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. |
9. 42x4811.294 x 206 |
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| 2. |
500+30+206/100+9/1000 |
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Answer» 500 + 30 + 206/100 + 9/1000 In decimal form 530 + 2.06 + 0.009 532.069 |
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| 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 -4sĺŚ12-15(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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Answer» 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 |
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| 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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Answer» 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 |
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| 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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Answer» 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.... |
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| 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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Answer» 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 |
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| 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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Answer» 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! |
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| 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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Answer» Area of rectangle = length × breadth 440 = 22 × breadth breadth = 440/22 = 20 m Perimeter = 2 ( length + breadth ) = 2 ( 20 + 22 ) = 2 × 42 = 84 m If you find this answer helpful then like it. |
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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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Answer» let the point of intersection be O .then vertically opposite angle are FOI and HOI , FOH and GOI |
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| 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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Answer» measure of other two angles are also same as 123° and 57° ... ( vertically opposite angles) |
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| 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 Ă 12.5. cm Ă 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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Answer» 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 |
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| 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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Answer» 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 |
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| 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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Answer» 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 |
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| 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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Answer» 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 |
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| 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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Answer» 1 2 your Answer is right |
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| 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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Answer» 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° Like my answer if you find it useful! |
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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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Answer» 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° |
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| 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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Answer» thanks 1 2 |
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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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Answer» 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. |
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| 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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Answer» 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 Like my answer if you find it useful! |
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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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Answer» 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 |
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| 42. |
erify the closure property of addition for the following integers:-5(a) and9(b) -19 and-1118 |
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Answer» -5+8/93/9 1/3 is the answer |
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| 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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Answer» Let the height be h.(perpendicular to)Distance=80 m(base)Angle of elevation=45°tan45°=perpendicular/base1=h/80h=80 m |
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| 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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Answer» 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 |
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| 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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Answer» 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 |
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| 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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Answer» Third question has a printing error |
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