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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. |
4 Form the differential equation of the family of curvesx = A cos nt + B sin nt, where A and B are arbitrary constants. [CBSE 2007] |
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| 2. |
is140Find the height of thevolume ise eylinder whose volume is 1.54 m3 and diameter of base |
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| 3. |
e diameter of the moon is approximately one fourth of the diameter of the eartlh.[NCERT]What fraction of the volume of the earth is the volume of the moon? |
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| 4. |
Nte - o ] |
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Answer» (n+1)P3 = 4*nP2 nPk = n!/(n-k)! using this formula (n+1)!/(n+1-3)! = 4*n!/(n-2)! (n+1)n!/(n-2)! = 4*n!/(n-2)! n+1 = 4 n = 4-1 n = 3 If you find this answer helpful then like it. |
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| 5. |
240 nt t g A |
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| 6. |
4. A wholesale dealer had a stock of 5698 shirt pieces. During the marriage season hesold away 3246 pieces. How many shirt pieces are still in the stock? |
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Answer» 1452 is the right answer The correct answer is 2452 5698-3246= 1452 is the answer the best answer is 2452 stock of shirt = 5698 sold = 3246 - = shil in stock = 1452 still left=5698-3246 =2452 |
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| 7. |
wali, Wilal LLCPU14. Monica cuts 46 m of cloth into pieces of 1.15 m each. How many pieces docs15. Mr Soni bought some bags of cement onHow many pieces does she get? |
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| 8. |
6A large pizza has toppings of capsicum and onion. In total there are 25pieces of both on the pizza. If there are 4 times as many onions piecesas capsicum pieces, how many pieces of each vegetable are there on thepizza?Butes ofA recipe needs a combined weight of 720 g of flour and sugar. If the recipe needs |
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Answer» 5 onion pieces and 20 capsicum pieces 5 onion pieces and 20 capsicum pieces 5 onion piece and 20 capsicum pieces |
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| 9. |
TUJUU ULIOSedCyclA company manufactures jeans. In 2014, the company made 30,000 jeans, out of them 400 jeans weredefective. The company sold fresh pieces at a profit of 750 and defective pieces at a loss of 200 perpiece. Find its profit/loss, ifit sold all pieces. |
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| 10. |
\sqrt [ 3 ] { - 2744 } = |
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Answer» as (14)³ = 2744so cube root of -2744 is -14 |
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| 11. |
[27 x (-2744)1/3 |
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| 12. |
1. [27 x (-2744)1/3 |
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| 13. |
The volume of a cube is 2744 cm^3.Its surface area is |
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Answer» Find the length: volume = length³ length³ = 2744 length = 14 cm Find the area of one side: Area = length x length Area = 14 x 14 Area = 196 cm² Answer: The length is 14 and the area is 196 cm². |
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| 14. |
2. The statistical data are of two types. These types area) technical data and presentation datac) primary data and personal datab) primary data and secondary datad) none of these |
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Answer» d)none of theseas statistical data is of two types that is numerical and categorical data |
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| 15. |
2x+3y=463x+4y=74Cross multiplication method. |
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| 16. |
cross multiplicationx-3y-3=03x-9y-2=0 |
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| 17. |
:-Solve the equations by cross- multiplication method.2x + 5y17, 5x + 3y14 |
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Answer» 2x+5y=17. .. ...(1) multiplied by 3 5x+3y=14. .....(2) multiplied. by 5 then 6x+15y=51 .......(1)25x+15y=70 .....(2) (1)-(2) we will getx=1 and y=3 |
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| 18. |
Solve by cross-multiplication method(i) 8x-3y =12; 5x= 2y +7 |
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Answer» 8x - 3y = 12.....(1)5x - 2y = 7.......(2) Multiply (1) by 2 and (2) by 3 then subtract 16x - 6y = 2415x - 6y = 21 After subtractionx = 24-21 = 3 Put value of x = 3 in eq (1)8*3 - 3y = 123y = 24-123y = 12y = 12/3 = 4 x = 3, y = 4 |
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| 19. |
solve the pair of linear equation by cross multiplication method x+ 2y = 2 x - 3y = 7 |
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| 20. |
The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34g/e. What isthe weight of the ball? |
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Answer» Diameter of the ball = 2.1 cm or radius = 2.1/2 = 1.05 cmVolume of the Spherical lead ball = 4/3πr³= 4/3*22/7*1.05*1.05*1.05Volume of the lead ball = 4.851 cu cmLead used for making lead ball = 11.34 g/cu cmSo, the weight of the lead ball = 11.34× 4.851Weight of the lead ball = 55.01 grams |
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| 21. |
The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/e: Whatisthe weight ofthe ball? |
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| 22. |
the diameter of a lead ball is 2.1cm.the density of the lead used is 11.34g/c^3 what is the weight of the ball. Answer : 5.5kg |
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Answer» Given : Diameter of the ball = 2.1 cm or radius = 2.1/2 = 1.05 cm Volume of the Spherical lead ball = 4/3πr³= 4/3*22/7*1.05*1.05*1.05 Volume of the lead ball = 4.851 cu cm Lead used for making lead ball = 11.34 g/cu cm So, the weight of the lead ball = 11.34× 4.851 Weight of the lead ball = 55.01 grams thankyou |
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| 23. |
The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/c3 . what is the weight of the ball ? |
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| 24. |
Solve the system of of equation using cross multiplication method:2x + 5y = 12x + 3y = 3 |
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| 25. |
16. In the Multiplications (I) and (II) given below, theletters a and b stand for the digits. Find the productof a and b.la73 a215762744Multiplication (I)A. 63C. 54E. of theseMultiplication (II)B. 28D. 72 |
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Answer» 1a7 × 81576 7×8=565 will be carry8×1=8last term is 15 so 7 must be the carry for 8×1so8×a+5=778a=72a=9now 3a2 ×b =2744b=2744÷392=7a×b=9×7=63 |
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| 26. |
solve using cross multiplication method x+2y equal to 7 and 2x-3y equal to 11 |
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| 27. |
Name the property under multiplication |
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Answer» Here law of associate multiplication is used |
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| 28. |
1/2 multiplication by 2 |
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| 29. |
show that log√8+log√27-log√125 =3/2________________________ log6-log5 |
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| 30. |
Evaluate log 2s 5 and log 1 811253V9(a)-2 , รู้ (b) 2,;,-2(d)-3,(c)2 |
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Answer» x = 1/3 |
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| 31. |
\frac{\log \sqrt{8}+\log \sqrt{27}-\log \sqrt{125}}{\log 6-\log 5}=\frac{3}{2} |
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| 32. |
7. Given log 2 a, log 3-b, express in terms of a and b(0) log 6(i) log 4.8(ill) log 125yin terms of a and b |
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Answer» log6=log(2.3)=log2+log3 log(4.8)=log(48/10)=log48-log10=log(2*2*2*2*3)-14log2+log3-1 |
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| 33. |
Show that:\frac{\log \sqrt{8}+\log \sqrt{27}-\log \sqrt{125}}{\log 6-\log 5}=\frac{3}{2} |
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| 34. |
\operatorname { log } _ { x } 25 - \operatorname { log } _ { x } 5 + \operatorname { log } _ { x } ( \frac { 1 } { 125 } ) = 2 |
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Answer» thnx |
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| 35. |
\frac{\log _{3} 125}{\log _{9} 25} |
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| 36. |
6.The sum of all values of0,"satisfying sinº 20 + cos20LJEE(Main)-Jan 19)(1)(2) |
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| 37. |
-19-1910x 10(1 - x)2(1-5 x 101912 |
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| 38. |
the diamrter of a lead ball is 2.1cm the density of lead used in 11.34g/c^3 what is weight of the ball |
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Answer» thank you |
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| 39. |
Determine the following products by suitable rearrangement:(i) 8 x 125 x 40 × 26(ii) 250 × 60 × 50 × 8 |
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Answer» 8 × 125 × 40 × 261000 × 40 × 2640000 × 261040000 |
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| 40. |
\left. \begin{array} { l } { \text { hich value of } p \text { is the solution of } 5 p - 1 = 2 p + 20 } \\ { \text { (1) } \frac { 19 } { 7 } } \\ { \text { (2) } \frac { 19 } { 3 } } \end{array} \right. |
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Answer» so 7 |
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| 41. |
( 2 x + 5 y ) ^ { 3 } = 8 x ^ { 3 } + 125 y ^ { 3 } + a x ^ { 2 } y + b x y ^ { 2 } |
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| 42. |
Calculate the mean for the following distribution.:x: 56 78 9y: 4 8 14 11 3 |
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| 43. |
16 x 125 x 8 x 625 |
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Answer» 16×125×8×625=10000000 16×125×8×625=10000000 |
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| 44. |
- 8 x53 x (-125)- (-41) x 102- 7 x (50-2)- (-57) x (-19)+ 57 |
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Answer» thanks |
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| 45. |
15. How many envelopes can be made out of a sheet of paper 72 cm by 48 cm, if each enveloperequires a paper of size 18 cm by 12 cm?(a) 4(b) 8(c) 12(d) 16 |
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| 46. |
substitutionmethod0 The difference between two numbers is 26 andone number is thrce timesthFind them |
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Answer» suppose two numbers are a and bso a-b=26now a=3bso 3b-b=26so b=13so a=39so two numbers are 13 and 39 thanks |
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| 47. |
3Thehe sum of Cproduct of twoof them is 24theoneproductofteam T orational numbers inand the other.154 m.1 |
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| 48. |
5/ How many envelopes can be made out of a sheet of paper 3 m 24 cm by 1 m 72om, foenvelope requires a piece of paper of size 18 cm by 12 cm? |
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| 49. |
5. How many envelopes can be made out of a sheet of paper 3 m 24 cm by 1 m 72 em, if eachenvelope requires a piece of paper of size 18 cm by 12 cm? |
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| 50. |
In 、 Joint stock laǐ shess A invensts Rs 1800 tav smepvott bath be sont |
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Answer» 1000 x 9 = 1800 x timeTime = 5 months |
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