Explore topic-wise InterviewSolutions in Current Affairs.

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.

6 Radha has a circular ring with a diameterof 49 cm. What is the circumference of thecircular ring?

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Diameter is 49 cmSo, radius is 49/2 cmCircumference is 2× pi × rcircumference = 2 × 22/7 × 49/2Circumference = 154 cm

2.

17. The weight of a ring is 2y gm, that of a necklace is (3y- 4) gm and that ofabangle is (2y+7)gm. Ifthe total weight is 45gm, find the weight of each

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3.

da + β)0, then sin (α-β) can be reduced toSe(b) cos 2ß(c) sin a. (d) sin 2o.(a) cos β(d) sin 2αANSW

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sorry but I cannot understand show another method sir

4.

ti (Construct a quadrilateral ABCD in which AB-54cmAC 4cm.)C45cm, CD = 4.3cm, DA+ 3.5cm and diagonal

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5.

s.Draw a circle with the help of bangle. And find the center of circle.

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1

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3

6.

5.Draw a circle with the help of bangle. And find the center of circle.

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1

2

3

7.

the rational nuolve for x, and3x + 2 / x = 2 /3

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9x+6=2x7x=-6so x=-6/7

8.

2. The given figure shows a circle with centreO. P is mid-point of chord ABShow that OP is perpendicular to AB

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9.

10. By selling 130 cassettes, a man gains an amount equal to the selling price of 5Find the gain per cent.

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10.

lling 130 cassettes, a man gains an amount equal to the selling price of 5 caseFind the gain per cent.

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Let the cost price of each cassette be rs.10cost price of 130 casts = 130*10=1300by selling , he gained sp of 5 cassettestherefore , cost price of 130 cassettes = selling price of 125 cassettessp of 5 cassettes = 1300/125×5= 52he gained Rs. 52 on selling 130 cassettesgain℅= gain/total cp * 100= 52/1300*100= 4℅

11.

By selling 130 cassettes, a man gains an amount equal to the selling price of 5 cassettes.Find the gain per cent.

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12.

(B) यदि p(9) =)-2 और qy)=2-1 त p) केज्ञात कीजिए।

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13.

3. The following figureshows a circle withcentre O.If OP is perpendicularto AB, prove that AAP = BP.

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to proof: AP = BPBy const.=draw radius from A to O and from B to OProof : In ΔAOP and ΔBOP AO = BO(radius of circle) angleAPO=angleBPO(each=90°) OP=OP (common) ΔAOP≈ΔBOP(by SAS) by cpct AP=BP

14.

px + qy = p - qqx - py + p+ q

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15.

Subtract the sum of a (a +2b + c) and-b(a-b+2c) from c (-a-b+c)

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16.

Q: If A + B = 2C and C + D = 2A, thenA) A + C = B+DB) A + C = 2DC) A+D = B+CDA+C = 2B

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A) A+C+= B+D is the right answer

17.

श् 3 टू पट हिटी _ \D QY— eAe . की के - A %

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10+15+30+55+68+38=235

18.

26.. State and prove the converse of the Pythagoras Theorem.onts from a noint X to the circle with

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STATEMENT :- In a triangle if the square of a side is equal to the sum of the squares of other two side then the angle opposite to the first side is a right angle..

19.

Fig. 8.31Vthat the line segments joining the mid-points of the opposite sides of aquadrilateral bisect each otherthrough the mid-noint M of hypotenuse AB

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20.

25. In the adjoining figure, PQR, is a right triangle, right angled at Q. X and Y aron PO and QR such that Px : xo 1:2 and QY: YR 2:1. Provee the pointsthat 9(PY2 XR) 13 PR2

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In triangle XRQ XQ²+QR²=XR²(2/3PQ)²+QR²=XR²4/9PQ²+QR²=XR² (1)BY Pythagoras THEOREM IN TRIANGLE PYRPQ²+QY²=PY²PQ²+(2/3QR)²= PY² (2)BY PYTHAGORAS PQ²+4/9QR²=PY²IN TRIANGLE PQRPQ²+QR²=PR² (3)PYTHAGORAS THEOREMAdding 1 and 24/9QR²+QR²+4/9PQ²+PQ²=XR²+PY²13/9 QR²+ 13/9 PQ ²= XR²+PY²1/9(13QR²+13PQ²) = XR²+PY²13(QR²+PQ²) =9(XR²+PY²) FROM 3 WE GET 13PR²=9(XR²+PY²)

21.

MATHEMATICAL RE1. Factorisation of xy- pq + qy -px is5.

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xy-pq+qy-px=xy-px+qy-pq=x(y-p)+q(y-p)=(x+a)(y-p)

hit like if you find it useful

22.

(B) Solve the following questions. (Any Two)1. The given figure shows the[4 Mameasures of a Joker's cap. Howmuch cloth is needed to make such21 cma cap ?10 cm

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Radius = 10 cm height = 21 cm

l =√r² +h²l =√10² + 21²l = √100 + 441l = √541l = 23.25 cm

cloth required = Curved surface area of one conical cap = πrl= (22/7)×(10)×23.25= 73.07 cm ²

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23.

10. By selling 130 cassettes, a man gains an amount equal toFind the gain per cent.ual to the selling price of 5 cassettes

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Let the cost price of each cassette be rs.10cost price of 130 casts = 130*10=1300by selling , he gained sp of 5 cassettestherefore , cost price of 130 cassettes = selling price of 125 cassettessp of 5 cassettes = 1300/125 * 5= 52he gained Rs. 52 on selling 130 cassettesgain℅= gain/total cp * 100= 52/1300*100= 4℅

24.

26. एक कम्पनी तीन तरह के क्रमिक बट्टे प्रदान करती है।company offers three types of successive discounts:(i) 25% और 15%(ii) 30% और 10%(iii) 35% और 5%

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option (ii) is the correct answer

ecvc🍈ufjyvj no CR DD hi Chris X

thanks jkr a Xdxd VCUXX schooluv j

option (iii) is the correct answer

ii) 30%&10% is the correct answer

option 2nd is the correct answer

2 is the right answer

25.

Ans.(1)4 If A and B are two sets such that n(A) 32,1n(B)- 28 and n(AB)-50, then n(A nB) will he(1n 10

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26.

56. A trader off[ Ans : 32901ers successive discounts of 20% & 15% what is the total discount a customer

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let the actual price be X

so, first discount is 20% , so the price becomes x-(20%ofx) => x-0.2x = 0.8x

now of the price of 0.8x , another 15% discount is there.. so, again the price drops to

= 0.8x -(15% of 0.8x) = 0.8x-0.12x = 0.68x

so, total discount on initial price is = (x-0.68x)×100%/x = 0.32*100 = 32%

27.

olve the following simul1. (a) x +y 10-у 12

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Adding the two equations2x=22 x=1111+y=10 y=-1

28.

6240 का 80% > 225 का 12% >> ?

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29.

Find x in terms of a, b and c:2cx#a,b,c

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given,a/(x-a) + b/(x-b) = 2c/(x-c)[a(x-b)+b(x-a)]/(x-a)(x-b) = 2c/(x-c) (x-c)[a(x-b)+b(x-a)] = 2c(x-a)(x-b)

ax^2 – 2abx + bx^2 - acx + 2abc – bcx = 2cx^2 – 2bcx – 2acx + 2abc ax^2+ bx^2 - 2cx^2 = 2abx – acx – bcx (a+b-2c)x^2 = x(2ab – ac – bc)(a+b-2c)x^2 - x(2ab – ac – bc) = 0 x[(a+b-2c)x - (2ab – ac – bc)] = 0 x = 0 or (a+b-2c)x - (2ab – ac – bc) = 0 x = 0 or (a+b-2c)x = (2ab – ac – bc) x = 0 or x = (2ab – ac – bc) / (a+b-2c) thus the 2 roots of the given equation are x = 0 and x = (2ab – ac – bc) / (a+b-2c)

30.

Find x in terms of a, b and cb 2c+, x#a,b,c13

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31.

olve the following:1) Find the value of x2xx6#x5 for x

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Value will bex⁽²⁺⁶⁾/x⁵x⁽⁸⁻⁵⁾x³

32.

(7) P(x)-x2+x-7 d P(-1)=

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P(x) = x^2 + x - 7

Then,P(-1) = (-1)^2 + (-1) - 7 = 1 - 1 - 7 = 0 - 7 = - 7

33.

asses1ve that the perpendicular at the point of contact to the tangent to a circle pthrough the centre.nnt from a noint A at distance 5 cm from the centre of the circle is 4

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34.

54. The centre of a circle is (2x- 1,3x + 1) and radius is 10 i.Find the value of xnasses through the noint (-1

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Given diameter of the circle = 20 units

Radius of the circle = d/2 = 20/2 = 10 units

Given center of a circle is (2x - 1, 3x + 1)

Given that circle passes through (-3, -1),

So, distance of (-3, -1) from the center of the circle should be equal to radius

of the circle

Hence, √(2x + 2)² + (3x + 2)² = 10

Squaring on both sides, we get

(2x + 2)² + (3x + 2)² = 100

13x² + 20x - 92 = 0

13x² -26x + 46x -92 = 0

13x(x - 2) + 46(x - 2) = 0

(13x + 46)(x - 2) = 0

Hence, x = 2 or -46/13

the diameter of the circle =20 units

35.

In the figure, XYZ is an isosceles triangle right angled atY. PQRY is a square of side 10 cm and S is the point ofintersection of XR and PZ. Find the length of XR + PZ.2015KO

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as this is an isosceles triangle, YP and RZ eill be the midpoint of XY and YZ respectively so the triangle is symmetric across the line QSY hence XR = PZ so XY=2PY, Applying Pythagoras theorem we get XR= 10√5, so XR+PZ=20√5

आपका उत्तर बरोबर हैमुझे अच्छा लगा

36.

जे विज लि रा6. Prove that the lengths of the tangents from an external point to a circle are equalR PR S W ७... ४...

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37.

b+c gtr y+2 apr5. c+ar+pz + x = 2 lb qya+b p+q x+y crz

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38.

or3. The following figureshows a circle withcentre 0.If OP is perpendicularto AB, prove thatAP = BP.ARC D is mid-noint of BC: AD

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first draw oa and ob as tringle

39.

Ahmed buys a plot of land for? 480000. He sellsshould he sell the remaining part of the plot to gain 10% on the whole?32of it at a loss of 6% At what gain per cent5

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the answer is incorrect

Buy of land =4800002/5*480000=192000 rsloss=6%100-6/100*192000=180480

total gain =10%100+10/100*480000=528000

sp=528000-180480=347520

cp of remaining part of land =3/5*480000=288000

gain% of remaining part of land= sp-cp/cp*100

<=>347520-288000/288000*100=20.6 %

40.

Ahmed's father is thrice as old as Ahmed. After 12 years, hisage will be twice that of his son. Find their present age.

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41.

56 A trader offers successive discounts of 20% &amp;[ Ans : 32%)5% what is the toresells it in Delhi for Rs 61,200. If then laipur, How muc0000 fipm jaipur and nfrom jaipur andeipur, How mh profit did he make?

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42.

G&amp;1 - &amp; (&amp;)el /uQ Nb\e Efif / 7v 521घट =&amp;}| कि % -

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1.

(x - 7)³

x³ - 7³ - 21x ( x - 7)

x³ - 343 - 21x² + 147x

43.

60% = 1280%nb:2090:c. 25%d. 2

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44.

olveenbcqutio2cx -a x-b x

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45.

(p+7)(p-4)

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(p+7) (p-4)P^2-4p+7p-28P^2+3p-28

46.

nleach of the fonodius 6 cm. From a point 10 cm away from its centre, construct the pairof tangents to the circle and measure their lengthsf rodiue d cm from a noint on the concentric circle of

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47.

a circle with the help of a bangle. Take a point outsıde the circlé. Consgents from this point to the circle.struct thepair

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48.

H.C.F. $ p^{2}, p^{5,} \&amp; p^{7} $ is

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First do factorization

p^2 = p*pp^5 = p*p*p*p*pp^7 = p*p*p*p*p*p*p

HCF is highest common factor which isp*p = p^2

49.

what is the diameterofthepond?6.6 cm15. A piece of wire is bent in the shape of a square of eachside 6.6 cm. It is rebent to form a circular ring. What is thediameter of the ring?[Hint. Perimeter of square Circumference of circular ring,i.e., π d 4x6516.6 cm6.6 cm06.6 cm

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50.

(ii). SP =1755, gain = 12*1/2%(iv) SP =5600, loss = 6*2/3%

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The u

Thanks for the answers

Thanks guys for the answers

Hi