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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. |
(-sqrt(5) %2B 5)*(sqrt(5) %2B 5) |
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Answer» 25-520 is a rational number is the correct answer of the given question To prove:(5+√5)(5-√5) is a rational number Solution:(5+√5)(5-√5)= 25-5√5+5√5-5=25-5=20 Here 20 is a rational numberHence (5+√5)(5-√5) is a rational number 20 is the right answer. |
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| 2. |
-2*(-3 %2B 5) %2B 5*((5*8)/2) |
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| 3. |
ises the rent of his house by 5% at the end of each year. If currently its rent2500 per month, how much will be the rent after 2 years? |
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Answer» percentage increase of rent in 1 year=5%so percentage increase of rent in 2 years=10%So rent after 2 years=2500+10% of 2500=2500+10/100x2500=2500+250=2750 for 1st yearrent=2500+5%of 2500=2500+5/100x2500=2500+125=2625for second year rent =2625+5%of 2625=2625+5/100x2625=2625+131.25=2756.25so this is the correct answer |
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| 4. |
フAman paysて500rent per month remains sameas rent for 3 months. How much does he hasto pay for a whole year, it the |
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| 5. |
Shaina pays 15000 as rent for 3 months. How much does she has to pay for awhole year, if the rent per month remains same? |
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| 6. |
6. Shaina pays7500asrentfor3months.Howmuchdoesshehastopayfor&tyear, if the rent per month remains same? |
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| 7. |
6. Shaina pays7500 as rent for 3 months. How much does she has to pay for a wholeyear, if the rent per month remains same? |
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| 8. |
IT X7-12mif 3x + 2y = 12 and xy = 6, find the value of 9x2 + 4y?onhoondoc form |
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Answer» S.B.S |
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| 9. |
P in the blank)a andbe two unit vectors and θ is the angle betweenthem. Then a +is a unit vector if :(A) θ -4(B) θ= π22π(Choose the correct one)Turn Over |
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| 10. |
-9 %2B 3 %2B 5*(5*9) |
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Answer» 3*1+9*5*5-9=3+45*5-9=-6+225=219 3+45*448*4192 is the answer |
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| 11. |
25. Simplify: cos (-870°) |
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Answer» cos(-870) = cos(870) = cos(720+150) = cos150 = cos(180-30) = -cos30 = -√3/2 |
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| 12. |
25. Simplify, |
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Answer» 8.0277 |
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| 13. |
Simplify 25 x 52 x t103x t4 |
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Answer» 25 * 5^2 * t^8/10^3 * t^4 = (5)^2 * (5)^2 * t^[8-4]/ (5*2)^3 = (5)^[2 + 2 - 3] * t^4/8 = 5t^4/8 |
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| 14. |
Simplify(i) (x2-5) (x +5) + 25 |
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Answer» x^3+5x^2-5x-25+25x^3+5x^2-5x |
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| 15. |
Choose the correct answer1. For two sets A and B, AUB=A if and only if(A) BsubsetA(B) AsubsetB(C) A!=B |
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Answer» If A∪B = A thenB⊆A, that is B is contained in A. A) option is correct |
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| 16. |
Find the distance between the following pairs of points |
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| 17. |
6 months. How much rent did it pay?14. The perimeter of an isosceles triangle is 42 cm and its base is 15 timeseach of the equal sides. Find (1) the length of each side of the triangle,(ii) the area of the triangle, and (iii) the height of the triangle. (Given,V7 = 2.64.) |
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Answer» I don't know iwant coins so let each equal sides be xcm base= 1½x=(3/2)xPerimeter=42cmx+x+(3/2)x=42cm(2x+2x+3x)=84[multiply by 2 both sides]7x=84x=84/7x=12 cm(i) length of each sides=xcm,xcm,(3/2)cm=12cm,12cm,(3/2)*12cm=12cm,12cm and 18cm.(ii)a=12cm,b=12cm,c=18cmSemiperimeter=(a+b+c)/2=(12+12+18)/2=42/2=21cmArea=√S(S-a)(S-b)(S-c)=√21(21-12)(21-12)(21-18)=√21x9x9x3=√3x7x9²x3=√3²x9²x7=3x9√7=27√7 cm²(iii) (iii)Area=27√71/2xbasexheight=27√71/2x18xheight=27√79xheight=27√7height=27√7/9 =3√7 =3x2.64 =7.92cm patani ki eda answer..... 27√7 cm^2 is the right answer 27√7 is the right answer |
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| 18. |
uitd 0l &glll lreular cone with radius 6 cm and height 8 cm.9. The base radii of two right circular cone of the same height are in the ratio 3:5. Find the ratioof their volumes. |
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Answer» Solution :- Let there be cone 1 and cone 2 respectively. Let the r and R be the radii of the two right circular cones respectively. Ratio of base radii = 3 : 5 Volume of cone = 1/3πr²h ⇒Volume of cone 1/Volume of cone 2 ⇒ (1/3*πr²h)/(1/3πR²h) ⇒ (1/3*π*3²*h)/(1/3*π*5²*h) = 3²/5² = 9/25 = 9 : 25 So, the ratio of their volumes is 9 : 25 thanks |
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| 19. |
Find the missing side of each triangle. Leave your answers in simplest radical form. State if the sidesform a Pythagorean triple.2)121)19v713 |
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Answer» Hyp² = opp²+adj² i) √19² = √7² + x² x² = 19-7 x² = 12 x = √12 ii) 13² = 12²+x² x² = 169-144 x² = 25 x = √25 x = 5 Like my answer if you find it useful! |
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| 20. |
. Find the distance between the following pairs of points |
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| 21. |
lve the following pairs of equati1.122y6 |
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Answer» 3y + 2x = 12 xy , (2y + 3x)6=6xy(13); 18x + 12y = 78xy, 9(2x + 3y = 12xy)= 18x + 27y = 108xy , 18x + 12y = 78 xy/ 15y =30; y = 30/25=2; 3y + 2x = 12xy ; 3(2) + 2x = 12(2)x; 6 + 2x = 24x ;; 6 = 22x, x = 22/6 |
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| 22. |
Compare the following pairs of rationalnumbers.[i)- an<d |
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Answer» 2/5 and 3/4 2/5 = 0.4 3/4 = 0. 75 So, 2/5 < 3/4 |
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| 23. |
x ^ { 2 } \times x ^ { 3 } \times x ^ { 4 } \div x ^ { 6 } |
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| 24. |
1. Add the following pairs of rational numbers:(11) 5'-9 |
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Answer» -2 1) -2/112)-1/9 is the correct answer of the given question |
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| 25. |
\begin{array}{ll}{\text { (i) }\left(a^{2}\right) \times\left(2 a^{2}\right) \times\left(4 a^{26}\right)} \\ {\left(-\frac{10}{3} p q^{3}\right) \times\left(\frac{6}{5} p^{3} q\right)} & {\text { (iv) } x \times x^{2} \times x^{3} \times x}\end{array} |
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| 26. |
\left(x^{3} \times x^{6}\right) \div\left(x^{4} \times x^{4}\right) |
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Answer» same base so power will be added x⁹ /x⁸x(⁹⁻⁸)x^1=x will be the answer tnx |
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| 27. |
\begin{array} { l } { \text { (viii) } \frac { 2 ^ { 8 } \times x ^ { 5 } } { 4 ^ { 3 } \times x ^ { 3 } } } \\ { \text { (xii) } \frac { 4 ^ { 5 } \times a ^ { 8 } b ^ { 3 } } { 4 ^ { 5 } \times a ^ { 5 } b ^ { 2 } } } \end{array} |
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| 28. |
(ĂźIf two circles with radii 8 em and 3 cm respectively touch externally,then find the distance between their centres. |
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Answer» The centres of the circles & the point of contact are collinear.Two circles touching internally at C.A & B are the centres.AC= AB+ BC (Betweenness property)AB is the distance between the centres.AC = 8 units (radius of the larger circle)BC= 3 units ( radius of the smaller circle)AC-BC= AB8–3 = AB5 = AB |
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| 29. |
\left \frac { 25 \times x ^ { - 4 } } { 5 ^ { - 3 } \times 10 \times x ^ { - 8 } } ( x \neq 0 )\right. |
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| 30. |
\frac{(\sqrt{x})^{\frac{3}{5}} \times(\sqrt{x})^{\frac{2}{5}}+2 \sqrt{x}}{2(\sqrt{x})^{\frac{2}{3}} \times(\sqrt{x})^{\frac{1}{3}}+\sqrt{x}} |
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| 31. |
3. Two concentric circles have radius 5 cm and 3cm respectively. Find the length of chord ofcircle which touches the smaller circle. |
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Answer» thank you |
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| 32. |
12. A cylinder and a cone have base radii 5 cm and 3 cm, respectively, and their respectiveheights are 4 cm and 8 cm. Find the ratio of their volumes. |
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Answer» Let the radius of Cylinder be r1 and height be h1 and the radius of cone be r2 and height of the cone be h2. Given = r1= 5 cm, h1= 4 cm r2=3 cm , h2= 8 cm Volume of Cylinder (V1)= πr1²h1 Volume of cone (V2) = 1/3πr2²h2 Volume of Cylinder(V1) : volume of cone(V2) πr1²h1 : 1/3πr2²h2 (5)²× 4 × 3 : (3)²× 8 25 ×12 : 9 ×8 25 × 12 : 72 25 : 6 Volume of Cylinder(V1) : volume of cone(V2)= 25:6 V1: V2 = 25:6 Hence, the ratio of volume of cylinder and volume of cone is 25:6 |
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| 33. |
25. Two circles with centres A and B of radii 3 cm and 4 cm respectively intersect at two pand D such that AC and BC are tangents to the two circles. Find the length of the common chord CDf n Ă P are SS, S1 respectively. Prove that: |
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| 34. |
dius of the circle is 6 cm. Theperpendicularefrom the centre of the circle to the chord.uhich is 8 cm in length, is(B) 3 cm(D) v7 cm2/5 cmO v5 cm |
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Answer» how we get root 20 into 2 root 5 |
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| 35. |
ult, lf the cost of 1 m? canvas is?70.of tarpaulin 3 m wide will be required to make conical tent of height & mdius 6 m? Assume that the extra length of material that will be required forrgins and wastage in cutting is approximately 20 cm (Use T-314) |
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Answer» Given a conical tent with Height ( h ) = 8m Radius ( r ) = 6m So, its length = l2= h2+ r2 = l2= ( 8 )2m + ( 6 )m2 = l2= 64m + 36m = l2= 100m = l = 10m Curved surface area of the cone = TTrl = 3.14 x 6m x 10m = 188.4 cm2 The extra length of material that will be required for stiching margins and wastages in cutting is approx 20 cm = 0.2m So, its length = l - 0.2m and its width = 3m area of sheet = CSA of tent so the area of the trapaulln = l x b = ( l - 0.2m ) x 3m = 188.4m2 = l - 0.2m = 62.8m2 = l = 63m Therefore, length of the trapaulin sheet required is 63m. |
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| 36. |
4. Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel Tdistance between ABand CD is 3 cm, find the radius of the circle. |
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| 37. |
22. Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively aremelted to form a single cube. Find the lateral surface area of the newcube formed. |
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| 38. |
Find the length of the side of a square where al eCISE-6.4-1MPind the one's digit of the cube of each of the following numbers.i) 3331 (ii) 8888 (iii) 149 (iv) 1005 () 1024 (vi) 77 (vii) 5022 (viii) 53f odd numbers using the |
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Answer» one's digits are following12954387 |
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| 39. |
90+80+100 |
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Answer» 90+80+10090+180270 answer |
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| 40. |
CD and GH are respectively the bisectorsof ZACB and LEGF such that D and H lieon sides AB and FE ofAABC and Δ EFGrespectively. IfAABC--A FEG, show that:CD ACGH FG(iii) Δ DCA~ ΔΗGF |
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| 41. |
9. Which term of A.P. 100, 90, 80, ... is zero? |
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Answer» Givena=100; d=90-100=-10an=0an=a+(n-1)d0=100+(n-1)-(10)0=100-10n+1010n=110n=110/10n=11 |
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| 42. |
11 )PA M P10. CD and GH are respectively the bisectorCB and LEGF such that D and H lieon sides AB and FE of Δ ABC and Δ EFGrespectively. IfAABC AFEG, show that:CD ACGH FG(iii) ΔDCA--AHGF |
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| 43. |
.10.CD and GH are respectively the bisectorsof ZACB and Z EGF such that D and H lieon sides AB and FE of Δ ABC and Δ EFGrespectively. If Δ ABC-AEG, show that:CD ACGH FG0)(iii) Δ DCA~ Δ HGF |
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| 44. |
ly tile bisectorsI ZACB and Z EGF such that D and H lieon sides AB and FE of Δ ABC and Δ EFGespectively. If Δ ABC-A FEG, show that:CD ACGH FGFig. 6.39 |
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| 45. |
of ZACB and ZEGF such that D and H lieon sides AB and FE of Δ ABC and Δ EFGrespectively. If Δ ABC-A FEG, show that:CD ACGH FG(ii) Δ DCB ~ Δ HGE(iii) Δ DCA ~ Δ HGF |
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| 46. |
10. CD and GH are respectively the bisectorsof ZACB and ZEGF such that D and H lieon sides AB and FE ofAABC and Δ EFGrespectively IfAABC AFEG, show that:CD ACGH FOG(ii) Δ DCB-AHGEFig. 6.39 |
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| 47. |
5. The radi of the internal and external surfaces of a hollouspherical shell are 3 cm and 5 cm respectively. If it is meltetand recast into a solid cylinder of height 22/3 cm, find the,die 'r, Ä°s melted and recast into a0wdiameter of the cylinder |
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| 48. |
. Some plastic balls of radius 1 cm were melted and cast into a tube. The thicknesslength and outer radius of the tube were 2 cm, 90 cm and 30 cm respectivelyHow many balls were melted to make the tube ? |
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Answer» sry wrong |
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| 49. |
sin x)Find the range of the function f(x) = |
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Answer» since the denominator is always +ve as [x] is squared. and [x] is always an interger.. and we know that sinnπ is always 0, where n is any interger. so sinπ[x] is always 0 so, the range is {0} right |
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| 50. |
dy= sin" x + sin-i VI-XFind-, when yах |
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Answer» Here |x| is due to the change in sign of X. |
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