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.
| 1901. |
In a cuboid the volume is 144, TSA is 192 , diagonal is 13 . Then find length , bredth, and height. |
| Answer» | |
| 1902. |
Find mean median mode |
| Answer» | |
| 1903. |
Sin Alfa =cos 135 what is the value of Alfa |
| Answer» | |
| 1904. |
Solve for x :-{1/(x-1)(x-2)} + {1/(x-2)(x-3)} = 2/3 |
| Answer» x=0,x=6 | |
| 1905. |
Which term of the sequence 48,43,38,33... is the first negative term? |
| Answer» | |
| 1906. |
What is the general term of an ap |
| Answer» an=a+(n-1)d | |
| 1907. |
What is the general form of quadratic equation |
| Answer» ax2 + bx + c = 0 | |
| 1908. |
What is work of median in triangle |
| Answer» Triangle is a three style | |
| 1909. |
If sinA=cosA. Find the value of A? |
|
Answer» If sina =cosa find the value a A=45 A= 45° |
|
| 1910. |
Prove that the tangents drawn at the ends of a circle make equal angles with the chord |
| Answer» Let NM be chord of circle with centre C.Let tangents at MN meet at the point O.Since OM is a tangent∴ MO ⊥ CM i.e. ∠OMC = 90°∵ ON is a tangent∴ ON ⊥ CN\xa0i.e. ∠ONC = 90° Again in ΔCMN , CM = CN = r\xa0∴ ∠CMN = ∠CNM∴ ∠OMC – ∠CMN = ∠ONC – ∠CNM⇒ ∠OML = ∠ONLThus, tangents make equal angle with the chord. | |
| 1911. |
Draw a pair of triangle to a circle of 5cm which are inclined to each other at an angle of 60 |
| Answer» Steps of construction:\tDraw a circle with centre O and radius 5 cm.\tDraw any radius OT.\tConstruct.\xa0{tex}\\angle TOT\' = 180^\\circ - 60^\\circ = 120^\\circ {/tex}\tDraw and\xa0{tex}TP \\bot OT{/tex}{tex}T\'P \\bot OT\'{/tex}. Then PT\'\xa0and PTare the two required tangents such that.\xa0{tex}\\angle TPT\' = 60^\\circ {/tex} Here, PT = PT\'. | |
| 1912. |
How to prepare for math exam |
| Answer» | |
| 1913. |
X2+6x-4=0 |
| Answer» X= 3+√13. | |
| 1914. |
Find the eleventh term from the last tera of the AP:27,23,19,....-56. |
|
Answer» Write the above A.P. in reverse order , we get\xa0-56,-52,-48..............a= -56 d= 4a11\xa0= a+(n-1)da11\xa0= -56+(11-1)*4a11\xa0= 56+10*4a11\xa0= -56+40a11 = -16 An = A + (n-1)DA= -56D= -4A11 = -56 + ( 11-1 ) -4A11 = -56 + (10) -4A11 = - 56 - 40A11 = - 96 -96 |
|
| 1915. |
Find the value of Q so that the quadratic equation px (X-3)+9=0 has two equal roots |
| Answer» we have,{tex}px^2 - 3px + 9 = 0\xa0{/tex}put a = p , b = -3p, c = 9 in\xa0{tex}b^{2}-4 ac{/tex}{tex}b^2 - 4ac = (-3p)^2 - 4(p)(9){/tex}{tex}= 9p^2 - 36p{/tex}for having equal roots\xa0{tex}b^2 - 4ac = 9p^2 - 36p = 0\xa0{/tex}{tex}9p^2 - 36p = 0\xa0{/tex}9p(p - 4)\xa0= 0\xa09p = 0 or p - 4 = 0\xa0p = 0 or p = 4 | |
| 1916. |
Can anyone solve this for me ..(1+tan^2A/1+cot^2A)=(1-tan A)^2/(1-cot A)^2=tan^2A |
| Answer» {tex}1+tan^2A/ 1+cot^2A{/tex}{tex}Sec^2A/Cosec^2A{/tex}1/cos2A .Sin2A/1Sin2A/Cos2ATan2A | |
| 1917. |
The circumfernce of two circles are in the ratio 2:3. Find the ratio of their areas?\xa0 |
| Answer» 2x+3x=1805x=180×=180÷5×=36 | |
| 1918. |
Pythagoras thorem |
| Answer» | |
| 1919. |
If 2^sinx+cosy=1and 16^sin^2+cos^2=4Then find some and cosy |
| Answer» | |
| 1920. |
ST PARALLEL TO RQ,PS=3,SR=4.FIND THE RATIO OF THE AREA OF TRIANGLE PST TO THE AREA OF TRIANGLE PRQ |
| Answer» PS = 3 cm, SR = 4 cm and ST || RQ.PR = PS + SR= 3 + 4 = 7 cmIn {tex}\\triangle{/tex}PST and {tex}\\triangle{/tex}PRQ{tex}\\angle{/tex}SPT\xa0{tex}\\cong{/tex}{tex}\\angle{/tex}RPQ (common angle){tex}\\angle{/tex}PST\xa0{tex}\\cong{/tex}{tex}\\angle{/tex}PRQ (Alternate angle){tex}\\triangle{/tex}PST\xa0{tex}\\sim{/tex}{tex}\\triangle{/tex}PRQ (AA configuration){tex}\\frac { \\text { ar } \\Delta P S T } { \\text { ar } \\Delta P Q R } = \\frac { P S ^ { 2 } } { P R ^ { 2 } } {/tex}{tex}= \\frac { 3 ^ { 2 } } { 7 ^ { 2 } } = \\frac { 9 } { 49 }{/tex}Hence required ratio = 9 : 49. | |
| 1921. |
What is circumvented |
| Answer» | |
| 1922. |
How to draw a perpendicular on opposite side from an angle by contruction |
| Answer» | |
| 1923. |
If alfa and byta are the zeroes of polynomial 2x^2-4x+5,find the value of alfa^2 and byta^2. |
| Answer» | |
| 1924. |
Find the number nearest110000 but greater than100000 which is exactly divisible by each of 8,15and21 |
| Answer» L.C.M of 8, 15 and 21.8 = 2 {tex}\\times{/tex}\xa02 {tex}\\times{/tex}\xa0215 = 3 {tex}\\times{/tex}\xa0521 = 3 {tex}\\times{/tex}\xa07L.C.M of 8, 15 and 21 = 23{tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa071= 840If we divide 110000 by 840, we will find out that it is not exactly divisible and we get 800 as remainder.Thus the number nearest to 110000 but greater than 100000 which is exactly divisible by 840 and also divisible by 8, 15 and 21 is = 110000 - 840 = 109200Hence 109200 is exactly divisible by 8, 15 and 21. | |
| 1925. |
Which is best publication book for board exam |
|
Answer» OK Ncert books Cbse sample paper |
|
| 1926. |
CSA of cone = 550 cm^2 , height = 24 cm. find volume = ?? |
| Answer» Height of the cone = 24 cmLet radius be r cm and slant height be l cm.{tex}\\therefore \\quad l = \\sqrt { h ^ { 2 } + r ^ { 2 } } \\Rightarrow l = \\sqrt { 24 ^ { 2 } + r ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad l = \\sqrt { 576 + r ^ { 2 } } {/tex}Now, curved surface area = 550 cm2{tex}\\Rightarrow \\quad \\pi r l = 550{/tex}{tex}\\Rightarrow \\quad \\frac { 22 } { 7 } \\times r \\times \\sqrt { 576 + r ^ { 2 } } = 550{/tex}{tex}\\Rightarrow \\quad r \\times \\sqrt { 576 + r ^ { 2 } } = 175{/tex}On squaring both sides, we getr2(576 + r2) = 30625{tex}\\Rightarrow{/tex}\xa0r4\xa0+ 576r2\xa0- 30625 = 0{tex}\\Rightarrow{/tex}\xa0r4\xa0+ 625r2\xa0- 49r2\xa0- 30625 = 0{tex}\\Rightarrow{/tex}\xa0r2(r2\xa0+ 625) - 49(r2\xa0+ 625) = 0{tex}\\Rightarrow{/tex}\xa0(r2\xa0+ 625)(r2\xa0-\xa049) =0{tex}\\Rightarrow{/tex}\xa0r2\xa0= -625 or r2\xa0= 49rejecting r2\xa0= -625we get, r2\xa0= 49{tex}\\Rightarrow r = 7 \\mathrm { cm }{/tex}Now, volume =\xa0{tex}\\frac { 1 } { 3 } \\pi r ^ { 2 } h{/tex}{tex}= \\frac { 1 } { 3 } \\times \\frac { 22 } { 7 } \\times 7 \\times 7 \\times 24{/tex}= 1232 cm3 | |
| 1927. |
8+root3+1 |
| Answer» 9root3 | |
| 1928. |
If angle between two tangents drawn from an external point is 60 degree then find the length of OP |
| Answer» We know that tangent is always perpendicular to the radius at the point of contact.So, ∠OAP = 90We know that if 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.So, ∠OPA = 12∠APB = 12{tex}\\times{/tex}60° = 30°According to the angle sum property of triangle-In ∆AOP, ∠AOP + ∠OAP + ∠OPA = 180°{tex}\\Rightarrow{/tex}\xa0∠AOP + 90° + 30° = 180°{tex}\\Rightarrow{/tex}\xa0∠AOP = 60°So, in triangle AOPtan angle AOP = AP/ OA{tex}\\sqrt 3 = \\frac{{AP}}{a}{/tex}therefore, {tex}AP = \\sqrt 3 a{/tex}hence, proved | |
| 1929. |
If Ais acute angle such that cosA =3/7find tan5tan25tan30tan65tan85 |
| Answer» Tan85tan5 tan65tan25 tan30Tan90. Tan90 1/root3Not defined not defined 1/root 3Not defined | |
| 1930. |
sinQ= |
| Answer» 1/cosecQ | |
| 1931. |
BL& CM are medians of a triangle ABC 90 degree at A .Prove that4(BL2+CM2)=5BC2. |
| Answer» BL and CM are medians of a {tex}\\triangle {/tex} ABC in which {tex}\\angle{/tex} A=900From {tex}\\triangle {/tex}ABC, BC2 = AB2 + AC2 ....(i)From right angled {tex}\\vartriangle {/tex} ABL,BL2 = AL2 + AB2i.e {tex}B{L^2} = {\\left( {\\frac{{AC}}{2}} \\right)^2} + A{B^2}{/tex}{tex}\\Rightarrow {/tex} 4BL2= AC2 + 4AB2 .....(ii)From right-angled {tex}\\triangle {/tex}CMA,CM2 = AC2 + AM2i.e {tex}C{M^2} = A{C^2} + {\\left( {\\frac{{AB}}{2}} \\right)^2}{/tex}[mid-point]{tex}\\Rightarrow {/tex}{tex}C{M^2} = A{C^2} + \\frac{{A{B^2}}}{4}{/tex}{tex}\\Rightarrow {/tex}{tex}4C{M^2} = 4A{C^2} + A{B^2}{/tex} .....(iii)Adding (ii) and (iii), we geti.e.4(BL2 + CM2) = 5(AC2 + AB2) = 5BC2 [From (i)] | |
| 1932. |
How many questions of 4 marks are come in paper? |
|
Answer» 6to8 answer\'s No |
|
| 1933. |
Find the value of Sec theta,if sin theta=3/5 |
|
Answer» Sin theta=3/5=P/HB=√(5)2-(3)2=√25-9=√16=4Sec theta=H/B=5/4 5/4 5/4 Sec theta =5/3Sec theta = hypotenuse/side opposite to angle theta |
|
| 1934. |
If 2cosA5=, find the value of 4 + 4 tan2 A |
| Answer» CosA =5/2=B/HP=√(5)2-(2)2=√25-4=√21TanA=P/B=√21/5=4+4*(√21/5)2=4+4*21/25=4+3.36=7.36 | |
| 1935. |
For any positive integer n, prove that n3-n divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) | |
| 1936. |
Abc is An isosceles triangle Right angled at C.prove that Ab square is equal to 2Ac square |
| Answer» Given: ABC is an isosceles triangle right angled at C.To prove: AB2 = 2AC2Proof : ABC is an isosceles triangle right angle at C.{tex}\\therefore {/tex} AC = BC........(1)and {tex}\\angle{/tex}C = 90o.......(2)In view of (2) in{tex}\\vartriangle {/tex} ABC,AB2 = AC2 + BC2........By Pythagoras theorem.AC2 + AC2 ........Frrom(1)= 2Ac2 | |
| 1937. |
if Sn the sum of first n terms of an AP is given by Sn =3n^2-4n,then find its nth term. |
| Answer» Sn = 3n2 + 4n.a1\xa0= S1 = 3(1)2\xa0+ 4(1) = 7a1\xa0+ a2\xa0= S2\xa0= 3(2)2 + 4(2)= 12 + 8= 20a2 = S2- S1\xa0= 20 - 7 = 13or, {tex}a + d = 13{/tex}or, {tex}7 + d = 13{/tex}{tex}\\therefore{/tex}\xa0{tex}d = 13 - 7 = 6{/tex}{tex}\\therefore{/tex}\xa0A.P. becomes 7,13,19,......\xa0Now, {tex}a_n =a + (n - 1 )d{/tex}{tex}= 7 + (n - 1)(6){/tex}{tex}= 7 + 6n - 6{/tex}{tex}= 6n + 1{/tex}or, an = 6n + 1 | |
| 1938. |
If the vertices (6,7),(4,-5),(x,2x) form triangle of area 3sq.cm2. Find x |
| Answer» | |
| 1939. |
where i get maths activities for maths lab? |
| Answer» Maths Lab Manual | |
| 1940. |
Sin0 +cos0 = √2 , then tan0 + cos0 = ? |
| Answer» | |
| 1941. |
What is a Pythagoras theorm |
| Answer» Pythagoras theorm is hypotenuse square = perpendicular square + base square | |
| 1942. |
First primeminister |
| Answer» Jawaharlal Nehru | |
| 1943. |
Circles theorem 10.2 |
| Answer» | |
| 1944. |
(a+b) ka whole square kya hoga? |
|
Answer» This is right . a ka square + b ka square + 2ab A square+Bsquare+2AB a squre + b square +2xaxb |
|
| 1945. |
Whatis the Eculid division |
|
Answer» Given a and b are two interges there exist unique integer\'s q and r satisfying.....a =bq +r a =bq+r |
|
| 1946. |
How to know that given experiment will give equally likely outcome? |
| Answer» | |
| 1947. |
Trignometry last ex full |
| Answer» | |
| 1948. |
Find the co-ordinate of the point on y -axis which nearest to the point(-2,5) |
|
Answer» (0,5 ) Use distance formula |
|
| 1949. |
Find the HCF of 1656 and 4050 by Euclid\'s division division algorithm |
|
Answer» Answer is 18 Answer = 18 |
|
| 1950. |
C bv gf n nh nn |
| Answer» | |