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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. |
The area of two similar triangles are respectively 9 cm2 and16 cm2. The ratio of their corresponding sides is: |
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| 2. |
tan 65cot 25°Evaluate |
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Answer» tan 65 / cot 25=tan 65 / cot (90 - 65) =tan 65 / tan 65 [Sincecot(90 - 65) = tanθ] = 1 |
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| 3. |
Example 10 Find the loliowing integrais10212 +6x5 |
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| 4. |
Example 10. Prove that(n !)2 s n" . (n !) < (2n)!for all positive integers n. |
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| 5. |
Rotation of Shapes1) Which one of the following is a 90° clockwise rotation ofabout the one |
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Answer» I don't know this answer sorry I don't know this answer sorry answer is figure number ccccccc (a) number option is answer of your question Option A is the right answer. A) is the right answer option A is the correct answer the answer is option ccccccc |
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| 6. |
\operatorname { tan } 5 ^ { \circ } \operatorname { tan } 25 ^ { \circ } \operatorname { tan } 30 ^ { \circ } \operatorname { tan } 65 ^ { \circ } |
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| 7. |
3 tan 25° tan 40 tan 50 tan 65 an 604 (cos229 cos 619)33. Evaluate: |
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Answer» Numerator = 3 tan 25° tan 40° tan 50° tan 65° - (1/2) tan² 60° = 3 ( tan 25 tan 65 ) ( tan 40 tan 50 ) - (1/2) ( √3 )² = 3 ( tan 25 cot 25 ) ( tan 40 cot 40 ) - (3/2) = 3(1)(1) - (3/2) = 3/2 ......................... (1) Denominator = 4 ( cos² 29° + cos² 61° ) = 4 ( sin² 61° + cos² 61° ) = 4(1) = 4 ..............................(2) From (1) and (2), Required Expression = ( 3/2 )/4 = 3/8 Ans. |
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| 8. |
EXAMPLE 10 Solve the following equations:(i) tan θ + tan 2θ + tan θ tan 2θ 1 |
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Answer» ok plse thanka acha pura karna tha |
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| 9. |
10ART-110 tsin? Itin tsin? 60in+sin 1067tsin? In190)4. Evalute cos 225º - sín 225º+tanuásº-cot 495oogie, 10720 |
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Answer» cos225-sin225+ tan 495- cot495=Cos (180-225)-Sin(180-225)+Tan(360-495)-cot(369-495)=Sin(-45)-cos(-45)+tan(-135)-cot(-135)=1/2-1/2+cot(-45)+tan(-45)=1/2-1/2-1+1=1/2-1/2=2/2=1 |
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| 10. |
100(100 â 40) |
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Answer» (40/(100-40) * 100)= (40/60)*100= (2/3)*100= 200/3= 66.66 |
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| 11. |
EXAMPLE 1Find the mean of the following distribution:1015f:1010 |
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Answer» (5×4+6×10+9×10+10×7+15×8)/(40)=(20+60+90+70+120)/40=360/40=9 |
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| 12. |
Example 10 The perimeter of a rectangular field is 100 m.NIf the length of the field is 30 m, find the area of the field |
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Answer» Perimeter = 2(l+b)=> 2(30+b) = 100=> 30 + b = 50=> b = 20m. Area = l*b = 30*20= 600 sq m. PLEASE HIT THE LIKE BUTTON Why take this much time to give answer |
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| 13. |
evalute (2x+1)^3 |
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| 14. |
Take Ď =, unless stated otherwise.1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder ofradius 6 cm. Find the height of the cylinder |
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| 15. |
The perimeter and the area of a square are k cm and k cm2 respectively. Find (a) the perimeter, (b) thearea, and (c) the length of a diagonal of the square. |
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Answer» Let the side of square be a perimeter of square is = 4a = k area of the square is = k² = a² => 4a = a²=> 4a-a² = 0=> a(4-a) = 0=> 4 -a = 0so, a = 4 cm hence perimeter = 4*4 = 16cm and area = (4)² = 16cm² length of diagonal is a√2 = 4√2cm |
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| 16. |
20. A metallic sphere of radius 4.2 cm is melted and recast intothe shape of a cylinder of radius 6 cm. Find the height of thecylinder. |
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| 17. |
1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder ofradius 6 cm. Find the height of the cylinder. |
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Answer» If a solid is recasted in any other solid the volume remains the same. so,vol.of sphere=vol. of cylinder 4/3πr cube= 2πr square ×h 4/3×22/7×42/10×42/10×42/10 = 2×22/7×6×6×h h= 98.784/36 H=2.744 |
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| 18. |
1.)A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder ofradius 6 cm. Find the height of the cylinder. |
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| 19. |
DALRLISE 13.322. unless stated otherwise.Take T =1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder ofradius 6 cm. Find the height of the cylinder.Metallica |
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| 20. |
at If the point P[2,2] is equidistant from the points A (2, k) and B (-2k, 3) find K. Also find thelength ofAP.nOR FAPOR such that P/RTS. Show that |
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Answer» thanks |
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| 21. |
Evalute 13[(-2)+1] |
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Answer» 13[-2+1] = 13 [-1] = -13 |
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| 22. |
evalute . tan 65 /cot 25 |
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Answer» tan(65°)-------------cot(25°) tan(90° - 25°)---------------------- cot(25°) cot(25°)-------------cot(25°) 1 |
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| 23. |
H Q RB are relas of Quadratispolynomial, p(x) dx24 be te thenevalute II |
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Answer» If alpha and beta are roots of ax^2 + bx + c = 0 Then,alpha + beta = - b/aalpha*beta = c/a....... (1) (alpha + beta)^2 = alpha^2 + beta^2 + 2alpha*beta alpha^2 + beta^2 = (b/a)^2 + 2c/a (alpha - beta)^2 = (b/a)^2 + 2c/a - 2c/a (alpha - beta) = b/a.......(2) Therefore, Value of (1/alpha - 1/beta)= (beta - alpha)/alpha*beta= - (alpha - beta)/alpha*beta= - (b/a) / c/a= - b/c |
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| 24. |
Evalute\frac{2 \sin ^{2} 63^{\circ}+1+2 \sin ^{2} 27}{3 \cos ^{2} 7-2+2 \cos ^{2} 73^{\circ}} |
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| 25. |
ŕ¤. > 100(100 â 40) |
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Answer» =(40/60) ×100=200/3= 66.6 |
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| 26. |
=4500(1+15/100)²-4500 |
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Answer» = 4500*[115/100]² - 4500= 4500*(23/20)² - 4500.= 5951.25 - 4500= 1451.25 Please hit the like button if this helped you. |
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| 27. |
Choose the correct ansteThe largest of the folloning is -a) o. 00ool b) I c) (o. 100) ² a) t = 0.1Iooo |
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Answer» b is the correct answer the answer is option d I am sorry the answer is b only options D is correct1/10÷0.1=0.1÷0.1=1 which is the largest number of the following number After solving all the option you will get 0.01 as the largest among all the numbers which is equivalent to option c . option D is correct answer option d is the correct answer of the given question |
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| 28. |
A colour TV is available for ₹26,880 inclusive of value added tax {VAT}.If the original cost of TV is ₹24000,find the rate of VAT. |
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Answer» tq |
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| 29. |
a colour Tv is available for₹26880 inclusive of VAT . if the original cost of TV is ₹24000. find the rate of VAT |
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Answer» can u resend it plz Original cost = ₹24000Selling price = ₹ 26880VAT price = ₹ 26880 - ₹ 24000 = 2,880VAT rate = (2,880/24000) × 100 = 12% to so much |
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| 30. |
(5+√7)(2+√5) |
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Answer» ( 5 + √7) ( 2 + √5) 10 + 5√5 + 2√7 + √35 if we write in explained |
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| 31. |
Amedicine capsule is in the shape ora cylinder with two--14m, m-hemispheres stuck to each of its ends. The length of thecapsule is 14mn. and the width is 5 mm. Find its surfacearea |
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| 32. |
Evaluate the Given limit: lim- |
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Answer» 5 is the correct answer |
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| 33. |
11. Evaluate the limitlim sex* Osin2x |
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| 34. |
Evalute the given limit : \lim _{x=3} \frac{x^{4}-81}{9 x^{2}-5 x-3} |
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Answer» thanks |
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| 35. |
Question 1. Define limit of a function. What are infinite limits andat infinity2 Make it clear.imits[Kanpur B.Sc. 20011 |
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Answer» The limit of a function at a point in its domain (if it exists) is the value that the function approaches as its argument approaches The concept of a limit is the fundamental concept of calculus and analysis. It is used to define the derivative and the definite integral, and it can also be used to analyze the local behavior of functions near points of interest. Infinite Limits. Infinite limits are those that have a value of ±∞, where the function grows without bound as it approaches some value a. For f(x), as x approaches a and what is infinity limit |
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| 36. |
lim sin x-sin ax→ ax -a |
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| 37. |
1. Find the remainder, if the polynomialx4 + x3-3x2 + 3 x + 1 is divided by :(ii) x(iv) 3+ 2x(iii) x + π(v) x |
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| 38. |
If(x + 1) and (x + 2) are the factors x^3 +3x2-3ax + b then find a and b |
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Answer» x+1 and x+2 are factors so if we put x=-1 and -2 we will get 0 as an answer. x*x*x+3x*x-3ax+b f(-1)=(-1)(-1)(-1)+3(-1)(-1)-3a(-1)+b = 0 -1+3+3a+b = 0 3a+b = -2 f(-2)=(-2)(-2)(-2)+3(-2)(-2)-3a(-2)+b = 0 -8+12+6a+b = 0 6a+b = -4 6a+b-3a-b = -4-(-2) 3a = -2 a = -2/3 3(-2/3)+b = -2 -2+b = -2 b = 0 |
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| 39. |
\lim _{x \rightarrow 3} x+3 |
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Answer» lim x->3 x+3by substituting x=3=3+3=6 |
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| 40. |
\operatorname { lim } _ { x \rightarrow 3 } \frac { \sqrt { 2 x + 3 } } { x + 3 } |
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| 41. |
1- 0050>(cosect âcotb) FoA AR 5 |
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| 42. |
\lim _{x \rightarrow 3} \frac{x^{2}-3}{x^{2}+3 \sqrt{3 x-12}} |
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| 43. |
\lim _ { x \rightarrow - 3 ^ { + } } \frac { x + 2 } { x + 3 } |
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| 44. |
1. Multiply the fo(i) 2xy, x |
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Answer» multiplication is2xy × x = 2x²y |
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| 45. |
2 MARKS QUESTIONFind the values using formulae:(i) 52^2 (ii) 77×83 |
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Answer» ii) (80+3)(80-3) = 80²-3² = 6400 - 9 = 6391 |
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| 46. |
\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3} |
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Answer» x^2-9 can be written as (X+3)(x-3)hence limits X tending 3 (X+3)(x-3)/(x-3)as x-3 will be get cancelledput in limits3+3=6 reply with step by step |
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| 47. |
Let's learn.Activity : Expand (x + + b) using formulae for areas of a square and a rectanglexxb |
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| 48. |
using distance formulae, show that points A (3,1),B(6,4), C(8,6) are collinear |
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Answer» love you prakat , can I have your WhatsApp number to take help during my exams, you are in 8 grade no for collinear AB +BC =AC ........ this is the basic formula of vector.... means they are in h same line |
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| 49. |
5. Find the square root of 625 and 441. Using these values, find the value of 6.25 x 70.0441. |
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| 50. |
7\lim _{x \rightarrow-3} \frac{2 x^{2}+7 x+3}{3 x^{2}+8 x-3} |
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Answer» by putting x = -3 we will get 0/0 form so using L'hospital rule lim x->-3 (6x+8)/(4x+7) = (-18+8)/(-12+7) = -10/-5 = 2 |
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