Saved Bookmarks
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. |
317+x=1 |
| Answer» | |
| 2. |
Page No.:Date :17का मान निकाल1133 - 28442 |
|
Answer» 5x is correct answer x=4 hoga. ,,,,,,,,,,,,,,,,,, x=1.333 is the correct answer x=1.333 is Wright answer 1.33 is correct answer 1.3333 is the correct answer X= 4/3 right Answer hai 1.333 is the right answer 1.333 is the bast answers x ka man = 1.33 is correct answer x=1.333 is the correct answer x=1.33 is the correct answer 1.33 is the correct answer |
|
| 3. |
317+x=1A< |
|
Answer» 3/7+X=17/7x=17/7-3/7=14/7=2 |
|
| 4. |
show that the Modulus Function fiKone nor onto, where lxl is , ifShow that the Signum. Function,f : R → R, givenbyx is positive or Omsk |
| Answer» | |
| 5. |
1) What is the value of 4/28 = 317 ? |
| Answer» | |
| 6. |
If matrix A =[1 2 3 ] , write AAT , where AT Is transpose of matrix A. |
| Answer» | |
| 7. |
. Find the area of sector whose radius is 7 em. and angle be 60" |
| Answer» | |
| 8. |
*(4x+3)4. If f(x) =- (6x-4)inverse of f?3show that fof(x) = x, for all x*-. What is theWIN |
| Answer» | |
| 9. |
the school ?An alloy consistscomposition of the alloy ?of 13 parts of copper to 7 parts of zinc and 5 parts of nickel. What is the percentage |
| Answer» | |
| 10. |
0. In the figure, if AC |I DE, then find the value of rcorD <) 55。28 B |
| Answer» | |
| 11. |
An alloy consists of 13 parts of copper to 7 parts of zinc and 5 parts of nickel. What is the percenngcomposition of the alloy?. |
| Answer» | |
| 12. |
An alloy consists of 30% zinc, 45% nickel and the rest copper. Find the weight of copper in 800 g of the alloy. |
| Answer» | |
| 13. |
(i) ) 103Ă107 |
|
Answer» the answer is 11021 of above question |
|
| 14. |
107 - 317 - 132 |
|
Answer» 10m²-31m-132=>m=5.5, -2.4 |
|
| 15. |
55. If the sum of p,q and r terms of an arithmetic progression is a, band c respectively, prove that |
| Answer» | |
| 16. |
0 1-1Aisan involutary matrix givenby a =14-3 43. -34|thenAl2the inverse of~will be |
| Answer» | |
| 17. |
(L.T.I.July 1998, Electrician, Electroplator)An alloy has 8 parts of copper, 5 parts orickel and 2 parts of cadmium.Find their percentages |
|
Answer» Percentage of copper= (8/15)*100= 53.4 % Percentage of nickel= (5/15)*100= 33.3 %.Percentage of cadmium= (2/15)*100= 13.3 % |
|
| 18. |
opees ow huch hust We pay Tor 12 kg onions?2. If 600 rupees buy 15 bunches of feed, how many will 1280 rupees buy?Exan |
|
Answer» 600 rupees buy 15 bunches of feed 1280 rupees can buy 1280×15/600 = 1280/40 = 32 bunches of feed If you find this answer helpful then like it. |
|
| 19. |
base of a triangular field is three times its altitude. If the cost ofsowing the field at Rs 58 per hectare is Rs 783,find its base and height. |
| Answer» | |
| 20. |
The base of a triangular field is three times its altitude. If the cost of sowing the field at rupees 58 per hectare is rupees 783, find its base and height? |
|
Answer» Area = 783 / 58 = 13.5 = 0.5 * b * h = 0.5 * 3h * h.h = 3b = 9 |
|
| 21. |
के छा कक.| 373, थी 1_ हा छांणा 600 6681 |
|
Answer» In given questionIn every column product of numbers in 1st, 2nd and 3rd row is written in 4th row For 2nd column = let missing no. is nThen 5*7*n = 105 n = 105/35 = 3 Missing number is 3 |
|
| 22. |
C Das or a u angle whose area is 90 cm' and height 2 cm5. The base of a triangular field is three times its height. If the cost of cultivating the field atR 1080 per hectare is 14580, find its base and height. |
| Answer» | |
| 23. |
17. रैखिक समीकरण युग्म कों, हल कीजिए :--~ t=3= 2.4S e373 |
|
Answer» Like my answer if you find it useful! |
|
| 24. |
103×107 |
|
Answer» 11021 is the answer because this is 11021 is the right answer 11021 is the right answer 11021 is the right answer. 103×107=11021 is the answer 103× 107=1102111021 is the right answer 103× 107 __________ 721 0000 10300____________ 11021 |
|
| 25. |
(o)) (55R Didign By |
|
Answer» 2tan^-1a=cos^-1(1-a^2/1+a^2)Now cos^-1(1-a^2/1+a^2)+cos^-1(1-b^2/1+b^2)=2tan^-1x2tan^-1a+2tan^-1b=2tan^-1xhencex=a+b/1-abHence proved |
|
| 26. |
103*107 |
|
Answer» (105-2) (105+2) =(105)²-2²=11025-4=11021 103×107=11021 is the best answer the correct answer is 11021 like kar please mera dost meri sister like 103×107=11021 like dost like sister like dear,please (100 +3) (100+7) =10000+2100+21=12121 the answer of the question is 11021 The answer of the question is 11021 |
|
| 27. |
a) I310. The last digit of the number (373)333 isb) 2c) 3al 1 |
|
Answer» 373¹=373373²=139129373³=51895117373⁴=19...............1373⁵=722................093 Hence the cyclicity of 373 is 4.So, when 333 is divided by 4 the remainder is 1, so 373 raised to the power 1 is 373, hence the last digit of the result is 373 raise to the power that is 3. Hence option c is correct. |
|
| 28. |
a)fs10. The last digit of the number (373)5s isd) 9a) 1b) 2 |
|
Answer» Suppose x is raised to the power y and we need to find unit digit of x^ySo, we will first divide y by 4 and find the remainder thus obtained. Since on dividing and integral number by four can yield only four remainders --> 0, 1, 2 and 3.This method applies for 4 values of x only, because for other values, you can split the values in the 4 values of x:-x = 2, 3, 7 and 8.For 2====If remainder [y/4] is 1,unit digit of = 2If remainder is 2, unit digit of = 4If remainder is 3,unit digit of = 8If remainder is 0,unit digit of = 6 For 3====If remainder is 1,unit digit of = 3If remainder is 2,unit digit of = 9If remainder is 3,unit digit of = 7If remainder is 0,unit digit of = 1 For 7====If remainder is 1,unit digit of = 7If remainder is 2,unit digit of = 9If remainder is 3,unit digit of = 3If remainder is 0,unit digit of = 1 For 8====If remainder is 1,unit digit of = 8If remainder is 2,unit digit of = 4If remainder is 3,unit digit of = 2If remainder is 4,unit digit of = 6 For memorising that of 3 and 7, you can check the unit digit of when 3 and 7 are raised to the power remainder [power = remainder]. For 2 and 8 this trend doesn't works.So in your question, Dividing 333 by 4, we get remainder 1.So using table, when remainder is 3, unit digit of 3 [because the unit digit of 373 raised to power something is because of its originalunit digit only] will be 3.Hence unit digit of will be 3. |
|
| 29. |
Parikshit makes a cuboid of plasticine of sides 5cm, 2cm, 5 cm.How many such cuboids Will be needed to form a cube? |
|
Answer» volume of cuboid =5*2*5=50cm^3he certainly no of cubes =volume of cube =volume of cuboidhence50=side^3side=50^1/3=25√2cm |
|
| 30. |
U = 110 voltsi = 4.55 ampsR =ohms |
|
Answer» V= IRR= V/I= 110/4.55= 24.17ohm |
|
| 31. |
k 40 and selling i a() 333u, e proĂa percentnge will ba(e) 20(a) 10ti roid ac t 360 and Chere is lon a |
| Answer» | |
| 32. |
In the following figures, find the values of x and y.1075TDC40"40107A |
|
Answer» X=50 is the right answer y=92 is the right answer x=50 y=92 is a answer |
|
| 33. |
4.In Fig. 6.31, if PQ 11 ST,2 PQR = 110° andRST-130o, find ZQRS.Hint : Draw a line parallel to ST through 110point R.]Fig. 6.31 |
| Answer» | |
| 34. |
Section-A1. Iftan 5_ :-V3 and θ is acute, then find the value of 2e2 |
| Answer» | |
| 35. |
43will terminate after how many places(CBSE 2009Q.7. The decimal representation ofc3tation of24.5of decimals? |
|
Answer» 43/2^4 5 ^3= 43 ×5/2^4 5 ^3 ×543 ×5/2^4 5 ^443 ×5/(2×5)^443 ×5/(10)^4215/100000.02154 places of decimal |
|
| 36. |
in the blanklo colo |
|
Answer» 1. >2. <. is the answers 1. >2. < is the best answer me < > is the best answer me please like my answer then I like your answer all off you i) 5/8 > 7/8v) 9/5 < 9/7 is the answer 1.>2.< is the right answer 1)5/8 < 7/82)9/5 > 9/7 5 7 1--: -- < -- 8 8 9 9 2--: -- > -- 5 7 |
|
| 37. |
2. Which is greatercolo |
|
Answer» 2/7 by multiplication method 5×7=352×8=16so 2/7 is greater 5/8 is greater than 2/7 |
|
| 38. |
SEC.. T ICN-Aemier14x1ニ6)id of triangle al ho.se vestices are AC-4,6) 丿e C2. -2) and c C2, s) |
| Answer» | |
| 39. |
-29,711, then find the-colovalue of x. |
|
Answer» Option c)15/2 is correct. |
|
| 40. |
5cm6cmIV1.5cm6.5cm IlcmIcm/III Icm2cm |
|
Answer» thanks my answer is crect 19.4cm2 |
|
| 41. |
1 मिनलिखित संख्याओं को अभाज्य गुणनखंडों के गुणनफल के रूप में व्यक्त कीजिए:| i) 140 (i) 156 (ii) 3825 (5 6 ) 7429। पर्णाकों के निम्नलिखित यमों के HCF और ICM जात कीजिए। इसकी जाँन == |
|
Answer» 5005 is the right answer for this question 5005 is the correct answer 5005 is correct answer 5005 is the correct answer. 5005 is the right answer 7429 is a anwser is a correct anwser |
|
| 42. |
5cm6cm:1.Sem81)086.5cmI1lemicm/ T lom2cm |
| Answer» | |
| 43. |
(2LA < ¢ ८.2e & -"b 0 |
| Answer» | |
| 44. |
EXERCISE 6.2t, InFig 6 17,0) and o, DI BC FindIC in () and AD in ()1.8 em1.sem icm7.2 cm3 em5.4 cm(oFig. 6.17 |
| Answer» | |
| 45. |
4. Find the number of sides of a regular pentagon whose each exterior angle has a measure of 45° |
|
Answer» Each exterior angle =360/n=360/45=8The number of sides is 8 |
|
| 46. |
DA5 cm and AC5cm.Radha made a picture of an aeroplane with coloured paper as shownthe total area of the paper usedwn in Fiz 1215.3.Scmocm15cm6 5cm2cm |
| Answer» | |
| 47. |
2.Find A BGven |
| Answer» | |
| 48. |
below14. Calculate the area of the shaded region in each of the figures gventi40 m |
|
Answer» tq brooo |
|
| 49. |
) सिद्ध कीजिए; £”! है - ००४6 X झा ‘COLA 4 cos A %‘q |
|
Answer» Lhs = (cotA - cosA)/(cotA+cosA)=(cosA/sinA - cosA)/(cosA/sinA + cosA) =cosA(1/sinA - 1)/cosA(1/sinA + 1) = (1/sinA -1)/(1/sinA+1)=(cosecA-1)/(cosecA+1) since 1/sinA= cosecA=RHS |
|
| 50. |
w Iftan A = , find sina colaIftan A=+COSA |
|
Answer» tanA=3/4; 1/sinA=1/3/5=5/3; 1/ cosA=1/4/5=5/4; 1/ sinA+1/ cosA=5/3+5/4==25/9+25/16=(25)^2/(25)=2/1=2 tanA=3/4; 1/ sinA=1/3/5=5/3; 1/ cosA=5/4; 1/ sina+1/ cosA=5/3+5/3=(25)^2/(25)=2/1 tanA=3;4sinA/cosA= 3/4sinA/3=cosA=/4=k(let)sinA= 3kcosA= 4k1/sinA+1/cosA= 1/3k+1/4k= 4k+3k/12k÷ 7k/12k= 7/12 |
|