Explore topic-wise InterviewSolutions in Current Affairs.

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1.

R17 If cosO+ sin0 = V2coso, show that cos 0 -sin 0 = 2

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cos o=/2 cos o-sinocoso-sino=/2coso_sino-sino =/2

It is correct answer

cosx + sinx=V2 cosx, then ( V2-1) cosx= sinx, multiplying both sides by V2+1, we get ( V2+1)( V2-1)cosx=(V2+1)sinx =cosx=V2sinx+ sinx; cosx-sinx=V2sinx, (V2+1)(V2-1)cosx=(V2+1)sin5x ;we know that; (a+b)(a-b)=a^2-b^2; ( V2)^2-(1)^2 cosx= ( V2+1)sinx; 2-1cosx=(V2+1)sinx, cosx=(V2+1)sinx;; cosx = V2 sinx + sinx; cosx-sinx= V2 sinx; cosx-sinx=+- sinx

it is correct answer

it is correct answer

2.

ㅣ요 In the given figure, PT = PS and TQ = SR. Prove that QS-RT.

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3.

101 (D) 14147 A SR5. & एके काम को 4 दिन में, छ 6 दिन में और ( 12 दिन मेंकरेंगे -() SL 3R

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PLEASE LIKE IT, IF YOU FIND THE ANSWER HELPFUL.

4.

21. Show thatcosec” O — tan“ (9 ) = sin 0 + sin (90° —0)

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5.

4 sin 0+3cose 607. If 4tan 0 3, then show that

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6.

6*x^2 - 7*x - 13=0*qquad - 78

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6x^2-6x+13x+13=0; 6x(x-1)+13(x-1)=(6x+13)(x-1); 6x=-13; x=-13/6; x=1

7.

\begin{array} { l } { \text { If } f ( x ) = x ^ { 2 } - 2 x + 3 , \text { then the value of } } \\ { \text { which } f ( x ) = f ( x + 1 ) \text { is } } \end{array}

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8.

\left. \begin array l \text If f ( x ) = \frac d x x %2B 1 , x \neq - 1 . \text Then, for what valu \\ \text is f ( f ( x ) = x \\ \text (a) \sqrt 2 \text (b) - \sqrt 2 \text (c) 1 \end array \right.

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a= -1

9.

\begin{array} { l } { \text { Using division algorithm, find the quotient and remainder } } \\ { \text { on dividing } f ( x ) \text { by } g ( x ) \text { where } f ( x ) = 6 x ^ { 3 } + 13 x ^ { 2 } + x - 2 } \\ { \text { and } g ( x ) = 2 x + 1 } \end{array}

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10.

1.Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then.Also, three years from now, I shall be three times as old as you will be." (Isn't thisinteresting?) Represent this situation algebraically and graphically.

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11.

. In the given figure, I l| m and t is a transversalIf 25 = 70°, find the measure of each of the angles41, 23. 24 and 28.

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12.

1. Aftab tells his daughter, "Seven years ago, I was seven times as old as you were thenAlso, three years from noy, I shall be three times as old as you will be." (Isa'tthinteresting?) Represent this situation algebraically and graphically

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13.

C2 The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum ofits first fifteen terms. (495)

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16th term = a + 15d3rd term = a + 2d

given

5(a + 2d) = a + 15d5a + 10d = a + 15d

4a - 5d = 0

4a = 5d

a = 5d/4

Also,

10th term = a + 9d = 41

a = 5d/4

so,

5d/4 + 9d = 41

5d/4 + 36d/4 = 41

5d + 36d/4 = 41

41d/4 = 41

41d = 164

d = 164/41

d = 4

a + 9d = 41

a + 36 =41a = 5

---------------------------------------------

n = 15a = 5d = 4

Sn = n/2(2a + (n-1)d)= 15/2 (10 + 14 * 4)= 15/2*66 = 495

14.

. In the given figure, I l|m and t is a transversal.If 15 70°, find the measure of each of the angles41, 23, 44 and 28.

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15.

2 Write 5 rational numbers which call Ue expioConvert each of the following into a fraction in its simplest form:a) 0.16b) 7.575c) 0.30l numhers:

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16.

given l=28,S=144,and there are total 9 terms.find a.

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Sum of n terms of AP= n/2(first term + last term)Given S = 144, l = 28, n = 9, a=?

Then144 = 9/2(a + 28)16*2 = (a + 28)a = 32 - 28 = 4

Value of a= 4

I got it tq

17.

x2,0Sx Sİ3x,35 x s10The relation fis defined by()The relation g is defined by g(x)-Show that /is a function and g is not a function.

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Observe the f have unique value for each x therefore a function.

But consider g at x = 2

it gives g(2) = 4

g(2) = 6

Therefore at x = 2 g gives two vaules, therefore not a function

18.

sing Lagrange's mean-value theorem, find a point on the curve ywhere the tangent is parallel to the line joining the points (1, 1) and (2,4),15. U[CBSE 2000C)

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thanks

19.

S) Draw a circle of radius 6cm. From a point 10 cm away from its centre, construct the pairof tangents to the circle and measure their lengths. Verify by sing Pythogoras Theorem.

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20.

6. Evaluaic Rt SR 5 Uo . If cosg +sing = cos then prove that cosg - sing = sing.. s gl ittt S

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21.

The relation f is defined by f(x) =3x,3 s x s 10The relation g is defined by g(x)x, 0 sxs2show that f is a function but g is not a function.

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22.

9. Is the function defined by22.0continuous at 1 Justify.

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wrong answer

23.

For what value of 2 is the function defined byf(x)-14x +1,if x >0

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24.

What is the mean of 1st ten prime numbers?

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First five prime numbers= 2, 3, 5, 7, 11, 13, 17, 19, 23, 29its mean= 2+3+5+7+11+13+17+19+23+29/10 =129/10= 12.9

25.

Write first four terms of the AP when the first term a and the common difference d aregiven as follows:(i) a 10, d 10(ii) a-2, d 0

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26.

22.In an AP, the sum of first ten terms is -150 and the sum of its next ten terms is550. Find the AP

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27.

he tUTTROL0 and the sum of its next ten terms is-550[CBSE 201045. In an A.P., the sum of first ten terms is-15Find the A.P.65. I

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28.

In an arithmetic progressionof 50 terms, the sum of firstten terms is 210 and the sum of last fifteen terms is 2565Find the arithmetic progression.

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Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P., an= a + ( n – 1)d

Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36= a + 35d

Sum of the last 15 terms = 15/2 [2a36+ ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

thank u sir

29.

The sum of the first five terms of an AP is 55 and the sum of the first ten terms of the sameAP is 235, find the sum of its first 20 terms.26.

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30.

7. In an arithmetic progression of 50 terms, the sum of first ten terms is210. And the sum of last fifteen terms of it is 2565. Find the

Answer»

Ans :-

Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P., an= a + ( n – 1)d

Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36= a + 35d

Sum of the last 15 terms = 15/2 [2a36+ ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

31.

28. In an AP, sum of the first ten terms is-150 arsum of its next ten terims is -550. Find the A

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32.

If the ratio of the 11th term of an AP to its 18h term is 2:3. Find the ratio of the sum offirst five term to the sum of first ten terms.

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answer was not match to the questions

33.

(iv) of theseIf a train covers 195 km in 3 hours, what distance will the train cover in 5 hours travelling at the same speed?

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thank you so much

34.

2. Find15th term of the AP with second term 11 and common difference 9.

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Given:a2=11d=9Formula; an=a+(n-1)da2=a+(2-1)911=a+9a=2a15=a+14da15=2+14(9)a15=128

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35.

bigger number and then count forward the smanumber. While adding 4 + 9 counting four numbers aeasier than counting nine numbers after 4.Let's practise Addition.the first number. Addition will be easy ifant forward the smaller9 four numbers after 9 is8+AS612+A15171213++2+6+4cटाटाऽauru N 61XAMMAMAMMA

Answer»

8+6 = 144+9=137+5=127+7=1412+4=1615+5=2017+2=1912+6=1813+4 =1711+7 =18

8+6=144+9=137+5=127+7=1412+4=1615+5=2017+2=1912+6=1813+4=1711+7=18

14131214162019181718is the correct answer

mark as best please and also leave a like

8+6=144+9=137+5=127+7=1412+4=1615+5=2017+2=1912+6=1813+4=17

36.

hesquare of first number is equal to the cube of second number. The numbers are) 9, 393) 8,4(d) 4, 2)6,3

Answer»

If you find this solution helpful, Please like it.

37.

If A= {3, 4, 5, 6, 7} B={ 1,6, 7, 8, 9) finds (A),n(B), n (AUB) and CanB) and establishthe relationship among these.

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38.

Find the median of first ten prime numbers.

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median

39.

3.Find the mean of first ten odd numbers.

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1+3+5+7+9+11+13+15+17+19this is an arithmetic serieswith the help of summation formula =n/2[2a+(n-1)d]10/2[2*1+(10-1)2]5[20]=100Mean of first 10odd numbers =sum of all the frequency/total number of frequencies =100/10=10

40.

Find the mean of the first ten natural numbers.

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41.

White first four terms of the AP, when the first term α and the common differece daregiven as follows(ii)a»-2,d-o(iii) a"4, d=-3() a-1.25, d--0.25Efferencc

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42.

,2: Find the median of the first ten natural numbers.

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find the median of first 10 natural numbers

43.

Prove that the sum of n terms of an AP in which first term acommon difference d and last termI, is given by

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Derivation for the Sum of Arithmetic Progression,SS=a1+a2+a3+a4+...+an

S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d]→ Equation (1)

S=an+a(n−1)+a(n−2)+a(n−3)+...+a1

S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n-1)d]→ Equation (2)

Add Equations (1) and (2)2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)

2S=n(a1+an)

S=n(a1+an)/2.

44.

A train is running at a speed of 75km/hr. In how much time will thetrain cover 350 km at the samespeed ?(JNV, 2002)(a) 2:30 hrs (b) 3:30 hrs(c) 4:40 hrs (d) 3:40 hrs

Answer»

the answer is (c) 4:40 hrs . see I have solved this question in your previous question .

option C is correct answer.

c is the best answer of options

45.

11. Find the distance between two points A and B on opposite sides of a river, a surveyor runs alonga line AC I AB. By measurement, he finds AC = 200 m and Z ACB = 45°.

Answer»

tanx=AC/AB; Tan45=AC/AB 45=200/AB; 45 AB=200; AB=200/45=200/1=200 cm

46.

finds in this equation

Answer»

2u+at×t=2s; at^2+2u/2=at^2+u=t

(2u+at)t=2s; u+at^2=s=at^2+u

s=ut+1/2at² is the correct answer

47.

man is standing on a railway bridge which is 180 mlong. He finds that a train crosses the bridge in 20 sec-onds but himself in 8 seconds. Find the length of thetrain and its speed.A. 36B. 54I. AC. 67D. 76

Answer»

Sol. Let the length of the train be x metres.Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 seconds.:. X/8 = (x + 180)/20 20x=8(x +180) x=120:. Length of train = 120m.Speed of train = (120/8) m/sec = 15 m/sec = [15 X 18/5 ] Kmph

= 54 Kmph.

48.

\begin{array}{l}{2 \frac{1}{8}+3 \frac{1}{6}=} \\ {\text { a) } \frac{127}{24}}\end{array}

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49.

I COULSLOWThe zeroes of the quadratic polynomial.? 99.7 + 127 are:(1) both negative

Answer»

Here zero of this polynomial will beLet given quadratic polynomial be p(x) =x2+ 99x + 127.On comparing p(x) with ax2+ bx + c, we geta = 1, b = 99 and c = 127

so

with the help of dharacharya formula-99+-√(99^2)-4*1*127/2

= -1.3 and -97.7

50.

(c) The ratio between the sum of n terms of twoarithmetic progression is (7n +1): (4n +27) Findthe ratio of their 11th terms.V: 70

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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)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].hence take m=11hence14*11-6/8*11+23=148/111