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.

Find the p dynomial whosearoes are 3253 and 3453

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2.

In Fig. 6.18, if LM | CB and LN | CD, prove thatAM ANAB AD

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In the given figure, LM || CB

By using basic proportionality theorem, we get,AM/MB = AL/AC ...(i)Similarly, LN || CD∴ AN/AD = AL/AC ...(ii)From(i)and(ii), we getAM/MB = AN/AD

3.

In Fig. 6.18, if LM || CB and LN | CD, prove thatAM ANAB AD

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In the given figure, LM || CBBy using basic proportionality theorem, we get,AM/MB = AL/AC ...(i)Similarly, LN || CD∴ AN/AD = AL/AC ...(ii)From(i)and(ii), we getAM/MB = AN/AD

4.

p(x).if a and B are zeroes of 2x247x-4.Find p(x) whose zeroes areI/a and 1/ß.D whereas 2 books and 3

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alpha and beta are zeroes that means they are solutions of the equation. To get that we have to compare equation with zero.2x^2+7x-4=02x^2+8x-x-4=02x(x+4)-1(x+4)=0so x=-4,1/2

now we have to find equation whose zeroes are 1/alpha and 1/betaso values are = -1/4,2so equation is (x+1/4)(x-2)=x^2-2x+x/4-1/2so equation is x^2-7/4x-1/2

5.

In the given figure LM || CB and LN || CDProve that AD AD18.AM AN

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6.

\begin{array} { l } { \text { In Fig. } 6.18 , \text { if } \mathrm { LM } \| \mathrm { CB } \text { and } \mathrm { LN } \| \mathrm { CD } \text { , prove that } } \\ { \frac { \mathrm { AM } } { \mathrm { AB } } = \frac { \mathrm { AN } } { \mathrm { AD } } } \end{array}

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LM || CBBy using basic proportionality theorem, we get,AM/MB = AL/AC ...(i)Similarly, LN || CD∴ AN/AD = AL/AC ...(ii)From(i)and(ii), we getAM/MB = AN/AD

thankyou soo much

7.

17. If p and q are the roots of the equation 3x2+6x+2=0show that the equation whose roots areandが)and LE) is(- p) is

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8.

12-İfthe polynomial (xt +2x3 + 8x2 +12x+18) 1s divided by anoth2x + 82 + 12x 18) is divided by another polynomial (x2+5) theremainder comes out to be (px+q) find p and c

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9.

(9) The difference between two acute angles of a rightangled triangle is . Find the angles in degrees.10

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10.

7. If the polynomiial x4 +2+8x2 +12 x +18 is divided by another polynomial x2 +5,the remainder comes out to be px +g. Then, find the values of p and q.

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11.

If the polynomial f(x)-x'-6x3 + 16x2-25x + 10 is divided by another polyncx2-2x +k, the remainder comes out to be x + a, find k and a.

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Given that the remainder is (x + a)⇒ (4k – 25 + 16 – 2k)x + [10 – k(8 – k) ] = x + a ⇒ (2k – 9)x + [10 – 8k + k2] = x + aOn comparing both the sides, we get2k – 9 = 1

⇒ 2k = 10∴ k = 5Also 10 – 8k + k2= a⇒ 10 – 8(5) + 52= a⇒ 10 – 40 + 25 = a∴ a = – 5

12.

6, In Fig. 6.44, the side QR of△ PQR is produced toa point S. If the bisectors of PQR andPRS meet at point T, then prove that

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13.

Fig. 6.42g.6.44, the side QR of A PQR is produced toa point S. If the bisectors of 4 PQR andZ PRS meet at point T, then prove that2Fig. 6.44

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14.

In which of the following situations, does the list of nombelprogression, and why?o The tasi fare after each km when the fare is t 1S for the first km and & for eachinder whm a yacuum pump removes of he

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For first km taxi fare = 15For each additional km taxi fare = 8

Then taxi fare for trip of n km= 15, 8, 8, 8, 8.........n

This is not Airthmetic Progression as in AP difference between any two successive term is same.

15.

g. 6.426. In Fig. 6.44, the side QR of A PQR is produced toa point S. if the bisectors of L PQR andZ PRS meet at point T, then prove thatSummary

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16.

6. In Fig. 6.44, the side QRofΔ PORİs produced toa point S. If the bisectors of PQR and< PRS meet at point T, then prove that2Fig. 6.44

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17.

छि% दि Mis 1% कप डि 18: है i) TMb bble 10 छह 1bD] Sl Mk % फंसb2 2k 2 Ikl 6L Mk Dblie b |93 OF ‘TI

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18.

Fig. 6.426. In Fig. 6.44, the side QR of PQR is produced topoint S. If the bisectors of PQR andPRS meet at point T, then prove that

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19.

e WMk e (q)

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20.

0०%] ७102 1५% bl Lb12B Mk bl B ks 20 b lojolieनि कर!

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आरुणि का शरीर सर्दी से अकड़ने लगा किंतु उसे तो एक ही धुन... गुरु की आज्ञा का पालन। बहता हुआ पानी कल-कल करता मानो आरुणि को कहने लगा-

मैं वर्षा का बहता पानी,

मेरी चाल बड़ी तूफानी।

उठ आरुणि अपने घर जा,

रास्ता दे दे, हट जा, हट जा।।

मगर गुरु भक्त बालक आरुणि का उत्तर था-

गुरुजी का आदेश मुझे है,

मैं रोकूंगा बहती धारा।

जय गुरु देवा, जय गुरु देवा,

आज्ञा पालन काम हमारा।।

21.

L) in Fig. 6. 13, lines AB and CD intersect at O. IfBOD 40°, findEAOC+ 4BOE< BOE and reflex70 andCOE.Fig. 6.13

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22.

13. In Fig. LM II CB; and LN II CD.Prove that:AM ANAB AD

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By Thales theorem or basic propositionality Theorem

IN∆ABCLM||BC so by B.P.TAL/AC=AM/AB-----------(1)

IN∆ADC

AL/LC=AN/AD------(2)from equation 1 &2 L.H.S is same so R.H.S will also same.soAM/AB=AN/AD

23.

TERNATIONAL OLYMPIAD UITIn the shown figure, POQ is a straight line.(2a + 12°) (3a - 17º)P0QBased on above information, choose the one which is correct?A) a = 38°(B) ZROQ = 96°-) Z POR = 84°407 ZROQ - Z ROP = 8° of thesethe given figure, it is given that OA|| EC and OBITED.

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24.

6. In Fig. 6.44, the side QR of A POR is produced toa point S. If the bisectors of PQR andZ PRS meet at point T, then prove thatFig. 6.44

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25.

3、rFig. 6.15. < POR =PQS = < PRTthen provet

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26.

22.In the given figure, if PQI| ST, Z PQR = 110° and Z RST = 130°. FindZ QRS:1100

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27.

28. In the figure the magnitude of <PQS is

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28.

l Exercise 9.2n Fig. 9.39 ABCD and AEFG are twoparallelograms. If<C= 55°, then tinFig. 9.39

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In parallelogram ABCD so,

angle A = angle C(opposite angle are equal)

angle A = 55°

Now in parallelogram AGFE,So,angle A = angle F (opposite angle are equal)

angle F = 55°

29.

of cementing its floor and wailis at lThe floor of a rectangular hall has a perimeter 236 m. Its height is 4.5 m. Find the costof painting its four walls (doors and windows be ignored) at the rate of 8.40 persquare metrelonath of 30 cmt a breadth of 20 cm and a height of 20 cm.

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Area of four walls = 2(l+b) * h

= 236m * 4.5m

= 1,062 m^2

Hence cost to do that at rate of ₹8.40 per square metre = ₹1062*8.40

= ₹8,920.8

30.

Fig. 6.426. In Fig.6.44, the side QR of A PQR is produced toa point S. If the bisectors of Z PQR andZ PRS meet at point T, then prove thatFig. 6.44

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31.

3. In Fig. 6.15, Z PQRPRQ, then prove thatf.Fig. 6.15

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32.

6. In Fig. 3.44, the side QR of A PQR is produced toa point S. If the bisectors of Z PQR andPRS meet at point T, then prove that2R s

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33.

20 A. B, C and D are four points on aand BD intersect at a point E suchBEC 130° and 4 ECD 20°. FindFig. 10.38n FigthatZBAC130Fig. 10.39

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34.

EXERCISE 6.2n Fig. 6.28, find the values of x and y and thenX show that AB CD13Fig. 6.28

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35.

In Fig. 6.39, sides QP and RQ of Δ PQR are produced to points S and T respectivelyIf angle SPR=135° andPQT = 1 10°, find < PRQ.

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36.

In Fig. $ 6.15 . \angle \mathrm{PQR}=\angle \mathrm{PRQ} $ , then prove that$ \angle \mathrm{POS}=\angle \mathrm{PRT} $ .

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37.

(4) In the figure, seg QS 1 side PR.seg RT u side PQ, P-S-R and P-T-QShow that (i) △PQS ~ △PRT.(ii) QSxTP=PS × RT.1

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38.

In Fig. $6.15, \angle \mathrm{PQR}=\angle \mathrm{PRQ}$ , then prove that$\angle \mathrm{PQS}=\angle \mathrm{PRT}$

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from 1 and 2 angle PQS = angle PRT

39.

P & Q are centers of circles of radii 9 cm and 2 cm respectively. PQ= 17 cm. R is the center of the circle of radius x cm which touches the above circle externally. Given that angle PRQ is 90°. Write an equation in x and solve it.

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40.

Fig. 6.15, < PQR = < PRQ, then prove that

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41.

(Uising (11. G3,(03) nd SAS nle).AD ACPCT)EXERCISE 7,2

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Given:

ΔABC is an isosceles∆ with AB = AC, OB & OC are the bisectors of ∠B and ∠C intersect each other at O. i.e, ∠OBA= ∠OBC & ∠OCA= ∠OCB

To Prove: i)OB=OC ii)AO bisects ∠A.

Proof:

(i)In ∆ABC is an isosceles with AB = AC, ∴ ∠B = ∠C [Since , angles opposite to equal sides are equal]

⇒ 1/2∠B = 1/2∠C [Divide both sides by 2]

⇒ ∠OBC = ∠OCB & ∠OBA= ∠OCA.......(1) [Angle bisectors]

⇒ OB = OC .......(2) [Side opposite to the equal angles are equal]

(ii)In ΔAOB & ΔAOC, AB = AC (Given) ∠OBA= ∠OCA (from eq1) OB = OC. (from eq 2)

Therefore, ΔAOB ≅ ΔAOC ( by SAS congruence rule)

Then, ∠BAO = ∠CAO (by CPCT)

So, AO is the bisector of ∠BAC.

42.

In Fig.9.23, E is any point on median AD of aABC. Show that ar (ABE) # ar(ACE).

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43.

7. It is given that in a group of 3 students, the probability of 2 students not having thesame birthday is 0.992. What is the probability that the 2 students have the samebirthday?l hallohall is drawn at random from the bag.

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44.

Express the following statements in terms of percentage.(a) 4 paise of 4 (b) 5 cm of im (c) 100 g of 2 kg (d) 3

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45.

EXERCISE 11.3In Fig. 11.23, E is any point on median AD ofaΔ ABC. Show that ar (ABE) = ar (ACE).

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46.

1. In Fig.9.23, E is any point on median AD of aABC. Show that ar (ABE)ar (ACE).

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47.

In Fig.9.23, E is any point on median AD of aA ABC. Show that ar (ABE) ar (ACE).

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48.

In Fig.9.23, E is any point on median AD of aA ABC. Show that ar (ABE)-ar (ACE).1.

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49.

1.) In Fig.9.23, E is any point on median AD of aA ABC. Show that ar (ABE) ar (ACE).

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50.

EXERCISE 11.21.Construct a triangle ABC in which BC = 7cm, LB = 75° and AB + AC = 13 cm.

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