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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. |
1. A cycle is bought for Rs.900 and sold forRs.1080, find the gain percent? |
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Answer» 20 percent he has gain gain is 1080-900=180. gain percentages =180/900×100=20% Any ANIRUDH RAVICHANDER fans? Please follow me if you love ANIRUDH gain is 1080-900=180.gain percentage=180/900=20% Gain = 1080 - 900 = 180 Gain % = (Gain/Cp) × 100 % = (180/900) × 100 % = 100/5 % = 20 % gain percent is 20 and the best answer. 20 percent is the correct answer 180/900x100=20% answer 20% is the right answer... |
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| 2. |
. The marked price of an article is1350 and theselling price is t 1080. Find the percentage discount. |
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Answer» MP=1350SP=1080discount=(1350-1080)/1350*100=20% |
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| 3. |
Find the whole quantity if 12% of it is 1080 |
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Answer» Given12% =1080 Then1%=1080/12=90 Whole quantity =100% So whole quantity =90x100 =₹9000 |
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| 4. |
find the whole quantity if no 12%of it is $ 1080 |
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| 5. |
1. Find the least number of four digits which is divisible by 4, 6, 810. Ans.- 1080 |
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| 6. |
I 6 Find the area of thetrapezoid belowq.5cm6cm5.5cm |
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Answer» Area of trapezoid = 1/2(sum of parallel sides) height= 1/2(9.5+5.5) 6=1/2(15) 6= 15×3= 45 |
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| 7. |
3.) Radha made a picture ofan aeroplane with coloured paper as shown in Fig 12.15. Findthe total area of the paper used.Sem6cmIV.5cm V1.5cm6.5cm |
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| 8. |
using long division method find f(x)=9x³-3x²+x-5,g(x)=x-2/3 |
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| 9. |
Question 3. Radha made a picture of an aeroplane with coloured papeas shown in figure. Find the total area of the paper used.5 cm6 cm1 cmUL1 cm2 cm |
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| 10. |
mcarlour is requireAkite in the shape ofsom and sides 6 cmbe of a square with a diagonal 32 cm and an isosceles triangle of base6 cm each is to be made of three different shades as shown inw much paper of each shade has been used in it?17. How much paper of |
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Answer» 256 The perimeter of a triangle is equal to the sum of its three sides it is denoted by 2S. 2s=(a+b+c) s=(a+b+c)/2 Here ,s is called semi perimeter of a triangle. The formula given by Heron about the area of a triangle is known as Heron's formula. According to this formula area of a triangle= √s (s-a) (s-b) (s-c) Where a, b and c are three sides of a triangle and s is a semi perimeter. This formula can be used for any triangle to calculate its area and it is very useful when it is not possible to find the height of the triangle easily . Heron's formula is generally used for calculating area of scalene triangle. ________________ ____________________ Solution: Let the kite is made with square ABCD & an isosceles ∆DEF. Given, sides of a ∆DEF are DE=DF= 6cm & EF= 8cm & Diagonal of a square ABCD= 32cm We know that, As the diagonals of a square bisect each other at right angle.OA=OB=OC=OD=32/2=16cm AO perpendicular BC & DO perpendicular BC Area of region I = Area of ∆ABC= ½×BC×OA [Area of right Triangle=1/2× base height] Area of region I= ½×32×16=256cm² Similarly area of region II = 256cm² For the III section, Now, in ∆DEF let the sides a=6cm,b= 6cm & c=8cm Semi perimeter of triangle,s = (6 + 6 + 8)/2 cm = 10cm Using heron’s formula, Area of the III triangular piece = √s (s-a) (s-b) (s-c) = √10(10 – 6) (10 – 6) (10 – 8) = √10 × 4 × 4 × 2 =√2×5×4×4×2 =√2×2×4×4×5 =2×4√5=8√5 = 8×2.24=17.92cm² [√5= 2.24...] Hence, area of paper of I colour used in making kite= 256cm² Area of paper of II colour used in making kite= 256cm² And area of paper of III colour used in making kite= 17.92cm² Hope this will help you..... |
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| 11. |
x^2 + 8x^2 + 21x +20 by (x+3) . divide using long division method |
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| 12. |
0.9find the square root of the followingdecimal numbers using long division method.b) 1.3456. |
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Answer» 1.16 is the right Answer 1.16 is the right answer 1.16 is correct answer |
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| 13. |
using long division method perform the following divisions (2y^3 - 12y^2 +16)÷(y^2 - 6y +8) |
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| 14. |
) 200 mL 2.5 litre and 4:50 |
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Answer» Please hit the like button if this helped you |
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| 15. |
1. Write the interval (3, 8) in set-builder form. |
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Answer» Set builder form A = { x | x ∈ R, 3 ≤ x ≤ 8 } |
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| 16. |
Write the following set as interval{x € R:1>= 2x-3>=0} |
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Answer» 1 > = 2x -3 > = 04 > = 2x > = 32 > = x > = 3/2 this is wrong answer |
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| 17. |
Write the following set as interval:\{x \in R : 1 \geq 2 x-3 \geq 0\} |
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Answer» 1≥ 2x-3 ≥ 0 => 4 ≥ 2x ≥ 3=> 2 ≥ x ≥ 3/2 so, value of x is = [3/2,2] |
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| 18. |
Fig 10.35line BC and EFIn the figure 10.36, the lines AB and ED are parallel. Also,Find : (i) DGC (ii) DEF2470°56G C8Fig. 10.37Fig. 10.36 |
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Answer» Since the corresponding angles formed by two parallel lines and their transversal are equal in measure,therefore i)∠DGC = 70° ii)∠DEF=∠DGC = 70° |
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| 19. |
F導6.isinFig. 6.16,ifx-y= w+z,then prove that AOBis a line4.Fig.616 |
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Answer» x+y=180(linear pair) simlarily w+z=180(linear pair from 1and 2 x+y=w+z |
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| 20. |
in Fig. 6. 16, ifx + y = w + z, then prove that AOBis a line.4.Fig. 6.16 |
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Answer» angle y =angle w ......1angle z =angle x ........2y+ x+z+w=360(complete angle)from 1 and 2y+y+z+z=3602(y+z) =360y+z =360÷2=180hence AOB is line |
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| 21. |
Fig. 8.31Show that the line segments joining the mid-points of the oppositequadrilateral bisect each other6.sides of a |
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| 22. |
Find the whole quantity if(a) 5% of it is 600. |
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Answer» thanks 5%of x is 6005÷100 × x =6005x=600×1005×=60000×=60000÷5×=12000 |
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| 23. |
Find the whole quantity if(a) 5% of it is 600 |
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Answer» if 5 percent is 600.then 1 percent is 600/5 = 120.so the 100 percent that is the whole is...120 x 100 = 12000. thanks just one more find the whole Quantity sum 5% of it is 600 |
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| 24. |
Find:||jar 15% of250find the whole quantity if64\(b)1% of 1 hour(c) 20% ofă2500(d) 75%of 1kg0 of itis 500km. |
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Answer» a)15% of 250 = 15/100 × 250 =37.5b) 1% of 1 hour = 1/100 × 60 min = 0.6 min = 36 secc)20% of 2500 = 20/100 × 2500 = 500 d)75% of 1 kg = 75/100 × 1000g = 750 gm |
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| 25. |
Find the whole quantity if(a) ,5% of it is 600.d) 70% of it is 14 minut |
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| 26. |
11 Find the whole quantity, if:a. 15% of it is Rs. 120)b, 7% of it isc, 40% of it is 200 kmlitre |
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Answer» it is right answer or Rong answer |
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| 27. |
be in A.Ρ. & hı, h,,h10 be in Н.Р. . If al-h1-2 & 47 h10-3 then athis.Let a1, a2,, a10 |
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Answer» Is it a4 x h7? |
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| 28. |
cami Page)icowing(J2 |
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Answer» i) yes Given : x² + 3x - 5 = 0 Sum of zeroes = coff of x²/coff of x = 3/1 = 3 So, sum of roots = 3 |
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| 29. |
Draw pin diagram of IC 741 |
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| 30. |
1. Find the HCF of the following using long division methoda. 92, 64 b. 90, 110 c. 216, 124 d. |
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Answer» 92: 2*2*2*364: 2*2*2*2*2*2GCF: 2*2 The Greates Common Factor (GCF) is: 2 x 2 = 4 |
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| 31. |
Using long division method, divide the following polynomials by a binomial andcheck your answer.2.5(0) 4p - 4p?+ 6p-by 2p-1(ii)x-6+ 15x2 by 2+ 3x |
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| 32. |
Let A4 5 6 1(i) What is the order of A?(ii) Write down the entries a31, a 4, a(ii) Write down AT31 24 |
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| 33. |
9. Let n(A) 4 and n(B) k. The number of all possible injections from A to B is 120 then k-b) 24c) 5d) 6a) 9 |
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| 34. |
4.In Fig. 6.17, if x + yw + z, then prove that POQ is a lineFig 6.17 |
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| 35. |
\begin{array} { c } { \text { Ex. } 12 . \text { Let } A = \{ \text { real numbers } } \\ { \text { Let } R = \{ ( a , b ) : a , b \in A \text { and } a - b < 5 \} } \\ { \text { Is } R \text { an equivalence relation? Justify your answer. } } \end{array} |
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| 36. |
(1m)known x in each of the following diagramsnach of the following diagrams |
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Answer» i) sum of angles in a ∆ is 180°so 50° + 2x = 180°2x = 130°x = 65° ii) supplement of 110° is 70°supplement of 120° is 60°so x + 70° + 60° = 180°x = 180° - 130° = 50° iii) 2x + x + 90° = 180°3x = 90°x = 30° |
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| 37. |
Find the values of a, b and c in the followingdiagrams |
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Answer» a = 180-49 = 131° is the answer supplementary angle so 49+a=180 so a=131 |
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| 38. |
11. आकृति में, यदि ZPOR=3DZPRO है, तो सिद्ध कीजिए कि ZPOS = ZPRT है।INCERT |
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| 39. |
३ आकृति 6.17 में, 800) एक रेखा है। किरण 00२ रेखा0 पर लम्ब हैं। किरणों 0? और 00२ के बीच में 05एक अन्य किरण है। सिद्ध कौजिए:1£ROS= — (£Q0S- ZPOS) |
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| 40. |
4. In Fig., POQ is a line. Ray OR is perpendicular toline PQ. OS is another ray lying between rays OPand OR. Prove that LROS 1/2(2QOS - ZPOS).0) |
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| 41. |
4. AABCis inscribed in a circle with centre O.If LAOB 140° and ZBOC 100, findZABC |
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Answer» angle ABC = 1/2*(360-angle AOB -angleBOC)=> angle ABC = 1/2*(360-140-100) = 60° thank-you |
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| 42. |
3. In given figure O is the centre of circle. IfLAOC 130° then find ZABC130 |
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Answer» ABC = half of AOC (angle substended by an arc is half the angle substended by an arc at the centre )=130÷2=65 |
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| 43. |
3)A tree is broken without separatingfrom the stene by the wind. Thetop touches the ground making anandle 30° at a distance of 12mend foot of the tree. Find the heightof the bee before breaking.Promo |
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A ladder leaning against a wall withefits top end at a height of 5m from ithe floor and its bottom end is at450 to the floor, the distance fromthe wall is(A) 4 m(B) 4.5 m |
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Answer» tan 45 ° =height/distance from the wall=>1=5m /distance from the wall=> distance from the wall=5m hit like if you find it useful |
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| 45. |
128×4-480+600÷3 |
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Answer» 128×4-480+600÷3128×4-480+200512-480+200512-280432 |
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| 46. |
2.Find the values ofthe unknowns Х'and 'y, in the following diagrams.R.5012090 |
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| 47. |
OR रेखा ।१(0 पर लम्ब है। किरणों (00/? और ZROS = दर (2 (005 - ८2205).हीe _ .'.. “**' |
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| 48. |
In thefigure ZROSis two timesZPOS.If ZPOS = ZQOR, find the measuresof all the angles.0 |
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Answer» let ANGLE POS = xangle QOS=xANGLE ROS= 2xx+x+2x=180angle on a st line4x= 180x= 180/4=45so angle POS= 45= ang QOSANGLE ROS= 2x= 2×45=90 |
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| 49. |
Find the values of the unknowns 'x 'and 'y 'in the following diagrams800450°120°50° |
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| 50. |
rmethe Cylinder thus formed.The thickness of a metallic tube is 1 cm and the inner diameter of the tube is12 cm. Find the weight of 1 m long tube, if the density of the metal is 7.8 g/em3cmo. |
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Answer» Inner radius,r=12/2=6cmthickness=1cmso,outer radius,R=6+1=7cmheight,h=1m=100cmvolume=pi×(R^2-r^2)×h=22/7×(7^2-6^2)×100=22/7×(49-36)×100=22/7×13×100286/7×100=28600/7since,density=7.7gm/cm^3also,weight=volume×densityweight=28600/7×7.7gm=31460gm or,31.46 kg Hit like if you find it useful |
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