1 |
| Answer» | |
| 26348. |
Find the points on X axis which are at a distance of 2√5 units from the point (7,-4) |
| Answer» (5,0) and (9,0) | |
| 26349. |
√105+√105= |
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Answer» 2√105 2√105 |
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| 26350. |
10 class datasheet Cbse |
| Answer» Go on CBSE.nic.com | |