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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. |
B0D Wants to coverside 0.5 m, find the number of square tiles needed to cover the floor.a room 3 m 50 cm wide and 6 m long by squared tiles. If each square tile is |
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Answer» Area of room= length × breadth=3.5×6 m²=21 m² Area of square=0.5×0.5 m²=0.25 m² Number of tiles required= 21/0.25= 84 tiles |
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| 2. |
21 aIn the given figure, ABCD is a rectangle of dimensions 21 em * 14 m. Asemicircle ,ส drawn with BC aa diameter Find the area and theperimeter of the shaded region in the figure |
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Answer» Area of shaded region = Area of rectangle - Area of semi-circle =(21 × 14) – ½ × π × 7 × 7 = 294 – 77 = 217 cm2 Perimeter of shaded region = 21 + 14 + 21 + π × 7 = 56 + 22 = 78 cm Like my answer if you find it useful! |
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| 3. |
NCERT)13. The curved surface area of a cylinder is 1320 cm2 and its base had diameter 21 em. Finheight and the volume of the cylinder. |
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Answer» thank you |
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| 4. |
Ex.6 A wall of length 10 m was to be buiät across an open ground. The height of the wall is 4 m and thickness8 em, thenof the wal is 24 cm. If this wall is to be built up with bricks of dimensions 24 em 12 cmfind the number of bricks which are required. |
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| 5. |
In a triangle the lengths of the two larger sides are 10 and 9.respectively. If the angles are in AP, then the length of thethird side can be(1987,2M) |
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| 6. |
,6.1 If Io lines interseer cach other then the verlically opposiangles are equal |
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| 7. |
3 m 4 dm - 1 m 8 dm |
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| 8. |
Theorem: Prove that:\sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}} |
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| 9. |
16. State the converse of Pythagoras theorem and prove 1 |
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| 10. |
L ‘.“,,..”.._a” का प्रहार कीजिए। |
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Answer» (3a - 7b - c)² 9a² + 49b² + c² + 2(-21ab - 3ac + 7bc) 9a² + 49b² + c² - 42ab - 6ac + 14bc |
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| 11. |
how that the elements on the main diagonal of a skew symmetric matrix are alzero.2. If a matrix A is symmetric as well as skew symmetric, then A -o3. If A and B are symmetric matrices of the same order, then show that AB issymmetric if and only if A and B commute, that is AB BA. |
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Answer» Crop only the question that you want a solution for. We will not be able to provide solutions to multiple questions. |
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| 12. |
Which of the following has a larger area and by how much ?A rectangle of length 24 cm and breadth 17 cm or A square of side 21 cm. |
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| 13. |
which of the following has a larger area and by how much?A rectangle of length 24 cm and breadth 17 cm or A square of side 21 cm. |
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| 14. |
Find the length of the longest rod that canbe fitted into a rectangular cuboidal box ofdimensions 6 dm (width), 8 dm (length) and7.5 dm (height). |
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Answer» Given length l = 8dmbreadth b = 6dm and height h =7.5dmLongest rod that can be placed in a room is nothing but its diagonal.Length of diagonal of a cuboid = √(l² + b² + h²)Length of longest rod = √(8²+ 6² + 7.5²) dm= √(64+ 36 + 56.25) dm= √156.25= 12.5dm |
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| 15. |
The larger triangle below has a base of 10.14 m; the gray triangle has an area of 40.325 m.40.325 m2- 10.14 mDetermine the area of the larger triangle If it has a height of 12.2 m.a. |
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Answer» Area of grey triangle40.325=0.5*12.2*bB=40.325/0.5*12.2=40.325/6.1=6.61mSo base of small triangle=10.14-6.61=3.53mSo,area of small triangle=0.5*3.53*12.2=21.53m^2 |
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| 16. |
If twice the area of a smaller square is subtracted from the area of a largersquare is equal io 14cm However, twice the area of the larger square isadded to 3 times the area of the smaller square the result is 203cmDetermine the sides of the two squares. |
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Answer» Let the side of the smaller square = a cm side of the larger square = b cm twice the area of a smaller square is subtracted from the area of the larger square =14 cm²b² - 2a² =14-----(1)twice the area of the larger square is add to three times the area of the smaller square =203 cm² 2b² + 3a²=203---(2) multiply (1) with 2 and subtract it from (2)2b²+3a²=2032b²-4a²= 14____________7a²=175 a²= 175/7a²=25a = 5 cm substitute a =5 in (1)b²-2a²=14b²-2*5²=14b²-50=14b²=14+50b²=64b=8 cmlarger square side = b =8cmsmaller square side = a = 5 cm |
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| 17. |
21) Find the length of the longest rod that canbe fitted into a rectangular cuboidal box ofdimensions 6 dm (width), 8 dm (length) and7.5 dm (height). |
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| 18. |
\sin \frac{\pi}{10} \cdot \cos \frac{\pi}{5}=\frac{1}{4} |
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| 19. |
n. The area of 4 walls of a cuboidal box is 256 cm2. If the length of the box is 5 times the heightHOTSand breadth is 3 times the height, find the dimensions of the cuboidal box. |
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Answer» Area of 4 walls = 256cm^2 Let the height be x Length = 5x Breadth = 3x To find - Length , Breadth , Height Calculation - As we know the area of 4 walls is given . Thus, it's lateral surface area of cuboid since it's area of 4 walls excluding the top and base. Lateral surface area of cuboid - 2h ( l + b) Thus, 2(x) [ 5x + 3x ] = 256 10x^2 + 6x^2 = 256 16x^2 = 256 Therefore x^2 = 256/16 = 16 => x = √16 = 4 Thus we got the value of x = height = 4 cm Length = 5x = 5(4) = 20 cm Breadth = 3x = 3 (4) = 12 cm thanks |
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| 20. |
The dimensions of a closed cuboidal box are in the ratio 5 : 4: 3. If the total surface area is0.94 m', find the dimensions of the box. |
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Answer» length 5xends 4x by 3x two sides 5x by 4x -->2*5x*4x = 40x^2top and bottom 5x by 3x --> 2*5x*3x = 30x^2two ends 2*4x*3x =24x^2so94 x^2 = .94x^2 = .01x = .1.5 * .4 * .3 |
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| 21. |
(iv) bases E lu2 Find the area of trapezium with base 15 cm and height 8 em, if the sidegiven base is 9 em long. |
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| 22. |
Q.6-The two sides of a triangle are 15 cm and 13 em. The length of third side cannot be more than orequal to. |
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Answer» According to triangle property sum of any two sides should always be greater than length of third side. Therefore for given triangle length of third side can not be more than or equal to(15 + 13) or 28 cm |
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| 23. |
pi=(4/5)*(x %2B 10) |
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Answer» the value x will be 40 the value of x after finding solution will be 40 the value of x after finding solution will be 40 |
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| 24. |
\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4} |
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| 25. |
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40square meters, |
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| 26. |
te tHustuii.1 A fet, the cap used by the Turks, is shaped like thefrustum of a cone (see Fig. 13.24). If its radius on theopen side is 10 cm, radius at the upper base is 4 emand its slant height is 15 cm, find the area of materialused for making it. |
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| 27. |
Find the median of the following frequencydistribution:Class-Interval Frequency40-4545-5050-5555-6060-6565-7070-75WOOOWN |
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Answer» thank u |
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| 28. |
A lawn of 20 m long and 7.5 m broad is to be paved with tiles each measuring 3 cm by 2.5 dm. Find thenumber of tiles required.8. |
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| 29. |
A lawn of 20 m long and 7.5 m broad is to paved with tiles each measuring 3 cm by 2.5 dm. Find thenumber of tiles required8. |
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| 30. |
HERON's FORMULA3. There is a slide in a park. One of its side walls has been painted in some colour with amessage "KEEPTHE PARK GREEN AND CLEAN" (see Fig. 12.10). If the sides of thewall are 15 m, 11 m and 6 m, find the area painted in colour.KEEP THE PARKGREEN AND CLEAN11 m15 m |
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Answer» Let the Sides of the triangular wall are a=15 m, b=11 m & c=6 m. Semi Perimeter of the ∆,s = (a+b+c) /2 Semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m Using heron’s formula, Area of the wall = √s (s-a) (s-b) (s-c) = √16(16 – 15) (16 – 11) (16 – 6) = √16 × 1 × 5 × 10 = √ 4×4×5×5×2 = 4×5√2 = 20√2 m² |
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| 31. |
There is a slide in a park One of sts side walls has been painted in sone coliour with amessage "KEEPTHE PARK GREEN AND CLEAN" (see Fig. I2.10). İfthe sides of thewall are 15 m 11 m and 6 m, find the area painted in colour |
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| 32. |
There is a slide in a park. One of its side walls has been painted in some colour with amessage "KEEPTHE PARK GREEN AND CLEAN" (see Fig. 12.10).If the sides of thewall are 15 m, 11 m and 6 m, find the area painted in colour.203ラ11 mKEEP THE PARKGREEN AND CLEAN15 m |
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| 33. |
3,There is a slide in a park. One of its side walls has been painted in some colour with amessage "KEEPTHE PARK GREENAND CLEAN" (see Fig. 12.10). If the sides of thewall are 15 m, 11 m and 6 m, find the area painted in colour.ツーKEEP THE PARKGREEN AND CLEAN15 mFig. 12.10 |
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| 34. |
\left. \begin{array} { l } { \text { sions for } m = 2 , n = - 3 \text { and } p = 1 } \\ { 6 m ( m + n ) + 10 } \end{array} \right. |
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Answer» m=2 n=-36(2)(2-3)+10=-12+10=-2 |
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| 35. |
find the length of the longest rod that can be fitted into a rectangular cuboidal box of dimensions 6dm width,8dm length and 7.5dm height. |
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Answer» 7.5 |
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| 36. |
find the length of the longest rod that can be fitted into the cubic having dimension 1m, 0.5m, and 0.75m as length, breath and height respectively |
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Answer» 1.34m approx is correct answer 1.34 approx is the answer for this question |
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| 37. |
he triangular side walls of a flyover have been used for advertiswalls are 122 m.'22 m and 120 m (see Fig. 12.9). The advertisementsaningoR5000perim peryear. Acompany hired one ofits walls for 3 months, Hch rent did it pay?122m22m120m |
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| 38. |
. Find the area of a triangle whose base is 8 cm and height is 11 cm. |
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Answer» The area of a triangle 44 cm square. area=11*8/211*4=44cm Area of triangle is 44cm2 The area of a triangle 44cm square. area of triangle=1/2bh=1/2×8×11=4×11=44cm |
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| 39. |
Find the area of a triangle whose base is 12 cm and height is 8 cm. |
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Answer» Area= 1/2* base * height= 1/2*12*8= 48cm^2 |
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| 40. |
3. Find the dimensions of a rectangle whose perimeter is 28 m and area is 40 sq.m. |
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Answer» Given :-Perimeter of rectangle = 28 2 (l + b ) = 28 l + b = 14 l = 14 - b --- (i) area = 40 l × b = 40 from (i) ( 14 -b ) × b = 40 (14 - b²) = 40 b² - 14b + 40 = 0 After spilliting the zeros you will get... (length) = 10(breadth) = 4 Area of a cube is 25 sq.m. what is its volume |
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| 41. |
2. Find the area of a rectangle whose perimeter is 47 m and length is 15 m. |
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| 42. |
\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\csc 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \cdot \csc 40^{\circ} |
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Answer» Taking L.H.S; sin 50/ cos 40 + cosec 40/ sec 50 - 4 cos 50 * cosec 40= sin 50/ cos 40 + cos 50/ sin 40 - 4 cos 50/sin 40[ since cosec A= 1/sinA and sec A= 1/cosA.]=cos(90-50)/cos 40+ sin(90-50)/sin 40 - 4 sin(90-50)/sin 40[since cos A= sin (90-A) and sin A= cos (90-A)] = cos 40/cos 40 + sin 40/sin 40 - 4 sin 40/sin 40=1+1-4*1=1+1-4=2-4=-2=R.H.S Hence,proved. sin(90–40)/cos40 + cosec(90–50)/sec50 –4cos(90–40).1/sin40= =cos40/cos40+sec50/sec50–4sin40.1/sin40= 1+1-4=-2 |
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| 43. |
lllilles produced and the cost of each article.iouuetlon on thatUT Ul artnciessions of a rectangle whose perimeter is 28 meters and whose area is 40square meters. |
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| 44. |
The dimensions of the room are 5mX4mx2m. Find the length of longest rod that can be placed in the roomIf a V2 + 1 then find (a - 1) |
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| 45. |
Q. 1. Write the expression which represents apolynomial. |
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Answer» nmathematics, apolynomialis anexpressionconsisting ofvariables(also calledindeterminates) andcoefficients, that involves only the operations ofaddition,subtraction,multiplication, and non-negativeintegerexponentsof variables. An example of a polynomial of a single indeterminate,x, isx2− 4x+ 7. An example in three variables isx3+ 2xyz2−yz+ 1. |
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| 46. |
find the area of each of the following parallelogram whose base is 340mm , height is 17cm |
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| 47. |
Find the area of each of the following parallelogram whose base 340 mm, height - 17 cm. |
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Answer» thanks |
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| 48. |
Fiad the areaea helowof the parallelogram whose base and corresponding altitude arethe19 em height 7 cm |
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Answer» area of parallelogram is base × height so here area will be 12× 7 = 84 cm² |
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| 49. |
1. Find the area of each of the following parallelogram whose base 340 mm, height 17 cm. |
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| 50. |
. Find the distance between the parallel lines of a parallelogram whose base is 12.5 cm and area is 75 sq.cm |
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Answer» they asked distance between parallel lines but you got height |
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