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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 75401. |
(2*tan(30^circ))/(tan(30^circ)^2 %2B 1) |
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Answer» tan 30 = 1/root(3) Then,2 tan 30/1 + tan^2 30 = 2* 1/root(3)/ 1 + (1/root(3))^2 = 2/root(3) / 1 + 1/3 = 2/root(3) * 3/4 = root(3)/2 root3/2 is the answer Bhai answer sin 60° hoga |
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| 75402. |
2tan 30Evaluate 1+tan230 |
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| 75403. |
ground atINCERT EXEMPLAR12.) The area of a circular playground is 22176 m2. Find the cost of fencing thisrate of50 per metre. |
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| 75404. |
to move once over to level a playground. Find the area of the playgroundu120Cl.Ittakes500 completein m2.Dillar is 50 cm in diameter and 3.5 m in height. Find the cost of paintingS Acyurved surface of the pillar at the rate of12.50 per m2. |
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Answer» 5)R = 25 cm = 0.25 mh= 3.5 mC.S.A = 2π r h= 2 ×22/ 7 ×0.25 × 3.5 = 5.5 m²cost of painting the curved surface of the pillar = 5.5× 12.50 = Rs 68.75 hit like if you find it useful |
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| 75405. |
Find the mean deviation about the mean for the following data.10-20 20-30 30-4040-5006070 70-30Marks obtainedNumber of students142 |
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| 75406. |
In the Given figure AB |ICD Find the values of x and y.7525° |
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| 75407. |
24. In the given figure, AB lI CD. If ZCAB-80°and LEFC = 25°, then <CEF = ?(a) 65°(c) 45°(b) 55°(d) 752580 |
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| 75408. |
(1) (5рео + 6b) |
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Answer» (5a'2 + 6b'2)25+36 add kar le àaaaaaaaàaaaaaaaa |
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| 75409. |
a+3b+6b=12 |
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Answer» a+3b+6b = 12 so, a+9b = 12=> a = 12-9b => a = 3(4-3b) |
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| 75410. |
v 1 L 312tan”' —+tan' —=tan™' —2 7 17 |
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Answer» formula.. plzz... you use in..?? |
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| 75411. |
(1+tan∅)²-sec²∅= 2tan∅ |
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Answer» Thank u so much |
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| 75412. |
2tan xProve that tan 2x =1-tan x |
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Answer» tan(2x) = (2 tan x) / (1 - tan²x) Take the LHS.tan(2x) = tan x = sin x / cos x.sin(2x) / cos(2x) = sin 2A = 2 sin A cos A.2 sin x cos x / cos(2x) = cos 2A = cos²A - sin²A.2 sin x cos x / (cos²x - sin²x) = Divide the numerator and denominator by cos²x.(2 sin x cos x / cos²x) / [(cos²x - sin²x) / cos²x] =[2 sin x(1) / cos x] / [(cos²x / cos²x) - (sin²x / cos²x)] =[2(sin x / cos x)] / [1 - (sin²x / cos²x)] = t tan x = sin x / cos x.(2 tan x) / [1 - (sin²x / cos²x)] = tan²x = sin²x / cos²x.(2 tan x) / (1 - tan²x) =RHS |
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| 75413. |
; 2tan 30° -1—tan?30° |
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Answer» tan30=1/√3Hence2*1/√3/1-1/3=3/√3=√3 |
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| 75414. |
0८% 1 कप Disjide DI 13l [ सकल को B ED Blo ek} W W £Ly(lage it 12 =¥ 128 & e का हा)7T Blnie T |
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| 75415. |
cos30=2tan 30/1+tan 30? |
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Answer» LHS = cos(30°) = √3/2 RHS = 2 tan(30°)--------------------1 + tan(30°) 2×(1/(√3))----------------- 1 + (1/√3) 2------------√3 + 1 So, RHS is not equal to LHS |
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| 75416. |
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painthe curved surface of the pillar at the rate of? 12.50 per m2 |
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| 75417. |
2.0, Ban ytes500 km from Delhi towards East direction. Bubly goes 200 km for Delhi towardWest direction. What is the distance Bubly has to cover to reach Bunty?rot |
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Answer» Bubly has to cover 500+200= 700 km in total 700 km distance bubly has to cover to reach bunty पटना की राजधानी कहाँ हैं |
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| 75418. |
the awenall the ange around a paunt ois egual cho 3605 |
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| 75419. |
(75*(25*25))/25 |
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Answer» 25 * 25 * 75 ÷ 25 Using BODMAS rule = 25 * 25 * 3 = 625*3 = 1875 25×25×75÷25=25×25×3=625×3=1875your answer is 1875 1875 is the best answer 25×25×75÷3=25×25×3=625×3=187t 25×25×75÷3=25×25×3=625×3=1875 1875 is right answer. 1875 is the correct answer 1875 is the best answer Given equation is 25 * 25 * 75 ÷ 25 = 1875. Using BODMAS rule 25×25×3 =1875 please like me 1875 is the answer which is correct 1857 is the correct answer....... |
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| 75420. |
ample 10Solve:- 22SolutionLCM of 4 and 6 = 12. therefore, multiply both sides of the equaWe have 2x3x 12** - 22 x 12 = 9x + 2x = 264 = 11x=000 = 24.Check: Left side - 3* 24+24= 18 + 4 = 22 = Right side.Self Practice 6BSolve each equation and check your answer.+ 18 = -246.157.61 + 1 - 11 - 2x) = -42+3I 32-110. - (8-61) =PROBLEMS BASED ON REAL LIFE SITUATIONSTo solve word problems, we first translate them into equations. Then we solveThe equations that we will make in this book are like those you have justshow how to turn word problems into equations.Sople in If 14 is added to a number, the sum is 35. Find theSolution Let the required number be x. When 14 is added to it.states that this sum is equal to 35.x + 14 = 35x = 35 - 14x =21The required number is 21.T umber is multiplied by 5 and 8 is subtracted |
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Answer» X/2+5=8X+10=16(multiply both sides by2)X=16-10X=6 x/2+5=8x+10=16(multiply by both sides)x=16-10x=6 x/2+5=8 = x+10 =16(multiply both sides by 2) = x = 16-10 = x = 6 is correct answer |
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| 75421. |
raus tho G |
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| 75422. |
Section-D3. Two water taps A and B together can fill a tank in 12 hours. Tap A takes 10 hours8x4=32less than the time taken by tap B to fill the tank separately. Find the time taken bytap B to fill the tank.OR |
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| 75423. |
Callbed. If the trihe scheduled timemore than the scheFi28 A train covered a certain distance at a30 km/h faster, it would have taken 2 hours less thansezone slower by 15 km/h, it would have taken 2 hours more thanFind the length of the journey.Hint Let the due speed of the train be x km/h and scheduled time hullength of journey = xy. Also (x +30) (y-2) = xy and (x - 15) (y + 2).- |
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Answer» Hi is Ka answer mujha Nahi pata ha yar sorry kya hova bol Bhai ha yar sorry I missed you |
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| 75424. |
Four times the area of the curved surface of acylinder is equal to 6 times the sum of the areas ofits bases. If its height is 12 cm, find its curvedsurface area |
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Answer» Curved Surface Area= 2πrh sq.cmArea of the base = πr² sq.cm.Sum area of the base= πr²+πr²= 2πr²sq.cmA.T.Q4(2πrh)= 6×2πr²8πrh=12πr²2h=3r2×12=3rr=8cm.Curved Surface Area=2πrh= 2×22/7×8×12=4224/7= 603.42 sq.cm |
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| 75425. |
5. The radii of two right circular cylinders are in the ratio 2: 3 and their heights arethe ratio 5: 4. Calculate the ratio of their curved surface areas and also the ratoof their volumes |
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| 75426. |
ISical value of\left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right) |
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Answer» thanks |
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| 75427. |
DTh each Pay aogls bloalchins angl and oile hm douneelown |
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Answer» Traingles need to provided for the solution |
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| 75428. |
\tan 70^{\circ}=\frac{\cos 25^{\circ}+\sin 25^{0}}{\cos 25^{\circ}-\sin 25^{\circ}} |
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| 75429. |
Example 2 : In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT areangle bisectors of Z POS and Z SOQ, respectively. If Z POS x, find Z ROT. |
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| 75430. |
a-ange also dvas the gooph ehe gaciph ctthe2 |
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| 75431. |
25 - 25 + 25 |
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Answer» 25 is the answer........... 25 - 25 + 25= 0 +25 = 25 25 is correct answer. -25 is the correct solution of given equation. 25165*5** |
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| 75432. |
A ain covered certain distance ar a uniform speed. If the train wouia have been 6 kn/hr faster,ld have taken 4 hours less than the scheduled time and if the train was slower by 6 km/lur, it wouldaken 6 hours more than the scheduled time. Find the length of the journey |
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| 75433. |
24. A man was driving a car with uniform speed 60km/h.write a linear a equation in two variabletime and distance .draw a graph |
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Answer» As the speed is uniform,so the graph will be a straight line .1. distance travelled in 5/2 hours= speed×timei.e. 60×5/2 = 150km.2.distance travelled= speed×time= 60× 1/2= 30 km. |
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| 75434. |
their helgub1their curved surface areas.6. The outer diameter of a metallic pipe is 18 cm andthe thickness of the pipe is 2 cm and the height is4 cm. Find the surface area of the metallic pipe.A rectangular sheet of paper of dimensions |
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| 75435. |
edge).The thickness of a holloof the wood required to make the cylinder, assuming it is open at either end.4.w wooden cylinder is 2 cm. It is 35 cm long and its inner radius is 12 cm. Find the volume |
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| 75436. |
15. Find the value of the expressions using identities.(a) 25x2 + 70x + 49 if x = -1(b) 49X2 – 84xy + 36y2 if x = 2, y = 3 |
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Answer» 15) a) 25x^2+70x+49; x=-1; =25(-1)^2+70(-1)+49= =25-70+49=74-70=4; b) 49x^2-84xy+36y^2; x=2, y=3; 49(2)^2-84(2)(3)+36(3)^2 =49(4)-84(6)+36(9) =196-504+324=520-504=16 |
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| 75437. |
eachfind9. In the figure given below, ray OS stands onine POQ. Ray OR and ray OT are angl rsPOSofand 2SOQ respectively. If 2POSx, findZROT |
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Answer» If angle POS = x degrees, then angle ROS = ½ x degrees. Angle SOQ is the supplement to angle POS, and thus it equals 180 - x degrees. Angle bisector OT causes angle SOT to measure (180 - x)/2 = 90 -½ x. Observe that angle ROT = angle ROS + angle SOT = ½ x + (90 -½ x) = 90o. |
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| 75438. |
(5x +7) (5x +7) is(A) 25x2 + 70x + 49(B) 25%-70x + 49(C) 5x2 + 70x + 7(D) 5x2-70x + 7 |
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Answer» (5x+7)(5x+7) = 25x²+35x+35x+49 = 25x²+70x+49 |
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| 75439. |
20+64x+70x |
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Answer» 20+134x will be the answer |
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| 75440. |
гvalog out accation of ).Example 18.12. An embankment, 5 m high, is made up of soil whose effectivestress parameters are d' = 50 kN/mʼand ' = 16°, and y = 16.2 kN/m . The pore pressureparameters as found from triaxial test are A=0.40 and B=0.92. Find the shear strengthof the soil at the base of the embankment just after the fill has been raised from 5 mto 9 m. Assume that the dissipation of pore pressure during this stage of constructionis negligible, and that the lateral pressure at any point is one-half of the vertical pressure.162 IN |
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Answer» the correct answer is 0.40 |
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| 75441. |
If ÎABC with sides 12cm, 5cm and 13cm revolved about 12cm side. Find volume and CSA of cone which generated by revolution. |
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| 75442. |
ine tiue sations no2. ) 3 and A |
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Answer» 1)The rational number between a and b is (a + b)/2 Hence rational number between 1 and 2 is (1+2)/2 = 3/2 Rational number between 1 and 3/2 is [1+(3/2)]/2 = 5/4 Rational number between 1 and 5/4 is [1+(5/4)]/2 = 9/8 Rational number between 3/2 and 2 is [(3/2) + 2]/2 = 7/4 Rational number between 7/4 and 2 is [(7/4) + 2]/2 = 15/8 Therefore five rational numbers between 1 and 2 are 9/8, 5/4, 3/2, 7/4 and 15/8 hit like if you find it useful |
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| 75443. |
s. A solid cube of metal each of whose sides measures 2.2 cm is melted to form a cylindricalwire of radius 1 mm. Find the length of the wire so obtained |
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| 75444. |
3. Each edge of a cubical box measures 160 cm. How many cubical packets of edge 8 cm can be packedin it? |
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Answer» no of packets = volume of box/volume of each packet = 160×160×160/8×8×8 = 20×20×20 = 8000 |
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| 75445. |
16. A solid cube of metal each of whose sides measures 2.2 cm is melted to form a cylindricalwire of radius 1 mm. Find the length of the wire so obtained. |
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| 75446. |
8.What is Condensation? |
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Answer» Condensation is the process by which water vapor in the air is changed into liquid water. I Condensationis the process by which water vapor in the air is changed into liquid water. In other words, the water in the air, a gas known as water vapor, from your hot shower cooled when it met the surface of the cold mirror. This caused the water vapor tocondense, or turn into its liquid form |
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| 75447. |
andiculaabebileeno ine Pa.asAnother ay |
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| 75448. |
ine Seomant Pa 12cm and Ri appsit |
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| 75449. |
Show that the volume of a cube, whose each edge measures 12 m, is 9 times the volumeof a cuboid of dimensions 8 m x 6 m x 4 m. |
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Answer» Volume of the cube with each edge measuring 12 m=(12×12×12) cubic metres =1728 cubic metres. Volume of the cuboid measuring (8 m ×6 m ×4 m )=(8×6×4) cubic metres. =192 cubic metres. |
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| 75450. |
Two coins are tossed simultaneously. Find the probability of getting :(i) exactly 1 head(ii) at most 1 head(ii) at least I head. |
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