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.

17. Sita has 40 more than 6 times that ofAwhat Gita has. If Sita has 286, howmuch money does Gita have?

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let geeta has x money,

Acc. to ques.

6x + 40 = 2866x = 286 - 406x = 246x = 246/6x = 41

2.

58,Sita deposited Rs. 5,000 at 10% simple interest for 2years. How much more money will Sita have in heraccount at the end of two years, If it is compoundedsemiannually

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3.

1e price in rupees of 1000 apples, when the cost of ascore is paise?Jow nmany houre will it tole totoe tly kmat 4 mn hour?

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4.

Find six ratiogàl numbers between 3 andl 4

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5.

J Radha tells his daughter Sita " Seven years agowas seven times as old as you were then. Alsothree years from now , I shall be three times asold as you will be". If the present age of Radha andSita are randy respectively, represent this situationgraphically.

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6.

Andl

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√2 is 1.414hence a rational number between 1 and √2 is 1.2

is this absolutely correct

7.

8.Find three different irrational numbers between the rational number andl

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8.

Obtain the volume of rectangular boxes with the following length, breadth andlrespectively

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9.

70. The angles of a triangle are (x – 40°), (x – 20°) and 2 -10°). Find the value of x and then theangles of the triangle.(C.B.S.E. 2015)

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10.

Which of the following is equal totan 80° × tan 70° × tan 60°x.X tan 10°?(A) 13 (B)/2(C) 1(D) 0

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This question is wrongno Option tan80= cot 10tan 70= cot 20now tan 60= √3now tan and cot will be cancelledand Answer = √3

This question is answered 0

11.

If $16^{n} P_{3}=13^{n+1} P_{3},$ find $n$

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12.

^{n} P_{4} :^{n+1} P_{4}=3 : 4

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nPr = n!/(n-r)!HencenP4 : n+1P4 = n!/(n-4)! : (n+1)!/(n-3)! = n!/(n-4)! : (n+1)(n)!/(n-3)(n-4)!Hencen-3/n+1 = 3/44n - 12 = 3n+3 n = 15

13.

6.In Figure, AB, CD and EF are three lines concurrent atFund the value of

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14.

2)=22-3реж). 05/8)) ) ))Cls d4)рез) |

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x^2 - 3 = 0

Using identity(a^2 - b^2) = (a - b)(a + b)

(x - root(3))(x + root(3)) = 0

x = root(3), - root(3)

Therefore,Zeroes of equation are root(3), - root(3)

-1/3 +2/5=2/5+-1-3 represent which of the following property

15.

AB, CD and EF are three concurrent lines passingthrough the point O such that OF bisects 2BOD.ZBOF 35°, find LBOC and LAOD.

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Tx fr giving soln...

16.

10 s0, rays AB and CD erDetermine y whenFig. 10.6010.61, lines AB, CD and EF intersect at O. Find the meavat es ofFig 10817.LDOE and ZBOF4E, CD and EF are three concurrent lines passing through the poist OLAOC LOE5.BOD. I LBOF 35, find ZBOCand LAOD.10.62, lines AB and CD intersect at O. IE LAOC BOE- 40° find LBOE and reflex LCOE.fretlines9. In Fig.LBOD 40a02DKFig. 10.62

Answer»

thanks for the answer

17.

9. In Fig 10.61, lines AB, CD and EF intersect at U.IDOE and /BOF.10. AB, CD and EF are three concurrent lines passing through the point O such that OFbisects BOD. If ZBOF-35°, find BOC and LAOD.tatements are true (T) and which are false (F)?

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18.

If attbytc-o, br-cy4a-0, crtaytb=0 are concurrent lines then 2 be. 2ca .2ab=

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tq

19.

If the median of the distribution 10, 12,13, 16, x, 20, 25, 30 is 18, then thevalue of x is

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Given data10, 12, 13, 16, x, 20, 25, 30

Median = 18

As total number of terms are 8 which is even. Hence median for given distribution= 1/2 [n/2 th term + (n/2 + 1)th term]

Hence,(16 + x) /2 = 1816 + x = 36x = 36 - 16 = 20

Therefore, value of x = 20

20.

(3) 70 x 10

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700 multiply krne pr

21.

4. Find x in the picture below7x-20 5x-103x 18

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angle FBE + angle EBD = 180°(linear pair)7x + 20° + 5x + 10° = 180°12x = 150°x = 12°

22.

60 x^{2}-70 x-30

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23.

he Jenath of a Countyardis 20 n-a92.scm andl breadth tocm ind

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24.

The following martk1 wene obldinethudann81,12,90, 86,65(

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Highest mark = 95Lowest mark = 62

Range = Highest mark - Lowest mark = 95-62= 27

25.

X and Y are the mid points of sides AC and AB respectively of ΔABC, QP BC and CYQ and BXP are straightlines. Prove that ar(AABP)ar(AACO)

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26.

apiup was 30 000.4 A dealer sold a motorcycle at a loss of 10%. If he had sold it motorcycle.for 4500 more, hemowould have made a profit of 5%. Find the selling price of the

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27.

ne two CisesA hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost ofin-plating it on the inside at the rate of ? 16 per 100 cm'

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28.

18. Show that the pair of straight lines ara + 2hry+by+298 +2fy+c=0 forma rhombus with the lines ar? +2hry+by = 0, if h (92- ) gf(a - b).

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hope it helps u

sorry I don't know please

29.

xampisy show that * : R × R → R given by a * b-> a + 2b is not associati

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Thanks a lot

30.

22. In the give figure, X and Y are the midpoints ofqAC and AB respectively, QPII BC and CYQand BXP are straight lines. Prove thatar(AABP) = ar(A ACQ).

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31.

LongAnswer Type Questions [4 Marks]10. In the given figure, X and Yare the mid-points of ACand AB respectively, QP II BC and CYQ and BXPare straight lines. Prove thatar(AABP ar(AACO)

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thanks

32.

210x92x = 10082.31+4|-71?1 216 7! 8!

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33.

The values of p for which the power of a point (2, 5) is negative with respect to a circlex^{3}+y^{2}-8 x-12 y+p=0 which neither touches the axes nor cuts them.

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34.

If both x-3 and 3x-1 are factors of px2+5x + r, show that p-r

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35.

(2) Sakshi prepared some jam at home and filled it in bottles. After giving away?the bottles to her friends, she still has 12 for herself. How many bones hasmade in all? If she filled 250g of jam in each bottle, what was the total weigthe jam she made?

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total bottles=19total weight=4750g

the right answer is 4750 gm

total bottel.19 total weight .4750g

Sakshi prepared a bottle of jam and she give to her friend = 7 jams bottle

She still have the bottels=12 bottle of jam

In each bottle 250g filled in bottle 12+7=1919×=

Sakshi prepared a bottle of jam and she gave her friend =7jams bottle

She still have the bottels=12bottle of jam

In each bottle 250g jam filled in bottle 12+7=1919×250=4750g

Ans. The total wieght ofthe jam ahe made 4750g

The right answer is 4750 gm

total botal=19total weight=4750g.

total weight = 4750 g is right

total bottles = 19total weight = 4750 gramis the best answer

total bottle=19.total weight=4750g

Total bottles =19 total weight =4750

correct answer is 4750gm

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4750 is correct answer of this question

total bottles=19 total weight=4750g

Total bottles =19total weight =4750g is the right answer for this questions...

4750 gm is the correct answer.

the correct answer is 4750gm

total bottles = 19total weight =4750g

35×12=420+12=432×250=107600gm = 107 kg and 600ymJam was made

hi bro mast Khel te hio

12+7=19250*19=4750gtotat bottles=19total weight=4750

total bottle es=19total weight=4750g

total bottles=19total weight=4750g answer.

the bottle were 19 with weight 4750g

36.

In $\Delta \mathrm{JAM},$ if $3 \angle J=4 \angle \mathrm{A}=6 \angle \mathrm{M},$ find all the angles.

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37.

ORIn figure ABCD In it, AD l l BC, LDAB = 90, AD-10 cm andcm. IfABE is a quadrant of a circle, find the area of the shaded region. [Take π = 1is a trapezium of area 24.5 sq, emcia647Section 'D

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38.

2x+3y=12;x-y=1

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39.

33. The area of the shaded region in thegiven figure is ISSC Constable, 20134545°(a) a"(π-1) sq units(b) (T-2) sq units(c) a (x -2) sq units(d) (t-1) sq units34. A street of width 10 m surroundsfrom outside a rectangular gardenwhose measurement is200 m x180 m. The area of thepath (in sq m) isSSC Constable, 2012](a) 8000 (b) 7000 (c) 7500 (d) 8200

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please give me a thanks

40.

(C)300%33. The area of the shaded region in thegiven figure is [SSC Constable, 2013145°45°(a) a (T -1) sq units(b) (T -2) sq units(c) a2 (n -2) sq units(d) ( 1 sq units34. A street of width 10 m surroundsfrom outside a rectangular gardenwhose measurement is200 m x180 m. The area of thepath (in sq m) isSSC Constable, 201(a) 8000 (b) 7000 (c) 7500 (d) 8200

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ABC is isosceles where BC=AC and ang.C=90°By Pythagoras theorem, BC=√2a.therefore,area = πa²/2 - 1/2x√2ax√2a= πa²/2 - a²=a²/2(π-2) sq units

ANSWER IS . OPTION NO IS 2.. 2 IS THE CORRECT ANSWER

41.

(3) The dimensions of a cuboid in em are 20 x 18 x 10. Its volume is(A) 7200 sq㎝(B) 3600 cm 3(C) 3600 sq cm(D) 1800 cm3

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42.

(3) The dimensions of a cuboid in cm are 20 x 18 x 10. Its volume is...(A) 7200 sq cm (B) 3600 cm^3 (C) 3600 sq cm (D) 1800 cm^3

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Volume of a cuboid is given as = length * bredth * height

Therefore 20 * 18* 10 = 3600 cubic units

volume=length×breadth×height=20×18×10=3600 cm^3

43.

60 x ^ { 2 } - 70 x - 30

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60x^2-70x-30=60x^2-90x+20x-30=30x(2x-3)+10(2x-3)=(30x+10)(2x-3)=10(3x+1)(2x-3)

44.

2.Consider an A.p. a, ay, az ....... such that as+as+ag=11 and a ta = -2, then the value of a +ar+a, is1)-82) 53) 74) 9

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let first term be aand common difference be da/c to que, a+2d+a+4d+a+7d=113a+13d=11anda+3d+a+d=-22a+4d=-2a+2d=-1so, a1+a6+a7=3a+11dputting the value of a=-5&d=2is 7

45.

Q 1: Decompose 32760 into prime factors.Q 2: Divide 24x2y2 - 4xy by 2xy.Q 3: a and are zeroes of polynomial 4x2 -2x + k +.7. Find the value ofk.sides has diagonals. If a polygon has 9 diagonals, find the number of sidesof the polygon.Q5: Simplify(1 + tan-6)(1-sin θ)(1 + sin θ)Q 6: IfdABC ~ ΔPOR ar(AABC)-16 cm? and ar(ΔΡ(R)-81 cm, AB-2 cm find PQ.

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46.

The coordinates of A, B, C and D are (6, 3), (-3, 5), (4, -2) and (x, 3x) respectively.ar(ADBC) : ar (AABC) = 1 : 2, find x.

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47.

ABCD is a parallelogram. P is a point on ABsuch that AP:PB-3:2 and Q is a point onCD such that CQ:QD-4:1. If PQ cuts thediagonal AC at R show that AR-7

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thx very much

48.

In the given figure, Pis a point on side BC of DABC such that BP: PC 1:2 and Q is a point on AP suchthat PQ': QA-2 : 3. Show that ar ( Δ AQC): ar (Δ ABC) 2 : S.

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49.

Q.4If both the roots of the quadratic equationx22kx +k2+k-5-0are less than 5, then k lies in the intgl4, 5](-co, 4)(5, 6]

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50.

(a) Prove that theequation2x2 -5xy+3y2-2x +3y 0 representstwo lines and find their point ofintersection.

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