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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. |
Solve the following quadratic equation for x andgive your answer correct to two decimal places(ICSE 2008)5x(x+2)-3 |
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| 2. |
2x +7Solve the equation:33x -1 4(2x + 7) , 4 =x+5 15 |
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| 3. |
What is the variable in the equation2x - 3 = 7? |
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Answer» x is the variable in it 2x is variable IN it and 3 X is the correct answer of the given question x is the variable in it variable is 2x as -3 and 7 are constant term and their value doesn't change 2x-3=72x=7+32x=10x=10/2x=5 2x is the variable in the given question |
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| 4. |
2x +73x 1Solve the equation:3x-1 4 |
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| 5. |
in the equation 2x-3y=7 if x=-1 then y= |
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| 6. |
Solve the equation 5(2x +7) 2(3x +15) 25 and verify the solution. |
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| 7. |
hl hat is this symbol Ď |
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Answer» it is used to denote any unknown quantity like the example you have provided in this it is denoting angle.it is pronounced as "phi" thank you |
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| 8. |
What will be the nature of roots of the quadratic equation 2x +4x-7-0 |
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Answer» 2x² + 4x - 7 = 0 Discrimanat = b² - 4ac = 4² - 4 × (2) × (-7) = 16 + 56 = 72 > 0 Roots are real and distinct |
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| 9. |
0. ABCDis a tapdistance between AB and CD isl, hind the cost of watering a trapezoidal field whose parallel sides are 10 m and 25/ el, Imli tance between them is 15pm and the rate of wateringsăŚ4permm respectively, the |
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| 10. |
A train covers a distance of 51 km in 45 minutes. How long will it take to cover 221 km? |
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| 11. |
7. A train covers a distance of 51 km in 45 minutes. How long will it take to cover 221 km? |
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| 12. |
find the cost of watering Trapezoidal field whose parallel sides are 10cm and 25 cm respectively the perpendicular distance between them is 15cm and they rate of wateringis?4 per month square |
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| 13. |
The length of a rectangular plotland is 50 m and its width is 30 mA triple fence has to be put along itsedges. If the wire costs 60 rupees permetre, what will be the total cost of thewire needed for the fence? |
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| 14. |
6, The length of a rectangular plot ofland is 50 m and its width is 30 m.A triple fence has to be put along itsedges. If the wire costs 60 rupees permetre, what will be the total cost of thewire needed for the fence? |
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Answer» Length of Reactangular plot = 50 mBreadth of Reactangular plot = 30 m Perimeter of Reactangular plot= 2(length + breadth)= 2(50 + 30) = 2*80= 160 m As triple fence is used so total length of wire required = 3*160= 480 m Therefore, total cost of wire required for fence = 480*60= Rs 28800 |
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| 15. |
=-X -+- +-13 47MIN+00winy |
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Answer» c) is the right answer of the following c) is the right answer of the following answer this question is a |
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| 16. |
(i) 18, 15 1/2 ,13,..,-47 |
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Answer» Thanks |
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| 17. |
the length and breathe of a rectangular ground are 90 m and 70 m respectively• find the cost of cultivating the ground at the rate of rupees 0•50 per sq m• |
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Answer» Total cost of cultivating rectangular field of 15 rupees=18330Therefore, area of the field=18330÷15= 1222Let the length of the field be xTherefore, l×b=1222= l=1222÷26Therefore, l= 47Now, perimeter=2(47+26)=146Therefore, cost of fencing the field=146×20 |
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| 18. |
Find the number of terms in the A.P. 18, 15 , 13,.., - 47 |
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Answer» Like if you find it useful |
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| 19. |
9. The total annual income of A, B and C is7.98 Lakh. They spend 90%, 87 % and 80%respectively of their incomes. If their annual savings are in the ratio of 12:17:16, find annual(Sagar, 2008) [Ans.äš79,800;äš99,750; 1,59,600]saving of each |
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Answer» Savings ratio = 12 : 17: 16Their savings are 12x, 17x and 16x respectively.It is given that A spends 90 %, B spends 87 1/2% and C spends 80 % of his income.So, their savings are,A's savings = 100 - 90A's savings = 10 %B's savings = 100 - 175/2B's savings = 25/2 %C's savings = 100 - 80C's savings = 20 % A's savings = 12x⇒ 10% of A's salary = 12x⇒ 25/100× A's salary = 12x⇒ A's salary = (100/10)*12x⇒ A's salsry = 120x B's savings = 17x⇒ 25/2% of B's salary = 17x⇒ 12.5/100× B's salary = 17x⇒ B's salary = (100/12.5)*17x⇒ B's salary = 136x C's savings = 16x⇒ 20 % of C's salary = 16x⇒ 20/100× C's salary = 16x⇒ C's salary = (100/20)*16x⇒ C's salary = 80x Total salary of A, B and C = Rs. 1800So,⇒ 120x + 136x + 80x = 7.9 L⇒ 336x = 7.9 L⇒ x = 2351 Salary of A = 120*2351A's salary = Rs. 282120B's salary = 136*2351B's salary = Rs. 319736C's salary = 80*2351C's salary = Rs. 188080 Now, savings of A, B and CA's savings = 10 % of 282120= (282120*10)/100A's savings = Rs. 28212B's savings = 12.5 % of 319736= (319736*12.5)/100B's savings = Rs. 39967C's savings = 20 % of 188080= (188080*20)/100c's savings = Rs. 37616 |
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| 20. |
2Find the number of terms in the A.P. 18, 15, 13,47 |
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Answer» Therefore 27 terms are there in the AP. |
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| 21. |
(ii)How many terms are there in the AP. I 8, 15, 13, ........-47 ? |
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Answer» There are 37 terms in the A. P. |
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| 22. |
Divide 3600 into two parts such that if one part be lent at 9% per annum and the other at10% per annum, the total annual income is 333. |
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Answer» à |
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| 23. |
Divide? 3600 into two parts such that if one part be lent at 9% per annum and the other a10% per annum, the total annual income is 333. |
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| 24. |
25. Divide? 3600 into two parts such that if one part be lent at 9% per annum and the other at10% per annum, the total annual income is333. |
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| 25. |
r Dvide? 3600 into two parts such that if one part be lent at 9% per annum and the other at10% per annum, the total annual income is 333. |
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| 26. |
3Ex.5. A man had Rs 8400. He lent a part of it at 8%S.I. and the remaining63% SJ. His total annual income was Rs588. Find the sum lentatat different rates. |
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Answer» please like if you find it useful |
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| 27. |
9, दि =, 1 cor” T बताइए । & |
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Answer» pratilom trikonmiti ke niyam se pls |
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| 28. |
Sinipleinterest?8,A sum of1550 is lent out into two parts, one at 8% and another one at 6%. If thetotal annual income is 106, find the money lent at eachrate. |
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| 29. |
COR A + 8 A |1+ sinकक; मल |
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| 30. |
The value of cosec" (2) - cor*(-13) is |
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Answer» I think this is right |
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| 31. |
If 3tan-x + cor x = 1, then x equals |
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Answer» j yutruiyrr8i6rrc7755334 |
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| 32. |
\left(-\frac{3}{5} i\right) \quad 2 \cdot i^{9}+i^{19} \quad 3 \cdot i^{-29} |
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| 33. |
R72. If cor' x + sin-', then x is equal to |
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Answer» We assumed x = 3 and proved it as RHS |
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| 34. |
l) parallelIf 9çąľ3y+12-0,2) coiriciden18x+6y+24-0, then the linear equationsare2.4) all the abor |
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| 35. |
1347 X ky is |
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Answer» 5205 is the answer yesssssssssss 5205 is the correct answerplz like my answer 5205 is absolutely correct answer the answer is 5025 and it's correct answer 5205 is ur answer nd this is the correct answer the answer is 5205 hope this help u 5205 is the correct answer |
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| 36. |
X2+18x+81=0 |
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Answer» Use (a+b)² = a² + b² + 2ab x² + 18x + 81 = 0x² + 2×9x + 81 = 0(x + 9)² = 0So,x = -9 |
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| 37. |
4००29, Solve: log(x +1)+log(x—1कक = [ g ¥)=logl1+2log333.E8 g SR |
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| 38. |
Ry shifting the origin to a suitable point, transformthe equation 5v90y+ 18x-9 -0 to theform _ _ _ = 1,(a > 0, b > 0)Hence find the co-ordinates of the point, where theorigin is shifted and the values of a and b.Y2 x2 . |
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| 39. |
वी ७० A+ cosec A+ (os A+ sec AY =7 + tan' A+ cor | |
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Answer» =(sinA+cosecA)²+(cosA+secA)² =sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA =sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA =1+cosec²A+sec²A+2+2 =5+(1+cot²A)+(1+tan²A) =7+tan²A+cot²A Identities used:1+tan²A=sec²A 1+cot²A=cosec²A sin²A+cos²A=1 cosecA=1/sinA secA=1/cosA |
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| 40. |
(ii)If y ะเ +tan 1, show that cor: d-y-2y-2da |
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| 41. |
10393910 10 10 1002610i。 |
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Answer» The answer of 70-26 is 44. |
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| 42. |
a=6, d=3,S27-?27S27=- [12 + (27-1)a]272= 27 × 45 |
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| 43. |
Express 27×27×27×2in exponential |
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| 44. |
a=6, d=3,S27-?Sn = 2 [D+ (n-1) d]S27 = [12 + (27-1) 미27272= 27 × 45=□ |
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Answer» like if you find it useful |
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| 45. |
12. Without actual division, prove that the polynomial 2x+4x x-34 is exactlydivisible by (x-2) |
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| 46. |
\frac { \operatorname { cos } 27 ^ { \circ } + \operatorname { sin } 27 ^ { \circ } } { \operatorname { cos } 27 ^ { \circ } - \operatorname { sin } 27 ^ { \circ } } = \operatorname { tan } 72 ^ { \circ } |
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| 47. |
14. For what value of x, is the mode of the following data 2725. 26, 27, 23, 27, 26, 24, x, 27, 26, 25, 25 |
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| 48. |
( a ) \quad \operatorname { log } _ { 2 } 5 + \operatorname { log } _ { 2 } ( x ^ { 2 } - 1 ) - \operatorname { log } _ { 2 } ( x - 1 ) ( b ) \quad 2 \operatorname { log } _ { 4 } 9 - \operatorname { log } _ { 2 } 3 \quad ( c ) \quad 3 ^ { 2 \operatorname { log } _ { 3 } 5 } |
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| 49. |
COWrite 2 log 3+3 log 5 -5 log 2 as a single logarithm2 |
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| 50. |
(ii) 9x +3y +12 = 018x + 6y+ 24 = 0 |
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Answer» the two lines are parallel so x and y has infinite value |
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