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.
| 2501. |
Pre board kb se hai |
| Answer» Jan 3 all India | |
| 2502. |
Triangles important questions |
| Answer» | |
| 2503. |
Raat Laura hai bore karegi |
| Answer» | |
| 2504. |
Is all congurent triangle are similar |
| Answer» Yes | |
| 2505. |
x2+5x-(a2+a-6)=0 find the value of x |
| Answer» Hkvhjnnvbncfh | |
| 2506. |
Sas Therom |
| Answer» Btao | |
| 2507. |
If alpha and bita are zeros of polynomial x^2-5+k such that alpha-bita=1 then . Find value of x |
| Answer» | |
| 2508. |
find the values of y for which the distance between the point p(2, -3) and q(10,y) is 10 untis |
| Answer» Using Distance formula, we have{tex}10 = \\sqrt {{{(2 - 10)}^2} + {{( - 3 - y)}^2}} {/tex}{tex} \\Rightarrow 10 = \\sqrt {{{( - 8)}^2} + 9 + {y^2} + 6y} {/tex}\xa0{tex}{/tex}{tex} \\Rightarrow 10 = \\sqrt {64 + 9 + {y^2} + 6y} {/tex}Squaring both sides, we get100 = 73 + y2 + 6y⇒ y2 + 6y - 27 = 0Solving this Quadratic equation by factorization, we can write⇒ y2 + 9y − 3y - 27 = 0⇒ y(y + 9) - 3(y + 9) = 0⇒ (y + 9)(y − 3) = 0⇒ y = 3, −9 | |
| 2509. |
Show that exactly one of the numbers n, n+2 , or n+4. is divisible by 3 |
| Answer» Let n =3kthen n + 2 = 3k + 2and n + 4 = 3k + 4Case 1: When n=3k ,n is divisible by 3 ............(1)n + 2 = 3k + 2or, n + 2 is not divisible by 3n + 4 = 3k + 4= 3(k + 1) + 1or, n + 4 is not divisible by 3Case 2:When n=3k+1, n is not divisible by 3\xa0n + 2 = (3k + 1) + 2=3k + 3 = 3(k + 1){tex} \\Rightarrow{/tex}\xa0n+ 2 is clearly divisible by 3..........................(2)n + 4 = (3k + 1) + 4= 3k + 5= 3(k + 1) + 2{tex}\\Rightarrow{/tex}\xa0n + 4 is not divisible by 3Case 3:When n=3k+2,n is not divisible by 3\xa0n + 2 = (3k + 2) + 2= 3k + 4(n + 2) is not divisible by 3x + 4 = 3k + 6 = 3(k + 2){tex}\\Rightarrow{/tex}\xa0n + 4 is divisible by 3........................(3)Hence, from (1),(2) and (3) it is clear that\xa0exactly one of the numbers n, n + 2, n + 4, is divisible by 3. | |
| 2510. |
If 3 tan 2A = underroot3 find the value of A |
| Answer» A= 15°. | |
| 2511. |
a=10 d=10 |
| Answer» AP = 10, 20, 30, 40... | |
| 2512. |
The lenght of a tangent from a point A at a distance 5 cm and 3 cm.find the radius of the circle |
| Answer» Answer will be 4cm. | |
| 2513. |
Yadi kisi sankya ka 3/5 uska 50% se 45 jyada ha to vo sankya ky h |
| Answer» Let the value will be xSo 3/5 of x be 3x/5Now, 3x/5 *x-50/100=45 3x/5*x/2=45 3x-x/10=45 3x-x=45*10 3x-x=450 2x=450 x=450/2 x=225 | |
| 2514. |
In abc Ab=Ac and D is point on side Ac such that (bc) whole saquare ac.cd prove Bd=Bc |
| Answer» To prove: BD = BCproof:BC2 = AC {tex} \\times{/tex} CD{tex}\\Rightarrow {/tex}{tex}\\frac{{AC}}{{BC}} = \\frac{{BC}}{{CD}}{/tex} ....(1)Also, {tex}\\angle{/tex} ACB = {tex}\\angle{/tex} BCD ....(2) .....[Common angle]In view of (1) and (2),{tex}\\triangle {/tex} ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}BDC .....[SAS similarity criterion]{tex}\\therefore {/tex}{tex}\\frac{{AC}}{{BC}} = \\frac{{AB}}{{BD}}{/tex} .......({tex}\\because {/tex} corresponding sides of two similar triangles are proportional)But AB = AC .....(Given)BD = BC | |
| 2515. |
Volume of frustum??? |
|
Answer» 1/3 pi h ( R2 + r2 + R1r2 ) One by three pi height 1/3^h(r1+r2+r1r2) |
|
| 2516. |
What is meant by maths |
|
Answer» Mathematics is the study of number,quantities and geometry. Mathematics is the study of number, quan |
|
| 2517. |
What is the median of first 10 natural Numbers |
|
Answer» 55/ 10 =5.5 5.5 |
|
| 2518. |
Study ho gya....?? |
| Answer» Kiska | |
| 2519. |
A chord of circle of radius 10cm subtent angle 60 find the area of major and minor segment |
| Answer» \tWe know that area of minor segment\t= Area of minor sector OAB - Area of\xa0ΔOAB\t{tex}\\because \\text { area of } \\triangle \\mathrm{OAB}=\\frac{1}{2}(O A)(O B) \\sin \\angle A O B{/tex}\t{tex}=\\frac{1}{2}(O A)(O B)\\left(\\because \\angle A O B=90^{\\circ}\\right){/tex}\tArea of sector = {tex}\\frac{\\theta}{360} \\pi r^{2}{/tex}\t= {tex}\\frac 14{/tex}(3.14) (100) - 50 = 25(3.14) - 50 = 78.50 - 50 = 28.5 cm2\tArea of major segment = Area of the circle - Area of minor segment\t= {tex}\\pi{/tex}(10)2 - 28.5\t= 100(3.14) - 28.5\t= 314 - 28. 5 = 285.5 cm2 | |
| 2520. |
difference between lateral and total surface area and volume |
| Answer» Hi.....Lateral Surface area of a cubiod or a cube is the surface area of its four adjacent sides. That is, area of the cube or cuboid excluding its top surface and bottom surface. I hope it helps...#BeCurious | |
| 2521. |
Maths ki optional exercise Final me aaigi Kya? |
| Answer» | |
| 2522. |
Aaj sunday ka khub fayda utha lo iss app pr roj - roj mauka thodi n milta h .. .. right iss.... |
| Answer» | |
| 2523. |
Express each no as a oroduct of its orime dactor 140 |
| Answer» So, the prime factors of 140 = 2 {tex}\\times{/tex}\xa02 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa07 = 22\xa0{tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa07. | |
| 2524. |
What is the value of sin90 |
| Answer» 1 | |
| 2525. |
Cbse pratical of sst kaise hoga |
| Answer» | |
| 2526. |
How to know curved surface area |
| Answer» | |
| 2527. |
Find the 10th term from the last of A.P. 7,14,21........343. |
|
Answer» 280 343-9*7= 280 |
|
| 2528. |
16x -10/x =27 by quadric formula |
| Answer» We have, {tex}16x - \\frac{{10}}{x} = 27{/tex}{tex}\\Rightarrow \\frac{{16{x^2} - 10}}{x} = 27{/tex}{tex}\\Rightarrow{/tex} 16x2 - 10 = 27x{tex}\\Rightarrow{/tex} 16x2 - 27x - 10 = 0{tex}\\Rightarrow{/tex} 16x2 - 32x + 5x - 10 = 0{tex}\\Rightarrow{/tex} 16x(x - 2) + 5(x - 2) = 0{tex}\\Rightarrow{/tex} (16x + 5)(x - 2) = 0{tex}\\Rightarrow{/tex} (16x + 5) = 0 or (x - 2) = 0{tex}\\Rightarrow x = - \\frac{{ 5}}{{16}}{/tex} or, x = 2Hence, the roots of given quadratic equation are 2 and\xa0{tex}-\\frac{{5}}{16}{/tex} | |
| 2529. |
AB parallel to DE and BD parallel toEF prove that DC^2=CF.AC |
| Answer» | |
| 2530. |
If tan (A+B)=roots and (A-B)=1 upon roots find A&B |
| Answer» tan(A + B) = {tex}\\sqrt 3{/tex}tan (A + B) = tan60°A + B = 60°. ...(i)tan (A - B) = 1tan (A- B) = tan 45°A - B = 45°.........(ii)Solving (i) and (ii), we getA =\xa0(52.5)° and B = (7.5)°.Hence, A = (52.5)° and B = (7.5)°. | |
| 2531. |
Pre board ho gay kya |
| Answer» | |
| 2532. |
How to solve elemenation method |
| Answer» | |
| 2533. |
2sin 2 sin theta + 15 cos theta find cot theta |
| Answer» | |
| 2534. |
Derive formula for TSA & CSA of CONE |
| Answer» TSA-22/7 r (l+r)CSA-22/7 rl | |
| 2535. |
Hai |
| Answer» | |
| 2536. |
CI. F. X 0-20 8. 1020-40 15. 3040-60. 20 50 60-80 p. 7080-100. 5 90 |
| Answer» | |
| 2537. |
??????⌚⌚ see watch and come fast .. ?? |
| Answer» | |
| 2538. |
What are concentric circles |
| Answer» Two or more circle from the same centre with different radius | |
| 2539. |
What is imagineary number |
| Answer» | |
| 2540. |
Hiiiii nidhi |
| Answer» | |
| 2541. |
Linear |
| Answer» | |
| 2542. |
All circle are what |
| Answer» | |
| 2543. |
If secA=x+1÷4x then prove that sec A+ Tan A= 2x or 1÷2x |
| Answer» | |
| 2544. |
SinA=√3|2and cosB √3|2 what is the value ofA+B |
| Answer» | |
| 2545. |
10 class chapter 8 exercise 8.4 solved |
| Answer» | |
| 2546. |
Why fireflies glow at night |
| Answer» They do even glow at daytime but because of the heat given out by sun kills those fireflies...and even if sone stay alive then they aren\'t visible because of sunlight. | |
| 2547. |
Find the probability of getting 53 of friday in a leap year |
| Answer» 2/7 | |
| 2548. |
Theorem BPT |
| Answer» | |
| 2549. |
What is the common difference of an ap in which a21 - a7-84? |
| Answer» Let {tex}a{/tex} be first term and {tex}d {/tex} be the common difference.{tex}a_{21} - a_{7} = 84{/tex}{tex}a + (21 - 1)d - a + (7 - 1)d = 84{/tex}{tex}a + 20d - a - 6d = 84{/tex}{tex}14d = 84{/tex}{tex}d = 6{/tex}Therefore, common difference is 6. | |
| 2550. |
Write the zeros of the polynomial x²-5x-6 |
| Answer» Let p(x) = x2 \x96 5x + 6= x2 \x96 3x \x96 2x + 6= x(x \x96 3) \x96 2(x \x96 3)= (x \x96 3)(x \x96 2)∴ (x \x96 3)(x \x96 2) = 0⇒ x \x96 3 = 0 or x \x96 2 = 0⇒ x = 3 or x = 2Thus 3 and 2 are the zeroes of the given polynomial Sum of roots (α+ β) = (3+2) = 5 = − (−5/1) = − (b/a) Product of roots (αβ) = 3 × 2 = 6 = (6/1) = (c/a) | |