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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. |
ooe crcie so formed.7. A piece of wire is bent in the shape of an equilateral triangle of each side 6,6 cm. Itform a circular ring. What is the diameter of the ring? |
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| 2. |
from A to this circle.7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pairof tangents from this point to the circle |
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| 3. |
2Ahmed buys a plot of land for? 480000. He sells-of it at a loss of 6%. At what gain per centshould he sell the remaining part of the plot to gain 10% on the whole? |
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Answer» Buys land at = 4800002/5*480000=192000 rsloss=6%100-6/100*192000=180480 total gain =10%100+10/100*480000=528000 sp=528000-180480=347520 cp of remaining part of land =3/5*480000=288000 Gain% of remaining part of land= sp-cp/cp*100 <=>347520-288000/288000*100=20.6 % thanks |
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| 4. |
Ahmed buys a plot of land for 96000. He sells of it at a loss of 6%. At what gain per centshould he sell the remaining part of the plot to gain 10% on the whole?25 |
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| 5. |
cp=25,sp=56 profit find it |
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Answer» Given : CP = 25 and SP = 56 As CP < SP There is profit Profit = SP - CP = 56 - 25 = 31 |
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| 6. |
). What happens to the volume of a cube when itsedge is (a) tripled (b) one third ? |
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Answer» Volume of cube is a^3. If it's tripled,Volume =(3a)^3 =27a^3So Volume becomes 27 times.If it's made one thirdVolume=(a/3)^3=a^3/27 So Volume becomes 1/27 times. |
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| 7. |
1. Find the surface area of a cube whose(7) edge is 8 cm,ii) perimeter of one face is 20 cm |
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Answer» 1) Edge of cube = 8 cmSurface area = 6*edge*edge = 6*8*8 = 384 cm^2 2) Perimeter of one face = 20 cmThen, 4*Edge = 20Edge = 20/4 = 5 cmSurface area = 6*edge*edge = 6*5*5 = 150 cm^2 |
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| 8. |
How many tangents can a circle have? |
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| 9. |
w many tangents can a circle have?il in the blanksA tangent to a circle intersects it inA line intersecting a circle in two points is called aAc-point (trcle can have parallel tangents at the mn) The common point of a tangent to a circle and the circle i |
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| 10. |
inu1weconsecutiveoddintegers,290sHow many terms to the A.P. : 9, 17,25,.... . must be taken to give a sumsum of whose squares isof 636? 419. If tangents PA and PB from a noint Pto cirgh centreare inclined to each other at |
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Answer» A.p= 9,17,25a=9d=17-9=8Sum of n terms=n÷2[2a+(n-1)d 636=n÷2[2(9)+(n-1)8]636=n÷2[18+8n-8]636=n÷2[10+8n]1272=10n+8n^28n^2+10n-1272=02[4n^2+5n-636]=04n^2+5n-636=04n^2+53n-48n-636=0n(4n+53)-12(4n+53)=0(n-12)(4n+53)=0n-12=0n=12 12 terms must be given to sum of 636 |
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| 11. |
Construction : To construct the tangents to a circle from a point outside it. |
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| 12. |
Consthe p7. Draw a circle with the help of a bangle. Take a point outside the circle.of tangents from this point to the circle. |
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| 13. |
Find the SP, ifi. CP= 1365, Profit = 300ii. CP=82200, Loss=ă330iii. CP=ăŚ1688, Loss= 144 |
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| 14. |
Draw a circle with the help ofa bangle, Take a point outside the cirecle. Construct the pairof tangents from this point to the circle measure them. Write conclusion. |
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| 15. |
|15. Draw a circle with the help of a circular solring. Construct a pair of tangents from a pointoutside the circle. |
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| 16. |
Prove that the tangents drawn from an external point to a circle are of equallength |
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Answer» To provePT=QT Proof: Consider the triangleOPTandOQT OP=OQ ∠OPT=∠OQT=90° OT=OT(common side) Hence by RHS the triangles are equal. HencePT=QT Hence Proved. |
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| 17. |
Prove that the tangents drawn at the ends of adiameter of a circle are parallel. |
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| 18. |
2. Find the selling price when:(i) CP =1650 and gain = 4%915 and gain26(11) CP =o/(iii)CP875 and loss = 12%(iv) CP =645 and loss-130 |
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| 19. |
(c)12000(d)16000, A trader sells his articles 20% above CP, andallows a discount of 10% on cash payment. Hisgain per cent is(a) 10%(c) 6%(b) 8%(d) 5% |
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Answer» thanks |
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| 20. |
3 Find the CPSP Profit/Loss CP(96-20% Profit !Pen |
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| 21. |
a) 7p + 5 = 19 [p=-2] |
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Answer» Given : 7p + 5 = 19 Put p = -2 7 × (-2) + 5 -14 + 5 = -9 But RHS is 9 so equality not holds so, p = -2 is not a solution of the given equation |
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| 22. |
a. Write (T) for true and (F) for talse(1) A cylinder has no vertexit) A cube has 6 faces. 12 edges and 8 verticesI) A cone has one vertex(lv) A sphere has one edge(v) A sphere has a curved surface. |
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Answer» 1-true2-true3-true4-false5-true |
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| 23. |
(1) (y2+ 10y + 24) รท (y + 4)(2) (p2 + 7p-5)รท(p + 3)(3)(3x + 2x2 + 4x3) รท(x-4) |
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| 24. |
(3q+7p²-2r³+4)-(4p²+2q+7r³-3) |
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Answer» 3q-2q+7p²-4p²+7r³-2r³+4-3= q+3p²+5r³+1 |
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| 25. |
Q. Find the value of r if 7P, 42.I7 |
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| 26. |
How many tangents can be drawn to a circleanswerfrom a point outside the circle ? Justify your |
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| 27. |
1. How many tangents can be drawn to a circle from a point on the same circlewhy? |
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Answer» Only one tangent can be drawn from a point because tangent is a line which touches the circle at one point only |
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| 28. |
13. Provethat the tangents drawn to a circle from a point in the exterior of the circleare congruent. |
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| 29. |
2906-30059 _ 1.| 24, % an 0 > “तो सिद्ध कीजिए कि। Lm0 b= ५ तो सिद्ध कोजिए कि e300 11 |
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Answer» PLEASE LIKE THE SOLUTION |
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| 30. |
प्रश्न 3. 5% वार्षिक साधारण व्याज की दर से 12000 का 1 माहका ब्याज ज्ञात कीजिए।(2014) | |
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Answer» 5/100×12000×1/12 =50 5/100*12000*1/12=10 |
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| 31. |
CP12000; SP = ?, Loss-73600 |
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Answer» When there is a Loss, SP = CP - Loss = 12000 - 3600 = 8400 |
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| 32. |
O is centre of the circle. PT is a tangent to the circle drawn from point P. PAB passes through the centre O of the circle. If PA = 3cm and PT = 6cm, find the radius of the circle |
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| 33. |
d. TIS Talse, ll Is trueThe solution of xy = y + 2Vy? - x? isdx |
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| 34. |
7p-10p |
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Answer» These are like terms. 7p - 10p = -3p |
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| 35. |
Solve the folkowing equations:(a) 10p= 100(b)10p+10 = 100 |
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Answer» p=100/10=1010p=100-10=90p=90/10=9 10p=100p=100/10ans.. p=10 thanks 10p+10=10010p=100-1010p=90p=90/10ans..p=9 |
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| 36. |
Page No:Date:Prove thetSee |
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Answer» Like if you find it useful |
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| 37. |
subtract 7p(3q+7p)from8p(2p-7q) |
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| 38. |
p2 - 10p +25 |
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| 39. |
P2- 10p +25 |
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Answer» p²-10p+25=p²-5p-5p+25=p(p-5)-5(p-5)=(p-5)(p-5) |
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| 40. |
EXERCISE 281. Write each of the following numbers in standard form(1) 57.36(iv) 168000000(il) 3500000(v) 46300000000002. Write each of the following numbers in usual form:(1) 3.74 x10(iv) 2.5 × 10 4(i1) 6.912 x 10(v) 5.17 x 103. () The height of Mount Everest is 8848 m. Write it(il) The speed of light is 300000000 m/sec. Express(i) The distance from the earth to the sun is 149600 |
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Answer» convert the following Hindu Arabic numerals to Roman numerals option 24 32 55 79 90 |
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| 41. |
Priyanka deposited6500 in a bank which pays 9% per annum interest. She withdrew1250 after theend of the first year. What amount will she get after three years? |
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| 42. |
2. Finthe othes HveThigohomenc uciih ngTacoh2 |
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| 43. |
In case of compound interest the principal everyyear |
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Answer» In case of compound principal changes every year. The principle for 2nd year will become (Principal+interest of first year). |
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| 44. |
9" If cos θ + sin θ = V2 cos θ t10. Using Euclid's division alenrithm ilhen show that cos θ-sin θ12 sin θ. |
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| 45. |
e300 upeesSeposited 12000 nipees at 9 p.epa. in a bank for some years, and withdrew hiaienenteriyeaz. At the end of the period, he had reseived aitogeter 17,400 rupesFor how many years had be deposited his money ? |
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Answer» Amount which is received by Javed is Rs. 17400 and he invested Rs. 12000 at 9% per annum. Interest he received is 17400 - 12000 = Rs. 5400 Using formula Pnr = 5400 12000× n× 9/100 = 5400 n = 5 years |
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| 46. |
(7p - 3р) (2p + 5р) |
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Answer» thanks |
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| 47. |
हट) 6 (006 5.prove thet सफल न घ0 |
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| 48. |
=-20 + 9p + 16-10p + 7 |
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Answer» -20 + 9p + 16 - 10p + 7 = 3 - p |
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| 49. |
Kabir takes a loan ofäš70,000 frorn Indu at arate of 2.5% pa. for two years compoundedannually. Find the amount he has to return. |
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| 50. |
oy 73 भिन, खांमन 100 का, मूदभत्र कोब़ 5%झाएन, मुदभत्र शंविमान कड ?(8) 3 O (b) 2 et(c) 1 b=t (त) (कांदरना््डि नया e |
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Answer» This is the answer 1 of the sum |
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