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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. |
LALI1. Three times a number decreased by 5 gives theresult 16. Find the number. |
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Answer» Let the no. be xA/q, 3x - 5 = 163x = 16 + 53x = 21x = 21/3x = 7 Let the no. be xA/q,3x - 5= 163x = 16 + 53x = 21x = 21/3x = 7 |
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| 2. |
1. Three times a number decreased by 5 gives theresult 16. Find the number.rhren consecutire multiples of 3 is |
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Answer» X= 7 is the correct answer of the given question 7 is the correct answer |
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| 3. |
Prove that (m2 -n2)4/mn |
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Answer» Tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθm-n=tanθ+sinθ-tanθ+sinθ=2sinθmn=(tanθ+sinθ)(tanθ-sinθ) =tan²θ-sin²θ∴, m²-n²=(m+n)(m-n)=2tanθ.2sinθ=4sinθtanθ4√mn=4√(tan²θ-sin²θ)=4√(sin²θ/cos²θ-sin²θ)=4√sin²θ(1/cos²θ-1)=4sinθ√(1-cos²θ)/cos²θ=4sinθ/cosθ√sin²θ [∵,sin²θ+cos²θ=1]=4sinθtanθ∴, LHS=RHS (proved) please like the solution 👍 ✔️👍 |
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| 4. |
Three numbers are given whose sum is 180and the ratio of the first two of them is 1:2If the product of the number is greatest, findthe numbers |
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Answer» 40, 60, 80 are the numbers. 40 and 80 after in the ratio 1:2. |
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| 5. |
If m+n-12 and mn=14 , find the value of (m2 + n2), |
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Answer» m^2+n^2=(m+n)^2-2mn=(12)^2-4(14(144-64=80 paagal pagal what is this I don't understand please tell clearly |
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| 6. |
5 y ^ { 2 } - 20 y - 8 z + 2 y z |
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| 7. |
|४ n का मान होगा, यदि "P:"P=2:1(a)5(c4(d) इनमें से कोई नहीं(b)6 |
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Answer» Option D is correct answer d is a right answere plz like thoko n!/(n-5)! divided by (n!/(n-3)!) = 2/1 n!/(n-5)! * (n-3)! / n! = 2n! cancel (n-3)! / (n-5)! = 2 Expand (n-3)! (n-3)(n-4)(n-5)! / (n-5)! = 2 (n-5)! cancel (n-3)(n-4) = 2n^2 - 7n + 12 = 2n^2 - 7n + 10= 2(n-5)(n-2) = 0n = 5 or n = 2, but we need n to be 5 or greater, son = 5 |
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| 8. |
\frac { d y } { d x } , \text { if } x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = a ^ { \frac { 2 } { 3 } } |
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| 9. |
1. Express the following percentages as fractions and decimals.a. 25%b. 125%C4c. 4% |
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Answer» a)25%=25/100fraction=1/4 decimal=0.25 b)125%fraction=125/100=5/4decimal=1.25 c)4(1/5)%=21/5%fraction=(21/5)/100=21/500decimal=0.042 25%25/1001/4 is right answer 125%125/100=5/4 is right answer |
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| 10. |
1 . 5. - L, ।।।।is for BC-DE250 = 2 BACआकृ-E।7. E F ! तुई हैं और ' का मट-लंद है।23 ८ रेखा के !! ही शरीर में बिंदुइस कार है कि / B12 - AB औरF- 048 हैं। देखिए आकृति :223।5 कि| | |-- Ha Lor( 41 = 13Lआ5 |
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| 11. |
4. Ray AX is the bisector of Z BAC and ray AYis the bisector of Z XAC. If L BAY 600, findВАС. |
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| 12. |
Q. 1. In the given figure, if Z BAC 340,LBCE-1630, then find the value of L ACB,LABC and LDBC.0340163 |
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Answer» ACB + BCE = 180° [Linear angles]ACB = 180° - 163°ACB = 17° ABC + ACB + BAC = 180° [Triangle sum property]ABC = 180° - 34° - 17° ABC = 129° DBC + ABC = 180°. [Linear angles]DBC = 180° - 129°DBC = 51° |
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| 13. |
of Ab, D ツCD, DA respectively, Giver thalet D be the midpoint of BC. If ZADB-450 and LACD 30070determine ZBAD.B) 20C)5D) noneA) 10 |
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| 14. |
12. The radius of a circle with centre at origin is 30 units. Write the coordinates of thepoints where the circle intersects the axes. Find the distance between any two suchpoints, |
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Answer» 0,0 is the points..... coordinates are, at X ( 30, 0 ) and at Y ( 0, 30 ), |
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| 15. |
s lope 7. Write the equation of each of thand hase equal slopes)(x-xines:(i) The X-axis and the Y-axis. |
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Answer» for x axis:y=0for y axis:x=0 |
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| 16. |
निम्न पर विचार(G) 3 x 33 |
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Answer» 3² × 3³ 3^(2+3) 3^5 3^2×3^3=3^(2+3)=3^5(ans) |
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| 17. |
e g USRS \mfl“33, सिद्ध करें कि 2(-3, 2), छि( ) e |
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| 18. |
16 \frac 2 3 \% \text of 600 g m - 33 \frac 1 3 \% |
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Answer» a)28% of 450 +45% of 280=(28/100)×450 +(45/100)×280=126+126=252 b)16(2/3)% of 600 gm -33(1/3)% of 180 gm=100-60=40 |
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| 19. |
| \forall \varepsilon - 8 z - | - | \varepsilon z - \angle S | |
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| 20. |
13. In the given figure, LADE: <ACB and <ABD-〈BAC. Prove that AD-BC. |
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| 21. |
RIANGLESABCD is a quadrilateral in which AD =BC andDAB CBA (see Fig.7.17) Prove that(i)ΔΑΒΙ)(ii) BD#ACABD-ABAC(iii)BAC.Fig.737diculars to a line |
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Answer» Given: In quadrilateral ABCD,AD = BC & ∠DAB = ∠CBA To Prove: (i)ΔABD ≅ ΔBAC (ii)BD=AC (iii)∠ABD = ∠BAC Proof: i)In ΔABD & ΔBAC, AB = BA (Common)∠DAB = ∠CBA (Given)AD = BC (Given) Hence,ΔABD ≅ΔBAC. ( by SAS congruence rule).(ii) Since, ΔABD ≅ΔBACThen, BD = AC (by CPCT) (iii)Since, ΔABD ≅ ΔBAC Then , ∠ABD = ∠BAC (by CPCT) hit like if you find it useful |
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| 22. |
8010 एक चतुर्भुज हैं. जिसमें ०0? « 80 और9:88 « ८ (8 हैं (देखिए आकृति 7.17)। सिद्धe किAABD= ABAC >) BD=AC) ZABD=Z£BAC |
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| 23. |
(ii) BD=AC(ii) ABD = L BACanrepfa 7.17 |
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| 24. |
liltereept on the x-axis.Poi() having its9. Find the equation of the straight line which passes through the point (4. 7) and hasinterceptson the axes equal in magnitude but opposite in sign. |
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| 25. |
, e Q R and S are respectively the midpoints of thedAB, BC. CD and DA of a quadrilateral ABCDShow thato POI AC and PO 2ACPOISR) PORS is a parallelogram. |
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| 26. |
ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that BCD is a right angle. |
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Answer» 1 2 |
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| 27. |
ΔABC is an isosceles triangle in which AB =ACSide BA is produced to D such that AD =AB (seefigure). Show that angleBCD is a right angle. |
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| 28. |
AABC is an isosceles triangle in whidh AB-ACSide BA is produced to D such that AD #AB(see Fig. 7.34)Show that Z BCDisaright angle |
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| 29. |
31. ABC is an isoscles right angled triangle with ZABC 90.Asemi-circle is drawn with AC as the diameter. IfAB = BC = 7 cm; findthe area ofthe shaded region (Take π=22)[ICSE 2012] |
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Answer» radius of the semicircles will be 7/2=3.5cmarea =πr^2/2as there are 2 semicircleshence2*πr^2/2=22/7**3.5*3.5=38.5cm^2 |
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| 30. |
21.AABC is an isosceles triangle in which AB AC. Side BA is producedto D such that AD AB. Show that 4BCD is a right angle.OR |
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| 31. |
29. If AABC is an isoseeles triangle such that AB-AC. Side BA is produced tooDsuch thatAD AB. Show that angle BCD is a right angle |
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Answer» tq tq |
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| 32. |
2. ABCD is a quadrilateral in which AD-BC andDAB = < CBA (see Fig. 7·17). Prove that(ii)BD=AC(iii)ABD = < BAC. |
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| 33. |
s in the given figure, ZABC 90 andBDLAC. If AB 5.7 cm, BD -3.8 cmand CD 5.4 cm, find BC |
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| 34. |
ABCD is a quadrilateral in which AD BC and2 DAB-CBA (see Fig. 7.17). Prove that2.(ii) BD=AC(iii) ABD=< BAC. |
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| 35. |
is a quadrilateral in which AD BC andCBA (see Fig. 7.17). Prove thatDAB = <(0(ii) BD=AC(iii) ABD = < BAC.119B. |
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| 36. |
GLESABCD is a quadrilateral in which AD BC andDAB CBA (see Fig.7.17). Prove that0) AABDEABAC(ii) BD=AC(ii) ZABD BAC. |
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| 37. |
2.ABCD is a quadrilateral in which ADBC andDABCBA (see Fig. 7.17). Prove that(ii) BD AC(iii) < ABD-2 BACFig. 7.17 |
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| 38. |
9. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. |
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| 39. |
drawn is neither a red card nor a queen.e points A(4.3) & B(x, 5) are on the circle with the center 0(2.3), find the value of X. |
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Answer» Number of non-red cards = 26.Number of non-red non-queen cards = 24Probability of non-red non-queen card = 24/52= 6/13. I want down answer plz |
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| 40. |
If the diagonals of a rhombus are 9 cm and 12 cm, find its sides |
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| 41. |
2. The radius of the incirele of a triangle is 4 cm and the segments into whichone side is divided by the point of contact are 6 cm and 8 cm. Determine theother two sides ofthe triangle. |
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Answer» Area Δ ABC = 1/2 . 4 . (AB + BC + AC) = root( s(s – a)(s – b)(s – c))i.e., 4 s = root( s(s – a)(s – b)(s – c) )16 s = (s – a) (s – b) (s – c)i.e., 16 (14 + x) = x X 6 X 8, i.e., x = 7Therefore, AB = 15 cm and AC = 13 cm. Samajh mein nahi aa raha |
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| 42. |
Two equal chords AB and CD of a circle when produced intersect at point P. Prove thatPB = PD |
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Answer» Given:AB=CDto prove:PB=PDconst: draw OE and OQ perpendicular on AB and CD respectivelyproof:given AB and CD are two equal chords of the same circle.OE=OQ(equal chords of a circle are equidistant from the center.)now in triangle OEP and OQP,OE=OQOP=OP(common)angle OEP = OQP =90 degree,by constructiontherefore triangle OEP = OQP (RHS congruency)EP = QP (CPCT)also AE=EB=1/2 AB and CQ=QD=1/2CD (the line joining the center of the circle is perpendicular to the chord and bisects the chord.)Now AB = AC implies AE = EB= CQ=QD ....(1)therefore EP-BE =QP - BEEP - BE = QP - QD (FROM 1)BP = PDhence proved |
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| 43. |
Q.IV. Solve the following.1. Two equal chords AB and CD ofa circle when produced intersectat point P. Prove that PB = PD. |
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| 44. |
In the figure, AB and CD are twoequal chords of the circle with centreO. Prove that:2) ZAOB COD |
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Answer» please like my answer if you find it useful |
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| 45. |
-m- DABCD is a quadrilateral in which ADLDAB = < CBA (see Fig. 7.17). Prove that(i) Δ ABD Δ BAC(ii) BD=AC2.BC andt(ii) Z ABD LBACFig, 7.17 |
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| 46. |
.26.Theradiusof the incircle of atriangle is 4 em and the segments intowhich one side is divided by the point ofcontact are 6 cm and 8 cm. Determinethe other two sides of the triangle.Sol.648 |
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Answer» Area Δ ABC = 1/2 . 4 . (AB + BC + AC) = root( s(s – a)(s – b)(s – c))i.e., 4 s = root( s(s – a)(s – b)(s – c) )16 s = (s – a) (s – b) (s – c)i.e., 16 (14 + x) = x X 6 X 8, i.e., x = 7Therefore, AB = 15 cm and AC = 13 cm. |
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| 47. |
16. AB and CD are two equal chords of a circle with centre OIf M and N are the midpoints of AB and CD respectivelyprove that (a) ZOMN-LONM, (b) LZAMN-LCNM.tO |
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Answer» 1 |
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| 48. |
2. The radius of the inefrete of a triangle is 4 cm and the segments into whichone side is divided by the point of contact are 6 cm and 8 cm. Determine theother two sides ofthe trianglesa |
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Answer» Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF bex.IntraingleABC,CF = CD = 6cm(Tangents on the circle from point C)BE = BD = 8cm(Tangents on the circle from point B)AE = AF =x(Tangents on the circle from point A)AB = AE + EB =x+ 8BC = BD + DC = 8 + 6 = 14CA = CF + FA = 6 +x2s =AB + BC + CA=x+ 8 + 14 + 6 +x=28 + 2xs= 14 +x meritnation se copy maar diya |
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| 49. |
is a quadrilateral in which AD- BC andCBA (see Fig. 7.17). Prove thatㄥDAB+i) BD AC |
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| 50. |
ही ही चतुर्भुज है, जिसमें pAB =4 CBA है (देखिए हा हिकीजिए कि शी() AABD=A BAC BD=AC(i) £ABD=~BAC |
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Answer» Given: In quadrilateral ABCD, AD = BC & ∠DAB = ∠CBA To Prove: (i)ΔABD ≅ ΔBAC (ii)BD=AC (iii)∠ABD = ∠BAC Proof: i) In ΔABD & ΔBAC, AB = BA (Common)∠DAB = ∠CBA AD = BC (Given) Hence,ΔABD ≅ΔBAC. ( by SAS congruence rule). (ii) Since, ΔABD ≅ΔBAC Then, BD = AC (by CPCT) (iv)Since, ΔABD ≅ ΔBAC Then , ∠ABD = ∠BAC (by CPCT) |
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