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. |
9. The wire needed, for four rounds of fencing around a rectangular grlength 235 m, is 3.2 km. What is the breadth of the ground? |
| Answer» | |
| 2. |
a rectangular piece of land measure 0.7 km by 0.5 km each side is to be fenced with 40 of wires what is the length of the wire needed? |
| Answer» | |
| 3. |
31, 9f sin 6 = दा तो (20 6 का मान है : |
|
Answer» Like if you find it useful |
|
| 4. |
(1) (20 + 6)â - (20 - |
|
Answer» a^3-b^3=(a –b) (a^2 + ab + b^2)hence(2a+b)^3-(2a-b)^3=(2a+b-2a+b)((2a+b)^2+(2a+b)(2a-b)+(2a-b)^2)=2b(4a^2+4ab+4a^2-b^2+4a^2+b^2-4ab)=2b(12a^2)=24a^2b |
|
| 5. |
3%,and 0.14?206. Which is largest in 6 |
|
Answer» 3/20 is the right answer this is correct answer is 3/20 3/20 is the correct solution 3/20 is the right answer for the given question Here, (6* 2/3) = (20/3) Now by multiplying by 100 we get, (20/3) × (1/100) = 1/15 = 0.06 ……. (1) Now solving for 3/20 = 0.15 ………. (2) And 0.14 ……….. (3) By comparing (1) (2) (3) we get, 0.15 > 0.14 > 0.06 So, the right answer is 3/20. |
|
| 6. |
2. Yamini covered a distance of 260 m by taking two rounds of ajrground of breadth 25 m. What is the area of the ground?1000 m (2) 900 m (3) 800 m2 (4) 625 m? |
| Answer» | |
| 7. |
fan 6 + tan 20 + tan 30 = tan 6 tan 20 tan 3. |
|
Answer» Tan x + tan 2x + tan 3x - tan x tan 2x tan 3x = 0tan x + tan 2x + tan 3x ( 1 - tan x tan 2x ) = 0tan x + tan 2x = - tan 3x ( 1 - tan x tan 2x )( tan x + tan 2x )/ ( 1 - tan x tan 2x ) = - tan 3xtan 3x = - tan 3x2 tan 3x = 0tan 3x = 03x = 2kpi where k = 0, 1,2,3,4,5,..............x = 2 k pi/3 where k = 0 , 1, 2 ,3 , 4 , 5 , 6..... Like my answer if you find it useful! |
|
| 8. |
4. Ifare in AP, prove that(b+c) (c+a) (a+ b)are in AP. |
| Answer» | |
| 9. |
10.ChordsAbandCDcutatPinside the circle : If AB7, AP-4, CP 2, then CD |
| Answer» | |
| 10. |
How to find the perimeter of a squareLike any polvon th |
|
Answer» Recall the formula for the perimeter of a square. For a square ofside lengthS, the perimeter is simply four times theside length:P=4s. Determine the length of one side, and multiply it by 4 to find the perimeter. |
|
| 11. |
___ 1. x + 3x2 + 3x + 1 को निम्नलिखित से भाग देने पर शेषफल ज्ञात कीजि। (i) x+1_(i) - (i) x_ (iv) x + ग ()(i)x+1(iv) x + Tt(v) |
|
Answer» 1 option is the right answer first option is rigjt The given polynomial isx3+3x2+3x+1x3+3x2+3x+1. By remainder theorem, ifp(x)=x3+3x2+3x+1p(x)=x3+3x2+3x+1is divided byx+1x+1then remainder is determined byp(−1)p(−1). p(−1)=(−1)3+3(−1)2+3(−1)+1=−1+3−3+1=0p(−1)=(−1)3+3(−1)2+3(−1)+1=−1+3−3+1=0 Thus, the remainder is 0. iiii The given polynomial isx3+3x2+3x+1x3+3x2+3x+1. By remainder theorem, ifp(x)=x3+3x2+3x+1p(x)=x3+3x2+3x+1is divided byx−12x−12then remainder is determined byp(12)p(12). p(12)=(12)3+3(12)2+3(12)+1=18+3×14+3×12+1=278p(12)=(12)3+3(12)2+3(12)+1=18+3×14+3×12+1=278 Thus, the remainder is278278. iiiiii The given polynomial isx3+3x2+3x+1x3+3x2+3x+1. By remainder theorem, ifp(x)=x3+3x2+3x+1p(x)=x3+3x2+3x+1is divided byxxthen remainder is determined byp(0)p(0). p(0)=(0)3+3(0)2+3(0)+1=0+1=1p(0)=(0)3+3(0)2+3(0)+1=0+1=1 Thus, the remainder is 1. iviv The given polynomial isx3+3x2+3x+1x3+3x2+3x+1. By remainder theorem, ifp(x)=x3+3x2+3x+1p(x)=x3+3x2+3x+1is divided byx+πx+πthen remainder is determined byp(−π)p(−π). p(−π)=(−π)3+3(−π)2+3(−π)+1=(−π)3−3π2−3π+1p(−π)=(−π)3+3(−π)2+3(−π)+1=(−π)3−3π2−3π+1 Thus, the remainder is(−π)3−3π2−3π+1(−π)3−3π2−3π+1. vv The given polynomial isx3+3x2+3x+1x3+3x2+3x+1. By remainder theorem, ifp(x)=x3+3x2+3x+1p(x)=x3+3x2+3x+1is divided by5+2x5+2xthen remainder is determined byp(−52)p(−52). p(−52)=(−52)3+3(−52)2+3(−52)+1=−1258+3×254−3×52+1=−125+150−60+88=−278p(−52)=(−52)3+3(−52)2+3(−52)+1=−1258+3×254−3×52+1=−125+150−60+88=−278 Thus, the remainder is−278−278. |
|
| 12. |
1. PQR is a triangle, right-angled at P. If PQ 10 cmand PR24 cm, find QR. |
| Answer» | |
| 13. |
solve for x and y 2x _ y = 6 and x_ y =2 |
|
Answer» 2x - y = 6.....(1)x - y = 2.......(2) Subtract eq(2) from eq(1)x = 6 - 2 = 4 Put value of x in eq(2)4 - y = 2y = 4 - 2 = 2 Therefore,Value of x = 4, y = 2 questions paper do u have Appu Ur there in Instagram or Facebook |
|
| 14. |
Prove that the function f given byf(x)-|x_ !! , x E R is not differentiable at x = 1. |
| Answer» | |
| 15. |
12. A line passes through (, y) and (h, k). I. If slope of the line is m,show thatk-y_{1}=m\left(h-x_{1}\right) |
| Answer» | |
| 16. |
A rectangular ground is 90 m long and 32 m broad. In the middle ofthe14 m32 mground there is a circular tank of radius 14 metrcs. Find theso strofturing the remaining portion at the rate ofき50 per lsquare metre.eo m |
| Answer» | |
| 17. |
274) The diagonal of a rectangular ground is 60 metres more than the breadth of the ground. If the length of the groundis 30 metres more than the breadth, find the area of the groundghr the destination 2 hours earlier If the |
| Answer» | |
| 18. |
The wire needed, for four rounds of fencing around a rectangular ground oflength 235 m. is 3.2 km. What is the breadth of the ground? |
| Answer» | |
| 19. |
8क्या 254 समांतर श्रेणी 4,9,14,19..का पद है? |
|
Answer» 254=4+(n-1)5254=4+5n-5254=5n-15n=255n=51 yes 245 samanter sari ka pad hi yes 245 समांतर है........ yes 51×5= 254 so answer is 'Yes' n= 51 is the correct answer yes 254 smanthar hai |
|
| 20. |
. Yamini covered a distance of 260 m by taking two rounds of a rectangularground of breadth 25 m. What is the area of the ground?(1) 1000 m2 (2) 900 m2 (3) 800 m 2 {4) 625 m2 |
|
Answer» distance covered=perimeterhence as two rounds so perimeter will be 260/2=130mnow perimeter=2(l+b)hence 130=2(l+25)65=l+25l=40mhence area is lxb=40*25=1000m^2 |
|
| 21. |
A rectangular ground is 90 m long and 32 m broad. In the middle ofthe ground there is a circular tank of radius 14 metres. Find thecost of turfing the remaining portion at the rate of 50 persquare metre.14 m32 m90 m |
| Answer» | |
| 22. |
254(a) 12(b) 15(c) 18(d)?206.[1] |
|
Answer» hit like if you find it useful |
|
| 23. |
e length of a rectangle is 16 cm and itsperimeter isequal to the perimeter of a squareun side 12.5 cm. Find the area of therectangle |
| Answer» | |
| 24. |
12 cm7.6 cmPQRS is a parallelogram. PM is theheight from P to SR and PN is theheight from P to QR. IFSR = 12 cmand PM= 7.6 cm.Š Find the area of the paral-lelogram PQRS 902 Find PN, if QR=8 cm. |
|
Answer» PQRS is a parallelogram, we know,area = b* h = SR * PM here SR = 12cm , PM = 7.6 cm so, area of PQRS = 12cm * 7.6 cm = 91.2 cm² 2). area = SR * PM = QR * PN so, 91.2 cm² = 8cm * PN 91.2/8 = PN PN = 11.4 cm hence, PN = 11.4 cm |
|
| 25. |
Find the 10th term from the end of the AP 4, 9, 14, .......254. |
| Answer» | |
| 26. |
4. If the perimeter and length of a rectangular hall are 84 m and 30 m respectively,then find the area of the hall. |
| Answer» | |
| 27. |
and PR 24 cm,find QR |
|
Answer» By Pythagoras theoremPQ²+PR²=QR²10²+24²=QR²100+576=QR²676=QR²Taking square root on both sides,√676=√QR²26=QRi. e QR=26 |
|
| 28. |
g"Δ ABC ~ △PQR and ar(AABC) = 4ar(APQR) . If BC = 12 cm, find QR. |
| Answer» | |
| 29. |
cosA sin (270° A) sin (270° - A)+ cos (180 A)(A) -1(C) 1(B) 0(D)野树君雨猿喃 |
| Answer» | |
| 30. |
x %2B 3*x^2 %2B 3*x^3 %2B x_ 4 %2B 1 |
|
Answer» X+1 is the factorx³+2x+X=0X=0,x²+2x+1=0(X+1) (X+1) is the factor x+1 IS THE CORRECT ANSWER |
|
| 31. |
tan ( 270 ^ { 0 } - \theta ) \cdot \operatorname { cot } ( 270 ^ { \circ } - \theta ) = \dots |
|
Answer» Sol:- tan*cot=1 |
|
| 32. |
Find the maximum and minimum values of x3-2x2+x+6. |
| Answer» | |
| 33. |
26.Find the maximum and minimum values of functionx) = sin 2x + 5. |
| Answer» | |
| 34. |
‘ g "“?’“"'*‘ ooकक.©°‘vf A G |
|
Answer» Sender's Address DateSubject Receiver's address I am-------(your name),secretary of sundarnagar colony.I would like to draw your attention towards the acute shortage of drinking water in our area.This problem has become a major issue how can anyone live without water.Our society has suffering a lot for one week.People have to buy drinking water.Our society receives very dirty water which is not adequate for any purpose.So,kindly take immediate steps for ensuring adequate water supply to the citizen because WATER IS LIFE. Thanking YouYours sincerely (Your name)Secretary of Sundarnagar(Delhi) |
|
| 35. |
ािसागपा4) सामवार3. निम्नलिखित प्रश्न में दिए गए विकल्पों मेंसे संबंधित अक्षरों को चुनिए :(1) 15-4MN : OL: : SH: ?/(3) 31-(1) VE (2) UF12.निम्नलिखि(3) UG (4) VFसे विषम4.निम्नलिखित प्रश्नों में दिए गए विकल्पों में (1) 8-से संबंधित अक्षरों को चुनिए :(3) 1213. निम्नलिSTOP : TRVT::?:?(1) MIND : IQLO(12) HAIL : PLCI(3) SAND : UDHS(4) BANK : CCQO5. निम्नलिखित प्रश्न में दिए गए विकल्पों मेंसे संबंधित संख्या को चुनिए :वालेEcho + |
|
Answer» questions 4 ,2 is correct answer 3 ka (1) hoga , M+2 = O, N -2 = l isi tarah,s+2 = v, H- 2 =E UFHAIL is correct answer |
|
| 36. |
\frac { 0 \cdot 891 \times 0 \cdot 642 + 0 \cdot 891 \times 0 \cdot 358 } { 0.9 \times 0.766 + 0.9 \times 0.234 } = ? |
| Answer» | |
| 37. |
\left( \cos 0 ^ { \circ } \times \cos 1 ^ { \circ } \times \cos 2 ^ { \circ } \times \ldots \times \cos 90 ^ { \circ } \right) |
|
Answer» as cos90 =0henceanything multiplied by 0 is 0 so the expression value will be 0 |
|
| 38. |
cos? o+ cos? (a+120) + cos? ( â120") = |
|
Answer» cos 120 = cos (90 +30) = cos90cos30 - sin90sin30 = -1/2sin 120 = cos90sin30+ sin90cos30 = sqrt3/2cos²( α + 120° ) =(-1/2*cosα -sqrt3/2*sinα)^2 = 1/4*(cosα + sqrt3*sinα)^2cos² ( α - 120° ) =(-1/2*cosα + sqrt3/2*sinα)^2 = 1/4*(cosα - sqrt3*sinα)^2 therefore given identity is= cos² α + 1/4*cos² α + 1/4*cos² α +3/4*(1-cos² α) + 3/4*(1-cos² α)= 3/2. |
|
| 39. |
0.746 \times 0.746 \times 0.746+0.254 \times 0.254 \times 0.254+3 \times 0.745 \times 0.254(0.746+0.254) is equal toA)1(B)10(C)1.920(D)0 |
| Answer» | |
| 40. |
In the given figure, AB|CD,ABE = 120°,LECD = 100° andLBEC = x。. Find the value of x.E1100-120° |
| Answer» | |
| 41. |
If ι and β are the roots of the equation Pr + qr + 1-0, find |
| Answer» | |
| 42. |
uct the right angled ΔPQR, where mLQ-909, QR1. ConstructPR 10 cm.8cm and |
| Answer» | |
| 43. |
The length of a rectangular plot is one and a half times its breadth, 20% of its pemeteris 20 m. What is the area of the plot? |
|
Answer» thanks |
|
| 44. |
POR is a triangle.right-angled at P1fPQ=10cmand PR = 24 cm, find QR.1. |
| Answer» | |
| 45. |
In the given figure, AB I| CD,ABE = 120°, <ECD = 100° and LBEC = x。.Find the value of x.Hint. Drau FEG || AB |CD.100120 |
| Answer» | |
| 46. |
4. The shape of a garden is rectangular in the middle and semi7 mcircular at the ends as shown in the diagram. Find the areae perimeter of the garden20 mHint: Length of rectangle 20 (3.5 + 3.5)] |
|
Answer» Mensuration: Mensuration is the branch of mathematics which concerns itself with the measurement of Lengths, areas & volume of different geometrical shapes or figures. Plane Figure:A figure which lies in a plane is called a plane figure. For e.g: a rectangle, square, a rhombus, a parallelogram, a trapezium. Perimeter: The perimeter of a closed plane figure is the total length of its boundary. In case of a triangle or a polygon the perimeter is the sum of the length of its sides. Unit of perimeter is a centimetre (cm), metre(m) kilometre(km) e.t.c Area:The area of the plane figure is the measure of the surface enclose by its boundary. The area of a triangle are a polygon is the measure of the surface enclosed by its sides. A square centimetre (cm²) is generally taken at the standard unit of an area. We use square metre (m²) also for the units of area. Circumference of a circle is the perimeterof a circle. In a circle the radius is half of thediameter. The approximate value of π( Pi) is= 22/7 |
|
| 47. |
Find the area of a rectangular field, if the length of the rectangular field is 164 m andbreadth is 20 m. |
|
Answer» Area of rectangle = l*b = 164*20 = 3280 m² Like my answer if you find it useful! |
|
| 48. |
5.A rectangular ground is 35 m long and 20 m broad. Find its perimeter. |
|
Answer» GIven:length=35mbreadth=20 mFormula: perimeter= 2(l+b)Solution:perimeter=2(35+20)=2*55=110Answer: perimeter= 110 m sq |
|
| 49. |
1510204080 . It is being given that 5 2236 and 10 3.162.Evaluate |
|
Answer» 15/(√10+√20+√40-√5-√80) 15/(√10+2√5+2√10-√5-4√5)15/(3√10-3√5) Put √5=2.236 and√10=3.162 15/(3(3.162)-3(2.236))5/(.926)=5.399 |
|
| 50. |
In the given figure, AB I CD,AABE = 120°, <ECD = 100° andLBEC = x。. Find the value of x.100120° |
| Answer» | |