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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. |
-4-19- 34 - 3=022 QA - 2 8compare TtLWA |
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| 2. |
8pare : -andEx. 1. Compare : andsol. =* = -8 andand |
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| 3. |
B - e o WA€= 2 -xaer-=£ 4 LySRRo |
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Answer» By elimination method x/2 + 2y /3 = - 1 ………..(1) x – y/3 = 3 ………..(2) Multiplying equation (1) by 2, we obtain x + 4y/3 = - 2 ………(3) x – y/3 = 3 ………..(2) Subtracting equation (2) from equation (3), we obtain 5y /3 = -5 Divide by 5 and multiply by 3 we get Y = -15/5 Y = - 3 Substituting in equation (2), we obtain x – y/3 = 3 ………..(2) x – (-3)/3 = 3 x + 1 = 3 x = 2 Hence our answer is x = 2 and y = −3 By substitution method x – y/3 = 3 ………..(2) Add y/3 both side we get x= 3 + y/3 ……(4) Plug this value in equation (1) we get x/2 + 2y /3 = - 1 ………..(1) (3+ y/3)/2 + 2y /3 = -1 3/2 + y /6 + 2y/3 = - 1 Multiply by 6 we get 9 + y + 4y = - 6 5y = -15 Y = - 3 Hence our answer is x = 2 and y = −3 Like my answer if you find it useful! |
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| 4. |
9. Quadratic polynomial having sum of its zeroes 5 and product of its zeroes -14 is(A) x2-5x -14 (B) 2-10x-14 (C) x5x 14 (D) -10x+ 14y- |
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Answer» Any quadratic polynomial having zeroes as p and q can be written as x² - (p+q)x + pq So, given p+q = 5 and pq = -14 So, equation is x² - 5x - 14 A) option is correct |
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| 5. |
6 \times 6 \times 6 \times 6 |
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Answer» 1296 is the answer 1,296 is the REAL answer |
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| 6. |
\frac { 1 } { 3 } \times ( 6 \times \frac { 4 } { 3 } ) \text { as } ( \frac { 1 } { 3 } \times 6 ) \times \frac { 4 } { 3 } |
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Answer» As product of real numbers is commutative so order of terms doesn't matter at all. Therefore both are equal.1/3 × (6×4/3) = 8/3(1/3×6)×4/3 = 8/3 |
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| 7. |
\frac{4+4 \times 18+6-8}{123 \times 6-146 \times 5}= |
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Answer» 4+4×3-8/738-730=8/8=1 (4+4×18÷6-8)/(123×6-146×5) = (4+12-8)/(738-730) = 8/8 = 1 |
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| 8. |
18 \div ( 25 - 10 - 6 ) %2B 5 \times 6 - 4 \times 8 |
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Answer» 0 is the answer of this question |
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| 9. |
A O 24 g sample of compound of oxygen and boron was foundby analysis to contain 0:096 g of boron and 0.144 g of oxygen.Calculate the percentage composition of the compound byweight. |
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Answer» Mass of compound = 0.24 g and, mass of boron = 0.096 g Percentage of boron in the compound = mass of boron / mass of compound * 100 = 0.096/0.24 * 100 = 40% Mass of oxygen = 0.144 g again, mass of compound = 0.24 g Percentage of oxygen in compound = mass of oxygen/mass of of compound * 100 = 0.144/0.24 * 100 = 60% Thus, the percentage of oxygen and boron in the compound is 60% and 40% respectively. Thank you friend |
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| 10. |
[ 3 \frac { 1 } { 4 } \div \{ 1 \frac { 1 } { 4 } - \frac { 1 } { 2 } ( 2 \frac { 1 } { 2 } - \frac { 1 } { 4 } - \frac { 1 } { 6 } ) \} ] |
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| 11. |
( - \frac { 7 } { 12 } ) ^ { 23 } |
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Answer» (-1)^23 (7/12)^23-(0.5833333333)^23-(0.004536)^2 (0.5833333333)^3- 0.00002025 x 0.1985-0.000004019625 |
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| 12. |
\frac { 7 } { 12 } + \frac { 5 } { 6 } - \frac { 2 } { 3 } |
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| 13. |
प्रश्न 14. यदि A+ B = है तो सिद्ध कीजिए कि4(1+ tan A) (1+ tan B)-2 |
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| 14. |
\frac { 4 } { 5 } x - \frac { 2 } { 3 } x = \frac { 7 } { 12 } |
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| 15. |
कै. कि कि एप As हलप्रश्न 14. समीकरण t\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-3}\left(\frac{8}{31}\right)को हलकीजिए। गा (2018, 17,12) |
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| 16. |
- ( 14 ) \frac { \operatorname { tan } \lambda } { 1 - \operatorname { cot } A } + \frac { \operatorname { cot } A } { 1 - \operatorname { tan } A } = \operatorname { sec } A \operatorname { cosec } A + 1 |
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| 17. |
19.Prove that the rectangle circumscribing a circle is a square.ndrant of a circle of radius 14 cm and a semicir |
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Answer» Given: A rectangle ABCD curcumscribe a circle which touches the circle at P, Q, R, S. To Prove: ABCD is a square. Proof:As tangents from external point are equal.AP = AS−−−−−−−−−−−(1)PB = BQ −−−−−−−−−−(2)DR = DS−−−(3)RC = QC−−−−−−−−−−(4) Add (1), (2), (3) and (4)AP+PB+DR+RC = AS +BQ+DS+QCAB +CD=AD+BC2AB =2BCAdjacent sides are equal so ABCD is a square. |
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| 18. |
e diameter of the circle circumscribing the rectangle ABCD is 4yOncoordinates of A and B are (-3, 4) and (5, 4) respectively, find the equation of thecircle.x+7. If the |
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| 19. |
using bodmas ; 5+3{-34-18-14}÷3[17+(-3)×4-2×(-7)] |
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Answer» the right answer is -1249 |
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| 20. |
7. How much will 50,000 amount to in 3 years,compounded yearly, if the rates for thesuccessive years are 6%, 8% and 10%respectively. |
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Answer» Principle amount P = 50000, n = 3 years, Amount = A Amount = 50000*(1+(6/100))*(1+(8/100))*(1+(10/100))= 50000*(1.06)*(1.08)*(1.10)= 62924 Therefore, After 3 years 50000 amount to Rs 62964 |
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| 21. |
25 \frac 7 8 %2B 8 \frac 7 12 |
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| 22. |
tan(-14*pi %2B theta) |
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| 23. |
14) Write the simplest form of\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) 0<x<\pi |
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| 24. |
Solve the following briefly : /2 marks eachl1. The area of a circle is (14+6 \sqrt{5}) \pi units, Find its radius. |
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| 25. |
Multiple Choice QuestionsChoose the Correct Option:I. Principal value of sin-1-=1/23TC4ĺ (d)none of these4 |
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| 26. |
Prove that a rectangle circumscribing a circle is a square. |
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| 27. |
ORFind the valbe s) of k for which the equati5kx16-O bas real and equal zots |
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Answer» x² + 5kx + 16 = 0For real and equal root discrimanat is 0b² - 4ac = 0(5k)² - 4×16×1 = 0(5k)² = 645k = √(64)5k = + or - 8k = 8/5 and -8/5 |
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| 28. |
Find the median of variates 5, 7, 8, 11, 13, 17 with frequencies 3, 7, 7, 12, 13, 10respectivelyhtends a right angle at the centre. What is th |
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Answer» Variates = 5, 7, 8, 11, 13, 17Frequency = 3, 7, 7, 12, 13, 10 Thusas we knowTotal Number of variates = 6 Thus Center two variates are 8 and 11And Frequencies are 7 and 12So Median = 7+12/ 2 = 19/2 = 95.5 |
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| 29. |
5. Convert the following into equivalent likefractions :3 1317 197 810 15 |
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Answer» 1,15/35,65/352,19/21,57/21 is the right answer c) is the correct answer |
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| 30. |
00288p)m(0(216p)m(d) none of theses The volume of a cube is 1000 cm' then its side is(a) 11 cm(b) 10 cm(c) 12 ctu |
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Answer» (a)^3 =1000 a= 10 cm |
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| 31. |
The circumference of the base of a cylindrical vessel is 132 cmco. How many litres ofwater can it hold? (1000 cm-11)1.buse radius and the sathat will help us, to seeeight is 252. The inner diameter ofoulnd |
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| 32. |
11. The HCF of the following fractions is 1734 an85317 3471703170313617(5) of these |
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Answer» Please like the solution 👍 ✔️ |
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| 33. |
Write two like fractionsfor each of the following.(a) 3/7(b) 4/11(c) 7/12(d) 9/13(e) 5/17 |
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Answer» a) 5,7 2/7 b) 7/11, 9/11 c) 3/12, 10/12 d) 4/13, 7/13 e) 8/17, 12/17 |
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| 34. |
5. 3, 4, 5, 7, 7, 10, 9,?,?r11, 112) 13, 113) 13, 134) 12, 15 |
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Answer» 3 4 3+2=5 4+3=7 5+2=7 7+3=10 7+2=9so missing terms are 10+3=139+2=11 |
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| 35. |
There is a cylinder circumscribingthe hemisphere such that theirbases are common. The ratio oftheir volume is |
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| 36. |
If the mass of 5 thin golden wires is250mg. How many wires would weigh1 kg?9. |
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Answer» 5 wires are of 250 mg now 1 kg = 1000000 mg Number of wires whose weight 1 kg are = (1000000×5)/250 = 20000 If you find this answer helpful then like it. |
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| 37. |
into an empty cylindrical vessel of radius 8 cm, find the hsvessel.A piece of ductile metal is in the form of a cylinder of diameterdrawn out into a wire of diameter 1 mm. What will be t1 cm and length 11 cm. It isso obtained?he length of the wireii melted to form a cylindrica |
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| 38. |
The circumference of the base of a cylindrical vessel is 132 cm and its height is25How many litres of water can it hold? (1000 cm-11)1. |
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| 39. |
Assume π--, unless stated otherwise.I.The circumference of the base of a cylindrical vessel is 132 cm and its height is 2sHow many litres of water can it hold? (1000 cm-1DActivity:1base radiusthat will h |
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| 40. |
0227The circumference of the base of a cylindrical vessel is 132 cmAssume π-, unless stated otherwise.1.How many litres of water can it hold? (1000 cmThe inner diameter of a cvlindrical11)2. |
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| 41. |
e circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cHow many litres of water can it hold? (1000 cm 11) |
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| 42. |
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cmHow many litres of water can it hold? (1000 cm3- 11) |
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| 43. |
Assumeunless stated otherwise.π =ctivity : Tr1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm that will helpase radius aHow many litres of water can it hold? (1000 cm3 11) |
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| 44. |
14. A rectangular vessel 22 cm by 16 cm by 14 cm is full of water. If the total water is pouredinto an empty cylindrical vessel of radius 8 cm, find the height of water in the cylindricalvessel. |
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| 45. |
12. A conical vessel has radius 10 cm and height 18cm and is completely filled with water which istransferred in a cylindrical vessel of radius5cm. Find the level of water (height) incylindrical vessel. |
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| 46. |
Two circles touch externally. The sum of their areas is 130Ď sq, cm and the distance betweentheir centres is 14 cm. Find the radii of the circles. |
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| 47. |
Two circles touch externally. The sum of their areas is 130Ď sq. cm and the distance betweentheir centres is 14 cm. Find the radii of the circles. |
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Answer» 1 |
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| 48. |
Two circles touch externally. The sum of their areas is 130T sq. cm and the distance betweentheir centres is 14 cm. Find the radii of the circles. |
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Answer» 1 |
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| 49. |
\frac { 1 } { 3 } \left[ \begin{array} { c c c } { - 2 } & { 1 } & { 2 } \\ { 2 } & { 2 } & { 1 } \\ { 1 } & { - 2 } & { 2 } \end{array} \right] |
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| 50. |
11.In the given figure, AB//CD. If LBAC = 75° and <CFE = 32",find the value of x75843°c. 48"b. 54d. 6332 |
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