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.

(vi)2 In the adjoining figure, show that ABCD isa parallelogram.Calculate the area of llgm ABCDD 5 cmA 5 cm B

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2.

. A 15 m tall boy is standing at some distance from a 30 m tall building. The angle ofclevation from his eyes to the top of the building increases from 30° to 60° as he walkstowards the building. Find the distance he walked towards the building.

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3.

EXERCISE 1.11. Use Euclid's division algorithm to find the HCF of:(ii) 196 and 38220135 and 225

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4.

Euclid's division algorithm to find the HCF of:135 and 225(ii)196 and 38220

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5.

Euclid's division algorithm to find the HC(1) 135 and 225(ii) 196 and 38220

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6.

JUse Euclids division algorithm to find the HCF of135 an 22() 196 and 38220

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7.

Use Euclid's division algorithm to find the HCF of:(i)135 and 225(ii)196 and 38220

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By euclid's division algorithm225=135×1+90135=90×1+4590=45×2+0hence the hcf is 45Please like the solution 👍 ✔️

As38220 = 196 x 195 + 0

H.C.F = 196

Please like the solution 👍 ✔️

8.

1. Use Euelid's division algorithm to find the HCF(1) 135 and 225(ii) 196 and 38220

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9.

17. In a triangle ABC, the incircle (centre O)touches BC, CA and AB at points P, Q and Rrespectively. Calculate(i) ZQOR (ii) ZQPR;given that LA = 60°.

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10.

1.Use Euclid's division algorithm to find the HCF(i) 135 and 225(ii) 196 and 38220

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11.

1. Use Euclid's division algorithm to find the HCF(i) 135 and 225(ii) 196 and 38220

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12.

16 in the above right sided figure, a circle inscribed in triangle ABC touches its sides AB, BC andAC at points D, E and F respectively. If AB 12 cm, BC 8 cm and AC 10 cm, then find thelengths of AD, BE and CF

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13.

7.ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively.Show that AXperimeter of AABC.

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We get thatAX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2axtherefore1/2perimeter of triangle =ax

14.

6. The 4tevill7. ABC is a triangle. A circle touches sides AB and AC produced and side BC at X. Y and Z respectivelyShow that AXperimeter of AABC

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AX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2axtherefore1/2perimeter of triangle =ax .

15.

2. The ratio of the corresponding sides oftwo similar triangles is 5:3. Then theratio of their areas is[ ]1) 5:3 2) 3:5 3) 6 : 10 49 25:9

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ans will be 25:9 formula- base×hight/2

16.

7.ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively.Show that AXperimeter of AABC.perim-2

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AX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2AXtherefore1/2perimeter of triangle =AX

please put the fig

17.

Obtain all other zeroes of $ 3 x^{4}+6 x^{3}-2 x^{2}-10 x-5, $ if two of its zeroes are $ \sqrt{\frac{5}{3}} $ and $ -\sqrt{\frac{5}{3}} $

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18.

19. The number of chairs left over when1000 chairs are arranged in rows con-taining 48 chairs each

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let x be the number of chairs left over 1 row contains 48 chairs 1000 chairs =48/1000 =20.833 rows so consider 20 rows which contain 48×20=960 chairstotally 1000 chair so 1000-960=40 chairs are kept left over

19.

2.If two equations 5x + 3y + 7 = 0 andlx + my +9=0 are inconsistent, thenwhat is l: m?(A) 3:2 (B) 2:3(C) 5:3 (D) 3:5

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20.

\begin{array}{l}{\text { Obtain all other zeroes of } 3 x^{4}+6 x^{2}-2 x^{2}-10 x-5, \text { if two of its zeroes }} \\ {\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}}\end{array}

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21.

7. If two adjacent vertices of a parallelogram are (3, 2) and (1, 0) and the diagonals intersect at(2, -5), then find the coordinates of the other two vertices.

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thank you

22.

2. The value of 1- tan1S:1+tan 2 30°

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1 - tan² 30°-----------------1 + tan²30°

1 - (1/√3)²----------------1 + ( 1/√3)²

1 - 1/3-----------1 + 1/3

2/3-----4/3

1/2

23.

The value of1 1s- 10

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Please hit the like button if this helped you

24.

In which quadrant the point (-2,-5) lies

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25.

what is the mirror image of the point 2, 5 in x axis

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P (2,5)After reflection on x-axis, P' (2, -5).

26.

If coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) anddiagonals bisect each other at (2,-5), find coordinates of the other two vertices.OR

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Let the coordinates of C and D be (a, b) and (c, d)

∴ (3 + a)/2 = 2

⇒ a = 1

and (2 + b)/2 = –5

⇒ b = –12

Also

(c + 1)/2 = –5

⇒ c = 3

and (d + 0)/2 = –5

⇒ d = –10

∴ Coordinate of C and D are (1, –12) and (3, –10)

27.

Example 13 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinatesof A and B are (3, -5, 7) and (-1, 7, - 6), respectively, find the coordinates of thepoint C.

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28.

A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB + CD =B+ DA

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29.

DatePage+ 3.2L+9一皿一3.to-tifらWhat value ed Xand y

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now substitute the value of x in any of the equation and find the value of y

thanks its right way and right ans

but equation 2-1,then value of x=200/7

30.

A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB +CD=BC+DA

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From the figure we observe that,

DR = DS (Tangents on the circle from point D) … (i)

AP = AS (Tangents on the circle from point A) … (ii)

BP = BQ (Tangents on the circle from point B) … (iii)

CR = CQ (Tangents on the circle from point C) … (iv)

Adding all these equations,

DR + AP + BP + CR = DS + AS + BQ + CQ

⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC

31.

A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB + CD BC+DA

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32.

A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB +CD=BC + DA

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33.

A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB + CDBC+ DA

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34.

1. 8 ball point pens cost & 64 What is the cost of 1pen 2

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cost of one pen is 64/8 = 8 rupees

35.

1f2-2x-1 = 4, then the value of xisよ

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x=3, so 3 ki power 3 ka answer hai 27

36.

4x + 3)7(4x +3)

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37.

IVALLILLE8. The total cost of 24 chairs is 9255.60. Find the9255.60. Find the cost of each chair.oh shirts can be m:

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38.

7.Can altitude lie outside a triangle?

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Analtitudeof atriangleis the perpendicular segment from a vertex of atriangleto the opposite side (or the line containing the opposite side). Analtitudeof atriangle can bea side or maylie outsidethetriangle.

39.

15. Does the point (-2.5, 3.5) lie inside, outside or on the circle y25?

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40.

15. Does the point (-2.,5, 3.5) lie inside, outside or on the circle xy 252

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41.

प्रश्न 4. सिद्ध कीजिए :९sinº - cost 8sin-a-cos-a-1.

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42.

8 pens cost rupees 144 how much the cost of 14 pens?

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If cost of 8 pens = 144cost of 1 pen = 144/8Then,Cost of 14 pens = 144*14/8 = 18*14 = Rs 252

43.

One apple cost 1 rs then what is the cost of 8 apples

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cost of 8 apples = 8 * cost of one apple = 8*1 = 8 rs.

44.

The cost of 8 chairs is 2,120. Find the cost of 25 chairs.

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Thanks

45.

Find the equation of the line passing through the point of intersection of the lnes4xED,-3 = 0 and 2x-3y + l = 0 that has equal intercepts on the axes.

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46.

= cosec A + cot A, using the identity cosec A=丨cof A.coS A + sin A -1

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47.

1+ cot Acosec A+ cot A using the identity cosec Acos A+ sin A 1

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48.

tan_TF cos A = m cos B!2 m-

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Given Cos A = m Cos B

⇒ Cos A / Cos B = m / 1. ( apply componendo and dividendo)

⇒ (Cos A + Cos B) / (Cos A - Cos B) = ( m + 1) / (m - 1)

⇒ {(2 Cos ( A+B) / 2 × Cos (A-B) / 2 } / {(- 2 Sin ( A+B) / 2 × Sin (A-B) / 2 } = ( m + 1) / (m - 1)

⇒ {(2 Cos ( A+B) / 2 × Cos (B - A) / 2 } / {( 2 Sin ( A+B) / 2 × Sin (B - A) / 2 } = ( m + 1) / (m - 1)

⇒ {( Cot ( A+B) / 2 × Cot (B - A) / 2 } = ( m + 1) / (m - 1)

⇒ {( Cot ( A+B) / 2 = { ( m + 1) / (m - 1) } Tan (B - A) / 2

Here x = ( m + 1) / (m - 1).

49.

| TFd-='HCF. (48,72). Eind-thevalue-of-01

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Factors of 48

List of positive integerfactors of 48 that divides 48 without a remainder.

1, 2, 3, 4, 6, 8, 12, 16,24

Factors of 72

List of positive integerfactors of 72that divides 72 without a remainder.

1, 2, 3, 4, 6, 8, 9, 12, 18,24, 36

So thegreatest common factor of 48 and 72is24.

50.

cos A -sin A + 1cA Cosec A + cot A, using the identity cosec A-1+ cot A.

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LHS:Cos A - Sin A + 1/ Cos A + Sin A - 1 dividing in numerator & denominator with Sin ACot A - 1 + Cosec A / Cot A - Cosec A + 1now putting 1= Cosec^2 - Cot ^2

= (Cot A + Cosec A) - (Cosec^2 A - Cot^2 A) / (Cot A - Cosec A + 1)

= (Cot A + Cosec A) [1 - Cosec A + Cot A) / (Cot A - Cosec A + 1)]

= (Cot A + Cosec A)

= RHS

Hence Proved