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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. The product of Virat Kohli's age 8 years agoand 6 years later is 680. Find his presentage. |
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Answer» Here is your answer :- Let the present age of virat kohli be x years Virat kohli age 8 years age = x-8 Virat kohli age 6 years later = x + 6 (x-8)(x+6) = 680 x² + 6x -8x -48 = 680 x² -2x -48-680 = 0 x² -2x -728 = 0 x² - (28-26)x - 728 = 0 x² -28x + 26x -728 = 0 x ( x -28) + 26( x-28) = 0 (x-28) (x+26) = 0 x = 28 |
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| 2. |
18. After 32 years, Rahim will be 5 times as old as he was 8 years ago. How oldisRahimtoday? |
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| 3. |
18.teAfter 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today? |
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Answer» thanks |
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| 4. |
Let n e N, if the value of c prescribed inRolle's theorem for the functionw) 2x(x-3)" on (0, 31 isthen nis equal to |
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| 5. |
6. Billy read 2 books. He read the first one in one week with 25 pages everyday. Heread the second book in 12 days with 23 pages everyday. What is the total numberof pages that Billy read? |
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Answer» in one week 7 days25*7=175 pagesnext one 12*23=276pagestotal number =276+175=451 pages |
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| 6. |
The population of a city is 78932541. Before ten years, it was 65364541Find the increase in the population of the city during the ten years. |
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Answer» Increase in population =78932451-65364541=13567910 |
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| 7. |
a-2-1+5Find the value of λ, such that the line--H3x -y-2z 7is perpendicular to the plane-4 |
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| 8. |
31. The solution of 2z-3-5 is |
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| 9. |
Find the equation of the plane through the points (1.-2, 4). (3,-4. 5) and perpendicular to the planes+ y-2z=6 |
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| 10. |
के “बडा ह G R W N R W e NSy7 3—X1611+ —x1496 =27e +“x |
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Answer» 7/9*1611+3/11*1496=1253+408=1661 the answer is 1661 🙂🙂🙂 |
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| 11. |
Ql—X-— XE 7<% .e W (S |
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Answer» this is wrong answer |
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| 12. |
(vi) {x|x=, n e W and n s 4) |
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Answer» the elements in the sets are.. { 0,1/4,2/5,3/6,4/7} |
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| 13. |
का ) W ,—Q\\Q_, सुर [कर वेट = 208 N Xe सिक Se \a s TS goee . 5\,*\\“\& carnses, ~%Lind e, e .: चार |
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| 14. |
1v(e “ i fl( A /_LU\"AL‘U (A,t,’fl/\,lu—€ थे o) जी +X 155R TN E WX५ |
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Answer» Derivative(d/dx)=d/dx(2x^4+X)=2*4*x^3+1=8x^3+1 |
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| 15. |
6. (2x + 6x + 2x +x-6) is di7. (4x°-12° +11x - 5) is divided by (2x - 1) |
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| 16. |
2. Solve the following equations.(0)(x | 2x + 6 0, x E Z(ii) (x 5x + 16 1, X E N(iv) (x | 4x - 25> 13, x E Z(ii) (x | 2x 3 7, x e WIlI) 1X(V) y 1-7 s 13, y is a prime number |
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Answer» 1) 2x +6 = 0 and x belongs to Z(integers) so solving 2x+6 = 0 => x = -3.2) 5x+16 = 1 => 5x=16-1 =15 => x = 15/5 = 3.3) 2x-3<7 , and x belongs to W so 2x<7+3 = 10 => 2x<10 => x < 5 therefore value of x is [0,5) 4) 4x-25> 13 => 4x> (25+13=38) => x > 38/4 => x > 9.5 , and since x=Z so rhe value is [10, infinity) 5) 5y/3 -7 <= 13 => 5y/3 <= 20 => y<= 12 and y is prime no. so values of y are 2,3,5,7 and 11. |
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| 17. |
Exerciseation tables to divg. 24 6I. 168q. 42 6ciated with dim, the number that |
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Answer» Use the tables of 5,6 and 8.15÷5=324÷6=416÷8=242÷6=7 b. 15/5=3g. 24/6=4l. 16/8=2q. 42/6=7 b.15/5=3g.24/6=4I. 16/8=2q. 42/6=7 |
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| 18. |
(text*((i*(s*(di*(v*(i*(de*(db*y)))))))*(x - 2)))*(-8*x %2B 3*x^4 - 6*x^2 %2B 2) |
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| 19. |
6. The diameter of the wheelof a car is 70 cm. How many revolutions will it make to travel1.65 km? |
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| 20. |
A copper wire is bent in the form of a square of side 44 cm. If it is re-bent in the form of a circle, what will be its radius?(Take pi=22/7) |
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Answer» copper wire is set as a square soarea of square = 44*441936cm^2perimeter= 4*44=176cmperimeter will be same so 2πr= 176r= 176/2π28cm |
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| 21. |
4z+3 = 6+2z2x-1 = 14-x |
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| 22. |
if k= x-y+2z where -2<=x<=1 and -1<=y<=2 and 3<=z<=6 then value of k |
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Answer» 9 0r 7 is right answer 9&7 is correct answer |
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| 23. |
9. A copper wire is bent in the form of asquare of side 44 cm. If it is re-bent in theform of a circle, what will be its radius?(Take n22 |
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| 24. |
Divide the following polynomials by a monomiat using longAlso, write the quotient and the remainder.ion mantnd(i) -8x + 6x-4+ 12x by 2x(ii) 5xto-gx8 -9x + 7x by x*(ii) 5z3- 622 + 7z by 2z |
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| 25. |
EXAMPLE 3.80Differentiate the following wrt z(a) sin(c) tan20 log (4x+3)](b) cos' (2z+ 1) |
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| 26. |
3. The coefficient of x in 4xy^2z is:(ii) 4y^2z(iv) 4yz4y^2(iii)4 |
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Answer» the answer is option 2 4y2 |
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| 27. |
3. Two adjacent sides of a rectangle are 8a -4. Subtract the sum of Sx 5y - 2z and -6x + 8y 4z from the sum of 6x +rtmie5b and 7a2 -4xy. Find its peand-6x + 8y-4z from the sum of 6x + 2y-4z and- 5y - 2z-2x + 4y -8z1tn get 4x2 -7xy - 4y +1? |
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| 28. |
solve the following equations and check your results.1. 3x = 2x + 18 2. 51-3=31-54. 4z + 3 = 6 + 2z 5. 2x - 1 = 14-X277x3. 5x + 9 = 5+ 3x6. 8x + 4 = 3x - 1)+75 26 |
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| 29. |
UĂłrethan2,000?IJ.IttheĹĄame astheper cent byEXERCISE 8.21. Aman got a 10% increase in his salary. If his new salary s?1,54,000, find hisoriginal salary. |
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| 30. |
Page No.Date:IAn Af consists of So teams of ushich 3nd toim isis lob Eind the onthten |
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Answer» Let a , d are first term and common difference of an A.P nth term = Last term = a + ( n - 1 )d an = a + ( n - 1 )d Now , It is given that , Third term = 12 a + 2d = 12 ------( 1 ) Last term = 106 a + 49d = 106 ---( 2 ) Subtract ( 1 ) from ( 2 ) , we get 47d = 94 d = 2 Substitute d value in equation ( 1 ) , We get a + 2 × 2 = 12 a = 12 - 4 a = 8 29th term = a + 28d a29 = 8 + 28 × 2 = 8 + 56 = 64 |
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| 31. |
EXERCISE 8.5Add:(ii) 15.6 + 7 + 9.25(10) 18.25 + 15.9 + 3.72(vi) 28.365 + 18.172 +9.275 +0.3694(6) 1.7 + 2.9 + 13.5(u) 16.75 + 10.25 + 9.6(iii) 6 + 1.5 + 2.5ibtract: |
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Answer» 1.7+2.9+13.5=18.115.6+7+9.25=31.356+1.5+2.5=1018.25+10.25+9.0=36.628.365+18.172+9.275+0.3694=56.1814 (i) 1.7+2.9+13.5=18.1(ii) 15.6+7+9.25=31.85(iii) 6+1.5+2.5=10(iv) 18.25+15.9+3.72=37.87(v) 16.75+10.25+9.6=36.60(vi) 28.365+18.172+9.275+0.3694=56. |
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| 32. |
u write more pairs in these examples?Can you write n-EXERCISE 1.2Wie down a pair of integers whose:(a) sum is -7(b) difference is -10(c) sum is0 |
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| 33. |
UB Bf their perpendiculatEXERCISE 11.2ng, give also the justification of the constructionpraw aents to the cirele and measure their lengths.circle of radius 6 cs u point 10 cm away from its centre, comstruct the pairI of tangent a tangent to a circle of radi |
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| 34. |
ORFind the dimensions of a rectangular park whose perimeter is 60 m andarea 200 m |
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Answer» Let the length of Reactangle be lLet the breadth of Reactangle be b Perimeter of Reactangle= l + b60 = 2(l + b)l + b = 30...........(1) Area of Reactangle= l*b200 = l*bl = 200/b............ (2) Put value of l in eq(1), we get200/b + b = 30b^2 - 30b + 200 = 0b^2 - 20b - 10b + 200 = 0b(b - 20) - 10(b - 20) = 0(b - 10)(b - 20) = 0b = 10, 20 If breadth = 10 m, length = 20 mIf breadth = 20 m, length = 10 m thanks |
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| 35. |
for the rice?4. Find the area of a rectangular park whose length is 45 m and breadth is 23 |
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Answer» L=4(4/5=24/5; b=3(3/4)=15/4;; Area=L x B=24/5 × 15/4=6/1×3/1=18;; correct answer is 18 Length = (5×4+4)/5 = 24/5Breadth = (4×3+3)/4 = 15/4 Area of Rectangle = L × B= 24/5 × 15/4 = 360/20 = 18 m^2 L=4(4/5=24/5;b=3(3/4)=15/4;Area=L×B=24/5×15/4=6/1×3/1=18 |
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| 36. |
with 66 cm long steps. In 400 steps, he makes a full rouildIfa wire is bent into the sha12.2. lfa wire is bent into the shape of a square, the area of the square is 49 cm2. When the wire is bent into a semia)IfatCircular shape, find the circumference ofthe semicircle.〔Take πig urhat is the ratio of their radii? |
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Answer» Area of square = 49 cma^2 = 49a= 7 cm Side of square = 7 cmPerimeter of square = Perimeter of semi circlePerimeter of semi circle = 4* 7 = 28 cm pi * r + 2r = 28=> r (2+ pi) = 28=> r (2+22/7) = 28=> r (36/7) = 28=> r = 28 * 7/36=> r = 7*7/9=> r = 49/9 cm Radius of semi circle = 49/9= 5.44 cm circumference = 2πr= 2*22/7*5.433.9cm |
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| 37. |
50. A piece of wire 11 cm long is bent into the form of an arc of a circle subtendingan angle of 45° at its centre. Find the radius of the circle. |
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Answer» 90 angle is correct answer 90 is the right answer |
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| 38. |
EXERCISE 1.4Find the union of each of the following pairs of sets:) X 1,3,5)(ii) A = [ a, e, i, o, u}(ii) A-r: is a natural number and multiple of 3).Y 1,2, 3)B = {a, b, c}B fr:ris a natural number less than 6A- r :x is a natural number and 1 <x s6)B = {x: x is a natural number and 6 < x < 10 }A(iv)(v){1,2,3},B |
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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. elements in set 1 but not in set 2 or not in set 1 but present in set 2 or present in both set 1&2 will make a place in union of two sets. try next question. if you can't i will explain. |
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| 39. |
A 44 cm long wire is bent to form a circle. Find the diameter of the circe. |
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Answer» Perimeter of Circle = 44 So, 2 × pi × r = 44 2 × 22/7 × r = 44 r = (7 × 44)/44 r = 7 cm So, diameter = 14 cm |
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| 40. |
A 44 cm long wire is bent to form a circle. Find the diameter of the circle |
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| 41. |
. A piece of wire in the form of a rectangle of 8.9 cm long and 54reshaped and bent in the form of a circle. Find the radius of the circle in mm. |
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| 42. |
7.A wire bent in the forn of a square encloses an area of 1936 sq. cm. If the same wire isbent into a circle, find the area enclosed by it. |
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| 43. |
८०538" ८०560 52"tan35° tan60° tan72° tan55° )I '. |
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Answer» good |
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| 44. |
Evaluate COS58+sn22evaluatecos58 sin22cos38 cosec52°sin32 cos68 tan18 tan35 tan60 tan72 tan55OR |
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Answer» cos 58° = cos(90 - 32°) = sin 32°sin 22° = sin(90 - 68°) = cos 68° cos 38° = cos(90 - 52°) = sin 52°tan 18° = cot 72°tan 35° = cot 55°[cos 58°/sin 32°] + [sin 22°/cos 68°] - [(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)] = [sin 32°/sin 32°] + [cos 68°/cos 68°] - [(sin 52°/sin 52°)/(cot 72° cot 55° tan 60° tan 72° tan 55°)] = [1] + [1] - [1/√3] (∵ tan 60° = √3) = (2√3 - 1)/√3 = (6 - √3)/3 (∵ Rationalise the denominator) |
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| 45. |
2 in a Î ABC, right angled at B, AB-24 cm, BC-7 cm . DetermineINCERTIINCERT(i) sin A, cosSA(ii) sin C, cos Cn 10 17 find fan Pand cot R. Is tan P- cot R? |
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| 46. |
ODUCTION TO IRIGONOMETRYEXERCISE 8.1In Δ ABC, right-angled at B, AB-24 cm, BC-7 cm. Determine :(i) sin A, cos A(Gi) sin C, cos Cn Fig. 8.13, find tan P-cot R.312 cm!\13 csin Acalculate cos A and tan A.4iven 15 cot A = 8, find sin A and sec A.13-, calculate all other trigonometric ratios.12Fig, 8.ensec θ |
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Answer» https://www.teachoo.com/amp/1760/529/Ex-8.1--2---In-fig--find-tan-P---cot-R---Chapter-8/category/Ex-8.1/Ex 8.1, 2 - In fig, find tan P - cot R - Chapter 8 Class 10 - Teachoo |
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| 47. |
radii i.e., (R-r).8. A piece of wire in the form of a rectangle of 8.9 cm long anisreshaped and bent in the form of a circle. Find the radius of the circle in mm.uare of side 11 cm. It is rebent to fo |
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| 48. |
. In the figure ABC is a right angled triangle, right angled at ZA. Find the area of the shadedregion, if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of the triangle ABC |
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| 49. |
2. There was a deserted land near a colony where peopleused to throw garbage. Colony people united to developa pond in triangular shape as shown in figure. The landis in the shape of a parallelogram ABCD. In the restof the portion medicinal plants are grown. Area of theparallelogram is 200 m2.(a) Calculate the area where medicinal plants are grown. |
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Answer» (i) Deserted land is in the shape of a parallelogram of area 200 m^2.Triangular portion of the land has the same base as the deserted land.If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.Therefore, area of the triangular portion where medicinal plants are grown is equal to 200/2 i.e. 100 m^2. |
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| 50. |
ORFind the dimensions of a rectangular park whose perimeter is 60 m andarea 200 m2 |
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Answer» Like if you find it useful |
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