Explore topic-wise InterviewSolutions in Current Affairs.

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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

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2.

coSI(9)log cos x(11) e(15) sin cos (sinx)

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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.

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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.

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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

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5.

ही. = __———.COtA _ 1+ secA cosecAl1—cotA 1-tanAs

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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.

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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.

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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

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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.

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10.

4. Given 15 cot A = 8, find sin A and sec A.

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U HAVE CLASS 9 SCIENCE OR MATHEMATIC QHESTION PAPER 2016 OR 2014

11.

4. Given 15 cot A8, find sin A and sec A.

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12.

In the given fig., find the value ofx and y.55 0 125°

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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)

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14.

In the given fig., find the value of x and y.55125

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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

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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.

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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

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18.

4. Given 15 cotA-8, find sin A and sec A.

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thanks

19.

which-98.term ofAP 12, 72,-3..-- is

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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

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21.

Show that a, a, , a6) a 3+4n. form an AP where a is defined as below:(ii) a 9-5m972

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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.

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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

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tq sis

24.

Given 15 cot A = 8, find sin A and sec A

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part 1

part 2

part 3

25.

Given 15 cot A 8, find sin A and sec A.

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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

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27.

63. Prove that: sinÂŽ (1 + tan ) + cos (1 + cot 6) i< = sec 0 + cosec 0. (CBSE2010] |

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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]

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29.

।. ड*-63+ 2-0 मगीकत़एनत वौजबाएसय(8) 2 दि. (८

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30.

sin® 63°+ sin” 27° .0 o 17° +cos” 73°

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31.

3. Evaluate:sin2 63+sin2 27°0 cos 17° + cos?7

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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

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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?

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34.

sin 47 + sin 61°-sin 11° - sin 25° is equal to

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(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

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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

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37.

4x?— 16x+ 15 =074

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38.

3. - 16x + 24x® – 12x® by 4x

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-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

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40.

-8x2 + 16x - 6 by 4x -2c.

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(-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

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41.

x²-16x+64-81

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(x-17) (x-1) is answer

(x-17) (x-1) is the best answer

42.

Vxim2x → 16 16x-xSP

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43.

The difference between the circumference and radius of a circle is37 cm. Using π , find the circumference of the circle. CBSE 2013

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44.

[x- 16x + 63 0

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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.

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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.

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47.

. Find the value of a in each of3 11 74 2 12

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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.

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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.

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50.

16x+14×=90 then x=?

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x=3 is the right answer