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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. |
Ahe cF of 3318 and lelth at-oliv," des12S ! jq377theand is28 learina emainders 1, 2 and srespect ive'destnumber. |
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| 2. |
(2xx+3= o EI=2 -+x+3 D] ल5; दिया e Bx=-3,x== |
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Answer» 2(2x-1/x+3)-3(x+3/2x-1)=5or, {2(2x-1)²-3(x+3)²}/(x+3)(2x-1)=5or, {2(4x²-4x+1)-3(x²+6x+9)}/(2x²+6x-x-3)=5or, 8x²-8x+2-3x²-18x-27=5(2x²+5x-3)or, 5x²-26x-25=10x²+25x-15or, 5x²-10x²-26x-25x-25+15=0or, -5x²-51x-10=0or, 5x²+50x+x+10=0or, 5x(x+10)+1(x+10)=0or, (x+10)(5x+1)=0Either, x+10=0or, x=-10Or, 5x+1=0or, 5x=-1or, x=-1/5∴, x=10,-1/5 Like my answer if you find it useful! |
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| 3. |
e b रत पा; ६ L005" w78 (m) » ——— S "एक.| ए 95 d kil & st kolhie i lel=E 2802-०5 |
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Answer» 140 = 2×2×5×7 156 = 2×2×3×13 3825 = 3×3×5×5×17 5005 = 5×7×11×13 If you find this answer helpful then like it |
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| 4. |
For what value of'p' the following pair of equations has a unique solutionlel2py5 and 3x +3y6 |
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Answer» quadratic equations2x + py = - 5 and 3x + 3y = - 6have unique solution Then,a1/a2 = b1/b2a1 = 2, b1 = p, a2 = 3, b2 = 3 2/3 = p/36 = 3pp = 6/3 = 2 Value of p = 2 |
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| 5. |
Find leLThe lengthit, along[Hint : Let the length-2x m and the breadth = X m.and breadth of a park are in the ratio 2: 1 and its perimeter is 240 m. Aits boundary. Find the cost of paving the path at Rs 3 per m2.path 2 mtunsThen, 2(2x + x )-24040] |
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Answer» nice |
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| 6. |
\left. \begin{array} { l l } { x y z = 24 , \text { is } } \\ { \text { (a) } 3 } \\ { \text { (c) } 90 } \end{array} \right. |
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| 7. |
The sides of a triangular board are 13 metres, 14metres and 15 metres. The cost of painting it atthe rate of Rs. 8.75 per m2 is |
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Answer» Sides of the triangle are 13m,14m, and 15m. semiperimeter of the triangle = (13+14+15)/2 = 42/2 = 21. Area of the triangle = √21(21–13)(21–14)(21–15) m² = √ 21*8*7*6 m²= √ 7056 m²= 84m² Rate of painting = ₹8.75/m² Cost of painting the board = ₹8.75*84= ₹735. |
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| 8. |
whenWhat will be he remainder left21815 is divided by 12 |
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Answer» 875/4=218 so remainder will be 3 so hence when 21 power 875 will be 3 |
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| 9. |
\left. \begin array l \text (A) \\ \text (C) 90 ^ \circ \end array \right. |
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Answer» Option ASin theta = cos theta It can be written as sin theta = sin(90 - theta) theta = 90 - theta 90 = 2 theta theta = 45 degrees. |
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| 10. |
3.Find the respective terms for the following APs.(1a, = 2, az = 26 find a, (b) |
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Answer» the answer is 14 because the common difference was 12 14 is correct answer your question |
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| 11. |
\left. \begin array l \operatorname cos ( 90 ^ \circ - A ) = \\ ( \Delta ) \quad \operatorname cot A \end array \right. |
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Answer» Answer : B) sinAExplanation : |
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| 12. |
\left. \begin{array} { l } { 180 ^ { \circ } - 30 ^ { \circ } = } \\ { 180 ^ { \circ } - 90 ^ { 6 } } \\ { 180 ^ { \circ } - 110 ^ { \circ } = } \end{array} \right. |
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Answer» (i) 150°(ii) 90°(iii) 70° |
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| 13. |
\begin{array} { l } { \text { Find the average of all prime numbers } } \\ { \text { between } 60 \text { and } 90 ? } \end{array} |
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Answer» Prime numbers bet. 60 and 90 are 61,67,71,73,79,83,89first we will add these no. and then divide it by 7 because the prime no. between.60to90 is 7 so the answer would be 7.42. |
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| 14. |
9. A boy spendsof his pocket money and then of the remainder is given to his sister. If he has & 40of the remainder is given to his sister. If he has 40left, what did he have at first? |
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Answer» Let pocket money be = xX -3/4x is amount remaining with him after spending money. X -3/4x= x/4 (on solving) now he gave 4/5of remaining amount to his sister. = x/4 - 4/5 of x/4= x/4 - x/5= x/20amount remaining with him is =40x/20=40x=800his pocket money is 800 |
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| 15. |
18. After spending 80% of hisincome and giving 10%,f the remainder in a charita46260 left with him. Find his income. |
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| 16. |
EXERCISE 9.41. The price of 3 metres of cloth is Rs 79.50. Find the price of 15 metres of such cloth. |
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Answer» 397.50rs is the price of 15m 3 m cloth = 79.50 rs.1 m cloth = 79.50 ÷ 3 = 26.50 rs15 m cloth = 26.50 × 15 = 397.50 rs.therefore price 15 meters of such cloth is 397.50 rs. 397.5 is the exact answer |
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| 17. |
Find the relationship between a and b so that the function f definebyax + 1 if x 3bx + 3 if x>3x) |
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| 18. |
IF x^2-1 is a factor of ax^4+bx^3-cx^2+dx+e=0.Show that a+c+e=b+d=0 |
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Answer» if x²-1 is a factor of ax⁴+bx³+cx²+dx+e=>x = +1 and -1 are the rootsof ax⁴+bx³+cx²+dx+e Now if f(x) = ax⁴+bx³+cx²+dx+ethan f(+1)=f(-1)=0so a+b+c+d+e=0 ….….….(1)and a-b+c-d+e = 0 ….…....(2) Hence from (1) and (2),a+c+e=b+d=0 |
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| 19. |
Findtherelationshipbetween a and b, so that the function f defined byf(x) = {ax + 1, x<=3; is continuous at xE bx 3, x> 1 |
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Answer» f(3) = 3a+1 also f(3) = 3b+3 for continuous , both the values should be equal so, 3a+1 = 3b +3 => 3a=3b+2=> 3(a-b) = 2 => (a-b) = 2/3 |
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| 20. |
16. If (x + 1) is the HCF of (ax2 + bx + c) and(bx? + ax + c), then the value of c is(a) 0(b) 2(d) 3(c) 1 |
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Answer» If (x + 1) is HCF of (ax^2 + bx + c) and (bx^2 + ax + c) Then,For x = - 1 both these equations should be equal to 0 Hence,a - b + c = 0..... (1)b - a + c = 0......(2) Add eq(1) and eq(2)2c = 0c = 0 (a) is correct option |
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| 21. |
Bx. 3 : Represent the following complex numbersin the polar formi) 1+i ii) 4+4V3-i iii) -2 iv) 31 v) -1- i |
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| 22. |
ar2 + bx + c = 03 |
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| 23. |
(c) 981 240 |
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Answer» (235440 ) is answer of this question 235440 is the correct answer of the given question ☺️235440 is your answer. BYE 235440 is right answer 981× 240 ________ 000 39240196200__________235440your answer is 2,35,440 235440 is the best answer 980×240=355440rait an. 235440 is correct answer 235440 is the correct answer 235440 is correct answer 235440 is correct answer. |
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| 24. |
2. A transA triangle having a perimeter 56 and its side are 2x, 2x + 3 and 2x + 5. Find the respective lengths ofthe sides of the triangle. |
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| 25. |
54. एक पत्थर को कुएँ में गिराया गया। पत्थर के पानीकी सतह पर टकराने के15 सेकण्ड बाद छपांक(splash) का शब्द सुनाई दिया। कुएँ की गहराई है-(मान लो ध्वनि का वेग वायु में 327 mls हैं)(a) 277m(b) 491 m(c) 660 m(d) 981 m. |
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Answer» because of acceleration the distance is 660 m your answer is because of acceleration the distance is 660m because of acceleration the distance is 660 m |
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| 26. |
8. The cost of 15 metres of a cloth is 981. What length of this cloth can be purchased forて1308? |
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| 27. |
\left. \begin array l 981 \$ 961 \\ 90 = 121 \end array \right. |
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Answer» Find the prime factorization of 186186 = 2 × 3 × 31Find the prime factorization of 196196 = 2 × 2 × 7 × 7To find the gcf, multiply all the prime factors common to both numbers: Therefore, GCF = 2 Like my answer if you find it useful! The answer is of H.C.F is The answer of H.C.F. is 2 |
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| 28. |
What is the remainder left after dividing Ol +11 +21+3981 |
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| 29. |
DULODOThe cost of 15 metres of a cloth is981. What length of this cloth can be purchased for1308? |
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Answer» 20 miteres clothesright ans like this answer 20 m is the right answer 20 meter is the right answer the cost of 15 meter of a cloth is ₹981.what length ofof cloth can be purchased for₹1308? 15 metres = 9811 metres = 981/15=65.4 rsfor rs 1308,1 rs = 1/65.4 metres1308 rs = 1/65.4× 1308 = 20 metre |
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| 30. |
A2. A ramp going towards a stage makes an angle of 48° with respect to the ground. Then, findthe angle made by the ramp with respect to the stage. |
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Answer» 180 = 90 + 48+ aa = 42° 180 = 90 + 48+ aa = 42° |
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| 31. |
7m+19/2=13 |
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Answer» 7m + 19/2 = 13 7m = 13 - 19/2 7m = (26-19)/2 7m = 7/2 m = 1/2 |
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| 32. |
The cost of 1 metre of cloth is 13. Find the cost of 19/2 m of this cloth. |
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| 33. |
If the cost of 8 toys is 192 What will be the cost of 14 such toys |
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Answer» cost of one toy=192/8=24rupeeshence of 14 toys will be 336rupees |
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| 34. |
The cost of13 toys is 117. Find the cost of 10 such toys.3. |
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Answer» cost of 10 toys = 10*117/13 = 90 rupees |
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| 35. |
If the cost of 9 toys is t 333, find the cost of 16 such toys? |
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| 36. |
हि27; - सिद्ध 'कीजिए-Sin i 1+ Cos0SNV = 26g(1) 1+Cos0 _ Sind R |
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Answer» =(sinA/1+cosA) + (1+cosA/sinA )= [sin^2A + (1 + cosA)^2]/sinA(1 + cosA)= (sin^2A + 1 + cos^2A + 2cosA)/sinA(1 + cosA) = (1 + 1 + 2cosA)/sinA(1 + cosA) [since,sin^2A + cos^2A = 1] = (2 + 2cosA)/sinA(1 + cosA) = 2(1 + cosA)/sinA(1 + cosA) = 2/sinA = 2 cosecA [since 1/sinA = cosecA] = R.H.S. hit like if you find it useful |
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| 37. |
10 1r0- 224 then 2 CotoC) 2cos2 cosD) 2 cosA)B) 2cos θ244 |
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Answer» it's option 3 |
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| 38. |
48. lfcose + sin θ :2cos θthen cose-sia) 2cos θb) '/2sin θc)sin θd) cos θ |
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Answer» CosA +sinA =√2cosAsquaring⇒(cosA + sin A)² = (√2cosA)²⇒cos²A + sin²A+ 2sinAcosA = 2cos²A⇒1-sin²A+ 1 -cos²A+ 2sinAcosA= 2cos²A⇒2 - 2cos²A =cos²A + sin²A -2sinAcosA⇒2(1 - cos²A)= (cosA - sinA)²⇒cosA - sinA =√[2sin²A]⇒cosA-sinA =√2sinA |
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| 39. |
: e cos0.cotf_.71âsecâd |
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| 40. |
If x +x-2cos 0, then x3 +xs. "4) 2cot 361) 2sin 302) 2cos 303) 2tan38t uhioh dividen the line segment joinj |
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Answer» tq |
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| 41. |
Page Nolot or seco IItsinotsectosino -1 tseco |
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| 42. |
Cote |
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| 43. |
tanÂŽ-cote = |
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Answer» a^2=tan^2x+cot^2xb=tanx-cotxso b^2=tan^2x-2tanxcotx+cot^2xso a^2-b^2=tan^2x+cot^2x-tan^2x+2-cot^2x=2Hence Proved |
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| 44. |
5JA kite is flying at a height of 60 m above the ground. The string attached to the kitetemporarily tied to a point on the ground. The inclination of the string with the gronis 60°. Find the length of the string, assuming that therę is no slack in the string |
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| 45. |
Q3. A kite is flying at a height of 60 m above the ground. The stringattached to the kite is temporarily tied to a point on the groundThe inclination of the string with theof the string, assuming that there is noground is 60°. Find the lengthslack in the string. |
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| 46. |
Q3. A kite is flying at a height of 60 m above the ground. The stringed to the kite is temporarily tied to a point on the ground.The inclination of the string with the ground is 60°. Find the lengthof the string, assuming that there is no slack in the string. |
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| 47. |
prove that: tane tseco +1= sec0+ tanø |
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| 48. |
3 cos 8 + 2 sin eQ. 13. If 2 tane = 1, find the value of3. If 2 tane = l.me te vare 2cos 9-sine |
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| 49. |
Express cos0 in terms of tane. |
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Answer» Tana=sina/cosasquare both sidestan^2a=(1-cos^2a)/cos^2atan^ 2 a=(1/cos^2a)-(1)tan^2a +1=(1/cos^2a)cosa=1/√[tan^2 a+1] |
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| 50. |
ashowthatcseco-tane- - singItsino |
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Answer» 1 - sinx/ 1+ sinx , 1- sinx / 1 + sinx × 1-sinx/ 1- sinx= (1- sinx)^2/(1+sinx)( 1- sinx)= ( 1 + sinx^2 -2sinx)/ 1 - sinx^2= (1 + sinx^2 -2 sinx)/ cosx^2 = 1/ cosx^2 + sinx^2/ cosx^2 -2sinx/ cosx^2 = secx^2 + tanx^2-2secxtanx = ( secx - tanx)^2 |
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