Explore topic-wise InterviewSolutions in Current Affairs.

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1.

9. 42x4811.294 x 206

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2.

500+30+206/100+9/1000

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500 + 30 + 206/100 + 9/1000

In decimal form

530 + 2.06 + 0.009

532.069

3.

48. यदि p=5-206 है, तो p+13 ज्ञात कीजिए।

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4.

EXERCISE 2.2d the zeroes of the following quadratic polynomials and verify the relationshipzeroes and the coefficients.12-2.x-8(i) 4s2 -4s妙12-15(ii) 6x-3-7x(vi) 3x2-x-4ith the uiven numhers as the sum and p

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5.

e S S B¢ e(o) Blbald 1 Q8 o) Be MBERD (0TI = OVe sl b= 0V ok ki) '3 nRhBRkk AOAV "8६८

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6.

ho givenOv2

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Step-by-step explanation:

∠PQR=90° (Angle in the semi circle is right angle)

Also,from triangle PQR, we have

∠RPQ+∠PQR+∠PRQ=180° (angle sum property)

∠RPQ+90°+∠PRQ=180°

∠RPQ=90°-∠PRQ (1)

Also, ∠BPR=90°(angle made on the tangent) (2)

Now, ∠BPR=∠RPQ+∠QPB

90°=∠RPQ+∠QPB

From (1), we get

90°=90°-∠PRQ+∠QPB

∠QPB=∠PRQ

Hence proved

7.

) % the natfo off the sumos the Kest 0 fewms ofw0 AP and & Grotn (५० नणे श rhen fTnd the xatloob-thoty . 9265.

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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

8.

Ftnd Prve as between 1 ond3 by mean method

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9.

breadth 2 m are enioueu. Tower bed is surrounded by a path 5 m wide. The diameter of flower bed is 65 m. What isthe area of the path?ith g 8 m long string. Find the area where the horse can graze.

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Step-by-step explanation:

Sol:) Radius of flower bed = 33 m (Diameter/2=radius)..

Radius of flower bed and path together = 33 + 4 = 37m.

So, Area of flower bed (path together )= 3.14 × 37 × 37 = 4298.66 m2.

Area of flower bed = 3.14 × 33 × 33 = 3419.46 m2

Area of path = Area of flower bed and path together − Area of flower bed = 4298.66 − 3419.46 = 879.20 m2....

10.

5. If length of a room is3 more than its breadth. Find the area of room if itsperimeter is 206 metre.

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Let the breadth be x so length is x+3perimeter = 2×(L+B ) = 206x+3+x = 1032x = 100x = 50So, breadth is 50m and length is 53m.

So, area is L×B = 50×53 = 2650 m²

thanks

11.

The area of four walls of a room is 51 m2.room is 5 m long and 3.5 m wide, find theheight of the roonm

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Area of four walls = LSA of room = 2 * h * (l+b) sq units

lenth=5m breadth = 3.2m,

Area of four walls = 2 *h * (5+3.2)

2*h*8.2=57.4m2

h=57.4/2*8.2

h=57.4/16.4

h=3.5 m

Cheers!

12.

(7) Find the breadth of a rectangular plot ofland, if its area is 440 m2 and the length is22 m. Also find its perimeter.

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Area of rectangle = length × breadth 440 = 22 × breadth breadth = 440/22 = 20 m Perimeter = 2 ( length + breadth ) = 2 ( 20 + 22 ) = 2 × 42 = 84 m

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13.

18. Prove that thecentre of a circle touching two intersecting lines lies on the angle bisector of thlinessin e-h cos θ = n, prove that az + 1,--m" + n-

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14.

Draw a pair of intersecting lines. Label andmeasure the vertically opposite angles.20.

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let the point of intersection be O .then vertically opposite angle are FOI and HOI , FOH and GOI

15.

Draw a pair of intersecting lines PQ andRS and mark the point where the twointersecting lines meet as O.3.

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16.

8. Two intersecting lines form angles of 123and 57°. What is the measure of the othertwo angles?

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measure of other two angles are also same as 123° and 57° ... ( vertically opposite angles)

17.

Find the angle between the two intersecting lines. Also find The condition ofPerpendicular and. Parallelism twolines,

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18.

13. Find the number of bricks, each measurincm × 12.5. cm × 7.5 cm requiredconstruct a wall 6m long 5 m heigh andm thick, while mortar occupies 5% ofvolume of the wall

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Volume of the wall = 6*5*0.5 =15 m3

1/20th of volume = 15*1/20 = 3/4 m3

So the volume to be covered by bricks =15−3/4=60−3/4=57/4m3

Volume of each brick =25/100*12.5/100*7.5/100=0.00234375m3

Number of bricks=57/4/0.00234375=6080

19.

angle,thenshoone of the four angles formed by two intersecting lines is a rightthat each of the four angles is a right angle.

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20.

13. Find the number of bricks, each measuringcm x 12.5 cm x 7.5 cm required toconstruct a wall 6m long 5 m heigh and o.5m thick, while mortar occupies 5% of thevolume of the wall

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Bricks Wall

l=25 cm l=6 m =600 cm

b=12.5 cm b=50 cm

h=7.5 cm h=5 m =500 cm

volume of wall=l x b x h

=600 x 50 x 500

=15000000 cubic cm

volume of mixture =1/20 x 15000000

=750000 cubic cm

volume left =15000000 - 750000

=14250000 cubic cm

required brick =14250000/25 x 12.5 x 7.5

=6080 bricks

21.

8. A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an aaof 22 m2, find the width of the varandah.

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22.

thehowelbe.. A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies anof 22 m2 find the width of the varandah.

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23.

the flower bed8. A room 5 m long and 4 m wide is surrounded by a verandah. Ifthe verandah occupies anof 22 m2 find the width of the varandah.

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24.

Problem 7.1The following concentrations wereobtained for the formation of NH, from Nand H2 at equilibrium at 500KIN,1 1.5 102M. [H2l 3.0 102 M andINHJ 1.2 10 M. Calculate equilibriumconstant.

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25.

inePHofa solution dependsconcentration.onITS| A) Hydronium ionsB) Hydride ionsC) 0xide ionsD) Hydroxyl ionsAnswer Key : AYour Response: Not AnsweredRaise Objection(Question No. 98)Question No. 99तीन पाइप P, Q और R, अलग-अलग एक टैंक को 10,15 और 20 घंटे में भर सकते हैं। P को सुबह 7 बजे, Qको सुबह 8 बजे और R को सुबह 9 बजे खोला गया। टैंककिस समय तक पूरा भर गया, यदि R एक बार में केवल3 घंटे की अवधि तक काम कर सकता है और उसके बादउसे 1 घंटे के लिए बंद करना पड़ता है?A) 12:30 बजे सायंB) 12:00 बजे सायंD) 12:12 बजे सायंc) 1:00 बजे सायंThree pipes P, Q and R can fill a tank in 10, 15 and20 hours separately. P was opened at 7 AM, Q at 8AM and R at 9 AM. At what time was the tankcompletely filled if R can work for only 3 hours at astretch and needs a 1 hour break?B) 12:00 PM| A) 12:30 PMC) 1:00 PMD) 12:12 PMAnswer Key : AYour Response: C (Wrong)Raise Objection(Question No. 99)Question No. 100परमाणु का द्रव्यमान, परमाणु केमें होता है।A) नाभिकB) इलेक्ट्रान क्लाउडC) रिक्त स्थानD) प्रोटॉन क्लाउडThe mass of an atom resides inof the

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26.

The sides of a triangular field are 51 m, 37 m and 20 m. Find the number of rose beds that can be prepared in the field, if each rose bed occupies a space of 6 sq.m

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Area of triangular field =a×b×c= 51×37×20m square=37740m squareNo. of flower beds =37740÷6=6290

The side of a triangular field are 41m,40m,and 9m. Find the number of rose beds can be prepared in the field ,each rose bed on an average needs 900cmsquare space

27.

Multiple Choice Questions (MCQs)17. Two small circular parks of diameters 16 m and12 m are to be replaced by a bigger circular park.What would be the radius of the new park if thenew park occupies the same space as the two

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The area occupied by both parks together would be(πx8²) + (πx6²) = 100π

So the new park should be of area 100π.π x (New Radius)² = 100π So, (New Radius)² = 100 [As we cancel π on both sides]Therefore, New Radius = 10m

28.

32x60°5yIn the given figure, two lines AB and CD intersect each other at a point E. Find the values ofx, y and z.

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1

2

your Answer is right

29.

3. In the given fig. lines XY and MN intersect at O. If POY 90°and a b 2:3, find c.

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30.

11.In the given figure, two straight lines AB and CD intersect at O. If LCOT60°, find a,b,c,2c4b60°

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Since ab and cd are straight lines,

4b+b+60 = 180°

5b = 180-60

5b = 120

b = 24°

Also, 4b+2c = 180°

2c = 180-4*24

2c = 180-96

2c = 84

c = 42°

Also, 2c+a = 180°

a = 180-2*42

a = 180-96

a = 84°

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31.

14. In the given figure, the two lines AB andCD intersect at a point o such thatZBOC = 125°. Find the values of x, y and z.А1250

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Angle BOA=180°ANGLE BOC=125°SO ANGLE AOC=X=180°-125°=55°ANGLE COB=ANGLE AOD=Y=125°(OPPOSITE ANGLES)ANGLE DOB=ANGLE AOC=X=Z=55°(OPPOSITE ANGLES)SO,X=55°Y=125°Z=55°

32.

meets thdistinctItis a linpoints. LQ'respe4. In the given figure lines XY and MN intersect Mat O. If LPOY 900 and a: b-2:3, find c.transvers

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33.

2. Find the area of a quadrilateral ABCD in which AB 3 cm, BC 4 cm, CD4cm.DA5 cm and AC5 cm.

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34.

żFind the area of a quadrilateral ABCD in which AB 3 cm, BC 4 cm, CD 4 cm,DA 5 cm and AC 5 cm.

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35.

Find the area of a quadrilateral ABCD in which AB-3 cm, BC-4 cm, CD 4 cmDA 5 cm and AC-5 cm.

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36.

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

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thanks

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37.

Find the area of a quadrilateral ABCD in which AB 3 cm, BC 4 cm, CD -4 cm,DA 5 cm and AC 5 cm.

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38.

closure property

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TheClosure Property statesthat when you perform an operation (such as addition, multiplication, etc.) on any two numbers in a set, the result of the computation is another number in the same set.

39.

There is internal choice for each quesiol.(A) Find the relation between x and y such that the point (x, y) is equidistant from points (7, 1)and (3, 5)14.(ORI

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40.

4 x 4 16MThere is internal choice for each question.(A) Find the relation between x and y such that the point (x, y) is equidistant from points (7,and (3, 5)14.

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By using distance formula

and squaring both sides

(x-7)²+(y-1)²= (x-3)²+(y-5)²

x²+ 49 -14x + y²-2y+1= x²+9 -6x + y²+ 25 -10y

50-14x-2y=34-6x-10y

25 -7x -y=17 -3x -5y

4x-4y=8

x-y=2

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41.

U1Vulpie CHOICE vuestungNumber of tangents that can be drawn to a circle is ...a. One b. Infinite c. At the most two d. of these

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number of tangent can be drawn to a circle is option B is rightonly one tangent can be drawn from a point of circle is one

Option C is the right answer

option 3 is a right answer

42.

erify the closure property of addition for the following integers:-5(a) and9(b) -19 and-1118

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-5+8/93/9

1/3

is the answer

43.

2. Verify associativity of addition of rational numbers i.e., (x + y) + 2 = x + (y + 2), when:

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44.

A person is standing at a distance of 80m from a church looking at its top.The angle of elevation is of 45°. Find the height of the church

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45.

Tuctice set 6.21. A person is standing at a distance of 80m from a church looking at its top.The angle of elevation is of 45°. Find the height of the church.

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Let the height be h.(perpendicular to)Distance=80 m(base)Angle of elevation=45°tan45°=perpendicular/base1=h/80h=80 m

46.

5) Other than those given as options15. What will be the approximate length of the diagonal of a square plot (in m) whoseperimeter is equal to perlmeter of the rectangular plot of length 23 m and breadth19 m ?

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Given,Length of Reactangle l = 23 mBreadth of Reactangle b = 19 mPerimeter of Reactangle Pr= 2(l + b) = 2(23 + 19)= 2(42) = 84

Let side of square be sPerimeter of Square Ps = 4*s

As per given condition Pr = Ps84 = 4ss = 84/4 = 21

Diagonal of square = s*sqrt (2)= 21*1.41= 29.61 m

47.

Prove that the commutative property holdsgood for the following products. Verify thefor the given values of the literals.64

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48.

6 Prove that the commutative property holdsgood for the following products. Verify the samefor the given values of the literals.44

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49.

mm) Internal choice 15 YIN U Call YouA) Prove that the distributive property holds for the rational numbers.Verify with example ?(OR)

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there are two types of distributatve property

1. left cancelation2. Right cancellation

1. (a O b) * c= (aoc) * (boc)2. (c O b) * a= (c O a) * (aob) is the correct answer

50.

6) Prove that the commutative property holdsgood for the following products. Verify the samefor the given values of the literals.4(ii)3p6XHq'); p =-1,q=-2

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Third question has a printing error