Explore topic-wise InterviewSolutions in Current Affairs.

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1.

Find LCM using division method.4, 8, 10

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2.

teiine hnof the lolowing polynomials has (x + 1asa factor(0 -r+ 1(i)+ 2x +2x +x+ 1Use the Factor Theorem to determine whether g(vis

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3.

L Fnd the zeroes of the polynomial.pe)-1 +30

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Splitting the middle term,x^2-15x-2x+30=0 x(x-15)-2(x-15)=0 (x-2)(x-15)=0 x=2 x=15

4.

ZDBC T समकोण हैADBC = AACRi(iv) CM = 7 AB.

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5.

1089 find the square root by division method

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thanos

thank u very much

6.

ハFind the common difference of the AP TA2aG,

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√7, √28, √63

√7, 2√7, 3√7 common difference = 2√7 - √7 = √7

7.

Find the square root of 5776 by division method.

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8.

Find the square root of 9216 by division method.

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9.

) Which of the following are Anthematic ProgressionFind common differenceV3. VE, Va, v12....

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Comman difference will not be thereas

No,They are not in AP because for being in AP the differences between two consecutive numbers should be same.But here, they are different.√6 - √3 = 0.717√9 - √6 = 0.551

10.

Find the common difference of the A.P 7 128,63

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hit like if you find it useful

11.

Find the common difference of the Arithmetic Progression (A.P.)

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Find the common difference a1,a2,a3hence difference =a2-a1

12.

2 AD is an altitude of an isosceles triangle ABC in which AB AC, Show that(j)AD bisects BC(ii) AD bisectsA.

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13.

3. elnAABC, bisector of < A intersects BC in D. If A-AC and BDAB-AC4.5 then find BC

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14.

2.AD is an altitude of an isosceles triangle ABC in which ABAC. Show that(1)AD bisects BC(ii) AD bisects Z A

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15.

AD is an altitude of an isoscejes triangle ABC in which AB AC, Show thatAD bisects AD bisects BC).

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16.

18. The 4th term of an AP is zero. Prove thatthe 25th term of the AP is three times itsCBSE 2016]11th term.

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17.

Find the LCM and HCF of the following pairs of integers and verify that IHCF product of the two numbers.(i) 26 and 91CM x(ii) 510 and 92(iii) 336 and 54

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18.

nd L.C.M.The H.C.F of two numbers is 12 and their difference is also 12. The numbersare:

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19.

8. The 4th term of an AP is zero. Prove thatthe 25th term of the AP is three times its[CBSE 2016]11th term.

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Hence 25th term is 3 times of 11th term

20.

nd the sum of their recipro5 The sum of two natural numbers is 9 and theis 호. Find the numbers.CBSE

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Let one number be xand other number be 9-x

a/q

1/x + 1/9-x = 1/2

9-x+x/x(9-x) = 1/29/x(9-x) = 1/2x(9-x)=18x^2-9x+18=0x^2-6x-3x+18=0x(x-6)-3(x-6)=0(x-6)(x-3)=0x=6,3when one=6, other=3

21.

In △ABC and △DBC are on the base BC. If AD intersects BC at 0 , provethatar(AABC)AOar(ADBC) DO

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22.

18 The 4th term of an AP is zero. Prove thatthe 25th term of the AP is three times its[CBSE 2016]11th term.

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Given a^4 = 0

That is (a + 3d) = 0

⇒ a = - 3d ........... (1)

nth term of AP is given by an = a + (n – 1)d

a^11 = a + 10d = – 3d + 10d = 7d [From (1)]

a^25 = a+ 24d = – 3d + 24d = 21d [From (1)]

= 3 x 7d

Hence a^25 = 3 x a^11

23.

29. The HCF and LCM of two numbers are 13 and 455respectively. If one of the numbers lies between 75and 125, then the number is(a) 78(c) 104(b) 91(d) 117

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24.

the given figure,AABC and ΔDBC are on the same base BC. If AD intersects BC atO. Prove thatar(AABC)AOar (ADBC) DO

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25.

first n terms.a. Show that a, a aform an AP where a, is defined as belov(ii) a, 9-5m(i) a" = 3 +4mAlso find the sum of the first 15 terms in each case.

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26.

Q21. Ifthe given figure, ABC andDBC are two triangles onthe same base BCProve that -ar(AABC) AOar(ADBC) DO

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You need to Proof the congruence of Triangle ABC and triangle BDC to Make AO AND OD equal

27.

the sum of the first n term of an AP is An=2n+n square .find common difference

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28.

3, In Fig 6.44, ABC and DBC are two triangles on thesame base BC. IfAD intersects BC at 0, show thatar (ABC) AOar (DBC) DO

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29.

If the perimeter of a semi conductor is 144 then what is the radius

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30.

3. In Fig. 6.44, ABC and DBC are two triangles on thesame base BC. IfAD intersects BC at O, show thatar (ABC) AOar (DBC) DO

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31.

the given figure, AABC and ΔDBC are on the same base BC. If AD intersects BC atO. Prove that ar(AABO AO(ADBC) DO

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32.

3. In Fig. 6.44, ABC and DBC are two triangles on thesame base BC. If AD intersects BC at O, show thatar (ABC) AOar (DBC) DO

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Thankyou

33.

Q15. In a quadrilatea quadrilateral ABCD ZA+ZD = 180° Does this mean AB | DC? Why? What special namedoes this quadrilateral have?

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Since, ABCD is a quadrilateral and ∠A+∠D= 180°.

From the figure, we can see that ∠A and ∠D form the adjacent angle pair and it is already known that the two adjacent angles always add to give their angle sum equal to 180°.

Also, adjacent angles are formed if the two lines are parallel which means that if two adjacent angles are given then the two ,lines must be parallel, therefore AB is parallel to CD.

If the two lines are parallel and also, the angle sum of two adjacent angles equal to 180°, then this type of quadrilateral is known as "Trapezium".

4 is the correct answer

Yes. If Angle A + Angle D = 180°,then definitely AB || DC.With this information, we can conclude that only one pair of opposite sides are parallel. Hence, a special name for this quadrilateral will be trapezium.See picture for the proof.

34.

If a pair of opposite sides of a cyclic quadrilateral are equal, prove thatother two sides are parallel

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35.

ABCD is a quadrilateral in which all four sides are equal. Show that both pairsopposite sides are parallel.

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ABCD is a quadrilateral with four equal sidesIf the sides are equal they will produce four equal anglesHence each angle will be 90 degree (Sum of interior angles is 360 degree)Hence ABCD is considered as a SQUAREAC and BD are joined which meet at OAC and BD are diagonals , which bisect interior angles.Angle BAD is bisected by ACAngles BAC = DACAngle BCD is bisected by ACAngles ACD = BCASince, BAD = BCD = 90°Angles BAC = DAC = Angles ACD = BCABAC = ACDThey are also interiorly alternate to each other.So by converse theorem,AB is parallel to CDSimilarlyAngles ABD = CBD = ADB = CDB (when BD is the bisector)CBD = ADBThey are also interiorly alternate to each other.So by converse theorem,AD is parallel to BCHence opposite pairs of sides are parallel to each other.

36.

40. Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its[CBSE 2012125th term.

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37.

40. Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find itsCESE 201225th term.

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38.

draw a cricle radius 4cm draw a tangent to circle making an angle of 30° with a line pass through the centre

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39.

9.If the sum of first 15 terms of an A.Ris 675 and its first term is 10,then find 25th term.

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a=10,s15=675,Sn=n/2(2a+(n-1)d)675=15/2(2×10+(15-1) d) 675 ×2/15=(20+14d)45×2=20+14d90-20=14d70=14d d=70/14d=5an=a +(n-1) dan=10+(25-1) 5an=10+24×5an=10+120an=130

40.

In an AP, th sum of first n term is \left(\frac{3 n^{2}}{2}+\frac{5 n}{2}\right) Find its 25th term.

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41.

A quadriatral is a parallelogram if a pair of opposite sides is equal and parallel.

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42.

us.rilateral.The distance between its pa

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43.

In an A.P., the sum of its first n terms is 6n -n2. Find its 25th term.

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44.

odd number

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The number of form 2n + 1 , where n is any natural number

45.

17. In an AP the sum of first n terms is +3n2 13nFind the 25th term.

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Sn = 3n²/2 + 5n/2

We know that ,

nth term = (sum of nth term) - (sum of (n-1)th term)

So , according to the question ,

25th term = (sum of 25th term) - (sum of (25-1)th term)= (sum of 25th term) - (sum of 24th term)= [3(25)²/2 + 5(25)/2] - [3(24)²/2 + 5(24)/2]= [1875/2 + 125/2] - [1728/2 + 120/2]= (2000/2) - (864 + 60)= 1000 - 924= 76

So , the 25th term of the AP is 76.

46.

Q7. Find the square root (by prime factorisation method)2352

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2352

/ \2 1176

/ \2588

/ \2294

/ \2147

/ \349

/ \77

/ \71

47.

L. 64b.By prime factorisation methodi. 576ii. 1225

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1)The prime factors are: 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

or also written as { 2, 2, 2, 2, 2, 2, 3, 3 }

Written in exponential form: 26x 32

2) The prime factors are: 5 x 5 x 7 x 7

or also written as { 5, 5, 7, 7 }

Written in exponential form: 52x 72

48.

Find the HCF of 72 and 84 using prime factorisation method

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2*2*2*3*32*2*3*7common is 3*2*2= 12

2×2×2×3×32×2×3×7HCF= 2×2×3HCF= 12 is the correct answer

49.

Find the LCM of 336 and 54 by prime factorisation method.

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336 = 2x2x2x2x3x7

54 = 2x3x3x3

LCM= 2x2x2x2x3x3x3x7 = 3024

50.

3. In quadrilateral ABCD, side AB is parallel to side DC. If ZA:LD 1 : 2 and C: B 4:5.(i) Calculate each angle of the quadrilateral(ii) Assign special name to quadrilateral ABCD

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