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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. |
£ Slफल p+AL |
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Answer» After cross multiplication21y+12= -4y-825y= -20y= -20/25= -4/5please like the solution 👍 ✔️ |
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| 2. |
o+l =3 बा,- Sl |
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| 3. |
cos30° *cos45° — sin30°* sin 45 |
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| 4. |
72*n - 52=71*n - 19 |
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Answer» 72x-71x= 52-19x= 33thanks |
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| 5. |
Is it right to say that cos(60° +30º = cos 60° cos30°-sin 60° sin 30°. |
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Answer» cos(60+30)= cos60cos30- sin60sin30; ((1/2)+(V3/2)) = =((1/2)(V3/2)-(V3/2)(1/2) ((V3/4)-(V3/4)) =0 Yes it is right to say |
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| 6. |
tEvehuale cos30 cos 45-sin 30 sin 45 |
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| 7. |
cos30tan60°sin45 COS30° |
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Answer» (1/3)(1/2)/(1/2)(1/√2) + √3/(1/2) = √2/3 + 2√3 thanks |
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| 8. |
(FRTC, 11-cos30°1 + cos30°ho = sec60° + tan60° |
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Answer» LHS : sqrt[(1 + cos30)/ (1 - cos 30)] = sqrt[(1 + cos30)^2/ (1 - cos^2 30)] = sqrt[(1 + cos30)^2/ (sin^2 30)] = (1 + cos30)/ sin 30) = cosec 30 + cot 30 = sec(90-30) + tan(90-30) = sec 60 + tan 60 = RHS Hence proved |
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| 9. |
. cos30-2cos40B esin 30 +2 cos 40 |
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| 10. |
72(vin, given an=4, d= 2, Sn=-14, find n and a.rt |
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Answer» an=a+(n-1)d=4or, a+(n-1)2=4or, a+2n-2=4or, a+2n=4+2or, a+2n=6or, a=6-2nSn=n/2[2a+(n-1)d]=-14or, n/2[2a+(n-1)2]=-14or, an+n(n-1)=-14or, (6-2n)n+n²-n=-14or, 6n-2n²+n²-n=-14or, -n²+5n=-14or, n²-5n=14or, n²-5n-14=0or, n²-7n+2n-14=0or, n(n-7)+2(n-7)=0or, (n-7)(n+2)=0either, n-7=0or, n=7or, n+2=0or, n=-2∵, n can not be negative ;∴, n=7∴, a=6-(2×7)=6-14=-8∴, n=7 and a=-8 Ans. Hit like if you find it useful |
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| 11. |
Evaluate the following.0 sin 45° +cos 45°sin 30°+tan45° -cosec 60°cot 45° +cos 60° -sec 30° |
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| 12. |
cos 30° cos 45° —sin 30° sin 45° |
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Answer» thnks |
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| 13. |
14. If one zero of a polynomial 3x2-8x + 2k + 1 is seven times the other, find the value of Ik.5 Solye the following pair of linear equations by substitution method: |
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Answer» 3x²-8x+2k+1 a=3, b=-8 ,c=2k+1suppose the zeroes of the equation are α and β . by the given condition α=7βas we know , α+β=-b/a7β+β=-(-8)/38β=8/3β=8/3×1/8=1/3αβ=c/a7β×β=2k+1/37×1/3×1/3=2k+1/37/9=2k+1/321/9=2k+121/9-1=2k12/9=2k4/3=2k4/6=k2/3=k |
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| 14. |
प्रश्न 2. x का मान ज्ञात कीजिए, यदि tan' (cr') = cot-4) |
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Answer» right answer is the2 the right ans of the following is +/- 2 |
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| 15. |
यदि(x+ik5 ए 2) =- (3.1; )5 x 3R y केमान ज्ञात।त कीजिएe Q \G |
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Answer» x + 1 = 3=> x = 2 y - 2 = 1=> y = 2 |
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| 16. |
x² - 8x + 16 |
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Answer» Use (a-b)² = a² + b² - 2abx² - 8x + 16x² - 2×(4x) + 4²(x-4)² |
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| 17. |
6. By how much must 3pq + 2p2 be increased to get5p9 - 3p2 +q2? |
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| 18. |
1. x> +8x+16 |
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| 19. |
sum of two integers is-30. If one of the integers is 15, determine the other.The difference of an integer a and-131s-4If the sum of-8 and a is-3, then find the additive inverse of aA submarine was 5827 m below sea level. If it ascends 2200 m,what is its new position?Keya opened a bank account by depositing 2500 in her account in May 2012. She deposited11 00 in the month ofJune 201 2 and withdrew乏1760 in the month of July 2012, Find her balance.. Find the value of a.TEGERS |
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| 20. |
S.T Ay wfe :(D SR i आर हरपी उधार 45% - आप 60१ - 1 वर 3 - ह बन्ल 45(%) § o 307+ 3 sin 607 - 2cosec’ 60° - 3 tan 30P (ii) sec 45° - cot* 45 |
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| 21. |
sec (45+)-cot (45-)Evaluatecos(45+)cosec(45°-oy1cos30"+sin Go+c05609+sin 30° |
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Answer» sec^2(45+A)-cot^2{90-(45-A)/cos^2(45+A) osec^2{90-(45-A) +√3/2+√3/2/1 + 1/2+1/2=sec^2(45+A)-tan^2(45+A)/cos(45+A)sec(45+A). + √3+√3/2 /2+1+1/2=1/cos(45+A)×(1/cos(45+A) +2√3/2/4/2=1/1+√3/44+√3/4 |
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| 22. |
= Then n72A) 45B) (45-y)D) 45 |
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| 23. |
\frac{\tan \left(45^{\circ}+\theta\right)-\tan \left(45^{\circ}-\theta\right)}{\tan \left(45^{\circ}+\theta\right)+\tan \left(45^{\circ}-\theta\right)}=\sin 2 \theta |
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| 24. |
3 Kg द्रव्यमान की एक वस्तु की गतिज ऊर्जाक्या होगी, यदि उसका संवेग 2 N S है\begin{array}{ll}{1 J} & {\text { (b) } \frac{2}{3} J} \\ {\frac{3}{2} J} & {\text { (d) } 4 J}\end{array} |
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| 25. |
sin30 + tan45 - cos 45 / sec30 + cos60-cot45 |
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| 26. |
Date'The third tenm ë¤ a GP.ěźproduct-ck14- |
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Answer» Suppose GP is a/(r*r) , a/r , a , ar ,ar*r,... Third term is 4 so a = 4 Product of first 5 terms = a/(r*r) * a/r * a * ar * ar*r = a*a*a*a*a = 3*3*3*3*3 = 243 |
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| 27. |
51—x+l,y—3—j=[3—'3—j ; तो + तथा ज्ञात कीजिए।g 3 |
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| 28. |
DHow much puse alcohal must be addedto 200 ml of a 15% solution to make itsStrength 32% |
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| 29. |
23. Find the ratio of the volume of a cube to that of a sphere which will fitinside it. |
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| 30. |
PA2328. Find the Coordinates of Point P on AB such thatwhere A(3,1) and B(-2,5) |
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| 31. |
Fit Cr4. Ifthe area of a circle is 50.24 m2, find its circumference.rumference is 22 cm. |
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| 32. |
1 the cumst nth man AP s qiventhe-Ap and-ーSn-sna-ndetermineind-ns |
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Answer» S1=3-1=2S2=3x4-2=10S3=3x9-3=24S4=3x16-4=44and so on the AP isa1=S1=2a2=S2-S1=10-2=8a3=S3-S2=24-10=14So the common diff is 6 and first term is 2the AP is 2,8,14,20,..the 25th term=a1+24d=2+24x6=146 |
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| 33. |
Subtract the sum of -1032 and 878 from - 34.at 124 from the cum conna 07 |
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Answer» 120 is the correct answer of the given question |
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| 34. |
2. varallel de U. CUM22). The'angles of a triangle are in the ratio 2:3:4. Determine the three angles2- |
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Answer» 2:3:4=180. so,2+3+4=9. 180/9=20 2*20=40. 3*20=60. 4*20=20.total sum is 180 the answer is total sum is 180. and20 |
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| 35. |
21. (k â 1)x - y =5,(k +1)x + (1 - k)y =(3k +1). |
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| 36. |
4 * ८०5 (4 + 2) - 2008 4+ ००8 दर को कथा 3. |
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Answer» sin(A-B) - 2 sin A + sin(A - B)/cos(A + B) - 2 cos A + cos (A-B) = 2sinA cosB - 2sin A/ 2cosA cos B - 2cos A = 2sin A(cosB - 1)/ 2cos A(cos B - 1 = tan A Hence proved |
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| 37. |
If \frac{13 k-1}{6}+\frac{2 k+5}{3}=\frac{5}{9}, then the possibel value of 'k' is _____ |
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| 38. |
(5) ( 4x - 5) + (3- 2x) j ug3u3. (2x +2,2x-2,3x- 3) |
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Answer» thanks |
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| 39. |
4. Determine the ratio of the volume of a cube to that of a sphere which will exactly fitinside the cube. |
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| 40. |
Percerrage10Ssculred45.1,550 and earns a profit of 81-%. Find the cost price of each radio.Tom sells 4 radios forfit of 1000 He also sells some copies of |
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Answer» SP=1550profit=8.33%so.CP=1550/1.0833=1430(approx)so per radio =1430/4=357.5 |
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| 41. |
anuualFind the hate dnterert |
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Answer» A = p(1+r/100)⁴ 256/81 p = p(1+r/100)⁴ 256/81 = (1+r/100)⁴ 1+r/100 = fourth root of 256/81 1+r/100 = 4/3 r/100 = 4/3-1 = 1/3 r = 100/3 r = 33.3% |
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| 42. |
कु 9४JF&E5rsb |
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Answer» A=1/2 b*hb=5cmh=4.5cmA=1/2 5*4.5A=2.5*4.5A=11.25cm² |
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| 43. |
A simple interest offrom a bank. Find the original principal, he invested for 6 years.(c)7.200 was given to Mr. Joshi with 8% pa. after 6 years |
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Answer» SI = ₹ 7200N = 6 yrsR = 8% SI = (PNR)/1007200 = (P × 6 × 8)/100 P = (7200 × 100)/6 × 8 = 15000Principle = ₹ 15000 |
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| 44. |
Sm DAGCD im which AB-5.6.com, Cum..LABS2B= 105 LD 80 |
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| 45. |
f = x2 y i-2xz j + 2yz k , find (i) div f ,(Kumaun 2008)e 15 |
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| 46. |
6) 7/11 of all the money in Hari's bank acccant is Rs. 77,000. How muchmoney does Hari have in his bank account? |
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Answer» 7/11×100=700/11=63.63 approximately |
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| 47. |
) In the figure, the sides of a rectangle aregiven. The lengths are in cm. Find the lengthand breadth of the rectangle.4x +272x+y+23x-2+32x+3y + 4 |
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| 48. |
4. The area of a rectangle is given by 6x^2y +4y^2x andand the width of the rectangle by 2xy. Find thelength of rectangle.ck |
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Answer» area is 6x^2y+4xy^2=2xy(3x+2y)width of 2xy so length is 3x+2y sir can you send photo of solution |
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| 49. |
In a class of 40 pupils, 35 % are girls. How many girls are there inthe class? How many of the pupils are boys? |
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| 50. |
13. Two equal circles intersect in P andQ. A straight line through P meetsthe circles in A and B. Prove that |
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