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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. |
The length and breadth of a park are in the ratio 2 : 1 and its perimeter is 360m. A path 2m wide runsinside it, along its boundary. Find the cost of paving the path at11·80per m2. |
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| 2. |
ratio 2:1 and its perimeieA path 2m wide runs inside it, along boundary. Find the cost of paving the path at3per m. |
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Answer» thank u |
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| 3. |
1. The length and breadth of a park are in the ratio 2: 1 and its perimeter is 360m. A path 2m wide runsinside it, along its boundary. Find the cost of paving the path at 80per m2. |
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| 4. |
T, b Ty ज़० n A WW?WM Pode sa o THEyoulie पा... o128 M fia/ )(एव 2 (7 7/:%&/"7 |
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Answer» Ratio=4.3Length=4x, breadth=3xArea=17284x*3x=172812x²=1728x²=1728/12=144x=12Length=48m, breadth=36m Perimeter=2(l+b)=2(48+36)=2(84)=168m Cost of 1 m=30RsCost of 168 m=30*168=5140Rs |
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| 5. |
pathThe length and breadth of a park are in the ratio 2: 1 and its perimeter is 240 m. A2 m wide runs inside it, along boundary. Find the cost of paving the path at ?5 perIn.2 |
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| 6. |
1.a)Solve: 11_=_1_-1X-aX-a+CX-b-cX-bhifactoriza. (2x272513x-2y-24- |
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| 7. |
(+ e ) SOD= {!ďŹ ÂĽ X) - |
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Answer» Sin(cot^-1(x+1)=cos(tan^-1(x) =>sin(sin^-1(1/sqrt(x^2+2x+2)))=cos(cos^-1(1/sqrt(1+x^2))) => 1/sqrt(x^2+2x+2)=1/sqrt(1+x^2) => x^2+2x+2 = 1+x^2 => 2x+1 = 0 Solving the above equation we get x=-1/2. HIT THE LIKE BUTTON! |
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| 8. |
quey ' छण्छय- न gS00°yS0d(d + V)sod |
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Answer» cos(a+b)=cosAcosB-sinAsinBhencedividing the given equation by cosAcosBhence1-tanatanb hence proved |
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| 9. |
What per cent of 10 kg is 250 g? |
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Answer» in 10kg the percent of 250 g is 25% 250g is what percent of 10kg? Let x be the required percentagex% of 10kg = 250g1kg = 1000g=> 10kg = 10000gx% 10000g = 250g(x/100) × 10000 = 250 x = 250/100 = 2.5 |
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| 10. |
at per cent of 10 kg is 250 g? |
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| 11. |
Datesap |
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| 12. |
8x-3=9_2x |
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Answer» here 8x-3=9-2x 8x+2x=9+310x=12x=12/10x=1.2 hope this will help you like my answer and MARK IT AS BEST ANSWER. 8x -3 = 9 - 2x8x + 2x = 9+310x = 12 x = 12/10 x = 1.2 8x-3=9-2x8x+2x=9+310x=12x=12/10x=1.2 answer. 8x-3=9-2x8x+2x=9+310x=12x=12/10x=1.2 8x + 2x = 9 + 310x = 12x = 1.2 X=1.2 is the right answer x=1.2is the best answer x=1.2 is the correct answer 8x-3=9-2x8x+2x=9+310x = 12X = 12/10X = 6/5X = 1.5 8x-3=9-2x One solution was found : x = 6/5 = 1.200 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 8*x-3-(9-2*x)=0 Step by step solution : Step1: Pulling out like terms : 1.1 Pull out like factors: 10x - 12=2•(5x - 6) Equation at the end of step1: Step2: Equations which are never true: 2.1Solve:2=0 This equation has no solution.A a non-zero constant never equals zero. Solving a Single Variable Equation: 2.2Solve:5x-6 = 0 Add6 to both sides of the equation:5x = 6Divide both sides of the equation by 5:x = 6/5 = 1.200 One solution was found : x = 6/5 = 1.200 Processing ends successfully 8x-3=9-2x One solution was found : x = 6/5 = 1.200 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 8*x-3-(9-2*x)=0 Step by step solution : Step1: Pulling out like terms : 1.1 Pull out like factors: 10x - 12=2•(5x - 6) Equation at the end of step1: Step2: Equations which are never true: 2.1Solve:2=0 This equation has no solution.A a non-zero constant never equals zero. Solving a Single Variable Equation: 2.2Solve:5x-6 = 0 Add6 to both sides of the equation:5x = 6Divide both sides of the equation by 5:x = 6/5 = 1.200 One solution was found : x = 6/5 = 1.200 10x=12x= 12/10is correct answer |
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| 13. |
Q10.Solve the pair of equations: 21x + 47y-1 10 and 47x + 21y = 162. |
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| 14. |
8x + 3 = 27 + 2x |
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| 15. |
yolve:1. 8x+3= |
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Answer» 8x + 3 = 27 + 2x 6x = 24 x = 4 |
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| 16. |
1. 8x + 3 = 27 + 2x |
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| 17. |
Solve:8x +3=27+2x |
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Answer» 8x-2x= 27-36x= 24x= 4 8x+3=27+2x8x-2x=27-36x=24X=24/68 8x+3=27+2x8x-2x=27-36x=24x=24/6x=4 |
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| 18. |
and 15cmrespectivelyClctethehcigt10. In Figure ABCD find the area of the shaded region.40 cm40 cm |
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Answer» ar of square = (side)^2 = (40 cm)^2 = 1600 cm^2ar of unshaded triangle = 1/2 (base)(height) = 1/2 (40)(40) = 800 cm^2ar of shaded region = 1600 cm^2 - 800 cm^2 = 800 cm^2 In a square all sides are equal |
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| 19. |
if 7 kg onions cost140 rupees how much must we pay for 12 kg onions |
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Answer» 7kg onions cost = 1401kg onion cost = 140/7 = 2012kg onions cost = 12*20 = Rs 240 thx |
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| 20. |
if 7 kg onions cost 140 rupees, then how many cost of 12 kg onions |
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Answer» Cost of 7 Kg Onion is ₹140Cost of 1 Kg Onion is ₹140/7= ₹20Cost of 12 Kg Onion is ₹ 20 × 12 kg = ₹240 |
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| 21. |
A heap of rice is in the form of a cone of base diameter 0.4 m and height 3.5 cm. Find the volume of rice. How much cloth is required to just cover the heap. |
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| 22. |
The quantity of rice used in a hostel for a dayis 10 kg 250 g. When a few more studentsoined, the consumption increased to12.3 kFind the extra quantity of rice required.9. |
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| 23. |
Substitution methodI=\frac{\sec ^{8} x}{\cos e c x} d x |
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Answer» Thanku so much 👍 |
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| 24. |
Integral(e^x, x)=C %2B e^x |
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| 25. |
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Findthe volume of the rice. How much canvas cloth is required to just cover the heap ? |
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Answer» Given, Radius r = 24/2 = 12 mHeight h = 3.5 m Volume of rice heap= 1/3*pi*r^2*h= 1/3*22/7*12*12*3.5= 22*4*12*.5= 88*6= 528 m^3 Slant Height l*l = 3.5*3.5 + 12*12l*l = 12.25 + 144 = 156.25l = 12.5 m Length of canvas cloth required= pi*r*l= 22/7*12*12.5= 471.4 m^2 |
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| 26. |
ORA heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Findthe volume of the rice. How much canvas cloth is required to just cover the heap ? |
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Answer» Volume = 1/3πr2h1/3 × 22/7 × 12×12×3.5= 528 m3 Canvas of cloth required = CSA Of conel = √144+12.25=25/2=12.5m CSA = 3.14×12×12.5=471 m^2 |
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| 27. |
. If 47x +31y 18 and 3Lx +47y 60, then find the value of x+y. |
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| 28. |
11) Find lim, sin 3x – 5x2 |
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| 29. |
27z(il e ) नg f,0y1p Lonceht of Llot (,0) ,(0b) wamd (Xy) e ८८:22: , ८८o 4 -८ that अदाथक |
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| 30. |
2(x) = x2-5x + 6(x) = 5x2 + 8x +3 |
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| 31. |
(ii)(-6x) x5x2 |
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Answer» (-6x) * 5x2 = -30x3-30x3 is right answer The right answer is = (-6x) × 5x^2= -30x^3 -30 x³is the answer (-6x)×5x^2 = -30x^3 is correct answer |
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| 32. |
p(x) = 5x2 + 8x3 |
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Answer» 5x² + 8x + 3 5x² + 5x + 3x + 3 5x ( x + 1) + 3 ( x + 1) (x + 1) ( 5x + 3) |
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| 33. |
Evaluate:5x2(x+1) (x2 + 4 |
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| 34. |
7:5x2 + 6x + 3x + 3 a3alloll |
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Answer» ( 5x^2+6x+3)/(x+3) |
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| 35. |
If35-5x2, then find the value of x. |
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Answer» it's linear equations in one variable... |
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| 36. |
\int e ^ { x } ( \operatorname { sec } x + \operatorname { sec } x \operatorname { tan } x ) |
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| 37. |
If sec= x+,x0, find (sec+ tan e).42 |
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Answer» x^2 +1+√x^4+x^2+1/x is the best answer answer this point Ted x2+1+√2x2+1/x |
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| 38. |
\int e ^ { x } \operatorname { sec } x [ 1 + \operatorname { tan } x ] d x |
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| 39. |
\frac{d y}{d x}-y \tan x=e^{x} \sec x |
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Answer» thank🙏💕 |
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| 40. |
\operatorname { sec } x + \operatorname { cos } e c x = 0 |
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Answer» u have to solve this equation.or find its general solution |
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| 41. |
⌠The ratio of the radii of two circles is 4: 5. Find the ratio of their arcas.d Find the area ove8. A horse is tied t |
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| 42. |
2e value ofsec xdxcosec" x2011 |
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Answer» thanks |
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| 43. |
( i ) \int e ^ { x } \operatorname { sec } ^ { 2 } ( e ^ { x } ) d x |
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| 44. |
If 8232A) 421x 31x 7z , then x + z-y-B) 5C) |
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Answer» Please hit the like button if this helped you |
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| 45. |
uDOImersect at a point, then find the value f3. If47x +31y w 1 8 and 31x+47y"60, then find the value of x+y. |
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| 46. |
yoidCalget:4 The exchange ratio of a certain foreign currency with the Indian rupee is 1:62.50. How muchof the foreign currency can be bought for 3125? |
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| 47. |
EXERCISE 8.1Find the ratio of:(a) 800 mL to 4.8 litres(c) 2g to 1 kDivide 336 among the thExpress the following as ra(a) 2596e) 5.5%press the following in pe(b2:25 |
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Answer» Ratio: c) 1l = 1000ml800ml: 4.8l= 800ml : 4800 ml= 800: 4800= 1:6d) 1 kg = 1000g2g: 1kg= 2g : 1000g= 2:1000= 1:500 Please hit the like button if this helped you |
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| 48. |
(ii) If sec 2A- cosec (A -27°) where 2A is an acute angle, find the measure of |
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| 49. |
1 + 2tan^2A /cos^2A = tan^4A + sec^4A |
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| 50. |
1. If 7 kg onions cost 140 rupees, how much must we pay for 12 kg onions? |
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