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.
| 6351. |
(O) In how many ways can 4 boys and 3 girls beseated in a row of 7 chairs if boys and girlsalternate ?(ii) In how many ways can 4 boys and 3 girls beseated in a row so that no two girls are together ?(ii) In how many ways can 5 girls and 3 boys beseated in a row so that no two boys are together ?(N.C.E.R.T. H.P.B. 2o11 |
| Answer» | |
| 6352. |
Akb z zas58. On the set of positive rationals, abinary operation * is defined bya* b = 2abgreetIf 2 * x = 3-1 then x =-10 minulgal2012 |
|
Answer» c is correct good b answer c) 5\12 is the correct answer of the following |
|
| 6353. |
40, 287 × 287 + 269 × 269-2 × 287 × 269 = ?(a) 534(b) 446 |
|
Answer» 287×287-2×269×287+269×269 = (287-269)(287-269) = 18×18 = 324 |
|
| 6354. |
8, Two co-ed school in your locality have 80 and 120 children respectively. In the first school, 60% andin the second 45% ofthe children are boys. what percent of the children in the two schoolsare boys? [Ans: 51%) |
|
Answer» 60*80/100 = 48 students 45*120/100 = 54 students total percent = 48+54/80+120 * 100 = 102/200 * 100 = 51% |
|
| 6355. |
287 x 287 + 269 x 269 â 2 x 287 x 269 =?) |
|
Answer» Let A = 287 B = 269 Given 287²+269²-2*287*269 We know that (a-b)² = a²+b²-2ab = (287-269)² = 18² = 324 Like my answer if you find it useful! |
|
| 6356. |
कि 2 उनe A IO—GES“— . 395 |
|
Answer» 1/4=2/86/10 is greater than 4/5 3/4 is less than 7/8 |
|
| 6357. |
χ-30 and ges acute angle..93-6 |
| Answer» | |
| 6358. |
5. 287 x 287 - 2 x 287 x 269 + 269 X 269 = ?(a) 556 (b) 446 (0) 354 (d) 3 |
|
Answer» 324 thanks u |
|
| 6359. |
A man of 1.5meter height is standing at a distance of 10V3 meter from a building. If the angle of elavationfrom his eyes to the top of the building is 60°. Find the height of the building, |
| Answer» | |
| 6360. |
p2 = a cos20 + b2 sin20 269 came ,p.+ dp-a262dO2p3 |
|
Answer» lnp^2=a^2+(b^2-a^2)sin^2x 2lnp=a^2+(b^2-a^2)sin^2x Now differentiate w.r.t.x 2/p(dp/dx)=(b^2-a^2)2sinx cosx=(b^2-a^2)sin2x dp/dx=p/2(a^2-b^2)sin2x Again defferentiate w.r.t x d^2p/d^2x=1/2(dp/dx)(a^2-b^2)sin2x+p(a^2-b^2)cos2x =p/4(a^2-b^2)^2sin^2(2x)+p(a^2-b^2)cos2x |
|
| 6361. |
7. If v2 is irrational number. Prove that 3-27 is also irrational.269 |
|
Answer» Lets assume that√7 is rational number.ie√7=p/q.suppose p/q have common factor thenwe divide by the common factor to get√7 =a/b were a and b are co-prime number.that is a and b have no common factor.√7 =a/b co- prime number√7= a/ba=√7bsquaringa²=7b² .......1a² is divisible by 7a=7csubstituting values in 1(7c)²=7b²49c²=7b²7c²=b²b²=7c²b² is divisible by 7that is a and b have atleast one common factor 7. This is contradictory to the fact that a and b have no common factor.This is happen because of our wrong assumption.√7 is irrationalHence an irrational number multiplied by 2 is irrational hence 2√7And a number added or subtracted to rational is irrationalhence 3-2√7 is irrational u |
|
| 6362. |
EXERDISE TE7ice a mumibver wihen decrased by 7 ges 45 Fme the umbe |
|
Answer» let the number be Xhence2x-7=452x=52X=26 thanks |
|
| 6363. |
01 =47 +27 ‘g=d+x Q.N 0 1% |
|
Answer» x+y= 5and 2x+2y= 10as Sol :Here the pair of linear equations arex+y = 5 2x+2y = 10(a1/ a2) = (b1/ b2)= (c1/ c2) =1/2the lines are coincident. |
|
| 6364. |
10. Two cones have their heights in the ratio 1:3 and the radii of their bin the ratio 3:1. Show that their volumes are in the ratio 3:1. |
|
Answer» thanks |
|
| 6365. |
15. In the given figure, O is a point in the interiorof square ABCD such that AOAB is anequilateral triangle. Show that AOCD is anisosceles triangle.seen atD, as showthe imageas the obje24 In the acne can |
|
Answer» Given:OAB is equilateral triangle. to prove:OCD is an isosceles triangle. construction:from point O, draw line from O to the points A,B,C,D. proof:line AC & BD are 2 diagonals equal. point O bisects all the diagonals. thus,AO=BO=CO=DO...CO is same as DO which proves that, OCD is isosceles. |
|
| 6366. |
(c) 7 m(d) none of theseThe radii of the bases of a cylinder and a cone are in the ratio 3: 4 and their heights are in the ratio: 3, then the ratio of their volumes is(a) 2:3(b) 9:8(c) 3:4 |
|
Answer» kaise aya or yaha b options me 9:6 malab 9:8 |
|
| 6367. |
of the base of a cylinder and a cone are in thThe radii of the base of a cylinder and a cone are in the ratio 3:4 and their heights ain the ratio 2: 3, then their ratio of volumes is -(A) 9:8(C) 3:e ratio 3:4 and their heighs ar(B) 9:4(D) 27:64 |
| Answer» | |
| 6368. |
two cylindrical jars have their diameters in the ratio 3:1 but height 1:3. Then the ratio of their volumes is |
| Answer» | |
| 6369. |
Two cylinders have their radii in the ratio 3 : 1but their heights are in the ratio 1 :3. Comparetheir volumes. |
| Answer» | |
| 6370. |
Two cylinder jars have their diameter in the ratio 3:1 but height 1:3.then find the ratio of their valumes? |
| Answer» | |
| 6371. |
3. Two numbers are in the ratio 3:5. If 7 is added toeach one, then the ratio becomes 11 : 16. Find thenumbers.Solution: |
| Answer» | |
| 6372. |
9 i अ०- il Ti नेT ©v W)97791 ‘01 ‘v dV का- et SR |
|
Answer» a1 = 6a2= 10d = 10-4 = 6 |
|
| 6373. |
One pot contains 8 litre of milk while other contains 750 milliliter. |
|
Answer» please post the complete question should find the ratio ans=32:3 but how . What is the process iam asking |
|
| 6374. |
10) If a tower 30m high,casts a shadow 10.3w 10V3 m fong(01)the angle of elevation of the sun?is 100 m. When the angle of elevation of Sun is 30, then(01)11) The height of the towerwhat is the length of shadow of the tower?line of sight with the horizontal when the object(01) |
| Answer» | |
| 6375. |
-3. Divide:(11 116v 7OBJE24 by 3(11) 6? by 11001(11) 5 by 3Mar!Mar(iv) 32 by 1Cico 01(w) 45 by 1(vi) 63 by 2 |
|
Answer» thanks |
|
| 6376. |
Write the name of universal donor blood |
|
Answer» Ans :- O negative In transfusions of packed redbloodcells, individuals withtypeO Rh D negativeblood are oftencalled universal donors. एल्कीन श्रेणी का सामान्य सूत्र लिखिए |
|
| 6377. |
1)Is 3456 divisible by 6? apply divisible rule. |
|
Answer» Divisibility by 6 is the nubmer is divisible by both 2 and 3 3456/2 = 1728 3456/3 = 1152 So, the number is divisible by 6 3456 is divisible by 6 because when the last number is even then the number will be divisible by 6 and answer is that when we divisible by 6 answer is 576 |
|
| 6378. |
1.2.Area of a circle is given by π. Find the area of a enArea of a triangle having sides a, b and c is given by the following expression,where s = a + b +cArea s(s - a)(s-b)(s -c)Find the area of a triangle having sidcs 3, 4 and 5 units.72 |
|
Answer» s=(3+4+5)/2 = 6area = root of [6×(6-3)×(6-4)×(6-5)]= root of (6×3×2×1)=root of(36)6 unit sq. |
|
| 6379. |
define 1. area 2. perimeter |
|
Answer» 1. Area is the size of a two-dimensional surface. 2. The continuous line forming the boundary of a closed geometrical figure. |
|
| 6380. |
28. Milk in a container, which is in the form of a frustum of a cone of height 30 cm and the radi ofwhose lower and upper circular ends are 20 cm and 40 cm respectively, is to be distrilbuted in acamp for flood victims. If this milk is available at the rate of Rs 35 per litre and 880 litres ofmilk is needed daily for a camp, find how many such containers of malk are needed for a cangand what cost will it put on the donor agency for this. What value is indicated through this bythe donor agency?imla triangles is equal to the square of the ratio of |
|
Answer» Given--Radius of lower end (R1)= 20cm Radius of upper end (R2)= 40 cm Height (H)= 30 cm So the volume of container= 1/3 pi H ( R1² + R2² + R1R2) = 1/3 * 22/7 * 30 ( (20)² + (40)² + 40* 20) = 1/3 * 22/7 * 30( 400 + 1600 + 800) = 1/3* 22/7 * 30* 2800 = 88000 cm³ or 88 litres. Amount of milk required = 880 litreNo. of containers needed will be= 880/88= 10 containers. Cost of 1 litre milk= Rs. 30So, Cost of 880 litre= 880 * 30 = Rs. 26400 |
|
| 6381. |
9, An electric pole is 10 m high. A steel wire tied to the top of the pole is affixed at at on the ground to keep the pole upright. If the wire makes an angle of 45째 withthe horizontal through the foot of the pole, find the length of the wire.poin |
| Answer» | |
| 6382. |
The sides of a triangle are 15 cm, 36 cm and 39 cm. Verify that it is a right angledand 14 m stand upright on a plane ground. li the distanc |
|
Answer» excellent thanks bro hello bro plz answer my questions so that I learn more for my exam iam house wife iam not getting time to solve my problem i will great ful to u |
|
| 6383. |
If the length of the arc of a quadrant of a circle is 11 cm, find itsarea2 |
| Answer» | |
| 6384. |
ittl tu 8ne decimalc pole is 10 m high. A steel wire tied to the top of the pole is affixed at aon the ground to keep the pole upright. If the wire makes an angle of 45 withpoint onontal through the foot of the pole, find the length of the wire. |
| Answer» | |
| 6385. |
From the top of a tower 50 m high, the angle of depression of thetop of a pole is 45° and from the foot of the pole, the angle ofelevation of the top of the tower is 609. Find the height of the pole ifthe nole and tower stand on the same plane. |
| Answer» | |
| 6386. |
List the possible outcomes, when a die is thrown once.die is thrown once. Find the probability of:a) getting a number > 3, andb) getting a number <3. |
|
Answer» Thanks |
|
| 6387. |
tan1°.tan2°.tan3°.tan4® ......... tan87° tan88°.tan89°के मान है |
| Answer» | |
| 6388. |
tan4 4(1 + tan24)? |
| Answer» | |
| 6389. |
4, x in quadrant 11= |
| Answer» | |
| 6390. |
in each of the followiFind the value of sin, cos and tantanx = - , x in quadrant 11 |
| Answer» | |
| 6391. |
dydx(11) If y = x), then --= |
| Answer» | |
| 6392. |
x^2 %2B 7*x %2B 1000 |
|
Answer» x^2+17x; x^2=17x; x=17 x2 + 17xx=17/2 this question is right answer x^2+7x+10x x^2+17x x (x+17) x=0 ; x+17=0x=-17 x(X+5x)+2x(X+5)=0now (X+2)+(X+5)=0ans X+2=0 x =-2 or ans X+5=0 x=-5both answer are in - that why no answer is there (x+5) (x+2 ) answer is ryt |
|
| 6393. |
Il) Maximum height is the maximum vertical distancetravelled by the projectile from the horizontal planousin0 |
|
Answer» the velocity in vertical direction is given by Vi = usin∅ so, using Newton's equation of motion at Highest point , the velocity is 0 => Vf²= Vi² - 2gH=> 0 = u²sin²∅ -2gH=> 2gH = u²sin²∅=> H = (u²sin²∅)/2g |
|
| 6394. |
\int \operatorname { sec } ( 11 + 12 x ) \operatorname { tan } ( 11 + 12 x ) d x |
| Answer» | |
| 6395. |
(2) tan 11+ tan+.. -o tan3 (b) tan() tan4-1-1(a) tan 3 (b) tan |
|
Answer» no wait... tan^-1(1) + tan^-1(1/2) + ....... = π/2 => π/4 + tan^-1(1/2) + ...... = π/2=> tan^-1(1/2) + ....... = π/2 -π/4 = π/4=> tan^-1(1/2) + tan^-1(x) = π/4=> (1/2+x)/(1-x/2) = tan(π/4)=> (1+2x)/(2-x) = 1=> (1+2x) = (2-x)=> 3x = 2-1=> x = 1/3 so, tan^-(1/3) is the answer. |
|
| 6396. |
( 11 ) \tan 18 ^ { \circ } + \tan 27 ^ { \circ } + \tan 18 ^ { \circ } \cdot \tan 27 ^ { \circ } = 1 |
|
Answer» Let us add 18 and 27 to get 45. So tan(18+27) = 1 = tan 18 + tan 27/[1 - tan 18.tan 27] or tan 18 + tan 27 = 1 - tan 18.tan 27, or tan 18 + tan 27 + tan 18.tan 27 = 1. Hence the answer is 1. |
|
| 6397. |
Gi12 24 aw 92 |
|
Answer» LCM of 18, 24 and 32 is288. Now, we have to get the multiple of288which also should be the samllest 4-digit number. Since288is divisible by 18, 24 and 32,1152is also divisible by all these numbers. Therefore,1152is the smallest four digit number divisible by 18, 24 and 32. |
|
| 6398. |
\frac { \cos 11 ^ { \circ } + \sin 11 ^ { \circ } } { \cos 11 ^ { \circ } - \sin 11 ^ { \circ } } = \tan 56 ^ { \circ } |
|
Answer» thanks |
|
| 6399. |
Find sinx, cosand tanin each of the following :48., xin quadrant 11tan x10.sinx =-, xin quadrant 114 |
| Answer» | |
| 6400. |
3y = tan-11dx13] |
| Answer» | |