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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.
| 66251. |
\left. \begin{array} { l } { 66 \div 3 } \\ { 97 \div 3 } \\ { 630 \div 5 } \end{array} \right. |
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Answer» thanks 1. 222.32.3333.126 |
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| 66252. |
1575 \div 21 \div 5 = \sqrt { ? } \times 6 |
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Answer» last step is wrong in previous answer 1575÷21÷5 = Root? × 6Let The Unknown Number Be X => 75 ÷ 5 = Root(x) × 6=> 15 = Root(×) × 6=> 15÷6 = Root(x)=> 5 ÷ 2 = Root(x) Squaring Both The Sides.... => (5÷2)^2 = x=> 25÷4 = x Value Of x = 25÷4= |
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| 66253. |
12 x y\left(9 x^{2}-16 y^{2}\right) \div 4 x y(3 x+4 y) |
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| 66254. |
(3)\left[(-4)^{12} \div(-4)^{8}\right] \div(-4) |
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| 66255. |
PORS is a parallelogram, T is the midpoint of PO (Fig. 11.23). ST bisectsZS. Prove that(a) QR=QT(b) RT bisects<R(c)LSTR=90°0r + 52) Find a |
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| 66256. |
\frac { ( 2 ^ { 5 } ) ^ { 2 } \times 7 ^ { 3 } } { 8 ^ { 3 } \times 7 } |
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Answer» thank you. |
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| 66257. |
9. In the given figure, O is the centre of acircle in which chords AB and CDintersect at P such that PO bisectszBPD. Prove that AB =CD. |
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| 66258. |
In the given figure, O is the centre of a circle inwhich chords AB and CD intersect at P such thatPO bisects ZBPD. Prove that AB CD.9.8 |
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| 66259. |
O is the centre of a circle inthe shords AB and CD intersect at P such thatbisects 4BPD. Prove that AB CDpo |
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Answer» thank you |
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| 66260. |
८५ 3-6 T कंकै 00 _ .40 Se©| =CoSOL 14030 N & xcmcAt -\ % |
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| 66261. |
SE 1.3fromA student27,36,879subtracted63,42,568 and then subtracted the differencefrom the smallest 9-digit number. What washis answer?11 produced |
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Answer» 96394311 is the correct answer of the given question |
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| 66262. |
2. In Fig. 10.11, if TP and TQ are the two tangentsto a circle with centre O so that POQ = 110°then Z PTQ is equal to(A) 60°(B) 70°(C) 80°(D) 90° |
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Answer» OPTION B 70 degree it is right answer in given question |
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| 66263. |
75% ofwhat number is 15?SE 8.2per cents.(G)140 |
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Answer» Percentage3/40 × 100 = 7.5% 2/7 × 100 = 28. 57% |
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| 66264. |
1. There are two parks A and B as shown. Ajaywalks around park A and Vijay walks aroundpark BPark APark B75 m75 ma) Who walks more distance?b) Which park has greater area? |
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Answer» both park are equal both are move equal distance and both park are same area |
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| 66265. |
Mr. Ajay went to market with 525, He bought fish for 150, oil for 90 andvegetables for?75. How much money was left with him? |
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Answer» Money left = Total - Fish - Oil - Vegetables =525-150-90-75 =525-315=210. |
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| 66266. |
SE2.75% ofwhat number is 15?9 is 25% of what number?EXERCISE 8.2Convert the given fractional numbers to per cents.40 |
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Answer» sendfash |
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| 66267. |
\frac { 441 } { 2 ^ { 2 } \times 5 ^ { 7 } \times 7 ^ { 2 } } |
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Answer» The deno is 2^3×7^2×5since it has 7 in deno and not in the form of 2^m ×5^n and the numerator ha 441=7×7×3×3when we simplify it we get 3^2 by 2^3×5hence it is a terminating decimal |
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| 66268. |
4 x ^ { 4 } \div 4 x ^ { 4 } = 0 |
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Answer» the given expression is wrong4x⁴ ÷ 4x⁴ = 1so LHS is not equal to RHS . |
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| 66269. |
500 \div 5 + 200 \div 4 |
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Answer» (500÷5)+(200÷4)=100+50=150 500÷5+200÷4=100+50=150 |
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| 66270. |
M is tăŞăźCM2 LABrn |
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Answer» tq |
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| 66271. |
sec 38/cosec52 + 2/root3 *tan17*tan38*tan60*tan52*tan73-3(sin^2 32 + Sin^2 58) |
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Answer» sec38/cosec52 + 2/√3*(tan17tan38 tan60tan52tan73) - 3(sin²32 + sin²58). ⇒sec(90 - 52)/cosec52 + 2/√3*{tan(90 - 73)tan73 tan(90 - 52) tan52 tan60} - 3{sin²(90 - 58) + sin²58}. we know :-sec(90 - Ф) = cosecФsin(90 - Ф) = cosФtan(90 - Ф) = cotФtan60 = √3let us apply here, ⇒cosec52/cosec52 + 2/√3 * (cot73 *tan73* cot52*tan52*√3) - 3(cos²58 + sin²58) [ sin²A + cos²A ] ⇒1 + 2(1/tan73 * tan73 * 1/tan52 * tan52) - 3(1) ⇒1 + 2 - 3 ⇒3 - 3 = 0 |
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| 66272. |
Evaluate eycos 58° . sin 22° \/' cos 38° cosec 52°sin 32° © ८०568" +[3 (tan 18° tan 35° tan 60° tan 72° tan 55°) |
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| 66273. |
\frac { \cos 58 } { \sin 32 } + \frac { \sin 22 } { \cos 68 } - \frac { \cos 38 \csc 52 } { \tan 18 \tan 35 \tan 60 \tan 72 \tan 55 } |
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Answer» cos58°= sin32° as cos(90-theta) = sin thetatan(90-theta)= cot thetasintheta= 1/cosec theta |
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| 66274. |
' 32) Find je' sin xdx . ‘( .R इक |
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Answer» Like my answer if you find it useful |
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| 66275. |
9. In the given figure, O is the centre of a circle inwhich chords AB and CD intersect at P such thatPO bisects ZBPD. Prove that AB CD |
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| 66276. |
9. In the given figure, O is the centre of a circle inwhich chords AB and CD intersect at P such thatPO bisects LBPD. Prove that AB- CD |
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| 66277. |
Ajay scored 75% in an exam. If his score was180, find the maximum marks of the exam. |
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Answer» let the maximum marks be xthen 75x/100=180;x=3600/15=720/3=240 |
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| 66278. |
âg uls puy s ZE =gQuel+Qo93s JI |
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Answer» secΘ +tanΘ = 2/3 1/cosΘ + sinΘ/cosΘ = 2/3 sinΘ + 1 = 2/3 cosΘ sinΘ = 2/3 cosΘ – 1 Θ = -0.39479 ± 1.176n Or Θ = -22.6198 degrees, which is in Quadrant 4 |
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| 66279. |
6, Ajay scored 75% in an exam. If his score was180, find the maximum marks of the exam 751 |
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Answer» Marks Ajay scored= 180% he scored= 75%let total be xso,75% of x= 18075/100*x = 1803/4 * x = 180x = 240. |
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| 66280. |
13. In an examination, 4examination, a student se 40% andfails by 10 marks if he scored 50%, he woupass by iS marks. Find the minimum marksrequired to pass the exama): 250b) 100c) 110d) 12529. IF Ais 2002 alter than nhw what percnt is 8 |
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Answer» answerrrrr issss (d) 125 |
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| 66281. |
ple 2: Two tangents TP and TQ are drawnto a circle with centre O from an external point TProve that PTQ 2 2 OPQ |
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Answer» Like if you find it useful |
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| 66282. |
\left. \begin{array} { l } { 3 \times 2 \div 2 } \\ { ( 21 \div 7 ) - 2 } \\ { 81 \div 3 \text { of } ( 2 \times 7 ) \times 5 \text { of } 6 } \end{array} \right. |
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| 66283. |
3 + [ 8 - 5 \div \{ 4 - 2 \div ( 2 + \frac { 8 } { 13 } ) \} ] |
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| 66284. |
6+8 \div 2-3 \times 2+10 \text { of } 2 \div 4 |
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| 66285. |
2 \div 2 + 2 |
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Answer» 2÷2+2=1+2=3follow VBODMAS |
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| 66286. |
54.The rn ultiplicative inverse ofs-19. If n--4. find 'm'm+n(1) 15(2) 23(3)-15(4) -23To io |
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Answer» Multiplicative inverse of 1/(m+n) is m+n i.e. -19 m + n = -19 when n = -4 m = -19 + 4 m = -15 (3) option is correct |
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| 66287. |
ix.Find the value of 5:9::x:72Eind the mean proportional's of 2 and 32 |
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| 66288. |
gelo)Find the domain and range of the function f (x) =" 2-sin 32 |
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Answer» #Hey there!!__________ ◆Given function is : f(x) = \frac{1}{2 - sin3 x} \\ ●For DOMAIN :----------------------- Function f(x) is not defined when => 2 - sin3x = 0 so , sin3x = 2 ---------(1) but we know that range of sinx € [ -1, 1 ] so maximum value of sin3x = 1 therefore sin3x ≠ 2 ( not possible) => 2 - sin3x ≠ 0 ------(2) so from equation (2) we can se tha function is defined for all values of x => Domain € R #FOR RANGE :----------------------we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\ now multiplying by -1 we get, => 1 ≥ - sin3x ≥ -1 adding 2 we get, => 2+1 ≥ 2 - sin3x ≥ 2-1 => 3 ≥ 2-sin3x ≥ 1 now taking inverse we get, \frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1 so Range € [ ⅓ , 1 ] _____________________________ ◆HOPE IT WILL HELP YOU like my answer and also make as expert |
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| 66289. |
Eind the value of\left(\frac{243}{32}\right)^{\frac{3}{5}} \times\left(\frac{2}{3}\right)^{-3} \times\left[\left(\frac{125}{27}\right)^{\frac{2}{3}}\right. |
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| 66290. |
io Chords Ab and CD cut at Pinside the eirele : If All7,AP 4, CP 2, then CD |
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Answer» please like my answer if you find it useful |
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| 66291. |
et G o 4 e |
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| 66292. |
URTwo tangents TP and TQ are drawn to a circle with centre O.from an external point T. Prove that PTQ = 2|OPQ |
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Answer» We know that, the lengths of tangents drawn from an external point to a circle are equal.∴ TP = TQIn ΔTPQ,TP = TQ⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)We know that, a tangent to a circle is perpendicular to the radius through the point of contact.OP ⊥ PT,∴ ∠OPT = 90º⇒ ∠OPQ + ∠TPQ = 90º⇒ ∠OPQ = 90º – ∠TPQ⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)From (1) and (2), we get∠PTQ = 2∠OPQ |
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| 66293. |
UrTwo tangents TP and Tthat < PTQ=2 < OPQ.Q are drawn to a circle with centre O from an external point. T. Prove |
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Answer» We know that, the lengths of tangents drawn from an external point to a circle are equal.∴ TP = TQIn ΔTPQ,TP = TQ⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)We know that, a tangent to a circle is perpendicular to the radius through the point of contact.OP ⊥ PT,∴ ∠OPT = 90º⇒ ∠OPQ + ∠TPQ = 90º⇒ ∠OPQ = 90º – ∠TPQ⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)From (1) and (2), we get∠PTQ = 2∠OPQ |
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| 66294. |
Example 2 :Two tangents TP and TQ are drawnto a circle with centre O from an external point TProve that ZPTQ 2 ZOPQ |
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| 66295. |
2 \frac{1}{2}+3 \frac{1}{8} \div 1 \frac{1}{4}+1 \frac{1}{4} |
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Answer» yes sir Sir please give me my therd and fourth questions answer thanks |
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| 66296. |
1 \div [ 1 + 1 \div \{ 1 + 1 \div ( 1 + \frac { 1 } { 3 } ) \} ] = |
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| 66297. |
3 \frac { 1 } { 5 } \div ( 2 - \frac { 1 } { 3 } ) + ( \frac { 2 } { 7 } \div 1 \frac { 1 } { 7 } ) |
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Answer» firstly solve the brackets2-1/3 = 6-1/3 = 5/3 2/7 ÷ 8/7 = 1/4 now , 16/5 * 3/5 + 1/4 48/25 + 1/4 =217/100 |
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| 66298. |
1 \div 1 = |
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Answer» 1÷1 is always equals to 0 |
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| 66299. |
1 \frac { 1 } { 4 } \times \frac { 1 } { 2 } \div 1 \frac { 1 } { 3 } |
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| 66300. |
15 - [ 4 \div 2 + \{ 3 \div \frac { 1 } { 4 } ( 1 + \frac { 1 } { 3 } ) \} |
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