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.

PRECIZE MATHEMATICFind the equation of line which passes through the following two points:(ii)(a, b), (a + r cos α, b + r sina).the trianglo u

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this is wrong

2.

MATHEMATICe9: Find the roots of 4x2 3x+ 5 0 by the method of completing the

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3.

A square and a rectangular field with measurements given in the figure have the same perimeter. Which field has a larger area?

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I know the answer that this aap is really work or not although thxx for answer

4.

L A square and a rectangular field withmeasurements as given in the figure have the sameperimeter. Which field has a larger area?

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5.

1. A square and a rectangular field withmeasurements as given in the figure have the sameperimeter. Which field has a larger area?

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6.

EXERCISE 11.1auare and a rectangular field withmeasurements as given in the figure have the sameerimeter. Which field has a larger area?

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7.

12abx2 -(9a2-8b2)x- 6ab=0

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Step-by-step explanation:

12abx^2-(9a2-8b^2)x-6ab

NOW WE SHOULD SPLIT THE MIDDLE TERM

12abx^2-9a^2x+8b^2x-6ab

USING FACTORIZATION METHOD

3ax(4bx-3a)+2b(4bx-3a)

(3ax+2b)(4bx-3a)

3ax+2b=0

3ax= -2b

x=-2b/3a

OR

x=3a/4b

8.

biological magnification define

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Biological magnification often refers to the process whereby certain substances such as pesticides or heavy metals move up the food chain, work their way into rivers or lakes, and are eaten by aquatic organisms such as fish, which in turn are eaten by large birds, animals or humans.

9.

Factorise the following expressions.(i) a² + 8a+16

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10.

Factorise the following expressionsa^2 + 8a + 16

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(i)a²+ 8a+ 16 = (a)²+ 2 ×a× 4 + (4)²

= (a+ 4)²

[(x+y)² =x²+ 2xy+y²]

11.

Factorise the following expressions.' ir a2 + 8a + 16 (i) p2-10p+25

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12.

Draw a rhombus whose side is 7.2cm and one angle is6O

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13.

Find the area of a rhombus whose side is 13 cm and altitude is 2 cm.

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14.

Findthe areaa of a rhombus whose side is 5 cm and the altitude is 6 cm.

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area=base × height=5×6=30cm^2

15.

Find the area of a rhombus whose side is of the length 5 m and one of its diagonals is of length 8 m.

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16.

Find the area of a rhombus whose side is of the length 5 m and one of its diagonals isof length 8 m

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17.

Find the area of the rhombus whose side is 6 cm and whose altitude is 4 cm. If cdiagonals is 8 cm long, find the length of other diagonal13.the

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18.

A truck requires 108 litres of diesel for covering a distance of 594 km. Howmuch diesel will be required by the truck to cover a distance of 1650 km

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this answer is same toe

19.

A truck requires 108 litres of diesel for covering a distance of 594 km. Howmuch diesel will be required by the truck to cover a distance of 1650 km?

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Amount of Diesel required by truck for covering a distance for 594 km = 108 litres

Amount of Diesel required by truck for covering a distance for 1 km = 108 ÷ 594

= (2/11)litre

So Amount of diesel required by truck for covering a distance for 1650 km = 1650 ×(2/11) = 300 litres

20.

ybananas can be purchased for2oht of 72 books isWhats ctÄąck requires 108 litres of diesel fur90?weightof 40suchbooks?tor coverng

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21.

9. A truck requires 108 litres of diesel for covering a distance of 594 km. Howmuch diesel will be required by the truck to cover a distance of 1650 km?

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22.

(L) 290 bt SRS A न£ 108. a’-3a+1=0626R, 8 + व 6 शा

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a^2-3a+1=0a^2+1=3a....(1)(a+1/a)^3=a^3+1/a^3+3a^2*1/a+3a*1/a^2a^3+1/a^^3=(a+1/a)^3-3(a+1/a)=1/a^3(a^2+1)^3-3/a(a^2+1)a^3+1/a^3=1/a^3(3a)^3-3/a(3a)=27-9=18

23.

BE 108 जद P. 5८८ 6 ने हिN3 B . WM PL\ ,\/\,,&n Pmmvc —th

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24.

니1) Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. lf one of the diagonals is 8-long, find the length of the other diagonal.

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25.

ample 1 For all n 2 1, prove that12+2+32+4%= n(n +1)(2n+1)+㎡

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For n = 1, the statement reduces to 12 =1 · 2 · 36and is obviously true.Assuming the statement is true for n = k:1^2 + 2^2 + 3^2 + · · · + k2 =k(k + 1)(2k + 1)/6, (1)we will prove that the statement must be true for n = k + 1:1^2 + 2^2 + 3^2 + · · · + (k + 1)^2 =(k + 1)(k + 2)(2k + 3)/6. (2)The left-hand side of (2) can be written as1^2 + 2^2 + 3^2 + · · · + k^2 + (k + 1)^2.In view of (1), this simplifies to:

1^2 + 2^2 + 3^2 + · · · + k^2+ (k + 1)^2 =k(k + 1)(2k + 1)/6+ (k + 1)^2=k(k + 1)(2k + 1) + 6(k + 1)^2/6=(k + 1)[k(2k + 1) + 6(k + 1)]/6=(k + 1)(2k^2 + 7k + 6)/6=(k + 1)(k + 2)(2k + 3)/6.Thus the left-hand side of (2) is equal to the right-hand side of (2). Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive integer n

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26.

11. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of isdiagonals is 8 cm long, find the length of the other diagonal

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27.

6.Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm.If one of its diagonals is 8 cm long, find the length of the other diagonal.to

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28.

Vample 1 Solve the following system of linear equations:4x – 3y = 8x-2y = -3.

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Multiply second equation be 4hence4x-3y=84x-8y=-12subtract both5y=20y=4hencex-2y=-3x-8=-3X=5

x=5,y=5 is the correct answer of this question

29.

ample 1 : Use Euclid's algorithm to find the HCF of 4052 and 12576.

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Use Euclid’s algorithm to find the HCF of 4052 and 12576.

Step 1: Since 12576 > 4052, apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

Step 2: Since the remainder 420 ≠ 0, apply the division lemma to 4052 and 420, toget

4052 = 420 × 9 + 272

Step 3: Consider the new divisor 420 and the new remainder 272, and apply thedivision lemma to get

420 = 272 × 1 + 148

Consider the new divisor 272 and the new remainder 148, and apply the divisionlemma to get

272 = 148 × 1 + 124

Consider the new divisor 148 and the new remainder 124, and apply the divisionlemma to get

148 = 124 × 1 + 24

Consider the new divisor 124 and the new remainder 24, and apply the divisionlemma to get

124 = 24 × 5 + 4

Consider the new divisor 24 and the new remainder 4, and apply the divisionlemma to get

24 = 4 × 6 + 0

The remainder has now become zero, so procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Also, 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) =HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052)

30.

€ 8 T=kih R (T 0 S{®RluE पाG L8

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31.

के न 2:22. gt — el कि N D L8 Bt eie blogx - e 4logxe3logx _ pllogx

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32.

g€ * दर कि ट्द - E—L जा स्कयुक 28 9

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33.

.A truck requires 108 L of diesel for covering a distance of 594 km. How much diesel willthe truck require to cover a distance of 1,650 km?

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For 594 km 108 L sufficient so for 1650 km required diesel = 1650×108/594 = 300 L

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34.

(व) 5 लॉटर (0) 7-5 लीटर L8 A gl तो नये9, U 35 लीटर मिश्रण में दूध और पानी का अनुपात 5:2 है. यदि इसमें 5 लीटर दूध और डाल 'दिया जायें ती नः ]मिश्रण में दूध और पानी का अनुपात कया होगा ?“५, 57: केश नि (2%

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3:1 aayega ishka answaer

35.

a’*+8a+ 16

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a²+8a+16= (a+4) ²

36.

+ 8a + 16

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a² + 8a + 16

a² + 2 × a × (4) + 4²

(a + 4)²

37.

a²+8a+16

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38.

1) a² + 8a + 16

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a^2 + 8a + 16 = a^2 + 4a + 4a + 16 = (a + 4)(a + 4)

a= -4. a=-4 is a best answer

39.

at+ 8a + 16

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a^2 +8a +16=a^2 +4a +4a +16=a(a+4)+4(a+4)=(a+4)(a+4)=(a+4)^2

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40.

8. Factorise : χ-64x3

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One more question

41.

a^2 + 8a+ 16

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a² + 8a + 16a² + 4a + 4a + 4²(a+4)²

42.

1/5th part of a drum is filled with milk.What is the capacity of drum if it require28 L more to fall the drum completely?

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Let capacity of drum is v

GivenMilk in drum = v/5

Then,v/5 + 28 = vv - v/5 = 28(4/5)v = 28v = 7*5 = 35

Capacity of drum is 35 litres

43.

8.Factorise each of the following:

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44.

What is translation ?

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Translation is a term used in geometry to describe a function that moves an object a certain distance. The object is not altered in any other way. It is not rotated, reflected or re-sized. ... In a translation, every point of the object must be moved in the same direction and for the same distance.

In English the process of translating words or text from one language into another is know as Translation.

45.

द जाकर ना वि.व 2. रथ व ervelidar के1 ०८2// 9 cm कि... ;7! ... क्र नि baost .Gar /',qn?jq|]29583:,'7>3°l

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Height (h) = 9 cm

Volume = 48π cm³

Let r be the radius of the base of the cone

Volume of a right Circular cone =1/3 πr²h

Volume ofa right Circular cone = 48π cm³( Given)

1/3 πr²h = 48π

1/3 ×r²× 9 =48

3r²= 48

r²= 48/3 = 16

r=√16

r = 4cm

[On taking positive square root]

Hence, the diameter of the base= 2×r= 2× 4= 8cm

46.

. ABCD is a field in the shape of a trapezium,ADIl BC, LABC-90° and ZADC 60° Foursectors are formed with centres A, B, C andD, as shown in the figure. The radius of eachDsector is 14 m. Find the followingtotal area of the four sectors,(i)i)area of the remaining portion, given that AD =55 m, BC-45 mand AB30 m.ICBSE 2013C]

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thanks

47.

DEC = 90°.EXERCISE 9B1. In the adjoining figure, ABCD is aparallelogram in which ZA -72°. CalculateZB, ZC and LD.72°

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48.

2. Prove that the adjacent angles of agm are supplementary

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Let ABCD be a parallelogram

Then, AD ∥ BC and AB is a transversal.

Therefore, A + B = 180° [Since, sum of the interior angles on the same side of the transversal is 180°]

Similarly, ∠B + ∠C = 180°, ∠C + ∠D = 180° and ∠D + ∠A = 180°.

Thus, the sum of any two adjacent angles of a parallelogram is 180°.

Hence, any two adjacent angles of a parallelogram are supplementary.

49.

1. In the adjoining figure, ABCD is a parallelogram inwhich <A = 72°. Calculate 2B,C and < D

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50.

13. In the given Mg. 12.11. ABCD is a quadrilateral, find

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ABCD IS A quardilaterAngle A =90* Angle B=105*Angle C=60*find angle DA.T.Q sum of quardilater =360*so A+B+C+D=360*90* +105* + 60* +D=360*255*+D=360*D=360*-255*D=105* ans.

x+90+105+60=360x=360-90-105-60x=105

x= 105 is right answer

105 (opposite angles are equal)

Sum of all angles of quadrilateral =360x+60+105+90=360x=105

x=105; answer is correct

A=90* le bhai answer dekhle

x=360- 255x=105 solved

x=105 is the correct answer

Given,ABCD(a quadrilateral)Angle A is 90°Angle B is 105°Angle C is 60°Angle D ??(x)We know,Sum of all the angles of a quadrilateral is 360°Therefore,Angle(A+B+C+D)=360° 90°+105°+60°+x=360°255°+x=360°x=360°-255°x=105°Angle D is 105°

105 degrees answer this question

value of x will be 105°

cerect answer is 105* ho ga

question the answer 105 .

ABCD is a quadrilateralsum of quadrilateral = 360*A+B+C+D= 360* 90*+105*+60*+x =360*255*+D=360*x=360*-255*x=105*

60+90+105+x=360therefore x=105

the value of angle D is equal to 105*

X _ 105 ( because opposite angle)

sum of angles in a quadrilateral =360then given angles = 105+60+90+x=360255+x=360x=360-255x=105

कोण D का माप 105 डिग्री है

105 is the answer of the ads

sum of quadrilater 360

sum of abcd 360angle à 90 angle b 105angle c 60angle d x 90+105+60+x=360x=360-(90+105+60)

angle [D] -?given -Angle (A,B,C,)angle A,B,C,D, are quardilater so, A+B+C+D=360 D=360-255 D=105

....🤗🤗...

60+105+90+x=360255+x=360x=105

right answer is 105 of angle d

the correct value of x is 105

the value of x is 75

cod ancda-90*b-105c-60d-105,

ABCD is a quadrilateral.angle A=90°angle B=105°angle C=60°angle D=?

Sum of all angles of the quadrilateral =360° (property)

So, A+B+C+D=360° = 90°+105°+60°+D=360° = 255°+D=360° = 255°-360°=D =105°

THIS IS THE FINAL ANSWER

x+90+105+60=360x+255=360x=360-255x=105 answer

105 is the correct answer