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. |
SECTION-B2Out of 500 car owners, 400 owned car A and200 owned car B. How many car owners haveboth car A and B? |
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Answer» number of persons who own board car A and B is equal to(400+200)-500=600-500=100 |
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| 2. |
72*x - 52=71*x - 19 |
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Answer» Here 72x-71x= 52-19x= 33thanks what do you mean by surveying |
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| 3. |
15. If ã and b are unit vectors and e is the anglebetween them, then the value of a+b is(a) 2000l)(D) 2 sin |
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Answer» (b) is a correct answer |
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| 4. |
log (sec x + tan xsec x + tan x(C) tan x |
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| 5. |
\begin{array} { l } { \text { 6. } \sqrt { 1 + 2 \tan x \cdot ( \tan x + \sec x ) } } \\ { 7 \cdot \tan x \cdot \tan 2 x \cdot \tan 3 x } \end{array} |
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| 6. |
\lim _ { x \rightarrow 0 } \frac { \tan x + 4 - \tan 2 x - 3 \tan 3 x } { x ^ { 2 } \cdot \tan x } |
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| 7. |
\tan ( x - y ) = \frac { \tan x - \tan y } { 1 + \tan x \tan y } |
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Answer» tan(x-y) = sin(x-y)/cos(x-y) = (sin x.cos y - cos x.sin y)/(cos x.cosy + sin x.sin y) Divide the numerator and the denominator by cos x.cos y tan(x-y) = (sin x/cosx- sin y/cos y) / (1 + sin x.sin y/cos x.cos y) = (tanx- tan y)/(1+ tan x.tan y) Why are u posting formulas what is ur question, are u asking for the proof of the formula |
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| 8. |
\lim _{x \rightarrow 0} \frac{\tan x+1-\tan 2 x-3 \tan 3 x}{x^{2} \cdot \tan x} |
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| 9. |
की हि + cosec20 = tan 0% “शक |
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| 10. |
\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \cdot \tan y} |
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| 11. |
13 x + 10 = - 3 x + 42 then the value of x |
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Answer» 13x + 10 = -3x + 4213x + 3x = 42 - 10 16x = 32 x = 2 |
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| 12. |
42. 12 मी० x 10 मी० के कक्ष की फर्श को पूरा ढकनेके लिए 10 से० मी० X 8 से० मी० की कितनीआयताकार पट्टियों की आवश्यकता है?(A) 12000 (B) 15000(C) 10000 (D) 18000 |
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Answer» 12x10=120मी2. 10x8=80सेमि120मी=12000सेमि120×100×100= 1200000÷80=15000B) 15000 12×10 =120 10×8=80 120=12000120×100×100=1200000÷80=15000 =15000 is answer the answer is 15000rectangular tiles |
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| 13. |
42. 12 मी० x 10 मी० के कक्ष की फर्श को पूरा ढकनेके लिए 10 से० मी० x 8 से० मी० की कितनीआयताकार पट्टियों की आवश्यकता है? |
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Answer» AREA =12*10=120m square=1200cm squarenow n* are an*80=1200n=1200/80=15 tiles क्षेत्र = 12 * 10 = 120 मीटर वर्ग = 1200 सेमी वर्गअभी वn * a हैंn * 80 = 1200n = 1200/80 = 15 टाइलें |
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| 14. |
1,31 1_1323. QďŹ.,4z+36+ X+23 |35 |
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Answer» 4 1/2 + 3 1/6 + x + 2 1/3 = 13 2/59/2 + 19/6 + x + 7/3 = 77/5x = 67/5 - (9/2 + 19/6 + 7/3) Take LCMx = (67*6 - 9*15 - 19*5 - 70)/30x = (402 - 135 - 95 - 70)/30x = (402 - 300)/30x = 102/30x = 17/5 = 3 2/5 |
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| 15. |
12. HCF of 3240, 3600 and a thirdnumber is 36 and their LCM is 24x 35 x 52 x 72. The third number is |
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Answer» 3600 = 2^4*3^2*5^23240 = 2^2*3^4*5H.C.F = 36 = 2^2 *3^2Since H.C.F is the product of the lowest power of common Factor, so the third Number must have (2^2 *3^2) as its factor.Since L.C.M is the Product of Highest power of Common Prime Factor, So the Third number must have 3^5 and 7^2 as its factor.Therefore, the third number = 2^2*3^5*7^2 |
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| 16. |
-1 %2B (3/11)*(5/6) %2B (5/13)*(6/15) |
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Answer» Kajal goal slaps hell PayPal well rappel woo slaps glacial |
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| 17. |
Find the difference?13÷(-6)-15÷(-7) |
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Answer» (-13÷6)+(15÷7)=30-91÷42-61÷42 |
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| 18. |
57/67 X 32/171 X 45/128 = ? |
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| 19. |
CAREER PATESURVIT-171me: 1:30 hoursONE STEP AHEAD.171 COORDINATE GEOMEM. Marks[5]The distance of the point (a, b) from origin is:(b) ? + b2(d) Vaz +6²Multiple choice questions:Que. (1) Multin(a)a + ba2 - b2 |
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Answer» option d is the correct option d is the correct answer |
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| 20. |
DatePage :5600171 |
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Answer» 3,274.8538011696 is correct answer two number=5600÷ 171=32.74853 |
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| 21. |
।!!! 1, तब ४ की माप बताइए।110°171१ |
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Answer» As l//m, x = 110° [corresponding angles] answer |
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| 22. |
Ifsin A = _ and cos B--, where 0 < A < 5,0 < B41, find the values of22() sin (A + B)(ii) sin (A -B)cos (A+B) ) cos (A-B)COS |
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| 23. |
Given that tanewhat is the value of5cosec 0-sec 0cosec20+ sec20 |
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| 24. |
Solve for x and y, given x0, y02 2 1 23 |
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| 25. |
fruits were ioSthobita bought 6.75 m fabric for her dress. She used 3.45 m for herself. How much fabric is lefther? |
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Answer» Total fabric left = 6.75 - 3.45 = 3.30 metres |
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| 26. |
Order Thinking Skills (HOTS)- -Gupta bought 45 metres of fabric. She used 38 CS to make 13. cumains and manraining to make four cushion covers? How such fabric i mares was used a custic |
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Answer» length of the fabric =45 1/4length of fabric she took tomake 12 curtains. = 38 1/3length of fabric used to make 4 cushions =45 1/4 - 38 1/3so, convert it into improper fractions = 181/4 - 114/3now find the LCM of the denominator of the fractions which is 12 now bring 12 as the denominator by multiply it by a number gives 12 and with the same number multiply the numerator alsolike this,=181×3/4×3 - 114×4/3×4= 543/12 - 456/12now both are like fractions so subtract it= 543-456/12=87/12so, 87/12 metre fabric is used to make 4 cushion,but the question also says that you have to find length of fabric used to make 1 cushion so we have to divide 87/12 from 4so, 87/12 ÷4 = 87/12÷4/1reciprocal = 87/12 × 1/4 =87/48now convert it into mixed fraction= 1 39/48therefore, 1 39/48 metre fabric was used to make 1 cushion |
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| 27. |
19. Prove that :(sin + cosec 0)2 + (cos ยง + sec 8)2 = 7 + tan? 6 + cot2 6. |
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| 28. |
19) Prove that (eusetu12 sin θ20) If cos θ + sin θy2 cos θ, show that cos θ-sin θ2 |
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| 29. |
point of Contac,then find the value of sin 9 + cos219/ If sin θ-cos θ |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 30. |
Today is Wednesday,what will be the day after 84 day? |
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Answer» 84 is exactly divisible by 7. So after 84 days, it'll be a Wednesday. |
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| 31. |
कि की पा अर अक <र्ु््िनि(DY 6 ६15 15% (7-०) ०० हू ६/फेदन्णकक e 4 20 () at =) |
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Answer» a = 2 +√31/a = 1/(2+√3)=2-√3a - 1/a = 2√3(a - 1/a)^3 =(2√3)^3=24√3 |
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| 32. |
19. If sin A--, cos B--, find A, B and sin 24. |
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Answer» SinA= sin30= 1/2 A= 30°Cos B= cos60° = 1/2 B= 60°Sin(2*30°)= sin60° = √3/2 |
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| 33. |
Draw a triangle ABC with AB-6 cm, BC-7 cm and CA- 8 om. Using ruler and compssalone, draw () the bisector AD of A and (i) perpendicular AL tróm Aon sc5Measure LAD |
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| 34. |
the roots of the following quadratic equations by factorisation.(0 x2-3x- 10 0() V2x+7x+ 5/2-0(v) 100x-20x+1 0(i) 2r+x-6 0(iv) 2x-x+01V |
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Answer» hand writting should be meet easy to understandable to answer it |
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| 35. |
(m) v2x(v) 100x-20+1-0mblems given in Ex |
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Answer» 100x^2 -20x + 1 = 0100x^2 -10x -10x + 1=010x(10x-1)-1(10x-1)=0(10x-1)(10x-1)=0x=1/10,1/10 |
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| 36. |
V2x +3Evaluate the limit: ;lim |
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| 37. |
Plot few points on the grample 6: Champa went to a Sale' to purchase some pants and skirts. When herfhends asked her how many of each she had bought, she answèred. "The number ofskirts is two less than twice the number of pants purchased. Also, the number of skirtsis four less than four times the number of pants purchased". Help her friends to findhow many pants and skirts Champa bought. |
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Answer» Very long |
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| 38. |
hu |
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Answer» Please post the full question tnx |
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| 39. |
(7)153-93 -631 Cu iei. |
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| 40. |
Amit has shirts and pants such that thetotal is less than 10. If the number of shirtsis 2 more than the number of pants, findthe maximum number of pants he couldhave |
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Answer» Let the number of shirt be x and pant be y => x + y < 10 now if x = y +2 then y+2 +y < 10=> 2y + 2 < 10=> 2y < 8 => y < 4 maximum number of pant can be 3 |
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| 41. |
diameter 3.5 cm, have been cut out. rmuarectangular ground is 90 m long and 32 m broad. In the middle ofthe14 m12. A32 mground there is a circular tank of radius 14 metres. Find thet ofcossquare metre.tering the remaining portion at the rate of 50 per90 m |
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Answer» thank you |
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| 42. |
1. Fand the total number of permutations of the letten of cach of the woreL, nd the total number of permutations of tho letterathe worsiven below) APTLE(iv) INITUTE) ARRANG(V)ENGINEERING(U) COMMEIC(vi) İNTERMEDIAIE |
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Answer» 1) 5!/2!=5*4*3=602)7!/2!=7*6*5*4*3=25203)8!/(2!*2!*2!)=(8*7*6*5*4*3)/(4)=50404)9!/(2!*3!)=(8*7*6*5*4)=67205) 11!/(3!*3!*2!*2!)=(11*10*9*8*7*6*5*4)/(6) =11088006)12!/(2!*2!*3!)=(12*11*10*9*8*7*6*5*4)/2 =39916800 |
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| 43. |
(рдкрд╛) 25274 L 29 Gk 2l0a 2% g S8 e N = |
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| 44. |
cos-1/n+5-5Et cosecha)AtsSoPovethat 'dy 20ce |
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| 45. |
(M.D.U. 1991)(Reproduce property 2, Arn. 10.10cro, ()- sinsin ()cos 019. IrShow that(MD.U. 1994; G.N.D.U. 1987)- sin nco. n- sin nCE.HU |
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| 46. |
) sin 60째 cos 30째 + sin 30째 cos 60째et |
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Answer» sin60=√3/2 cos30=√3/2sin30=1/2 cos60=1/2hence√3/2*√3/2+1/2*1/2=3/4+1/4=4/4=1 |
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| 47. |
- G |sin 27째e et scos 63째)cos 63째sin 27째 ) |
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Answer» (sin27/cos63)*(sin27/cos63) + (cos63/sin27)*(cos63/sin27) = (sin27/cos(90-27))*(sin27/cos(90-27)) + (cos63/sin(90-63))*(cos63/sin(90-63)) = (sin27/sin27)*(sin27/sin27) + (cos63/cos63)*(cos63*cos63) = 1 + 1 = 2 If you find this answer helpful then like it. |
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| 48. |
If Sn is denoted the sum of first n terms of an AP , proved that S12 =3(S8 -S4) |
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Answer» like if you find it useful Thanks |
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| 49. |
(3) Determine the value of x +y if\left[ \begin{array}{cc}{2 x+y} & {4 x} \\ {5 x-7} & {4 x}\end{array}\right]=\left[ \begin{array}{cc}{7} & {7 y-13} \\ {y} & {x+6}\end{array}\right] |
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Answer» thank you |
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| 50. |
1. The marked price of a pant is 1,250 and the shopkeeper allows a discount of 8%on it. Find the discount and the selling price of the pant. |
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