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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.
| 36451. |
Class 10 cbse introduced 2 types of levels ...1.basic 2.standard |
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Answer» Lekin yaar agle saal walo k maze h ? Jo bhi ho dena to padega na exam Yeah .... Fr next year I think it\'s fr next year |
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| 36452. |
The radii of buket of height 24 cm are 15 cm and 5 cm find metal sheet used to make it |
| Answer» Firstly,you have to find the slant height of the bucket, "l"=26cm.Then, find the CSA of the bucket.Ans-816.4cm sq.units | |
| 36453. |
Sir board me kitna year old paper aata hai |
| Answer» Bhai ncert hi kr le usme se aayega . | |
| 36454. |
9x2-9p(p+a)x+(2p2+5pq+2q2) solve equation x |
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Answer» iss se karle bhai bahut jaada bada ha d= b2 -4ac 2p+q/3 and p+2q/3 |
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| 36455. |
Show that the reciprocal of irrational number is irrational .... |
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Answer» If any irrational no. Will be there instead of pi then... We know that the irrational numbers are always terminating or non repeating . So take the example of pi= 22÷7 = 3.1285714 and it\'s reciprocal 7÷22 = 0.31818818 which are also irrational.. |
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| 36456. |
Find the 25th term of A.P when the given A.P is 11,16,21...... |
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Answer» 131 Hello. Honey hamare comments gayab ho gaye yaar.. a= 11d= 5 a25= a+ 24d11 + 24×5 a25 = 11+120 a25= 131 131 |
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| 36457. |
What is the zero of the polynomial? |
| Answer» Zero of a poly. is defined as the trm when inserted in the poly. =0 and it satisfies it | |
| 36458. |
The length of the shadow of a man is equal to the height of man |
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Answer» Angle of elevation is tan45° Angle of elevation is 45. |
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| 36459. |
solve x^2_45x+324=0 |
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Answer» x2 - 45x + 324 = 0x2 - 36x - 9x + 324 = 0x ( x - 36) - 9 ( x - 36) = 0(x - 36) (x -9) = 0x - 36 = 0 and x - 9 = 0x = 36 and x = 9 36,9 |
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| 36460. |
Cos^3+sin^3/cos+ sin. +. Cos^3- sin ^3/cos- sin=1. Please solve it |
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| 36461. |
Find the roots of the following 2 |
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| 36462. |
What is the sum of first \'n\' odd positive integers? |
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Answer» n2+2n N square |
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| 36463. |
If sec thitta =x .write the value of tan thitta ? |
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Answer» X²-1 Root x^2-1 |
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| 36464. |
what is square root of 2 |
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| 36465. |
7sin^[email\xa0protected]+3cos^[email\xa0protected]=4Show that [email\xa0protected]=1/root3 |
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| 36466. |
Maths chapter 2 questions no 3 ka solution . Page no 36 ncert class 10 . Take divisor 3x2 - 5 |
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| 36467. |
theorems of chapter circles |
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Answer» tangents make 90 degree with the diameter tangents drawn from external point are equal |
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| 36468. |
If triangle def is similar to triangle abc, area of triangle def=4/9(area of abc) then find bc /ef |
| Answer» Area ofΔabc/area of Δdef=9/4=(bc)²/(ef)²√9/√4=bc/ef3/2=bc/ef. Ans | |
| 36469. |
Find a quadratic polynomial whose zeroes are 5+√2 and 5- √2 |
| Answer» X square - 10x+23 | |
| 36470. |
Ch 3 ex 3.5 Q 2 1 explain krdoo Koi ncert ka |
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| 36471. |
Show that 12n cannot end with the digit0 for any natural number n |
| Answer» For any digit ending with zero must have both 5 and 2 as factor but 12 have only 2 as factor not 5 so it cannot end with 0 for any natural no n | |
| 36472. |
Sin70/cos20+cosec20/sec70_2cos70 cosec20=0 |
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| 36473. |
If (cosec theta - sin theta) =m³ and (sec theta - cos theta) = n³ then prove that m⁴n² + m²n⁴ =1 |
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| 36474. |
If the sum of m terms of an AP. is the same as that sum of its n terms, show |
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Answer» m/2 (2a+(m-1)d)=n/2 (2a+(n-1)d) 2am+(m-1)md=2an+(n-1)nd 2a(m-n)+d (m^2-n^2-m+n)=0 2a+d (m+n-1)=0 Show that the sum of its (m+n) terms is zero |
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| 36475. |
Express 156 as the product of its prime factors |
| Answer» 2×2×3×13 | |
| 36476. |
Evalute sin219°+sin272° |
| Answer» sin219+sin272=sin(180+39)+sin(270+2)= -sin39-cos2=0.63-0.99=-1.62 | |
| 36477. |
In an ap of 50 terms the sum of 10 terms is 210 and the sum of 15 term of 2565 find ap |
| Answer» Let a be the first term and d be the common difference of the given AP. Therefore, the sum of first n terms is given by{tex}S _ { n } = \\frac { n } { 2 } \\cdot \\{ 2 a + ( n - 1 ) d \\}{/tex}{tex}\\therefore{/tex}\xa0S10 = {tex}\\frac{{10}}{2}{/tex}{tex}\\cdot{/tex}(2a+9d) {tex}\\Rightarrow{/tex}5(2a+9d)=210{tex}\\Rightarrow{/tex}2a+9d=42. ...(i)Sum of last 15 terms = (S50 - S35).{tex}\\therefore{/tex}\xa0(S50\xa0- S35) = 2565{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{{50}}{2}{/tex}(2a+49d)- {tex}\\frac{{35}}{2}{/tex}(2a+34d)=2565{tex}\\Rightarrow{/tex}\xa025(2a+49d)-35(a+17d)=2565{tex}\\Rightarrow{/tex}\xa0(50a-35a)+(1225d-595d)=2565{tex}\\Rightarrow{/tex}\xa015a+630d = 2565 {tex}\\Rightarrow{/tex}\xa0a + 42d = 171 ...... (ii)Therefore, on solving (i) and (ii), we get a=3 and d=4.Hence, the required AP is 3,7,11,15,19..... | |
| 36478. |
Find the point on the x-axis which is equidistant from (2,5) (-2,9)?? |
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Answer» Yogita ingle plz explain about balkan crisis.. (-7,0) Let the point of x-axis be P(x, 0)Given A(2, 5) and B(-2, 9) are equidistant from PThat is PA = PBHence PA2 = PB2 → (1)Distance between two points is\xa0√[(x2 - x1)2 +\xa0(y2\xa0- y1)2]PA =\xa0√[(2\xa0- x)2\xa0+\xa0(5\xa0- 0)2]PA2\xa0= 4 - 4x +x2 + 25 =\xa0x2 - 4x + 29Similarly,\xa0PB2\xa0=\xa0x2\xa0+ 4x + 85Equation (1) becomesx2\xa0- 4x + 29 =\xa0x2\xa0+ 4x + 85- 8x = 56x = -7Hence the point on x-axis is (-7, 0) |
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| 36479. |
Up board ka question no. 1 |
| Answer» Sorry its cbse board | |
| 36480. |
In a acute triangle ABC, AD is a median. Prove that 4AD^2 + BC^2 = 2AB^2 + 2AC^2 |
| Answer» \xa0Given:\xa0In ∆ ABC, AD is the medianConstruction:\xa0Draw AE ⊥BC\xa0\xa0Now since AD is the median∴ BD = CD =BC ....... (1)\xa0In ∆ AEDAD2\xa0= AE2\xa0+ DE2 (Pythagoras theorem)⇒ AE2\xa0= AD2\xa0– DE2 ......... (2)\xa0In ∆ AEBAB2\xa0= AE2\xa0+ BE2\xa0= AD2\xa0– DE2\xa0+ BE2\xa0(from (2))= (BD\xa0+ DE)2\xa0+ AD2\xa0– DE2\xa0(∵ BE = BD + DE)= BD2\xa0+ DE2\xa0+ 2BD·DE + AD2\xa0– DE2= BD2\xa0+ AD2\xa0+ 2·BD·DE\xa0In ∆ AEDAC2\xa0= AE2\xa0+ EC2= AD2\xa0– DE2\xa0+ EC2 (from (5))= AD2\xa0– DE2\xa0+ (DC – DE)2= AD2\xa0– DE2\xa0+ DC2\xa0+ DE2\xa0– 2DC·DE= AD2\xa0+ DC2\xa0– 2DC·DE\xa0Adding (3) and (4) we get\xa0AB2\xa0+ AC2\xa0=BC2\xa0+ AD2\xa0+ BC·DE\xa0+ AD2\xa0+BC2\xa0– BC·DE⇒ 2 (AB2\xa0+ AC2) = BC2\xa0+ 4AD2⇒ 4AB2\xa0+ BC2\xa0= 2AB2\xa0+ BC2 | |
| 36481. |
Draw a triangle ABC with side < BC=5, |
| Answer» Draw kiwe kria atha | |
| 36482. |
How to draw a tangent to a circle from any poiy |
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Answer» U can get the answer by ncert R u asking this question with regard to the chapter\'\' construction \'\'.. If yes then u can get the solution in Google because I don\'t know how to post pic here.. Draw an radius and name the point as A .Then construct 90° at A and hence a tangent is formed. ☺ Point* |
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| 36483. |
A guadrilateral adcb is drawn to circumscribe a circle prove that Ab+Cd=Ad+Bc |
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Answer» Ncert ka hai Rd sharma ka questio h or bahut easy h.Or tu ise online bhi search kr sakta h? |
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| 36484. |
Sample paper buy nhii ho rahe hai |
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| 36485. |
2√5+√5 |
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Answer» 3√5 It will be 3√5 Yrr kya question puch rahe ho |
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| 36486. |
If the ratio of the sums of first n terms of is 7 n + 1: 4n + 27 find the ratio of the ninth terms |
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Answer» 24:19 use n=2m-1where m is the required term 24:19 |
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| 36487. |
1:- if nth term of an AP is (2n+1) ,what is the sum of its first three terms |
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Answer» 15 is ans . put n=1 u will get first term similarlly find 2nd and 3rd term and then find sum 15 |
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| 36488. |
The sum of the first n terms of an AP is 3n2+6n .find the nth term of this AP |
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Answer» Sry, 6n+3 6n-3 Usme n=1 phir n=2 put karke a,d find karle |
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| 36489. |
2+89 |
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Answer» 2+89=90+1=91 Silly question 91...... why are yu asking such silly questions ? 90+1 |
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| 36490. |
All formula for 14 chapter statics |
| Answer» Check out in this app....... | |
| 36491. |
What are coprimes |
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Answer» *numbers Minders which do not have factors other than 1 nd itself. Numbers which do nt have any common factor other than 1...... Whose hcf is 1 |
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| 36492. |
4q+1, 4q+3 are all positive number |
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Answer» yes Ya , its positive odd number Sry ya it is positive odd no. No |
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| 36493. |
Find the rational number which the y-axis divides the segment joining (-3,6)and(12,-3) |
| Answer» Ratio would be 1: 4 ; y= 21/5 Hope this would help you | |
| 36494. |
Practice sums |
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Answer» Kr lenge paii practice hum ..saadi chinta ni kr tu..?? Ji hum practice karenge sum ki ? What? |
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| 36495. |
Derivation of bpt and converse of bpt |
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Answer» O hello itni akad h to que puchne se phle book pad lo that will be beneficial for u or ye koi notebook nhi jo pura description likhu answer ka If you know just help me out Refer ncert |
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| 36496. |
If tan A = cot B , then prove that A+B =90°. |
| Answer» tan A = cot B. also tan A = cot(90-B).cot A= cot (90-B).A=90-B. A+B=90 | |
| 36497. |
prove :- sinA-2sin cube A / 2cos cube A-cosA= tan A |
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Answer» L.H.S.= sinA-2sin cube A / 2cos cube A - cosA =sinA(1-2sin square A) / cosA(2cos square A -1) =tanA[1 - 2sin square A / 2(1 - sin square A) - 1] = tanA[1 - 2sin square A / 2 - 2sin square A - 1] = tanA[1-2sin square A / 1-2sin square A] = tanA×1 = tanA = R.H.S. Ncert is useful Ncert .. chapter trignometry ka exercise 8.4 question 5th ka (VII) Ncert ka h ex 8.4 |
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| 36498. |
Tptal surface area of frustum |
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| 36499. |
what is constant polynomial |
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| 36500. |
Sample paper buy nhi ho rahe hai ise app ke |
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