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} { l } { \text { If } f ( x ) = \frac { x - 1 } { x + 1 } , \text { then } f ( a x ) \text { in terms of } f ( x ) \text { is equal to } } \\ { \text { (a) } \frac { f ( x ) + a } { 1 + a f ( x ) } } \end{array} \right. |
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| 2. |
If z varies directly as x and inversely asincrease of 12% in x and a decrease of 20% in y.Find the percentageasy. Find the percentage increase in z due to anIfx+1 men will do the work in x + 1 days, find the number of dayfinish the same work.I days, find the number of days that (x+2) men can |
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Answer» Given z∝ (x/y)Hence z = (x/y) assuming proportionality constant as 1.x increase by 12% and y decreases by 20%Hence z value becomes, z = 1.12x/0.8y = 112x/80yChange in z value = (112x/80y) - (x/y)= 32x/80yIncrease in z% = (32x/80y)/(x/y) x 100= (32/80) x 100 = 40% |
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| 3. |
If the equation 3x2 2(a + b + c)x+ a2+b2 +c0 has real roots thenSelect an answerA a bD of these |
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| 4. |
17. Find total number of prime factors ofexpression (6)15 (5) * (9)" |
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Answer» =2^15×3^15×3^11×3^11×5^11=2^15×3^37×5^11 2^15×3^15×3^11×3^11×5^11=2^15×3^37x5^11 |
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| 5. |
thnewtienu4. If x varies directly as y, then find the missing entry.1560 |
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Answer» where the four comes ok thanks |
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| 6. |
lf.), varies directly as x2 and y = 8 when x=2, findy when x = 6.l 10 |
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| 7. |
2. दर्शाइए कि कोई भी धनात्मक विषम पूर्णांक 64+1, या 64+3. या 6q+ 5, के रूप का होता है ।जहाँ कोई पूर्णाक है। |
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| 8. |
4. Divide 340 in the ratio 2:3. |
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| 9. |
If 5" 125, then 5625(64)5052 of these |
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| 10. |
64. 0.2x -53.5x-3 5 |
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Answer» Thank you |
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| 11. |
24*(text*(60*(f*o))) %2B 220 - 11 |
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Answer» the correct answer is1649 220+24 of 60 -1089÷99220+24*60-11220+1440-111649 |
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| 12. |
2*(470/235) %2B 220 |
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| 13. |
. If d varies directly as t, and ifd=4 when t = 9, find d when t=21. |
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Answer» d is proportional to td = kt4 = k × 9k = 4/9So when t = 21then d = (4×21)/9 = (4×7)/3 = 28/3 |
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| 14. |
-e ) ;.”".f?"f हक.कि जे |
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Answer» (1-x)tan(pie*x)/2 = (1-x)/cot(pie*x)/2 lim x->1 (1-x)/cot(pie*x)/2 if we will put x=1 we will get 0/0 form so using L'hospital rule lim x->1 (-1)/(-cosec(pie*x)/2*cosec(pie*x)/2) = (-1)/(-1) = 1 your answer is wrong |
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| 15. |
18) x varies directly as y, when x = 5, y - 30. Find the constant of variation and equation of variation |
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Answer» option a is the right answer option a is the right answer option a . is the correct answer |
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| 16. |
if p varies directly as q and p is equal to 282,when q =5.1. if q=6.8,then what is p |
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| 17. |
5.y varies directly as square root of x. When x = 16, y = 24. Find the constant olvariation and equation of variation. |
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Answer» It is given that , y varies directly as square root of x . x = 16 , y = 24 y = k × √x ( k is a constant ) 24 = k × √16 24 = 4k 24/4 = k 6 = k Therefore ,Required constant = k = 6 And Equation y = 6√x |
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| 18. |
.2: Find the area of the sector of ciich subtends an angle of 120° at the cente radius of the circle is 6 cnm. |
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Answer» Area of sector = (120/360)*π*r²= (1/3)*(22/7)*(6²)= 37.71 cm² Please hit the like button below |
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| 19. |
BSelect a figure from the options which will complete the patterFigure (X) |
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Answer» Sarah, Leah, Haley, and Kim bought555bottles of water to share among themselves. They divided each bottle of water into444equal portions. Sarah took111portion from each bottle of water. Which equation represents how much of a bottle of water Sarah took? |
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| 20. |
Q27 Two dice are thrown simultaneously. What is the probability that the total scoprime number? u cne ae o |
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Answer» first hav to write prime numbers below 12 because 6,6 sum ends with 12,so there ll be 15 terms as like 1 1,2 1,1 2,.........6 5,so 15/36=5/12 can u tell it in a simple way |
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| 21. |
लॉग,Wi< |
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Answer» Ans :- (4/7) ÷ (-5/7) = (4/7) × (-7/5) = -4/5 PLEASE LIKE THE SOLUTION |
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| 22. |
S. D, E and F are respectively the mid-points of0) BDEF is a parallelogram.(ii) ar (DEF)ar (ABC)the sides BC, CA and AB of a ΔABC. Show that^ ar (ABC.(ii) ar (BDEF) - |
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Answer» Proof (I) and (ii) Proof (III) |
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| 23. |
2. If E,FG and H are respectively the mid-points ofthe sides of a parallelogram ABCD, show thatD EFigar (EFGH)--ar (ABCD).dAn resnectively |
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| 24. |
Express each of the ratios in the simplest form:(i) 6 and 50 p to? 4m 60 cm to 5 m |
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Answer» i) 1 rupee= 100 paise.₹6.5 : ₹4= 6.5:4= 13:8ii) 1m = 100cm.3m 60cm = 360cm.5m = 500cm.3m 60cm : 5m = 360:500= 18:25 Please hit the like button if this helped you thanks you |
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| 25. |
645.) 1 km to 600 m in simplest form is:(a) 5:3(b) 100:6 |
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Answer» 1 km = 1000 m So, ratio is 1000 m : 600 10 : 6 5 : 3 a) Option is correct |
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| 26. |
(iii)SP. =340 and gain =ăŚ20 |
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Answer» Gain= sp-cp20= 340- CPCP= 320 ruppes |
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| 27. |
20 - 85 %2B 220*340 |
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Answer» 340×220-340÷4+2074800-85+2074800-6574735 |
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| 28. |
450+340=? |
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Answer» 790 is the correct answer... |
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| 29. |
4. Divide 340 in the ratio 2:3 |
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Answer» Let the parts be 2x and 3x2x + 3x = 3405x = 340x = 68 340 can be divided in 2:3as 136, 204 |
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| 30. |
4. Divide 340 in the ratio 2: 3. |
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Answer» Total parts = 3+2= 5now we have to divide it 3/5*340= 204and second part2/5*340= 136 |
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| 31. |
IfE F.G and H are respectively the mid-points ofthe sides of a parallelogram ABCD, show thatar (EFGH)ar (ABCD)2 |
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| 32. |
Examine the tile pattern shown at right.a. On graph paper, draw Figure 4b. How many tiles will Figure 10 have?ć ć 1-15.Figure 1Figure 2FigureHow do you know? |
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Answer» In figure 10 tiles will be 6+12+16+20+24+28+32+36+40+44.... these are the sequence of tile which will formed ... at 10th place there will be 44 tiles It would be an AP with a= 6 and d=4Figure 10 will have6 + 9*4 = 42 |
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| 33. |
Q7, i Define regular poly[Draw any figure of polygon)any figure of polygon) Find xy.z, w in the given司Find , w z wint eg eachfigure(2 mark each)120 |
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Answer» Apolygonisregularwhen all anglesareequal and all sidesareequal (otherwise it is "irregular"). This is aregular pentagon(a 5-sidedpolygon). |
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| 34. |
2.IfE,FG and H are respectively the mid-points ofthe sides of a parallelogram ABCD, show that"ar (EFGH)ar (ABCD) |
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| 35. |
Figure A is a scale image of figure B.Figure AFigure BFigure A maps to figure B with a scale factor ofWhat is the value of x? |
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Answer» the answer is,the value of x is 12 how |
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| 36. |
है ~ A yescdical pudo 4 loasgle G onCusiea ol p कि. लि. |
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Answer» Length of the vertical pole = 6m (Given) Shadow of the pole = 4 m (Given) Let Height of tower =hm Length of shadow of the tower = 28 m (Given) In ΔABC and ΔDEF, ∠C = ∠E (angular elevation of sum) ∠B = ∠F = 90° ∴ ΔABC ~ ΔDEF (By AA similarity criterion) ∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional) ∴ 6/h= 4/28 ⇒h= 6×28/4 ⇒h=6 × 7 ⇒h= 42 m Hence, the height of the tower is 42 m. |
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| 37. |
2, If E,F,Gand H are respectively the mid-points ofthe sides of a parallelogram ABCD, show that]ar(EFGH) = - ar (ABCD) , |
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| 38. |
21 ABCD is a parallelogram. E and F are points on side AB such thatAE-EFzfB. Show that arDAE) =-ar (ABCD).A4 |
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| 39. |
2.94 (-€((७न का + iol = न 3 दि3 P |
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Answer» p = tanФ + secФp -secФ = tanФp² + sec²Ф - 2 p secФ = tan²Ф= sec²Ф - 1So secФ = (p² + 1) / 2p so cosФ = 2p/(1+p²)sinФ = √[1 - cos²Ф ] = (p²-1)/(1+p²) Samagh nahi aa rha |
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| 40. |
s. D, E and F are respectively the mid-points of the sides BC, CA and AB of a AABC.Show thatG) BDEF is a parallelogram.(iii) ar (BDEF) ar(ABC)(i) ar (DEF4 ar (ABC)2 |
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| 41. |
3, 4 weeks3 daysdays |
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Answer» 1 week=7 days4 weeks=7×4=24 days4 weeks 3 days=24 days +3 days= 27 days thank |
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| 42. |
ratios in simplest form 1m 5 cm: 63 cm |
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Answer» 1 m = 100 cm 1 m 5 cm = (100+5) cm =105 cm 105 : 63 = 5 : 3 |
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| 43. |
Which of the following ratios is the largest?(A) 7:15(B) 15:23(C) 17:25(D) 21:29 |
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Answer» 21/29 is largest in all the given ratio |
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| 44. |
3, 4 weeks 3 daysdays |
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Answer» 4 weeks=4*7=28 days1 week=7 days28+3=31 days |
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| 45. |
TR U P WP L OL = वितन० हे विद न ० |
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Answer» nPk = n!/(n-k)! (n+2)P3 : (n+1)P3 = 10 : 7 (n+2)!/(n+2-3)! : (n+1)!/(n+1-3)! = 10 : 7 (n+2)!/(n-1)! : (n+1)!/(n-2)! = 10 : 7 (n+2)(n+1)!/((n-1)(n-2)!) : (n+1)!/(n-2)! = 10/7 (n+2)/(n-1) = 10/7 7n + 14 = 10n - 10 3n = 24 n = 8 If you find this answer helpful then like it. |
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| 46. |
340 days into months, weeks and days |
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Answer» Months= 340/30 = 11 monthsWeeks = {340 - (11*30)}/7= 10/7= 1 week and 3 days. |
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| 47. |
MPLE 27 The angle of elevation of a jet fighter from a point A on the ground is60°. After a flight of 15 seconds, the angle of elevation changes to 30If the jet is flying at a speed of 720 km/hour, find the constantheight at which the jet is flying. [Use 3 1.732.] [CBSE 20080 |
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Answer» Like if you find it useful |
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| 48. |
ICBSE 2010]a rectangle.19. In the given figure, ABCD is aparallelogram, AM CN. Prove thatBNDM is a parallelogram.riren figure, ABCD is a |
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| 49. |
23. In the given figure, ABCD is aparallelogram. L is the mid-point of AB.Prove that ACBM is a parallelogram. |
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Answer» Construct:ACandMBIn∆ALMand∆BLCAL=BL(LismidpointofAB)∠ALM=∠BLC(VerticallyoppositeAngle)∠AML=∠LCB(Alternateangles)Therefore,In∆ALM≅∆BLC(byASA)AM=BC(byCPCT)Hence,ACBMisaPararllelogram(Apairofoppositesidesisequalinparallel) |
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| 50. |
Divide (5x-4x+3x +18) by (3-2x +x2). |
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