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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. |
D-15ByDivide (12-17-5x) by (3-5x) and verify the division algorithm |
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| 2. |
mple 10: Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form aportion? |
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| 3. |
ला woluo: 2o छा Wllo o o Pशा 27. नालi : ol |
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| 4. |
7.Divide (6+ 19x +x-613) by (2+ 5x- 3a2) and verify the division algorithm. |
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| 5. |
1. Use the long division method to divide thefollowing:(a) x2 + 5x + 6 by x + 3 |
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| 6. |
17. 3 weeksdays |
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Answer» 1 week=7 days3 weeks=3×7 days 3 weeks=21 days Thank you |
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| 7. |
nd the area ol tno quPT Usha runs around a circular p20 rounds at the speed of 16 km/hr. |
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Answer» Area if circular path=πr²=754600 m²r²=754600*(7/22)r²=240100r=√240100r=490 m Total distance covered in 20 rounds=20×490 m=9800 m=9.8 km Time taken=Distance/speed=9.8/16 hr=0.6125 hr |
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| 8. |
The ratio between the curved surface and total surface of a cylinder is 1:2. Find the volume of the cylinder, given that its total surface area is 616 cm^3 |
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Answer» k h yo ratio of curved surface area(c.s.a) and total surface(t.s.a) of a right circular cyclinder = 1:22πrh/2πr(r+h) = ½h/(r+h) = ½(2πrh on the both sides will be cancelled)2h = r+h (by cross multiplying)2h-h = r therefore h=r therefore height of cylinder = radius of cylinder = 1:1 volume = π r^2hTSA = 2πr(r+h)r= hso 2πr(2r)4πr^2= 616r^2= 616/4πr= 7then volume = π×7*7*722*491078 cm^3 |
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| 9. |
17. The ratio between the curved surface and total surface of a cylinder is 1:2. Find thevolume of the cylinder, given that its total surface area is 616 cm2cost |
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| 10. |
Find the area of polygon ABCDEF, if AD18cm, AO 14cm, AP 12 cm, AN 8cm,AM 4 cm, and FM, EP QC and BN areperpendiculars to diagonal ADcm i5 cm Ncm!4 cm |
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| 11. |
The simple interest on a certain sum of money for two years at 51% is 6,600,What will be the compound interest on that sum at the same rate for the sameime period?2 |
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| 12. |
18. The given figure shows parallelogram ABCD.Points M and N lie in diagonal BD such thatDM BN. |
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Answer» hit like if you find it useful |
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| 13. |
The angle of elevation of clouds from a point 60m above a lake is 30and angle of depression of the reflection of clouds in the lake is 60 findthe height of clouds |
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Answer» Let AO=HCD=OB=60mA'B=AB=(60+H)m In triangke AOD,tan30'=AO/OD=H/OD H=OD/√3OD=√3 H Now, in triangle A'OD,tan60'=OA'/OD=(OB+BA')/OD√3=(60+60+H)/√3 H =(120+H)/√3 H=>120+H=3H 2H=120 H=60m Thus, height of the cloud above the lake = AB+A'B =(60+60) = 120m |
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| 14. |
51, 15 गी, 17 रो, मी लग्ये था 2 मी. 2 सेमी चौड़े फर्श परबिकने के लिये ग से । तिने वर्गाकार टाइल्सों कीजरूरत होगी।(a) 840 (b) 841(c) 820(d) 814. |
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Answer» please answer is 820 because he is a simple question Length of the room = 15 m 17 cm = 15 *100 +17 = 1517 cm.Breadth of the room = 9 m 2 cm = 902 cm.The HCF of the 1517 and 902 will be size of square tiles.HCF 1517 and 902 = 41 cm.Area of the room = length * breadth = 1517 * 902 cm2.Area of tiles = 41 *41 cm2 So, no. of tiles required = (1517 *902) /(41*41) = 814 tiles. |
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| 15. |
\left. \begin{array} { l } { \text { 4. Divide } 55 ( x ^ { 4 } - 5 x ^ { 3 } - 24 x ^ { 2 } ) \text { by } 11 x ( x - 8 ) } \\ { \text { Divide } x ( 4 x ^ { 2 } - 5 x ^ { 3 } - 24 x ^ { 2 } ) \text { by } 11 x ( x - 8 ) } \end{array} \right. |
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Answer» thank you |
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| 16. |
24. Divide 3x5 - 8x2 + 3x + 2 by x2 – 3x + 2 and verify the division algorithm. |
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| 17. |
mple 15 : Divide 8v15 by 23 |
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| 18. |
Divide44\left(x^{4}-5 x^{3}-24 x^{2}\right) \text { by } 11 x(x-8) |
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| 19. |
x+x129. Divide 5x 3_13x2+2) -14 by (3-24+x2)and verify the division algorithm. |
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| 20. |
jessie joined a course on 7th july. she attended the course for three weeks. on what date she completed the course? |
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| 21. |
ime the Vave ol p dor which |
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Answer» Given , Points = ( p+1 , 2p-2 ), ( p-1 , p) and (p-3 , 2p-6) For the given points ( x₁ , y₁ ) , ( x₂ , y₂ ) and ( x₃ , y₃) to be collinear then [ x₁ ( y₂ - y₃ ) + x₂( y₃ - y₁ )+ x₃( y₁- y₂ )] = 0 Here, x₁ = p + 1 y₁ = 2p - 2 x₂ = p - 1 y₂ = p x₃ = p - 3 y₃ = 2p - 6 Substituting the values in the formula , ( p + 1 ) ( p - ( 2p - 6 ) ) + ( p - 1 ) ( 2p - 6 - ( 2p - 2 ) ) + ( p - 3 ) ( 2p - 2 - ( p ) ) = 0 ( p + 1 ) ( p - 2p + 6 ) ) + ( p - 1 ) ( 2p - 6 - 2p + 2 ) ) + ( p - 3 ) ( 2p - 2 - p ) = 0 ( p + 1 ) ( -p + 6 ) + ( p - 1 ) ( -4 ) + ( p - 3 ) ( p - 2 ) = 0 - p² - p + 6p + 6 - 4p + 4 + p² - 3p - 2p + 6 = 0 - 4p + 16 = 0 4p = 16 Dividing both the sides by 4 4p / 4 = 16 / 4 p = 4 Hence, For the points to be collinear , p = 4 |
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| 22. |
. Find the sum of oder integers from ime |
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Answer» odd numbers=1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99,101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 135, 137, 139, 141, 143, 145, 147, 149, 151, 153, 155, 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 177, 179, 181, 183, 185, 187, 189, 191, 193, 195, 197, 199 odd integers from 1 to 2001=1,3,5,7,9,11,...........,2001a=1 d= 2 n =? tn =2001tn = a+(n-1)d2001= 1+(n-1)22001=1+2n-22001=2n-12n=2001+12n=2002n=2002/2n=1001sum of the integerssn= n/2 (a+tn) =1001/2 × (1+2001) =1001/2 ×2002 =1001×1001 =10,02,001 is the answer |
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| 23. |
anel5 mutt onowits flooif each ime |
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Answer» area of room=5*7=35m² area required by 1 child=70cm*50cm=3500cm²=.35cm² let n be no. of studentsthen area of room=n*area required by 1 student35=n*.35n=35/.35n=100 |
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| 24. |
4. Which term of the GP. ime Color2,8, 32, ..... is 512 ? |
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| 25. |
A rectangular sheet of paper is l 2-crn long and 103 cm wide.ime |
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| 26. |
10 x763A circle passes hrough A,B,C and D as shown in the figure. IfBAD = 63°,find x. |
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| 27. |
is7 The product of HCF andy(x > y)of HCF and LCM of two positive integers and(A) x+y(B) rJx×y(D) x |
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Answer» option c This is an identity |
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| 28. |
find the value of tan 63° x tan 27° |
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Answer» tan63=tan(90-27)=tan27=tan^227 |
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| 29. |
400/841 |
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Answer» √400√841400841 Simplify thenumerator. Tap for more steps... 20√84120841 Simplify thedenominator. Tap for fewer steps... Rewrite841841as292292. 20√29220292 Pulltermsout from under the radical, assuming positive real numbers. 20292029 The result can be shown inmultipleforms. Exact Form: 20292029 Decimal Form: 0.68965517… |
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| 30. |
(1)841 (2) 1369 (3) 2116 (4) 3249 |
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| 31. |
INCERT13. An open box is made of wood 3 cm thick. Its external length, breadth and height ae1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of 250 pesg. metre |
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| 32. |
INCEREbox is made of wood 3 cm thick. Its external length, breadth and heigh a2.48 m 1.16 m and 8.3 dm. Find the cost of painting the inner surtace of t S0 p13. An opensg metre |
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| 33. |
4 10 40Question 13. Assume that the chances of n patient bavinattack is 40%. Ireduces the risk of heart attack by 30% and prescription of eereduces its chances by 25%. Attwo options with equal probabilities. It is given that after going throuone of the two options the patient selecteattack. Find the probability that the patient followed a cmeditation and yogaPt is also assumed that a meditation and yoga coursertain drugpatient can choose any ore of thetimeghd at random suffers a heart |
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| 34. |
24 x94x4Simplify using laws of exponents 3x42x27 |
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| 35. |
ion 13. Assume that the chances of a patient having a heartQuestattack is 40%. It is also assumed that a meditation and yogareduces the risk of heart attack by 30% and prescription of certain dreduces its chances by 25%. At a time a patient can choose any one of thetwo options with equal probabilities. It is given that after going throughone of the two options the patient selected at random suffers a heartattack. Find the probability that the patient followed a course ofmeditation and yogai?rug |
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| 36. |
Se area or the quadrilaterae adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is4 m, find the distance between the shorter sides. |
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Answer» thank you so much... |
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| 37. |
lakes.various human actfvlttlements have nowAreas of Triangles and QuadrilateraIPLE 2 Find the perimeter and area of a triangle whose sides are of lengths51752 cm, 56 cm and 60 cm respectively. |
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| 38. |
PSQR is a parallelogalXERCISE 8.11. The angles of quadrilatera are in the ratio 3:5:9:13. Find all the angleIf the diagonals of a parallelogram are equal, then show that it is a rectangle.ofgquadrilateral.hiect each other at right angles, the |
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| 39. |
I. Solve the following. Simplify using laws (10x (10) |
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| 40. |
Using laws of exponents, simplify:(3^-5*10^-5*125)/(5^-7*6^-5) |
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Answer» thank you sir |
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| 41. |
1. Using laws of exponents(i) 32 x 34 x 38 |
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| 42. |
.By using distributive laws, find: 257 x 1007 |
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| 43. |
.1 Solve using the laws of exponents:- 63 x 63 |
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Answer» 6^3 * 6^3 = 6^[3 + 3] = 6^6 |
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| 44. |
To stitch a shirt. 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched andzuch cloth will remain unused? |
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| 45. |
(iii) x4+64 |
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Answer» x4+ 64 =x4+ 16x2- 16x2+ 64x4+ 16x2- 16x2+ 64 =x4+ 16x2+ 64 - 16x2x4+ 16x2+ 64 - 16x2 = (x2+ 8)(x2+ 8) - (4x)2= (x2+ 8)2- (4x)2(x2+ 8)2- (4x)2= (x2+ 8 + 4x)(x2+ 8 - 4x), |
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| 46. |
numbers, then verifyLCM (p, q) x HCF (p, q)pq |
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Answer» take lcm and hcf which are common p = a²b³ q = a³b HCF ( p,q ) = a²b [∵Product of the smallest powerof each common prime factors in the numbers ] LCM ( p , q ) = a³b³ [∵ Product of the greatest power of each prime factors , in the numbers ] Now , HCF ( p , q )× LCM ( p , q ) = a²b× a³b³ = a∧5b∧4 --------( 1 ) [∵ a∧m× b∧n = a∧m+n ] pq = a²b³× a³b = a∧5 b∧4 ---------------( 2 ) from ( 1 ) and ( 2 ) , we conclude HCF ( p , q )× LCM ( p ,q ) = pq |
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| 47. |
If p and q are primes, then what will beHCF (p, q)? |
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Answer» If p and q are both primes, then their factors are 1 and the number itself. So the only common factor is 1. Hence HCF will be 1. Not only for two primes, the HCF of any number of prime numbers is always 1. |
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| 48. |
If 0.254 expressed as (HCF of p, q is 1) then the value of p+ q is |
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| 49. |
(iii) x4 +64 |
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| 50. |
x4 -5x2 +4 |
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Answer» hit like if you find it useful |
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