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.

show that the lines x-1\2=y-2\3=z-3\4and x-4\5=y-1\2=z-0\1 intersect

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2.

41. 115 + 57 ~ 1 3 + 1का मान क्या है?4=(2) 139784ॐ| (4) 1189420

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Take l.c.m of 5,7,3,4=420So,after addition answer =1189/420(4) is the right answer

1189/420 is right answer

1189/420 is the right answer

3.

3. Convert each of the following into a decimal:...2516(115)(1) 156liv) 3524100100(iv) 1000(vi) 17(vii) 25into libre derimals:

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4.

i. Let Δ ABC . A DEF and their areas be, respectively, 64cm2 and 121 cm2. If15.4 cm, find BC.2. Diagonals of a trapezium ABCD with AB lI DC intersect each other at the poIfAB-2 CD, find the ratio of the areas of triangles AOB and COD.

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5.

The straight lines x +y-0, 3x + y-4-0, x+3y-4-0from aAJísosceles(D) none of these(B) equilateral(C) right angledints then locus of the po

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Since none of the lines are perpendicular therefore it is not a right angled triangle.

now find the intersection points of x+y=0, 3x+y-4=0, x+3y-4=0, we get(2,-2),(-2,2) & (1,1)

let A(2,-2) B(-2,2), C(1,1)

By distance formula we get AC=AB but not equal to AB

Therefore it is a isoscels triangle.

6.

4,-Find the angle between the lines xa and by+(10.

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The line x = a is a straight vertical line, and the line by + c = 0, if rearranged, is y = -c/b. Since there is no x variable in y = -c/b, it is also a straight line, but a horizontal straight line. If one line is going straight up and down, and the other left and right, and the angle between them is 90°

7.

In the given figure, Q> R. PS is the bisector of ZOPR, PM L QR. Find CMPS80°Q M s

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15° will be tha answer right

not sure what time you going to the gym today and tomorrow is a good time to go back to the gym now I'm going 🔙 to you later on tonight at work today and it is not a problem

8.

4.Find the angle between the lines x,, a and by + c-0

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9.

The diagonals of a parallelogram PQRS are along the lines x +3y 4 and 6x -2y 7. ThenPQRS is

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10.

X+2x+4

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3x+4 is the answer to this

and if you equate the equation to zero, you get the value of x as -4/3

11.

Solve:x-3=27 + 2x4,) 9x + 5 = 4(x-2) + 8

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1) 8x + 3 = 27 +2x6x = 24x = 42)9x + 5 = 4(x-2)+85x = -5x= -1

12.

In the given figure, PQ is a diameter of a circlewith centre O. If 4PQR 65, SPR 40and ZPQM 50° find ZOPR, Z0PM andPRS.40°659O 50

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13.

Fi 10 37,4 POR 100,whre P.sd are fpoints on a circle with centre O. Find zOPR

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14.

11. In the given figure. LPQR =120°, where P,Q and R are points on a circle with centreO. Then ZOPR is(A) 20°(B) 10°(C) 30°(D) 40°P120

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thank

15.

ejWhich angles are adjacent angles?ZNMK and LOPMNMK and LLMPZNMK and ZOPRZNMK and ZLMK

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angle NMK and angle LMK are adjacent angles

(4) option is correct

16.

हर्टc DX 8 C

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What is called Japan in Japanese language

17.

1. Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm2 and 121 cm2.15.4 cm, find BC.

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18.

Let Δ ABC ~ Δ DEF and their areas be respectively 64 cm2and 121 cm2. If EF 15.4 cm. Find BC.

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19.

Ans. ZIMC 1151? in the given Fig. 47, ZOPR LPOR and M and N are points respectively on sides QR and PR of APOR such thatOM- PN. Prove that OP 0Q. where O is the point of intersection of PM and ON

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20.

ADEF and their areas be respectively 64 cm2 and 121 cm2. If EF 15.4 cm.213.Letfind BC.AABC

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21.

2. In Δ ABC, DEIBC., If DE3 BC and area of Δ ABC-81 cm2 , find the area of Δ ADE.

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In triangle ADE and triangle ABC

angle A = angle A ( common )

angle ADE = angle ABC ( corresponding angles, DE is parallel to BC)

therefore, triangle ADE is similar to triangle ABC ( by AA corollary of AAA)

area of triangleADE / area of triangle ABC = (DE)^2 / (BC)^2

area of triangle ADE / 81 = (2/3 BC)^2 / (BC)^2

area of trangle ADE / 81 = 4BC^2 / 9 / BC^2

area of triangle ADE = 4 / 9 * 81

area of triangle ADE= 36 cm^2

it's not answer

22.

tan(32 )14. If x-200s 0-cos 2 e and y-2sin e-sin 2 θ, prove that aye-sin 26,provethatartan |30tCBSE2013]

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23.

26. Using integration, find the area of theregion in the first quadrant enclosed by theX-axis, the line y = x and the circlex" + y" = 32

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24.

1. Find the area of the region bounded by the curve y- z and the linesx 4 and the x-axis in the first quadrant.

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25.

Starting from ()*5, write various producteshowing some pattern to show(-1)x(1)-1

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26.

ली1 है? क्यों अथवा क्यों नहीं?8- का प्रतिलोम 0.3 हैक्‍या 8: का IR

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27.

PInFig. 10.37, < PQR =100, where P, Q and R arepoints on a circle with centre O. Find ZOPR.

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28.

2. Find the area of the region bounded by y9x, x 2x 4 and the x-axis inthefirst quadrant

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29.

Find the slope of a line (i) which bisects the first quadrant angle (iikie direction of /-axis measured anticl

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30.

LAB C ~ Δ PQR M is the mid-point of BC and B is the mid-point of QR. The area ofAABC-t 100 sq. cm and that of ΔPOR . 144 sq. cm. If AM cm, then find PN.

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31.

chord of a circle is equal to the radius of thecircle-. on the minor arc and also at a point on theFi. 10.37, PQR- 1009, where P. Q and R areP185angle subtended by the chord atmajor ar100ircle with centre O. FindOPRpoints on aFig. 10.37ㄣ. 10.38,L ABC = 69 , < ACB = 31°, findn FigZ BDCFig. 10.38 .

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32.

Find the area of the region bounded by xfirst quadrant.4y, y2, y4 and the y-axis in the

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33.

ar (PQR).ABCD is a rectangle. If AB 20 cm and ar (ABCD) 100 cm2find BC.7. 1AnsmFind the length

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area of rectangle ABCD = base × height =AB ×BC100 cm^2= 20 cm × BCBC= 5cm

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34.

2' 2(C)し22)2The equation of the circle which touches both the axes and the line+1 and lies in the first3 4quadrant is (x-c)2 + (y-c)2弌2 where c is(A) 1(B) 2(C) 4(D) 6

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35.

^ { 41 } C _ { 40 } ^ { 8 } C _ { 5 } , 11

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1

2

3

36.

a ^ { 3 } - b ^ { 3 } + 8 c ^ { 3 } + c

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(a)³ +(-b)³ +(2c)³ - 3(a)(-b)(2c) = (a-b+2c)(a²+b²+4c² +ab +2bc -2ca)

37.

< ig uRl pue g WIS AR[NOEd * 7 न 6509 भर!

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Cos theta = 3/5 means base= 3 and height is 5by pythagoras theoremP^2 + B^2 = H^2P^2 = 25 -9P = Root 16P= 4sin theta = 4/5tan theta = 4/3

38.

1. Write various uses of concave mirror in everyday life.

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Large concave mirrors are used to focus sunlight to produce heat in the solar furnace. They are also used in solar ovens to collect a large amount of solarenergyin the focus of the concave mirror for heating,cooking, melting metals, etc. These were some common uses of concave mirror in daily life.

39.

3 pusie 1 By be b 8 RET IR By pae) LT

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Ans :- बिन्दु की X-अक्ष से दूरी को उस बिन्दु का Y-निर्देशांक याकोटिकहते है।

40.

v3+!3 +12 का- हो, तो x2 +43.और yयदि x ==3-13+1मान है-

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x=√3+1/√3-1*√3+1/√3+1=3+1+2√3/3-1=2+√3and y=√3-1/√3+1*√3-1/√3-1=3+1-2√3/2=2-√3and x^2+y^2=4+3+4√3+4+3-4√3=7+7=14

41.

θ lies in the first quadrant and 5 tan θ-4, then

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angle is in first quadrant5tan=4so tan=4/5that means sin=4/root(41)so (5sin-3)/(5sin+2)=(5*4/root(41)-3)/(5*4/roo(41)+2)=(20-3root(41))/(20+2root(41))

42.

E is the mid-point of side QR of a ΔΡ0RD is the mid-point of PE when producedmeets PR in G. Prove that PG =-PR.3

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43.

3. Find the area of the region bounded by ' 4y, y 2, y 4 and the >-axisfirst quadrant.

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The area can be found out by the integration of x = 2√y , from y = 2 to y = 4

so the integration is 2y^(3/2)/(3/2) = 4/3y^(3/2)

now putting the limit we get area = (4/3)[4^(3/2) -2^(3/2)] = (4/3)[8-√8] = (32-8√2)/3 sq.units

44.

In figure, the line segment ST is parallel to side PR of ΔPOR and it divides the triangle into two partsPSPQofegual area. Find the ratio-

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45.

98037+ 9Sl BUE 5 [ elव.6 500९ न छ पाधटु &

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5 tan theta=4

So tan theta=4/5

therefore, sin theta/cos theta = 4/5

Hence we can say that,

sin theta= 4x , cos theta= 5x

(5*4x-3*5x)/(4x+2*5x)

5x/14x

= 5/14

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46.

जगावान काना 5 6 (रू) "सन हिरू2,6]3 L o NG . - N कि RS A (Sl S T

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Thank you

47.

L é-’y\,g-{ e atif o~ O{ o ESZL(Q;‘(J WShose $2deकं श्ग1y तिल -

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Area of square = (side)²Area of square = (20 m)² = 400 m²

48.

Fig. 5.382. In figure 5.39, □PORS and □MNRL arerectangles. If point M is the midpoint of side PRthen prove that, (i) SL LR, (ii) LNSQFig. 5.39

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49.

20% of a number is 1 more than 79. Findthe number

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let number be XhenceX*20/100+1=790.2x+1=790.2x=80x=400

thans

50.

~ 4 [Prove dted oppa; E Sl gL omSen 6 coodo | Cebele G R/ *‘a#/(;"’“: at Lo oenh9 el .

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