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. |
\left. \begin{array} { c c c } { a ^ { 2 } + 2 a } & { 2 a + 1 } & { 1 } \\ { 2 a + 1 } & { a + 2 } & { 1 } \\ { 3 } & { 3 } & { 1 } \end{array} \right| = ( a - 1 ) ^ { 3 } \cdot 2 |
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| 2. |
if α and β are the roots of the equation 3x2+x-10=0 then find the value of 1 upon α+ 1 upon β is..? |
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Answer» If alpha and beta are roots of quadratic equation Sum of roots (alpha + beta) = - b/aProduct of roots (alpha*beta) = c/a For 3x^2 + x - 10 = 0 Sum of roots (alpha + beta) = - b/a = - 1/3Product of roots (alpha*beta)= c/a = - 10/3 1/alpha + 1/beta= (alpha + beta) /alpha*beta= (- 1/3)/(-10/3)= 1/3 ans |
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| 3. |
2 =32, find a.(A)-2(B) 4(C) -6(D)-4 |
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| 4. |
324. Arrange - - and 17 in thedescending order of their magnitudes.Also, find the sum of the lowest and thelargest of these rational numbers. Express theresult obtained as a decimal fraction correctto two decimal places. |
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Answer» 17/32, 5/8, -1/4, -3/16 |
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| 5. |
A rexiangular ground is 90m long and 32m broad. In the middle of the ground there is a circular tank of radiusja remaining portion at the rate of *50 per square metremeeres. Find the cost of turfing the14 m32 m90 m |
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| 6. |
5. Find the equations to the altitudes of the triangle whose angular poins art ioining the points ,A (2, -2), B (1, 1) and C (-1, 0). |
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| 7. |
45 Show that P (1,-1) is the centre of the drcle circumscribing the triangle whoseangular points are A (4, 3), B (-2. 3) and G (6, -1) |
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| 8. |
: The co ordinates of angular point of a triangleABC are A(O, 1), B(2, 0) and C(-1, -2). Find theeqvation of side AB of the triangle.. |
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| 9. |
The co-ordinates of angular point of a triangleABC are A(O, 1), B(2, 0) and C(-1,-2). Find theequation of side AB of the triangle. |
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Answer» thank you mam |
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| 10. |
1 The area of a n equilateral triangle is49 J3 cm. Taking each angular point as centre, circles are drawnwith radius equal to half length of the side of the triangle. Find the area of triangle not included in the circles |
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Answer» sun konsa lesson ka h |
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| 11. |
Sin2f ney=12and my=32 find utty.. |
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Answer» 80 is the correct answer of the given question x+y=12, x-y=32x^2+y^2=(x+y)^2-2xy =(12)^2-2×32 =144 -64 =80 |
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| 12. |
2. Ifone of the roots of the quadratic equation x -10x + 2k 0 is (h +2V6), findthe values of h andk. |
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Answer» will you check your sum once...there is a mistake in it... |
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| 13. |
Two numbers p and q are such that the cosum of roots and as the product of roots, then find theequation py' + 3x t20 hp and |
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| 14. |
The area of an equilateral triangle is 49/3 cm2. Taking each angular point as centre, circlesare drawn with radius equal to half the length of the side of the triangle. Find area of triangle notincluded in the circles. [Take V3 1.73 |
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| 15. |
how many bricks,each 25 cm by 15cm by 8cm are required for a wall 32m long.3m high,40cm thick? |
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| 16. |
How many bricks, each measuring 25cmx 11.25cm6cm will be needed to build a wall 8 m long, 6 m highand 22.5 cm thick? |
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| 17. |
An ohjeet falls fhom the top of the tower travela 2,5 m in the Inst second hefore sriking theground, Find theind the value of acceleration du to araviu a theeight of tho tower yVA) |
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Answer» It is given that distance travelled by the object in last second = 24.5 m let the height of tower = h Sn = u + 1/2a(2n-1) 24.5 = 0 + 10/2(2n-1) 24.5 = 5(2n-1) 24.5 = 10n - 5 19.5 = 10n 1.95 = n Distance travelled in last second is 1.95 But distance travelled in (n-1) th second (Sn) = u + 1/2g(2n-1) Sn = 5 [ 2(n-1) -1] Sn = 5 [ 2n -2 -1 ] Sn = 5 [ 2n - 3 ] Sn = 10n - 15 Sn = 10n - 15 24.5 = 10n -15 9.5/10 = 0.95 Total time taken = 1.95 + 0.95 = 2.9 seconds height of tower 1/2gt² 5(2.9)² 5 * 8.41 = 42.5 m |
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| 18. |
6. Find the square roots of the following numbers by prime factorization method.а)729b)1764c)9604 |
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Answer» a) 729 = 3×3×3×3×3×3 √729 = √(3×3×3×3×3×3) = 3×3×3 = 27 b) 1764 = 2×2×3×3×7×7 √1764 = √(2×2×3×3×7×7) = 2×3×7 = 42 c) 9604 = 2×2×7×7×7×7 √9604 = √(2×2×7×7×7×7) = 2×7×7 = 98 If you find this answer helpful then like it. |
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| 19. |
square root of 230 by prime factorization method |
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| 20. |
Find the square roots of 441 and 144 hauare roots of 441 and 144 by the method of repeated subtractionFind the square roots of the following numbers by the prime factorization method.() 441(ii) 324(11) 324(iii) 54756 (iv) 2025 (v) 6561 |
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Answer» 441 is answer of this question |
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| 21. |
factorization method =x 2+ 3x+ 40=0 |
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Answer» 5x=-40x=-8 answer. |
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| 22. |
Find cube root of 175616 by factorization method. |
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Answer» 175616=2³×2³×2³×7³=56 very very thank you |
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| 23. |
A field is in the shape of a trapezium having parallel sides 90 m and30 m. These sides meet the third side at right angles. The length of thefourth side is 100 m. If it costs 5 to plough 1 m2 of the field, find thetotal cost of ploughing the field. |
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Answer» Given Parallel sides of trapezium= 90 m and 30 m,So ,Let AB = 90 m ,CD = 30 mAndAB | | CD And these sides meet third side at right angles , So let third side = ADAndLength of fourth side ( BC ) = 100 m Now we draw a perpendicular from point C at AB that meet AB at E , SOHeight of Trapezium ABCD = AD = ECAndAE = CD = 30 m So,BE = AB - AE = 90 - 30 = 60 m Now we apply Pythagoras theorem in∆BEC , As : BC²= BE²+ EC² 100²= 60²+ EC² EC²= 10000 - 3600 EC²= 6400 EC = 80 m ( This also height of trapezium field ) We know Area of Trapezium = 12( Sum of parallel sides )×( Height )So,Area of Trapezium ABCD =12( 90 + 30 )×( 80 ) Area of TrapeziumABCD=120×40 Area of TrapeziumABCD= 4800 m² And ,costs Rs 4 to plough 1m² of the field . Total cost to plough the whole field = 4800× 5 = Rs. 24000 |
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| 24. |
find the HCF of 84,98 using prime factorization method |
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| 25. |
Find the root of the quadratic equation by factorization method |
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| 26. |
1. Write four rational numbers equivalent to each of the following rational numbers |
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Answer» Equaling to 2/9 4/8 , 6/21 , 14/ 63 , 10/45 equivalent to2/92/9×2/2=4/182/9×3/3=6/272/9×4/4=8/362/9×5/5=10/45therefore2/9's equivalent numbers are 4/18,6/27,8/36,10/45 |
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| 27. |
(e) Which of the following rational numbers is the smallest?on whichthe following rational numbers is the smallest? |
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Answer» first find the absolute value then take out lcm then make the de nominates equal the compare which is the smallest |
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| 28. |
ce four rational numbers equivalent to each of the following rational numbersWIN |
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| 29. |
ボA = {1,2,3} .訂計37,, (1,2)すtiqut FROTI a -Let A = ( 1, 2, 3) thennumber of equivalence relation containing(1, 2) is (a) 3 (b) 4 (c) 1 (d) 2 |
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| 30. |
Write four equivalent rational numbers to each of the following rational numbers14.2.(o -518(iii) 15-19(ii)(iv)17-23166j |
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Answer» 1) -10/36, -15/54,-20/72,-25/90.2) 14/34,21/51,28/68,35/853) -30/-38, -45/-57, -60/-76, -75/-954) 28/-46, 42/-69,56/-92,70/-115 hii muskan do you need any help |
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| 31. |
Write four rational numbers equivalent to each of the following rational numbers(1) 15(1) -11 |
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| 32. |
(b) Three numbers preceding the Roman numeral D |
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Answer» D means 500 so preceding value ia 499 |
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| 33. |
ustration 4.59 Prove thatthe numbers 49, 4489,444889obtained by inserting 48 into the middle of the precedingnumbers are square of integers. |
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| 34. |
The houses in a row are numbered consecutively from 1to 49. Show that there exists a value of X such that sumof numbers of houses preceding the house numbered Xis equal to sum of the numbers of houses following X |
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| 35. |
1. The houses of a row are numbered consecutively from 1 to 49. Show that there is a valueof x such that the sum of the numbers of the houses preceding the house numbered xi(NCERT)equal to the sum of the numbers of the houses following it find x. |
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Answer» thanks |
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| 36. |
e houses of a row are numbered consecutively from 1 to 49. Show that there is a valueof x such that the sum of the numbers of the houses preceding the house numbered x isequal to the sum of the numbers of the houses following it. Find this value of x. |
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Answer» another way to solve it |
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| 37. |
9. If a rope with length of 750 m is used tofence three sides of a garden as shown in thefigure fourth side wall of the house is withlength of 50 m, what is the length of sides ofgarden?DD50 m |
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Answer» as shown in question 3sides=750mwhich means 1side carries 250m |
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| 38. |
a 68 cm long rope is used to make a rhombass on graund the distance between theair of opposite corner is 16cma. what is the distance between the other two cornerb. what is the area of the ground bounded by the ropeENG 1049 PMOType here to search |
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Answer» Perimeter of rhombus = length of rope. Let side length of rhombus is aLet diagonals of rhombus are d1 and d2 Then,4a = 68a = 68/4 = 17 cm (1) Given d1 = 16 cm Using pythagoras theoram a^2 = (d2/2)^2 + (d1/2)^2 17^2 = d2^2/4 + 64 d2^2 = (289 - 64)*4 d2^2 = 225*4 d2 = 15*2 = 30 cm Therefore distance between other two corners = 30 cm (2) Area bounded by rope = d1*d2/2 = 16*30/2 = 240 cm^2 thanks |
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| 39. |
On an algebra test, the highest grade was 56 pointsmore than the lowest grade. (In statistics, thedifference between the largest number and thesmallest number is known as the range.) The sumof the two grades was 128. What were the highestand lowest grades on the test? |
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Answer» let lowest grade=xhighest = x+56A/Sx+x+56=1282x= 128-56= 72x= 72/2=36so lowest grade=36highest grade= 36+56= 92 |
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| 40. |
me numerator of a fraction is 6 less than the denominator3 is set to theumerator, the fraction becomes equalCONFind the original fraction |
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| 41. |
The numerator of a fraction is 6 less than the denominator. If 3 is added to theumerator, the fraction becomes equal to 3 Find the original fraction.2 |
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| 42. |
3. The numeratorumerator of a fraction is 6 less than the denominator. 1 3 is added to therator, the fraction becomes equal to Find the original fraction.numerator, the fra |
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| 43. |
EXERCISE 1A-35-42981. Expressas a rational number with denominator(i) 20(ii) -30(iii) 35(iv)2. Expressas a rational number with denominator 7.. |
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| 44. |
uSsamixedfraction as an impr(Whole x Denominator) + NumeratorDenominatorEXERCISE 7.2w number lines and locate the points on them2 31 2 37i fractions: |
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| 45. |
2The area of an equilateral triangle is numerically aqual to.its perimeterFind a side of the trangle [Take v3 1-731 |
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Answer» Like if you find it useful |
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| 46. |
at the rate ol 250 Ppel10. Find the area of an equilateral triangle each of whose sides measures (ü) 18 cm, (i) 20 cm[Take V3 = 1.731 |
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Answer» thank you |
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| 47. |
The height of an equilateral triangle measures 9 cm. Find itcorrect to 2 places of decimal. (Take v3 1.732.)s area, |
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| 48. |
8. From the top of a 60 m high building, the angles of depression of the top and the bottomof a tower are 45° and 60° respectively. Find the height of the tower. (Take V3 = 1.732) |
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| 49. |
The angles of elevation of the top of a rock fros thetop and fool of a 60 m high tower are 45" and strespectivelg Find the height of the tocktake, v3 1.7321 |
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| 50. |
40. If three circles of radius a each, are drawn such that each touches theother two, prove that the area included between them is equal toa2[Take v3 1.73 and Tt 3.14.]425 |
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