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. |
5 The sides BA and DC of a quadrilateral ABCD15.are produced as shown in the figure. If AB| DC,then(a) atx bty (b)a+bx+y(c) aty btx (d)axby |
| Answer» | |
| 2. |
coSI(9)log cos x(11) e(15) sin cos (sinx) |
|
Answer» Mark the appropriate question to be solved. No-7 No-7 Is it integral or derivation? If it is derivation, then this is the solution using Taylor expansion Yes Tq |
|
| 3. |
If the ratio of the sum of the first n terms of two A.Ps is (7n +1): (4n + 27),then find the ratio of their 9th terms. |
|
Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)We can consider the 9th term as the m th term.Let’s consider the ratio these two AP’s m th terms as am: a’m→(2)Recall the nth term of AP formula, an= a + (n – 1)dHence equation (2) becomes,am: a’m= a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].now substitute the value of m as 9so the answer becomes120/95 |
|
| 4. |
If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n+27nthen find the ratio of their 9th terms. |
|
Answer» Answer. Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27) Let’s consider the ratio these two AP’s mth terms as am : a’m →(2) Recall the nth term of AP formula, an = a + (n – 1)d Hence equation (2) becomes,am : a’m = a + (m – 1)d : a’ + (m – 1)d’ On multiplying by 2, we get am : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’] = [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1 : S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)] = [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23] Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23]. So, the ratio of 9th term=14(9)-6: 8(9)+23120:95=24:19 hit like if you find it useful |
|
| 5. |
ही. = __———.COtA _ 1+ secA cosecAl1—cotA 1-tanAs |
|
Answer» Like my answer if you find it useful! |
|
| 6. |
If the ratio of the sum of the first n terms of two A.Ps is (7n+ 1): (4n+27),then find the ratio of their 9th terms. |
|
Answer» hit like if you find it useful |
|
| 7. |
Q.28) If the ratio of the sum of the first n terms of two A.Ps is (7n + 1):(4n+27),then find the ratio of their 9th terms. |
|
Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27) We can consider the 9th term as the m th term. Let’s consider the ratio these two AP’s m th terms as am: a’m→(2)Recall the nth term of AP formula, aₙ= a + (n – 1)d Hence equation (2) becomes,aₘ: a’ₘ= a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getaₘ: a’ₘ= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].now substitute the value of m as 9so the answer becomes120/95. |
|
| 8. |
y - 9 = 72 |
|
Answer» y= 81y-9=72y= 72+9=81 y-9=72y=72+9y=81 This is the correct answer y=81y-9= 72y=72+9=81 y=81 is the right answer y= 81 is the correct answer of the given question y-9=72y=72+9y=81this is best anst y=81 is the right answer Y = 81 is the answer |
|
| 9. |
4. Given 15 cot A 8, find sin A and sec A. |
| Answer» | |
| 10. |
4. Given 15 cot A = 8, find sin A and sec A. |
|
Answer» U HAVE CLASS 9 SCIENCE OR MATHEMATIC QHESTION PAPER 2016 OR 2014 |
|
| 11. |
4. Given 15 cot A8, find sin A and sec A. |
| Answer» | |
| 12. |
In the given fig., find the value ofx and y.55 0 125° |
|
Answer» Sum of angles in a triangle is 180° 55° + x + y = 180° The angles in a linear pair is 180° 125° + y = 180° So, y = 55° Substituting in the first equation, we get 55° + x + 55° = 180° x = 180° - 110° x = 70° |
|
| 13. |
In the given fig, find the value of x.(x +20) |
| Answer» | |
| 14. |
In the given fig., find the value of x and y.55125 |
|
Answer» y=180°-125°y=55°x =180°-(55°+55°) =180°-110° =70° |
|
| 15. |
17. From the given fig., find the value of20(a) sin LABC(b) tan x - cos a + 3sin x912 |
|
Answer» thanks for sending the answer still I have kept questions see and send me the answers I will keep some questions you can send me the answers |
|
| 16. |
The sides BA andDC of quad. ABCD areproduced as shownin the given Fig. Find the value of x. |
| Answer» | |
| 17. |
18. From the given fig., find the value of3(a) 4 sin x(b) 6 tan x(c) 4 cos x 15 sin y -tan xbn |
| Answer» | |
| 18. |
4. Given 15 cotA-8, find sin A and sec A. |
|
Answer» thanks |
|
| 19. |
which-98.term ofAP 12, 72,-3..-- is |
| Answer» | |
| 20. |
The sum of sh and 9th terms of an AP is 72 and sum of 7th and 12th terms is 97. Find the APA DRC |
| Answer» | |
| 21. |
Show that a, a, , a6) a 3+4n. form an AP where a is defined as below:(ii) a 9-5m972 |
|
Answer» i) a(n) = 3+4na(1) = 7a(2) = 11a(3) = 15so common difference is 4 , an AP therefore. ii) a(n) = 9-5na(1) = 4a(2) = -1a(3) = 9-15 = -6So common difference is -5, an AP therefore |
|
| 22. |
If 15 cotA -8 then find the value of sin' A. |
|
Answer» cot A = 8/15cosec^2 A = 1+ cot^2 A = 1 + (8/15)^2 = 1 + 64/225 = 289/225sin^2 A = 1/ cosec^2 A = 225/289 |
|
| 23. |
value of Sin4 +cos Asin A -cos AGiven, 12 cotA - 8, the value ofis1) 74) 17 |
|
Answer» tq sis |
|
| 24. |
Given 15 cot A = 8, find sin A and sec A |
|
Answer» part 1 part 2 part 3 |
|
| 25. |
Given 15 cot A 8, find sin A and sec A. |
|
Answer» Tnkuuu |
|
| 26. |
oo chords AB and eD o length 6om and 12em vespectively of acle ave lir s the ditance betuween As and CD3cm find thebetuweerncrrcle are lYradius ob the cirele |
| Answer» | |
| 27. |
63. Prove that: sinÂŽ (1 + tan ) + cos (1 + cot 6) i< = sec 0 + cosec 0. (CBSE2010] | |
| Answer» | |
| 28. |
If the radii of the circular ends of a bucket, 45 cm high are 28 cm and 7 cm (as shown in given fig),find the capacity of the bucket.IAI-2004] |
| Answer» | |
| 29. |
।. ड*-63+ 2-0 मगीकत़एनत वौजबाएसय(8) 2 दि. (८ |
| Answer» | |
| 30. |
sin® 63°+ sin” 27° .0 o 17° +cos” 73° |
| Answer» | |
| 31. |
3. Evaluate:sin2 63+sin2 27°0 cos 17° + cos?7 |
|
Answer» We know,cos x = sin (90 - x)sin x = cos (90 - x)sin^2 x + cos^2 x = 1 Then,sin^2 63 + sin^2 27/ cos^2 x 17 + cos^2 73= sin^2 63 + cos^2 (90 - 27)/ sin^2 (90 - 17) + cos^2 73= sin^2 63 + cos^2 63/ sin^2 73 + cos^2 73= 1/1= 1 Therefore its value = 1 |
|
| 32. |
3.114αν τα.Given 15 cotA= 8, find sin A and sec A |
|
Answer» CotA = 815 Or BP=815 Let B=8K and P=15k So in a right angle triangle with angle A P^2+ B^22=H^2 Or H=17K Sin A = P/H = 15/17 Sec A = H/B = 17/ cotA=815 is correct |
|
| 33. |
Which term of the AP 72, 63, 54,is 0? |
| Answer» | |
| 34. |
sin 47 + sin 61°-sin 11° - sin 25° is equal to |
|
Answer» (sin47 - sin25) + (sin61 - sin11) (2cos36 x sin11) + (2cos36 x sin25) 2cos36 ( sin11 + sin25) 4cos36 x sin18 x cos7 sin18 = (root5 - 1)/4 and cos36 = (root5 + 1)/4 so on solving we get, cos7. |
|
| 35. |
Find the coordinates of the point, where theline x-y 5 cuts Y-axis.CBSE 2013 |
|
Answer» if cuts y axis, x is 0 thenhence =0-y=5y=-5 coordinates are (0,-5) thanks |
|
| 36. |
2. In Fig. 8.13, find tan P- cot R.12 cm13 an. 3.. If sin A= 4, calculate cos A and tan A4, Given 15 cotA 8, find sin A and sec A.1311 the triconometric ratios.Fig. 8.13 |
| Answer» | |
| 37. |
4x?â 16x+ 15 =074 |
| Answer» | |
| 38. |
3. - 16x + 24x® – 12x® by 4x |
|
Answer» -4+6x^4-3x^2 is the correct Answer!! -4+6x⁴-3x² is the correct answer of the given question -16x+24x^5-12^3/4x-4+6x^4-3x^2 ans |
|
| 39. |
ty Tthe dis tonce betuweentheparallel |
| Answer» | |
| 40. |
-8x2 + 16x - 6 by 4x -2c. |
|
Answer» (-8x*x + 16x - 6 )/(4x-2) = -2(4x*x-8x+3)/(4x-2) = -2(4x*x-6x-2x+3)/(4x-2) = -2(2x-1)(2x-3)/(4x-2) = -2(2x-1)(2x-3)/(2*(2x-1)) = 3-2x If you find this answer helpful then like it. |
|
| 41. |
x²-16x+64-81 |
|
Answer» (x-17) (x-1) is answer (x-17) (x-1) is the best answer |
|
| 42. |
Vxim2x â 16 16x-xSP |
| Answer» | |
| 43. |
The difference between the circumference and radius of a circle is37 cm. Using π , find the circumference of the circle. CBSE 2013 |
| Answer» | |
| 44. |
[x- 16x + 63 0 |
| Answer» | |
| 45. |
20. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute[CBSE 2013]hand in 5 minutes. |
|
Answer» thank |
|
| 46. |
circle7. Achord of length 30 cm is drawn at a distance of 8 cm from the centre of[CBSE 2013]a circle. Find out the radius of the circle. |
| Answer» | |
| 47. |
. Find the value of a in each of3 11 74 2 12 |
|
Answer» 3/4 + 11/2 = 7/12 + a (3 + 22)/4 = 7/12 + a 25/4 - 7/12 = a (75 - 7)/12 = a a = 68/12 |
|
| 48. |
If 2 \cos ^{2} \theta+11 \sin \theta=7, then find the principal value of theta. |
| Answer» | |
| 49. |
36. If the radii of the circular ends of a bucket 28 cm high, are 28 cm andCBSE 2011)7 cm, find its capacity and total surface area. |
| Answer» | |
| 50. |
16x+14×=90 then x=? |
|
Answer» x=3 is the right answer |
|