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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Aare shrushti maine pucha tu online naahi hai??? |
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| 2. |
RIMSHA gud ni8 dear ab kal wait krunga aapka |
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| 3. |
RIMSHA........ |
| Answer» | |
| 4. |
Can 2 no.s have 16 as their hcf and 380 as their lcm? Give reason |
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Answer» LCM completely No bcz HCF divides LVM cmetely |
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| 5. |
write both answer -5^2 and (-5)^2 |
| Answer» Its 25 !! | |
| 6. |
In an equilateral traingle altitude is drawn. Prove ad2=3bd2 |
| Answer» Wrong question | |
| 7. |
What is the LCM of 1_3 +1_2+1_4 |
| Answer» What?? | |
| 8. |
Solve for x: 1/2a+b+2x=1/2a+1/b+1/2x , x not equal to 0 |
| 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 | |
| 9. |
If sin=3/4 then find 4+4cot squareA |
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| 10. |
7/75 is terminating or non terminating number |
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Answer» Non terminating repeating number Non terminating |
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| 11. |
TanA-sinA÷tanA+sinA=secA-1÷secA+1 |
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| 12. |
The ordinate of a point A on y axis is 5 and B has a coordinates (-3,1). Find the length Ab |
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Answer» Root 65 Point A (5,0) bcz of y axis ...now find. LENGTH AB BY distance formula Length of Ab will be 5 |
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| 13. |
7/75 is terminating or non terminating |
| Answer» Non terminting | |
| 14. |
If the ratio of the roots of the equation tx^2 +nx +n is p:q then prove that √p\\q +√q\\p +√n\\t =0. |
| Answer» Roots of given equation {tex}lx^2 + nx + n = 0{/tex} are\xa0{tex}\\frac { - n + \\sqrt { n ^ { 2 } - 4 n l } } { 2 l }{/tex}\xa0and\xa0{tex}\\frac { - n - \\sqrt { n ^ { 2 } - 4 n l } } { 2 l }{/tex}{tex}\\therefore{/tex}\xa0{tex}\\frac{p}{q}{/tex}\xa0=\xa0{tex}\\frac { - n + \\sqrt { n ^ { 2 } - 4 n l } } { - n - \\sqrt { n ^ { 2 } - 4 n l } }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt {\\frac{p}{q}} {/tex}\xa0+\xa0{tex}\\sqrt {\\frac{q}{p}} {/tex}\xa0+\xa0{tex}\\sqrt {\\frac{n}{l}} {/tex}={tex}\\sqrt { \\frac { - n + \\sqrt { n ^ { 2 } - 4 n l } } { - n - \\sqrt { n ^ { 2 } - 4 n l } } }{/tex}\xa0+\xa0{tex}\\sqrt { \\frac { - n - \\sqrt { n ^ { 2 } - 4 n l } } { - n + \\sqrt { n ^ { 2 } - 4 n l } } } + \\sqrt { \\frac { n } { l } }{/tex}={tex}\\frac { - n + \\sqrt { n ^ { 2 } - 4 n l } - n - \\sqrt { n ^ { 2 } - 4 n l } } { \\left( \\sqrt { - n - \\sqrt { n ^ { 2 } - 4 n l } } \\right) \\left( \\sqrt { - n + \\sqrt { n ^ { 2 } - 4 n l } } \\right) } + \\sqrt { \\frac { n } { l } }{/tex}={tex}\\frac { - 2 n } { \\sqrt { n ^ { 2 } - n ^ { 2 } + 4 n l } } + \\sqrt { \\frac { n } { l } }{/tex}={tex}\\frac { - 2 n } { 2 \\sqrt { n l } } + \\sqrt { \\frac { n } { l } }{/tex}\xa0=\xa0{tex}\\frac { - \\sqrt { n } } { \\sqrt { l } } + \\sqrt { \\frac { n } { l } }{/tex}\xa0{tex}= 0{/tex}Hence proved. | |
| 15. |
585445884*78888 |
| Answer» 4.6184655e+13 | |
| 16. |
What is the general form of every position odd integer in terms of some integer p? |
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Answer» 2p+1 A+bn |
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| 17. |
Cot theta + Tan theta = x Sec theta - Cos theta = yThen find (Xsquare Y)-(X Ysquare)=? |
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| 18. |
If √3sinx find the value of sinx.tanx(1+cotx)/sinx+cosx |
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| 19. |
Who was pythagorus ? |
| Answer» Pythagoras was\xa0a Greek\xa0mathematician,\xa0philosopher, and mystic. He wrote nothing himself, so his ideas survive through the writings of others, including\xa0Aristotle. Many people are familiar with him as the\xa0mathematician\xa0who formulated the Pythagorean theorem in geometry that relates the lengths of the sides in a right triangle. | |
| 20. |
2√45÷2√5 |
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Answer» 3 3 9 15 3 15 |
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| 21. |
Find the HCF of 1365 and1560 by prime factorization method hence find their LCM |
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| 22. |
2y-x=8,5y-x =14,-x+y/2=1/2 are vertices of triangle. Determine graphically the vertices of triangle |
| Answer» {tex}2y - x = 8{/tex}{tex}5y - x = 14{/tex}{tex}y - 2x = 1{/tex}Now, {tex}2y - x = 8{/tex}{tex}x = 2y - 8{/tex}When {tex}y = 2{/tex} then {tex}x = - 4{/tex}When {tex}y = 5{/tex} then {tex}x = 2{/tex}Thus, we have the following table giving points on the line {tex}2y - x = 8{/tex}\tx-42y25\tNow, {tex}5y - x = 14{/tex}{tex}x = 5y - 14{/tex}When {tex}y = 2{/tex}, then {tex}x = 1{/tex}When {tex}y = 3{/tex} , then {tex}x = 1{/tex}Thus, we have the following table giving points on the line {tex}5y - x = 14{/tex}\tx-41y23\tWe have,{tex}y - 2x = 1{/tex}{tex}x = \\frac{{y - 1}}{2}{/tex}When {tex}y = -1{/tex}, then {tex}x = 1{/tex}When\xa0{tex} y = 3{/tex} , then {tex}x = 1{/tex}Thus, we have the following table giving points on the line {tex}y - 2x = 1{/tex}\tx-11y-13\tFrom the graph of the lines represent by the given equation, we observe that the lines taken in pairs intersect at points A(-4, 2) B(1, 3) and C(2, 5).Hence the vertices of the triangle are A(-4, 2) B(1, 3) and C(2, 5) | |
| 23. |
The sum of three numbers in AP is -3 and their product is 8.find the numbers |
| Answer» -4, -1, 2... | |
| 24. |
Exterior angle bisector theorm |
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| 25. |
If points (1,2),(-5,6)and(a,-2) are colliner.Find\'a\'. |
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Answer» a=7 Use the formula of area of triangle and equate the equation with zero...you will get the answer |
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| 26. |
6(ax+by) =3a+2b , 6(bx-ay)=3a-2b |
| Answer» 6(ax + by) = 3a + 2b6ax + 6by = 3a + 2b.........(i)6(bx - ay) = 3b - 2a6bx - 6ay = 3b - 2a.........(ii)Multiplying (i) by a and (2) by b,So, 6a2x + 6aby = 3a2 + 2ab .......... (iii)And 6b2x - 6aby = 3b2 - 2ab ......... (iv)Add (iii) and (iv), we get6a2x + 6aby + 6b2x - 6aby = 3a2 + 2ab + 3b2 - 2ab⇒ 6a2x + 6b2x = 3a2\xa0+ 3b2⇒6 (a2x + b2x) = 3(a2\xa0+ b2)⇒{tex}x = \\frac { 3 \\left( a ^ { 2 } + b ^ { 2 } \\right) } { 6 \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 3 } { 6 } = \\frac { 1 } { 2 }{/tex}Substituting\xa0{tex}x = \\frac 12{/tex}\xa0in (i),we get{tex}6 a \\times \\frac { 1 } { 2 } + 6 b y = 3 a + 2 b{/tex}⇒ 3a + 6by = 3a + 2b⇒ 6by =3a + 2b -3a⇒ 6by = 2b⇒ {tex}y = \\frac { 2 b } { 6 b } = \\frac { 1 } { 3 }{/tex}Hence, the solution is\xa0{tex}x = \\frac { 1 } { 2 } , y = \\frac { 1 } { 3 }{/tex} | |
| 27. |
I\'m new here..who wanna frd |
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| 28. |
(TanA+CosecB)²-(CotB-SecA) ²=2TanA CotB(CosecB+SecA) |
| Answer» So easy ?? | |
| 29. |
Show.Cosec^2 theta - tan^2 (90-theta) = ssin^2 theta + sin (90-theta) |
| Answer» So easy ?? | |
| 30. |
How to find "A" in assumed - mean method |
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| 31. |
Samruddhi? |
| Answer» | |
| 32. |
Facebook pe kon kon h? |
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Answer» Dear once trust on me M hu na lokendra nathawat |
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| 33. |
Have anyone given NTSE exam |
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| 34. |
14.1 3rd question |
| Answer» Refer this app | |
| 35. |
Yaha par koi DIXITA VERMA hai |
| Answer» So easy ?? and bye | |
| 36. |
Date sheet aa gyi kya... |
| Answer» No but it will come probably this week | |
| 37. |
Kya koi bataega ki board exams ki date kya h |
| Answer» It will be start after holi | |
| 38. |
2sin68/cos22 - 2cot15/5tan75 - 3tan45tan20tan40tan50tan70/5 |
| Answer» | |
| 39. |
RIMSHA........???????? |
| Answer» Absent | |
| 40. |
The surface area of sphere is 616 cm^2.find its radius |
| Answer» 7cm | |
| 41. |
If sec theta. Sin theta=0 find theta |
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Answer» =0 0 |
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| 42. |
What is the datesheet for 10th boards |
| Answer» Not yet announced | |
| 43. |
Is this site is cbse affiliate d |
| Answer» | |
| 44. |
tanA/1-cotA+cot A/1-tanA =1+secAcosecA prove this question |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 45. |
A horse is tied to a pole with 28 m long string . Find the area where the horse can graze |
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Answer» 3.14*28^2 Ncert easiest question |
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| 46. |
Explain why 7×11×13+13 & 7×6×5×4×3×2×1+5 are composite numbers. |
| Answer» Numbers are of two types - prime and composite.Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.It can be observed that7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)= 13 × (77 + 1)= 13 × 78= 13 ×13 × 6The given expression has 6 and 13 as its factors.Therefore, it is a composite number.7 × 6 × 5 × 4 × 3 × 2 × 1 + 5= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)= 5 × (1008 + 1)= 5 ×10091009 cannot be factorized furtherTherefore, the given expression has 5 and 1009 as its factors.Hence, it is a composite number. | |
| 47. |
If A+B=90 and sec A =5/3, then find the value of cosec B |
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Answer» Given A+B=90° i.e,A=90°-Band secA=5/3sec(90°-B) =5/3 (because A=90°-B) CosecB=5/3(because sec(90°-A)=cosecAHence proved A +B=90 A=90-B --------1 SecA = 5/3Sec 90-B =5/3 ( from 1)Cosec B = 5/3 Hence solved Same |
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| 48. |
Hey! Frnds I\'m new here |
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| 49. |
Hlo jiiii |
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| 50. |
RIMSHA plzzz yrrr |
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