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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. |
in the given figure, AD 13 cm, BC 12 cm,AB 3 cm and angle ACD angle ABC 90Find the length of DC. |
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Answer» from triangle ABC we can find ACAC^2 =AB^2 +BC^2 AC^2=3^2 +12^2AC^2 =9+144AC^2=153now in triangle ADCAC^2 +DC^2 =AD^2153+DC^2 =13^2DC^2=169-153DC^2 =16DC=4cm |
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| 2. |
a) 16 cmu 8.5 m19. The perimeter of a square is 28 cm. Find its area and the length of its diagonal. |
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Answer» area of square = 196cmdiagonal=14√2 Area of square=196cmdaigonal=14√2 area of square =196cmdiagonal=14√2 area of square =196cmdiagonal =14/2 |
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| 3. |
9. Mother gave Jyoti 14 piece of orange and 1 piece of orange to Meena. What was the total amount oforanges given to both of them? |
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Answer» Answer of your question is 11/4 Total part= 5/4+3/2LCM ° = 8= 10+12/822/8 part( in total she gave ) |
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| 4. |
\frac { 9 ^ { 3 } \times 27 \times 81 ^ { 4 } } { 3 ^ { - 2 } \times 3 ^ { 4 } \times 81 ^ { 2 } } |
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Answer» can you take a clear picture it is not much visible |
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| 5. |
A piece of cardborad in the shape of a trapeziumABCD and AB IlDC, LBCD = 900, quarter circleBFEC is removed. Given AB- BC3.5 cm, D2 cm. Calculate the area of remaining piece ofcardborad. |
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| 6. |
Math Sumetan t° Jan 220 tom 60 tan 67 tan 8+Lot 54 t sin 20 sec 30 -2dan 36 |
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| 7. |
\frac 2 ^ 3 \times 3 ^ 4 \times 4 3 \times 32 |
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| 8. |
aSimplify(1) SUME E VIVEiv75 xbox 363140 x 20042837 |
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| 9. |
35. Find the quadrantic polynomial withthe given numbers at the sume andproduct of its zero v2, |
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Answer» Let the polynomial be ax^2 + bx + c, and its zeroes be α and ß α + ß = √2 = 3√2/3 = -b/a αß = 1/3 = c/a If a = 3, then b = -3√2, c = 1 Therefore, the quadratic polynomial is 3x^2 -3√2x +1. Like my answer if you find it useful! |
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| 10. |
। (28+1) +52 =4, (x -3) |
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| 11. |
So)eveeL = ihe |
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Answer» (4/7) / ( 3 1/3 + 2 4/5 )= (4/7)/[ (10/3)+(14/5)]= (4/7)/(92/15)= (4*15)/(7*92)= 15/161 Please hit the like button if this helped you |
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| 12. |
स्वाध्याय [ध्छी2 (0810 भा,ही 40 खने 28 थे. ८ 6 नी. दिभाऊ छे, (खुखो 2u5(d 7.16.)नित 5सो 3 2 280 52 280,खने. 90 विशे तमे. शुं 5डी शी ? |
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| 13. |
ind the coefficient of range of the followingdistribution:40, 52, 47, 28, 45, 36, 47, 50 |
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| 14. |
(ix) given as,U, u() given /-28, S-144, and there are total 9 terms. Find a. |
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Answer» We know, S = n/2 × ( a + l) 144 = 9/2 × ( a + 28) (144 × 2)/9 = a + 28 32 = a + 28 a = 32 - 28 a = 4 thank you so much for your answer Please hit like button if you like my answer |
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| 15. |
11. The adjacent sides of a parallelogram are8 cm and 9 cm. The diagonal joining theends of these sides is 13 cm. Calculatethe area of the parallelogram. |
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Answer» Let tqo adjacent sides are a=8cm, b=9cmdiagonal d=13cmthese three sides show one triangles= (a+b+c)/2=(8+9+13)/2=30/2=15s-a=15-8=7s-b=15-9=6s-c=15-13=2area of the triangle = √[s(s-a)(s-b)(s-c)= √[15*7*6*2area of the parallelogram =2*area of triangle=2*[√(15*7*6*2)]=2× 20.4939015319=40.98 cm^2 |
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| 16. |
length of the ribbon they boughtNaina was given 1 piece of cake and Najma was given 1the total amount of cake was given to both of them |
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| 17. |
Find the diametercircle of the circle whose circumference 11 cm. |
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| 18. |
3.Naina was given 1 2 piece of cake and Najma was given 1 5 piece of cake. Findthe total amount of cake was given to both of them. |
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| 19. |
In the given figure find:(i) ZACD5.(ii) ZAED1540° |
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Answer» angle ACD = 40+15 = 55° angle AED = 55+52 = 107° |
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| 20. |
Naina was given 15 piece ofcake and Najmathe total amount of cake was given to both of them.3.was givern1;2 |
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| 21. |
piece of cake and Nagma was givenMoni got 4of cake given to both of them.z piece of cake and Nagma was given 1 piece of cake. Find the total amount7. |
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| 22. |
The ratio between the curved surface and total surface of a cylinder is 1:2. Find thevolume of the cylinder, given that its total surface area is 616 cm2. |
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| 23. |
Find the circumference of a circle whose radius isa.(ii) 10.5 cm(ii) 3.5 mFind the circumference of a circle whose diameter is(i1) 35 cmi) 10.5 m |
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| 24. |
find the radius of circle whose circumference is 57.2 cm |
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Answer» thank you |
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| 25. |
. Find the equations of two straight lines passing through (1, 2) and making an angleof 60° with the line x +y 0. Find also the area of the triangle formed by the three[BTE 2004]lines. |
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| 26. |
find the radius of a circle whose circumference is 57.2 CM |
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| 27. |
what is cellular respiration |
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Answer» Cellular respirationis a set of metabolic reactions and processes that take place in the cells of organisms to convert biochemical energy from nutrients into adenosine triphosphate (ATP), and then release waste productsPlease change the subject here.It is a biology question so make sure the subject indicates biology and not mathematics. |
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| 28. |
38. Find the equation of the straight lines passing through (2,-1) and(P. U. 67).making an angle of 45 with the line 6x + 5y 8 |
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| 29. |
. In Fig. 1, AD = 4 cm, BD = 3 cm and CB = 12 cm. Find cot θ |
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| 30. |
7. AABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB-5 cm.Fig 11.24BC = 1 3 cm and AC = 12 cm. Find the area of AABC. Also find the length ofADb cmD 9 cm13 cmFig 11.25Fig 11.26 |
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| 31. |
Construct a quadrilateral:1. ABCD in which AB = 4.5 cm, BC = 3.5 cm, AD = 3 cm. AC-5 cm and BD-45 cm. |
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Answer» diagram |
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| 32. |
I. In Fig. 6.17.0 and (ĹŠ), DE BC. Find EC in (0 and AD in (i).1.5 cm 1 cm1.8 cm7.2 cm3 cm54 cn |
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Answer» i) AD/DB = AE/EC => 1.5/3 = 1/EC => EC = 2cm ii) AD/DB = AE/EC => AD/7.2 = 1.8/5.4=> AD = 7.2/3 = 2.4 cm |
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| 33. |
2. In the given figure, DE || BC.(i) If AD-3.6 cm, AB = 9 cm and AE = 2-4 cm, find EC.AD 3DB 5(ii) Ifand AC = 5.6 cm, find AE.=(iii) If AD = x cm, DB = (x-2) cm, AE = (x + 2) cmand EC = (x-1) cm, find the value of x. |
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| 34. |
In figure 3.8, <ACD is an exterior angle of Δ ABC.ZB 40°, ZA 70°. Find the measure of ZACD |
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Answer» thanks ji |
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| 35. |
Construct a quadrilateral:1. ABCD in which AB-4.5 cm, BC- 3.5 cm, AD 3 cm, AC- 5 cm and BD 4.5 cm. |
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Answer» thank you |
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| 36. |
For ABC, find the measure of angle ACD. |
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Answer» angle ACD = angle CBA + angle BAC= 66+45 = 111° |
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| 37. |
2. Show that any positive odd integer is of the form 6q+1, or 64 +3. or 69 +5, where issome integer |
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Answer» Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder. according to Euclid's division lemmaa=bq+r a=6q+r where , a=0,1,2,3,4,5then,a=6qora=6q+1ora=6q+2ora=6q+3ora=6q+4ora=6q+5 but here,a=6q+1 & a=6q+3 & a=6q+5 are odd. |
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| 38. |
/2,Showthatany positive odd integer is of the form 6q+ 1,or 6q + 3, or 6g + 5, where gissome integer. |
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| 39. |
| — I shoo jtgddto« (}ucem Mw— "/ |
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| 40. |
2.Show that any positive odd integer is of the form 6q +1, or 6q +3, or 6q + 5, where g issome integer |
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| 41. |
चर नल Shoo Hab 2 log(r))= 4 तन वीक:i % |
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Answer» x² + y² = 7xy Add 2xy on both sides x² + y² + 2xy = 9xy ( x + y)² = 9 xy Take log on both sides 2log ( x + y) = log3² + log (xy) 2log(x+y) = 2log3² + log x + logy |
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| 42. |
2. Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q issome integer |
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| 43. |
Practice set 3.11. In figure 3.8, ZACD is an exterior angle of A ABC.4-40°, 2A= 70°. Find the measure of ACD.In A POR,<P = 70°、20-65° then find <R. |
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Answer» Angle ACD=angle B + angle A(exterior angle property) Angle ACD=40+70=110 |
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| 44. |
Sides BC, CA and AB of a triangle ABC are produced in an order,forming exterior angles ACD, BAE and LCBF. Show that\angle A C D+\angle B A E+\angle C B F=360^{\circ} |
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| 45. |
In right angle ABC,LB=90, AB=6cm, BC=8Cm, Find AC |
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Answer» Thanks |
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| 46. |
Simplify and express each of the following in exponential form:0 23 324())(s)xs)s(253x32 |
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Answer» ((5^2)^3*5^4)÷5^7(5^6*5^4)÷5^75^10÷5^75^3as according to law of indicesx^a*x^y=x^a+y |
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| 47. |
साई 34 syle (1)(2x+y)(2x-y)—(2x + y)? |
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Answer» (2x+y)(2x - y) - (2x+y)^2 Taking (2x + y) common,(2x+y) ( 2x - y - 2x - y)= (2x+y)(-2y)= -4xy - 2y^2 PLEASE HIT THE LIKE BUTTON |
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| 48. |
Simplity andexpress each of the following in exponential form03x323x7 x1I2Ix11'3(vi) 2+3+43 x31a) (2'x2true or false and justify your answer |
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| 49. |
(A) x> 1(C) x<1The solution of the inequation 4(B) x21(D) xs124is :(A) χ>1(c) xs 1(B) x2 1(D) xs1 |
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| 50. |
28. Solve for x:x-1 2x +12x +1 x-1 |
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Answer» Please hit the like button if this helped you |
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