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. |
(i) 300 rupees (11)laved deposited 12000 rupees at 9 p.c.p.a. in a bank for some years, and withdrew hiinterest every year. At the end of the period, he had received altogether 17,400 rupeeFor how many years had he deposited his money? |
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Answer» Amount which is received by Javed is Rs. 17400 and he invested Rs. 12000 at 9% per annum. Interest he received is 17400 - 12000 = Rs. 5400 Using formula Pnr = 5400 12000× n× 9/100 = 5400 n = 5 years |
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| 2. |
hen prove thet Jano v23-COS 2 8tan p |
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| 3. |
intersect each other at O. If LPOT = 75°, find thevalues of a, b and c4b[Board Term L, 2016, Set-JQ22L5C] |
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| 4. |
Evaluate :() 7PDyoue thet 10p |
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Answer» 7P4=7×6×5×4=42×20=840 |
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| 5. |
. Prove thet Stinc cosotSelco - tano |
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| 6. |
23Given Acompute A-1 and show that 2A-1 91-A.-4 7 |
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| 7. |
. Kabir invested a sum of r 12,000 at 5% per annum at compound interest. Heafter n years. Find the value of r1 interest of & 410 in 2 years |
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Answer» Formula to find the amount in case of compound interest: A= P * {1+ (r/100)}^n; where A is the amount, P is principal,r is rate,n is the number of years. Putting these values in the formula: 13230 = 12000 * {1 + (5/20)}^n=> 13230/12000 = (105/100)^n Simplifying the equation, we get: 441/400 = (21/20)^n(21/20)^2 = (21/20)^nTherefore, n=2 Number of years = 2 |
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| 8. |
(2) Find the quadratic equation whose roots are the reciprocals of the rootsof + 4x - 10 = 0.et c.&B be the |
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Answer» x=-5, x=-1 is the best answer x^2+4x-10=0as per shridhar acharyax= -b+-√b^2-4ac/2ahere a= 1, b= 4, c= -10x= -4+-√(4)^2-4×1(-10)/2×1 = -4+-√16+40/2 = -4+-√56/2 = -4+-2√14/2 = 2(√14-2)/2= +-(√14-2)roots are (√14-2) & (2-√14)reciprocal of root =1/√14-2 & 1/2-√14sum of root= 1/√14-2 + -(1/√14+2)= √14+2-√14+2/(√14)^2-(2)^2=4/14-4=4/10=2/5product of root= (1/√14+2){-(1/√14+2)=-{1/(√14)^2-(2)^2= -1/14-4= -1/10eq=x^2-x(sum of root)+(product of root)=0x^2-x{2/5)+(-1/10)=010x^2-4x-1=0 X=-5. X=1 is the answer of this question |
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| 9. |
22. In the adjoining figure, OPQR is a square. A circle oxdrawn with centre O cuts the square in X and Y. YProve that QX QY. |
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| 10. |
A = \left( \begin{array} { c c c } { 1 } & { 2 } & { 2 } \\ { 2 } & { 1 } & { 2 } \\ { 2 } & { 2 } & { 1 } \end{array} \right) , \text { show thal } 1 ^ { 2 } - 4 \Lambda |
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| 11. |
Prove that 7 \log 16 / 15+5 \log \frac{25}{24}+3 \log \frac{81}{80} = log 2 |
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| 12. |
*29. Show thal1 4 7then prove that (A + B)= A' + BIfA=1258]4 21 |
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| 13. |
find -ha leaf no twhich must be achsd to leaskgisqvare. find tls sq root o4 thal ra |
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Answer» thus 489 to be added to 100000 to make it a perfect square |
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| 14. |
If α, β are roots of the equation x2 + 5x + 5 0, thenwrite on quadratic equation whose roots are α + 1 and |
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| 15. |
G hetween them is 18 cm. Find the length of each of the parallel sides.r than the other by 6 cm, find the two parallel sides.sides ishe area of a trapezium is 180 cm2 and its height is 9 em. If one of theongn. If one of the parallelea the nther. If the area of the |
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| 16. |
The equation x2 + 4x + 3=0 has roots α and β. Findthe equation whose roots are (1 + α/β ) and (1+β/α) |
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| 17. |
(i)J(sinx + cos x) dx |
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Answer» thanks sin2x/2+cos2x hota hai |
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| 18. |
4. In each of the following assumethat the base is 10. Prove(i) log()+10g(j)+loger log()-loge,-0.(ii) log360 3log2 2log3 + log5(iii)log ( 5) = log2-2log5-10g3-210g7.(iv) logi I 0) log(100)ă10g( 1 000) + log( 10000) 10.(v) 510g3-log9-log27. |
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Answer» please like my answer if you find it useful |
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| 19. |
F Gnza . (ena=zCnat®ieaSive. . नस_&S_a_l__‘[ (o a I l ‘[fl na |
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Answer» Please like the solution 👍 ✔️👍 |
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| 20. |
18. If log 2 0-3010, and log 3 0 4771, find the value of() log 6 () log 24 (i) log 5 |
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Answer» part 3 log 5 ?? |
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| 21. |
tay44If ω is imaginary cube root ofunity, then sin! (ω 13 +ω 20)π +-t is equal to |
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Answer» thnks |
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| 22. |
\frac { 3 x } { 4 } = 21 - 3 |
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Answer» 3x/4 = 183x= 18*4x= 72/3=24 |
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| 23. |
EXAMPLE 12 The diagonals of a parallelogram ABCD intersect at O A line through Ointersects AB at Xand DC at Y. Prove thal OX - OY |
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| 24. |
7. If β is perpendicular to both a and where α kand y 2i + 3j + 4k, then what is ß equal to?(a) 3i 2j (b)-3i +2j ( 21-3 (d)-21 +3 |
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| 25. |
-6x+21=-3 |
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Answer» -6x + 21 = -3-6x = -3 -21-6x = -24x = (-24)/(-6)x = 4 |
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| 26. |
0.04 0.080.08-0.120.12-0.160.16-0.200.20-0.2499240.90.20 7 20,1 니9. Find the coordinates of the centroid of a triangle whose vertices are (x,y(ち,与). |
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| 27. |
21/3 |
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Answer» 21/3 =7. hoga because 21 ko 3 se devide karenge to 7 aayega |
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| 28. |
| यदि 0 इकाई का काल्पनिक मूल है तो निम्नलिखित का मान निकालें ।[If o is an imaginary cube root of unity, then evaluate the follow[Hint: I+0+0=0]| - 0टिनाला ओटी में है जो तिलिवित का मान निकाले ।। |
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Answer» | 1 w w² || w w² 1 || w² 1 w | R(1) --> R(1) + R(2) + R(3) | 1+ w+ w² 1+ w+ w² 1+ w+ w²| | w w² 1 || w² 1 w | 0 0 0|| w w² 1 || w² 1 w | So, 0 is answer |
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| 29. |
x ^ { 3 } - 9 x ^ { 2 } + 23 x - 15 = 0 \text { whose roots are in } A.P. |
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| 30. |
. If sinx cOS Xcos x is equal to |
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| 31. |
45. Evaluate:log 8x log90) log 27 |
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Answer» Please hit the like button if this helped you |
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| 32. |
ea n |
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Answer» x1^3+4x1-80=0compare it with ax^2+bx+c=0so D=root(b^2-4ac)=root(16+320)=root(336)= 4*root(21)x1=(-b+root(D))/2a=(-4+4root(21))/2=-2+2*root(21)orx1=(-b-root(D))/2a=(-4-4root(21))/2=-2-2*root(21) |
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| 33. |
5. Evaluate : log 27. |
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Answer» Answer is 6Use property of logRoot 3 can be written as 3 power 1/2 and 27 can be written as 3 power 3Use property u will get the answer |
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| 34. |
andmintheena-thand also itsea |
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Answer» Area= 1/2 x diagonal 1x diagonal 2= 1/2*16*12= 12*8= 96cm^2 AREA=d1×d216×12×1/28×12=96 it's area |
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| 35. |
i chin ea |
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Answer» An integer (pronounced IN-tuh-jer) is a whole number (not a fractional number) that can be positive, negative, or zero. Examples of integers are: -5, 1, 5, 8, 97, and 3,043. Examples of numbers that are not integers are: -1.43, 1 3/4, 3.14, .09, and 5,643.1. The set of integers, denoted Z, is formally defined as follows: Z = {..., -3, -2, -1, 0, 1, 2, 3, ...} In mathematical equations, unknown or unspecified integers are represented by lowercase, italicized letters from the "late middle" of the alphabet. The most common are p, q, r, and s |
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| 36. |
। ताह11% 20 1 0D L kb |
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Answer» उत्तर : वाहन चलाते समय आपातकालीन स्थिति में प्रतिक्रिया करने से रुकने तक की दूरी "रुकने की दूरी" कहलाती है। |
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| 37. |
Convert ea3/8 |
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Answer» 0.375 is the answer of 3/8 3/8 = 0.375 |
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| 38. |
cos ω(x-pcos o)xcos ω + ysin ωp(sino + cosh,orsin ωorxcos (o + y sin ωp.Hence, the equation of the line having normal distance p from the origin and anglewhich the normal makes with the positive direction of x-axis is given byExample (i1) Find the equation of the line whose perpendicular distance from theorigin is 4 units and the angle which the normal makes with positive direction ofx-axiis 150Solution Here, we are given p 4 andω-50 (Fig 10.18).cos 150 V3 +1 |
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| 39. |
If o is an imaginary cube root of unity, then (1+w-)?equals-[2002] |
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| 40. |
Evaluate:tan 35°cot 55°cot 55ta3Sec 40tan 35°tan 35cosec 50。220 |
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| 41. |
) Evaluate :/а.xdx-Ń |
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| 42. |
10. Evaluate ſxcos xdx. |
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Answer» Plzz like my ans🙏🙏 |
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| 43. |
2 tan A8. Given: sec A= 29 , evaluate Asia1, evaluate Asin A-tan Acosec A |
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| 44. |
Evaluate log xdx |
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| 45. |
correctlyest form:4/5 x 3/7 |
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Answer» 4/5 * 3/7 = 4*3/5*7 = 12/35 Aap hum ko apna whatsapp numbers dijiay please |
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| 46. |
PoyawkolpollaD-SHT=Kb+kqD-2=19+kb |
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| 47. |
1gb=?kb |
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Answer» the answer of your question is 1gb=1000kb |
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| 48. |
1. Find the value of (-1)2002 (-1)2003x |
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Answer» this is ringm |
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| 49. |
Solve $ \sin 2 x-\sin 4 x+\sin 6 x=0 $ |
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| 50. |
tan(theta)^2 %2B cot(theta)^2 %2B 2=sec(theta*(cosec^2*theta))^2 |
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