Saved Bookmarks
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. |
5. Prove that the perpendicular at the point of contact to the tangent to a circle passesthrough the centre. |
| Answer» | |
| 2. |
Q.5Prove that thecontact to the tangent to acircle passes through thecentre. |
| Answer» | |
| 3. |
√64/³√64 |
|
Answer» 64=8*864=4*4*4hence√64=8and 3√64=4hence8/4=2 |
|
| 4. |
Prove that out of all chords which passesthrough any point circle, that chord will besmallest which is perpendicular on diameterwhich passes through that point. |
|
Answer» Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle. Let CD is another chord passes through point M. We have to prove that AB < CD. Now join OM and draw OL perpendicular to CD. In right angle triangle OLM, OM is the hypotenuse. So OM > OL => chord CD is nearer to O in comparison to AB. => CD > AB => AB < CD So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point. |
|
| 5. |
\cos 64^{\circ} \cos 19^{\circ}+\sin 64^{\circ} |
|
Answer» cos(A) cos(B) - sin(A) sin(B) = cos(A-B)so cos(64°) cos(19°) - sin(64°) sin(19°) = cos(64 - 19) = cos(45°) = 1/√2 |
|
| 6. |
The value of 64+64 |
| Answer» | |
| 7. |
x^3/64 - 64/x^3 |
|
Answer» x^3/64 - 64/x^3 = (x/4)^3 - (4/x)^3 Using identitya^3 – b^3 = (a – b) (a^2 + ab + b^2). = (x/4 - 4/x)(x^2/16 + x/4*4/x + 16/x^2) = (x/4 - 4/x)(x^2/16 + 1 + 16/x^2) |
|
| 8. |
\frac { x ^ { 3 } } { 64 } - \frac { 64 } { x ^ { 3 } } |
|
Answer» plzz ono nioymay dau na ai tai hocha na |
|
| 9. |
Prove that the perpendicular at the point of contact to the tangent to a circle passesphrough the centre. |
| Answer» | |
| 10. |
j)25(u(す) 100(3) 1922(3) 15(3) 950(a) 1× 1000(B) 1650 30(5) 94 × 100(c) 25(B) 5235 × |
|
Answer» like kro agar uttar se satisfied ho to |
|
| 11. |
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre |
| Answer» | |
| 12. |
^{20} C_{5}+\sum_{j=2}^{5} 25-j_{C_{4}} |
| Answer» | |
| 13. |
1 ~ 161 + 64 |
|
Answer» l² - 16l + 64 l² - 8l - 8l + 64 l(l - 8) - 8 ( l - 8) ( l-8)² |
|
| 14. |
0,1, 12-13, 1+161+19yo |
|
Answer» sum= 10+11+12+13+14+15+16+17+18+19= 145total= 10 mean = 145/10= 14.5 |
|
| 15. |
1. A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end ofthe diameter through Ais(A) (y-q) 4px (B) (x-4py (C) (y -p)4qx(D) (x-p4qy |
| Answer» | |
| 16. |
1. Evaluate each of the following(i) 161/2 |
|
Answer» 16 raised to the ½ power is equivalent to saying the square root of 16. The square root of 16 is the number, which, when multiplied by itself equals 16. 4 multiplied by 4 equals 16. Therefore, the square root of 16 is plus or minus 4. |
|
| 17. |
11.12. J25 - let's find its value.NE - let's find the value.J |
| Answer» | |
| 18. |
5. 2x-3y + 13=0,3x-2y + 12 = 0equatigns |
|
Answer» 2x-3y=-133x-2y=-12 6x-9y-6x+4y=-39+24-5y=-15so y=32x-9=-132x=-4so x=-2 |
|
| 19. |
with thED, prove that AB = cirstand Mandip arelaving a game by standing on a circle of radius3m drawn in a park. Reshma throws a balltoSalma, Salma to Mandip, Mandip to Reshma. Ifthe distance between Reshma and Salma andbetween Salma and Mandip is 6m each, what isThree girls Reshma. Salmathe distance between Reshma and Mandip? |
| Answer» | |
| 20. |
Example Observe Fig. 6.30 and then find PIR7.63.60Сре12 |
| Answer» | |
| 21. |
The slant height of a frustum of a cone is 4 cm and the perimeter (circumference) of its circular ends are 18cm and 6 cm. Find the curved surface area of the frustum. |
|
Answer» Given slant height(l)=4cm Circumference1(C1)=18cm 2 pie r1 = 18cm 2*22/7*r1 = 18cm therefore, r1 = 63/22cm similarily C2 = 6cm so, r2 = 21/22cm now CSA of frustum of cone = pie l (r1+r2) 22/7*4(63/22+21/22) 84/22 * 4 * 22/7 =48 sq. cm.------------Ans. |
|
| 22. |
of 12 and 36.. Find the H.C.Fwhat do you observe ? |
|
Answer» the HCF of 12and 36is 12 12................... |
|
| 23. |
A variable circle having fixed radius 'a' passes throughorigin and meets the coordinate axes in point A and B.Locus of centroid of triangle OAB, 'O, being the origin, is(a) 9(x2 +y?) 4a2(c) 9(x2 + y?) 2a2(b) 9(x2 + y?) a2(d) 9(x2 + y?) 8a2 |
| Answer» | |
| 24. |
09.r = 161161= 99161..1.62 = 20Try THESEI. Find the decimal values of the following:1.iii.v1.vii.IX+Observe the following decimals |
|
Answer» 0.142857 is the correct answer |
|
| 25. |
In a AABC, 4B-90°, AB-5cm and (BC+ AC)-25 cm. Findthe values of sin A, cos A, cosec C and sec C. |
| Answer» | |
| 26. |
Find the ratio in which the line seghent J11. Solve: 99x + 101y499;101x + 99y= 501 using suitabat civon that VE is irrational. |
| Answer» | |
| 27. |
99x+101y=499;101x+99y=501 |
|
Answer» Consider 99x+101y=499_______(1) 101x+99y=501 ________(2)multiplying equation (1) by -101 we get-9999x-10201y=-50,399_________(3)multiplying equation (2) by 999999x+9801y=49,599___________(4)adding (3) and (4) we get-400y=-800y=2substituting y=2 in equation (1) we get99x+202=49999x=499-20299x=297x=297/99x=3 |
|
| 28. |
\begin{array} { l } { \text { Solve the following pairs of equations } } \\ { \frac { 1 } { 2 x } + \frac { 1 } { 3 y } = 2 } \\ { \frac { 1 } { 3 x } + \frac { 1 } { 2 v } = \frac { 13 } { 6 } } \end{array} |
| Answer» | |
| 29. |
\begin{array}{c}{\text { Solve the following pairs of equations }} \\ {\text { (i) } \frac{1}{2 x}+\frac{1}{3 y}=2} \\ {\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}}\end{array} |
| Answer» | |
| 30. |
x2÷x5+×4+x3 |
| Answer» | |
| 31. |
orresponding sides of the AABCProve that:fProve that 2cos A - coSAsin A-2sin A28. The diameters of the lower and upperends of a bucket in the from of a frustum of a cone are 10 cm and 30 cmrespectively. If its height is 24 cm, find:The area of the metal sheet used to make the bucket.(0(ai) Why we should avoshserved from the top of a 100 m high light house from the sea-level, the angles of depression of two ships areid the bucket made by ordinary plastic ? [Use r 314in Ä°s exactly behind the other on the same side of the lighthouse, find the distance between |
| Answer» | |
| 32. |
23. If A 3x + 8 and B x2 +2 then AB?3x3 +8x2 +6x +163) x3 +8x2 +12x +162) x3 +8x2 +6x +84) of these. |
| Answer» | |
| 33. |
5. The diagonals of a rhombus are 7.5 cn and 12 cm. FindObserve thatName them. Whiits area |
| Answer» | |
| 34. |
7) Two-Dimensiona! FiguresName each polygon by its mumber of sides. Then 3)classify it as convex or concave and regular or area of each figureirregulor. Round to the nearest3 cm4) Find the perimeter er circumference ond area 5) Find the perimeter or circunference andof each figure. Round to the neorest tenttenth6) Graph eech figure with the given vertices ond identify the figure. Then find the perimeterand orea of the figure |
| Answer» | |
| 35. |
99x + 101y = 499; 101X + 99y = 501 |
| Answer» | |
| 36. |
(v) Solve 99x + 101y = 499101x +99y = 501 |
|
Answer» 99x + 101y= 499 .......1101x + 99y = 501......2By subtracting 1 and 2101x + 99y = 501-99x - 101y =-499______________2x - 2y = 22 (x-y) = 2x-y =1_____3By adding 1 &2101x+99y=50199x+101y=499_________________ 200x+200y = 1000200(x+y)=1000x+y = 5______4By adding 3 and 4x-y=1x+y=5_______2x=6x=3 x-y=13-y=1-y= -2y=2 |
|
| 37. |
write" - AndC hanged at detAb undANis+A-= courtABcotA + tan Al(sin A-Cost) = sinna A- Cos A cotAABC right anged at 8he opposite side andigonometric ratio ofAB cos 1908 - A-)ACBC sec 15A)AB |
|
Answer» (1 + cotA + tanA)(sina-cosA)= ( 1 + cosa/ sinA + sina/ cosa)( sinA- cosa)= ( sinacosA+ cosa^2 + sina^2/ sinA cosa + ( sina- cosa) = ( sinacosa + 1) + (sina/ cosa sina + cosa/ cosa sina) = Sina cot a- cosatana L.H.S=(1+cotA+tanA)(sinA-cosA)=sinA-cosA+cotAsinA-cotAcosa+tanAsinA-tanAcosA=sinA-cosA+cosA/sinA×sinA-cotAcosA+tanAsinA-sinA/cosA×cosA=sinA-cosA+cosA-cotAcosA+tanAsinA-sinA=sinAtanA-cosAcotA LHS (1 + cota + tana)( sinA - cosa)= (1+ cosx/ sinx + sina/ cosa)( sina-cosa)= =( cosxsinx + cosx^2+ sinx^2)/ sinxcosx ( sinx - cosx) = (cosxsinx+1 /sinxcosx) ( sinx - cosx) = cosxsinx^2- cosx^2sinx- sinxcosx/ sinxcosx = cosx(1- cosx^2) - sinx(1-sinx^2) = cosx^3 - sinx^3/sinxcosx = cosx^2 / sinx - sinx^2/ cosx = cotx/cosecx - cosecx/ secx^2 |
|
| 38. |
square each of these and show that they call be lewlu. Use Euclid's division lemma to show that the cube of any positive integer is of theform 9m. 9m+ 1 or gm+ 8. |
|
Answer» Let a be any positive integer and b = 3a = 3q + r, where q ≥ 0 and 0 ≤ r < 3Therefore, every number can be represented as these three forms. There are three cases.Case 1: When a = 3q,Where m is an integer such that m = Case 2: When a = 3q + 1,a3= (3q +1)3a3= 27q3+ 27q2+ 9q + 1a3= 9(3q3+ 3q2+ q) + 1a3= 9m + 1Where m is an integer such that m = (3q3+ 3q2+ q)Case 3: When a = 3q + 2,a3= (3q +2)3a3= 27q3+ 54q2+ 36q + 8a3= 9(3q3+ 6q2+ 4q) + 8a3= 9m + 8Where m is an integer such that m = (3q3+ 6q2+ 4q)Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8. |
|
| 39. |
YRoO'=Sl 1.1पुणेठा वते:2: ~Gar135 M3 225 (07196 73 38220' ४.०. ey Sकि ¥£/ देठां सिडीगां मघिशाहां रा W R 2 (H.C F) U3 aua w8L H.2 (HC |
|
Answer» (i) 135 ਅਤੇ 225 225> 135 ਤੋਂ ਲੈ ਕੇ ਅਸੀਂ ਡਿਵੀਜ਼ਨ ਲੇਮਮਾ ਨੂੰ 225 ਅਤੇ 135 ਤੇ ਪ੍ਰਾਪਤ ਕਰਦੇ ਹਾਂ 225 = 135 × 1 + 90 ਬਾਕੀ 90 ≠ 0 ਤੋਂ, ਅਸੀਂ ਡਿਵੀਜ਼ਨ ਲੀਮਮਾ ਨੂੰ ਪ੍ਰਾਪਤ ਕਰਨ ਲਈ 135 ਅਤੇ 90 ਤੇ ਲਾਗੂ ਕਰਦੇ ਹਾਂ 135 = 90 × 1 + 45 ਅਸੀਂ ਨਵੇਂ 90 ਡਿਵੀਜ਼ਨ ਅਤੇ ਨਵੀਂ ਬਾਕੀ 45 ਨੂੰ ਵਿਚਾਰਦੇ ਹਾਂ, ਅਤੇ ਪ੍ਰਾਪਤ ਕਰਨ ਲਈ ਡਿਵੀਜ਼ਨ ਲੇਮਮਾ ਲਾਗੂ ਕਰਦੇ ਹਾਂ 90 = 2 × 45 + 0 ਕਿਉਂਕਿ ਬਾਕੀ ਦਾ ਸਿਫਰ ਹੈ, ਪ੍ਰਕਿਰਿਆ ਰੁਕ ਜਾਂਦੀ ਹੈ. ਕਿਉਂਕਿ ਇਸ ਪੜਾਅ 'ਤੇ ਭਾਗੀਦਾਰ 45 ਹੈ, ਇਸ ਲਈ, 135 ਅਤੇ 225 ਦੇ HCF 45 ਹੈ. (ii) 196 ਅਤੇ 38220 38220> 196 ਤੋਂ ਲੈ ਕੇ ਅਸੀਂ ਡਿਵੀਜ਼ਨ ਲੇਮਮਾ ਨੂੰ 38220 ਅਤੇ 1 9 6 ਤਕ ਲਾਗੂ ਕਰਦੇ ਹਾਂ 38220 = 196 × 195 + 0 ਕਿਉਂਕਿ ਬਾਕੀ ਦਾ ਸਿਫਰ ਹੈ, ਪ੍ਰਕਿਰਿਆ ਰੁਕ ਜਾਂਦੀ ਹੈ. ਕਿਉਂਕਿ ਇਸ ਪੜਾਅ 'ਤੇ ਭਾਗੀਦਾਰ 196 ਹੈ, ਇਸ ਲਈ, 1 9 6 ਅਤੇ 38220 ਦੀ ਐਚ ਸੀ ਐੱਫ 196 ਹੈ. please like the solution 👍 ✔️ |
|
| 40. |
2Lend the value-odcand. |
| Answer» | |
| 41. |
find the sum of 256 cand its addictive inverse |
| Answer» | |
| 42. |
d. 12. If a Matrix is of order 4 x2, then the Number of elements isĐ°. 4b. 2a.8 |
|
Answer» Given : order of matrix is 4 × 2 Numbers of elements = 8 option (d) is correct thank u |
|
| 43. |
2. Arrange the following in ascending order42473355 |
|
Answer» L.C.M of the denominators=354/7×5/5=20/352/5×7/7=14/354/35×1/1=4/358/7×5/5=40/35ascending order means small nos. to big nos.=4/35<14/35<20/35<40/35=4/35<2/5<4/7<8/7 ans. |
|
| 44. |
A triangle with vertices (4, 0), (-1, -1), 3, 5) is |
| Answer» | |
| 45. |
27. The circumcentre of the triangle with vertices (4, 0),(O, O) and (O, 6) is.IS |
|
Answer» let A (4,0) be (x1,y1) and B (0,0) (x2,y2) and C(0,6) (x3,y3) find circumcentre (x,y) by centroid formula x=x1+x2+x3/3 and y=y1+y2+y3/3x=4+0+0/3 and y=0+0+6/3x=4/3 and y=6/3therefore, circumcentre (x,y) is (4/3,2) Like my answer if you find it useful! |
|
| 46. |
9.Find the sum of:-73wand -1-(0) 21:14(b) and(c)and(d) |
| Answer» | |
| 47. |
7 18 369 -14 21 |
|
Answer» 7/9*18/-14*36/211/1*2/-2*36/21=-36/21=-12/7 Refer to the above attachment * this one |
|
| 48. |
\frac { 4 } { 21 } \div \frac { 9 } { 14 } |
|
Answer» 4/21 ÷ 9/14=4/21 × 14/9=4/3 × 2/9=8/27 |
|
| 49. |
27. What type oflyuregehcarbonatepowder?graph is obtained between current flowing through a conductor and potentialsignificance.a esperiment to verify Ohn's law. Draw the graph and write it |
|
Answer» When we draw a curve between potential difference and current through conductor then according to Ohm's law (Potential difference = current flowing x R) we can see that the equation is of the format : y = mx +c where m is the slope and c is the point on y-axis ,i.e, when x=0. Hence the graph is a straight line with m = R and c = 0. Here R is the resistance. |
|
| 50. |
Line segments AB and CD intersect at O such that AC | DB.If < CAB = 35° and < CDB = 55°, then BOD- |
|
Answer» Tanks |
|