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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. |
Find all zeroes of the polynomial p(x) = 2x4-3x3-9x2 + 15x-5, if two of its zeroes are5 and-5 |
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| 2. |
cot 250tan 65耳何何ATTeTQ : (cos? 20° + cos? 70°) +cot 5° cot 10° cot 60cot 80° cot 85° |
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Answer» cos²20+cos²70+cot25/tan65+cot5cot10cot60cot80cot85 =cos²20+(sin20)²+tan65/tan65+cot5cot10cot60tan10tan5 =since cos²x+sin²x=1 1+1+cot60 2+1/√3 Like my answer if you find it useful! |
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| 3. |
8 %2B 88 %2B 888 %2B 8888 |
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Answer» 98760 iska uttar hoga 98760 you are answer 98760 is right answer |
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| 4. |
Find the sum to n terms of the sequence, 8, 88, 888, 8888... |
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| 5. |
compute and give the answer in simplest form 2√72×5√32×3√50 |
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| 6. |
A closed cylindrical tank of height 1.4m and radius of the base is 56 cm ismade up of a thick metal sheet. Howmuch metal sheet is required ? |
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| 7. |
(ix) |
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| 8. |
AB 1PQ 3ar Δ ABCGiven Δ ABC ~ Δ PQR, if, then find ar A PCRar Δ PQR |
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Answer» The answer may be 1: 9 because the area of two similar triangles is equal to ratio of sum of square of there corresponding side |
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| 9. |
AB 1ar Δ ABCGiven Δ ABC ~ Δ PQR, if PQ3, then find ar A POR |
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Answer» As triangle ABC is similar to triangle PQRBy Area Theoremar(ABC)/ar(pqr)=(AB/PQ)^2=(BC/QR)^2=(AC/PR)^2 =(1/3)^2 =1/9 |
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| 10. |
ix) 3Vi331= |
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| 11. |
ix) 3v1331= |
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| 12. |
correspoesment28. If the median of aABC intersect atshow that ar(AGB) ar(AGC) - ar(BGC -1/3 ar(ABC) |
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Answer» given ,AM , BN & CL are mediansto prove -ar. ΔAGB = ar.ΔAGC , ar.ΔAGB = ar.ΔBGC & ΔAGB= 1/3ar.ΔABCproof ,inΔAGB &ΔAGC AG is the median∴ar. ΔAGB = ar.ΔAGCsimilarly , BG is the median∴ar.ΔAGB = ar.ΔBGCso we can say thatar. ΔAGB = ar.ΔAGC =ΔBGCnow ,ΔAGB +ΔAGC +ΔBGC = ar.ΔABC1/3ΔAGB +1/3ΔAGC + 1/3ΔBGC ( they are equal in area)=ar.ΔABCso we can say that , ΔAGB = ar.ΔAGC =ΔBGC = ar 1/3ΔABC ( PROVED ) |
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| 13. |
™=Sys-31:फिरA ARGl |
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Answer» (5ᵐx5³x5⁻²)/(5⁻³)=5¹²5ᵐ⁺³⁻²⁺³=5¹²5ᵐ⁺⁴=5¹²m+4=12m=8 |
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| 14. |
and C(-3, -1) are the vertices of AABC. Find the length of median AD.-1)arA(5, 1); B(1, 5)何 -可 |
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| 15. |
16, AABC and ABDE are two equilateral trianglessuch that D is the midpoint of BC. Then,ar(ABDE) : ar(AABC) =?1 : 4(d) 3:4(c) /3:2 |
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| 16. |
\frac { 2 ^ { 3 } \times 3 ^ { 5 } \times 16 } { 3 \times 32 } |
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Answer» (2^3×3^5×2^4)/(3×2^5)=(2^2×3^4)=4×81=324 answer kaisa aaya h mujhe smjh mai nahi aaraha h plz thoda acha se smjha dye |
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| 17. |
\frac { 15 ^ { 3 } \times 18 ^ { 2 } } { 3 ^ { 5 } \times 5 ^ { 4 } \times 12 ^ { 2 } } \quad ( v ) ( \frac { 6 } { 15 } ) ^ { 3 } \div ( \frac { 25 } { 32 } ) ^ { 2 } \times ( \frac { 45 } { 16 } ) ^ { 3 } |
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Answer» please do 5th question also |
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| 18. |
Find the equation of the straight line which cuts off an intercept -sfrom the y-axis and makes an angle of sin 13 with the x-axis4.12 |
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Answer» accha se likh ke Bhejiye |
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| 19. |
( \frac { 243 } { 32 } ) ^ { - \frac { 3 } { 5 } } \times ( \frac { 2 } { 3 } ) ^ { - 3 } \times [ ( \frac { 125 } { 27 } ) ^ { \frac { 2 } { 3 } } ] |
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Answer» thank you |
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| 20. |
1. State the universal law ofgravitation. |
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Answer» universal law of gravitation states as 1 directly proportional to the product of the masses of the objects and 2 inversely proportional to the square of the distance between them The universal law of gravitation states that the force of attaction between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.This law is applicable on any two objects anywhere in the universe. |
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| 21. |
(ix)6xy - 4y +6-9x |
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Answer» 6xy - 4y + 6 - 9x2y ( 3x - 2) -3 (3x - 2)(2y - 3)(3x - 2) |
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| 22. |
11. The length and the breadth of a rectangle are 36 cm and 28 cm. If its perimeter isequal to the perimeter of a square, find the side of the square. |
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Answer» Reactangle length = 36 cmReactangle breadth = 28 cm Perimeter of Reactangle= 2(length + breadth)= 2(36 +28)= 2*64)= 128 cm Let length of square is sThen Perimeter of Square = 4*s Given,4*s = 128s = 128/4 = 32 Side of square = 32 cm |
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| 23. |
28 cmHow many bricks, each of size 25 cmx13.5 cmx6 cm, will be required to build a walland 33 cm thick?8 m long, 5.4 m high |
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| 24. |
A rectangular sheet of paper is 33 cm long and 32 cm wide. It isalong its length to make a cylinder of height 32 cm. A circular shepaper is attached to the bottom of the cylinder formed. Find theof the cylinder formed |
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Answer» Given : Length of the rectangular sheet = 33 cm And, Width of the rectangular sheet = 32 cm As we know that, length of the rectangular sheet = circumference of the circle Circumference of the circle = 2πr ⇒ 33 cm = 2*22/7*r ⇒ r = (33*7)/(22*2) ⇒ r = 231/44 ⇒ r = 5.25 cm Capacity or volume of the cylinder =πr²h ⇒ 22/7*5.25*5.25*32 ⇒ 22/7*27.5625*32 ⇒ 19404/7 ⇒ 2772 cm³ Hence, the capacity or the volume of the cylinder is 2772 cm³ Like my answer if you find it useful! |
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| 25. |
ch side of a square piece of cardboacm. What is the area of the cardbomany square pieces of side 3 cmn be cut out of it?ow |
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| 26. |
Each side of a square piece of cardboard is30 cm. What is the area of the cardboard ?How many square pieces of side 3 cmcan be cut out of it? |
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| 27. |
i.Converting a Rectangle into a SquareOn a thick sheet of paper, draw a rectangle of dimensions 5 cm x 2 cm.2 cmUsing three straight cuts, divide the rectangle into 5 pieces such that these pieces whenrcarranged give a square. |
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Answer» 2(I+b)=2(5+2)=2(7)=14 |
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| 28. |
Evaluate by using formula 61^2-39^2 |
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| 29. |
Maths Project WORKÝ, Do These loWeite zure equivalent practionsfor 13 (1) |
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Answer» add 1/2 to4/7 u will get another five equivalent fraction (i) 3/5=6/10 =9/15 =12/20 =15/25 =18/30(ii) 4/7=8/14 =12/21 =16/28 =20/35 =24/42 3/5×2/2=6/10; 3/5×3/3=9/15; 3/5×4/4=12/20; 3/5×5/5=15/25; 3/5×6/6=18/30; (ii)4/7×2/2=8/14; 4/7×3/3=12/21; 4/7×4/4=16/28; 4/7×5/5=20/35; 4/7×6/6=24/42; 4/7×7/7=28/49 |
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| 30. |
. The capacity of an open cylindrical tank is 2079 m3 and the diameter of its base is 21m.Find the cost of plastering its inner surface at 40 per square metrelindar of height 1.21 m and diameter 28 cm is melted and recast |
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Answer» please like the solution 👍 ✔️ |
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| 31. |
It is required to make a closed cylindrical tank of height 1 m and base diameter 140cmfrom a metal sheet. How many square metres of the sheet are required for the same |
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| 32. |
Using the principle of mathematical induction prove that:. nintD(nt 5) is a multiple of 6 See Solved ix |
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Answer» Assume P(k) is true: K(k+1)(k+5)=6m [multiple of 6] Consider P(k+1)=( k+1) (k+2)( k+6) =k(k+1)(k+2)+6(k+1)(k+2)………split (k+6) =6n+6(….)…………………………………………….k(k+1)(k+2) product of 3 consecutive no’s is divisible by 6 therefore it is true for all natural nos Thnx |
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| 33. |
It is required to make a closed cylindrical tank of height 1 m and base diameterfrom a metal sheet. How many square metres of the sheet are required for the140 cm |
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| 34. |
an A. P., lfp. if o term is andPm term isă.term isprove that the sum of finst pand c"erms is pg +1), where p gfa certain number of terms of the A.P.25,22, 19,.. is 16. Find the |
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Answer» let the first term be a and common difference be dSo, pth term= 1/q a+ (p-1)d = 1/qqth term= 1/p a+(q-1)d = 1/pfrom these two equations, d= 1/pqa= 1/pq sum of first pq terms = (pq/2)(2/pq + (pq-1)*1/pq)= (pq/2)*(2/pq + 1- 1/pq)=(pq/2)*(1+1/pq)=1/2*(1+pq) |
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| 35. |
15 Ea2fnd the value of15 Iia 3-2,2, find the value of a |
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| 36. |
17¡1nte figure i, Ill, and at iia'. Find the value ofx,矸-15 |
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Answer» Please hit the like button if this helped you |
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| 37. |
s gCos © s/B o (एलW Coeded iia था |
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| 38. |
(2) It is required to make a closed cylindrical tank of height 1 m and base diameter 140ăfrom a metal sheet. How many square metres of the sheet are required for the same |
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| 39. |
8. In the given Fig. 1, AB BC and BP BQ, Show that AP CQ |
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Answer» Like my answer if you find it useful! |
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| 40. |
oftheThe area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimetertriangle (π ,ふ1.73) |
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Answer» thankyou |
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| 41. |
Find the rato or area onnages Iia diameter AB of a circle with centre O bisects each of the two chords CD and EF as shown 2in the figure, Prove that the two chords are parallel |
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| 42. |
7x+3-4x=10 |
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| 43. |
H+4X+103 L. |
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Answer» X + 4X + 10= 5X + 10 If X + 4X + 10 = 05X + 10 = 05X = -10X = - 2 |
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| 44. |
| 26.Solve the following differential equation :निम्न अवकल समीकरण को हल कीजिए :xcos) =ycos(१) +3 |
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| 45. |
\begin{array} { l } { \text { Using finst principle, find the derivative } } \\ { \frac { 1 } { 2 x + 3 } } \end{array} |
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Answer» d/dx(2x+3)^(-1)=-1(2x+3)^(-2)d/dx(2x+3)=-2/(2x+3)^2 what is (<)this sign in between |
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| 46. |
, where p and q are1. Is zero a rational number? Can you write it in the formand q #02 |
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Answer» no 0 is not a rational number 0 can not be written in p/q form yes zero is a rational no. it can be written in the form of p/q = 0/1 the question is not visible prooerly and the answer is 0 is a rational number and it can be written in the form of p_ q as 0÷1 No, 0 is not a rational number. It cannot be written in te form of p/q. |
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| 47. |
usveWeitetheresoes of70303y po) |
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| 48. |
3... निम्नलिखित व्यंजकों में से प्रत्येक व्यंजक को सरल कौजिए:o (3+5)(2++2) (3 - \/5) (3 = \fi) |
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| 49. |
\frac { 3 } { 4 } + \frac { 4 } { 13 } \div \frac { 3 } { 52 } - 6 |
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| 50. |
52:3.9::3:4 |
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Answer» 5.2 : 3.9 = 3 : 4 5.2/3.9 = 1.3×4 / 1.3×3 = 4/3 so this ratio is not same. |
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