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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. |
X -alirn . _ = 9, find all possible values of a.Ex. |
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Answer» thnxxx |
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| 2. |
The digit at ten's place of a two digit number is 2 more than twice the digitat unit's place. If the product of digits is 24, find the two digit number. |
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| 3. |
\frac 5 \frac 1 2 - \frac 2 3 \text of \frac 4 5 2 \div \frac 1 6 - ( \frac 3 5 %2B 2 ) = ? |
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Answer» 5 1/2 - 2/3 * 4/5/2 ÷ 1/6 - (3/5 + 2) = 11/2 - 8/15/ 12 - (3 + 10)/5 = (11*15 - 16)/30 / (60 - 13)/5 = (165 - 16)/30 / (47/5) = 149/30 *5/47 = 149/(6*47) = 149/282 |
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| 4. |
s Iftan A tan B-x and, cot B-cot A -y, prove that cot (A-B)+ |
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| 5. |
23. If tan A + tan B = a and cot A + cot B = b, prove that: cot (A + B) =. |
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Answer» TanA + tanB = A --------(I) & cotA + cotB = B-------(II)or from (I) 1/cotA + 1/cotB =Aor { cotA + cotB}/cotA cotB = Aor B/cotA cotB = Aor cotA cotB = B/A-------------------------------------...so cot( A+B ) =cotA * cotB -1/ cotA+ cotB= { B/A -1}/B= 1/A - 1/B |
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| 6. |
Between I and 3I, m numbesequence is an A. P. and the ratio of 7h and (m-1)t numvalue of m.rs have been nserted in s |
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Answer» Since m numbers are inserted between 1 and 31 then the total number of terms in the A.P. = n =(m+2).The first term = a = 1The last term = = a+(m+2-1)d= 31or, 1+(m+1)d=31or, (m+1)d=30or, d=30/(m+1) ---------------------(1)Now, according to the question,[a+7d]/[a+(m-1)d]=5/9or, (1+7d)/{1+(m-1)d}=5/9or, 9+63d=5+5(m-1)dor, 63d-5(m-1)d=5-9or,(63-5m+5)d=-4or, (68-5m){30/(m+1)}=-4or, 30(68-5m)=-4(m+1)or, 2040-150m=-4m-4or, -150m+4m=-2040-4or, -146m=-2044or, m=(-2044)/(-146)or, m=14 Ans. hit like if you find it useful |
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| 7. |
L Problem: Prove that cot A +cot B+ cot Ca +b+ c4A |
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| 8. |
(-(-sqrt(5) %2B 1)^2/4 %2B ((1 %2B sqrt(5))/2)^2)/sqrt(5) |
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Answer» 1÷√5 is the correct answer (1+ √5)^2 - (1-√5)^2 = ( 1+ √5 + 1-√5)( 1+√5 - 1+√5) = 2 ( 2√5) = 4√5 So 1/√5. ( 4√5/4) = 1/5 this is the perfect... hi babe u r sooo cute babe 1 is the correct answer |
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| 9. |
I 8 workes can buid a wall in150 hours How many workers will bereguleed to do the same workin 20 hours. |
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| 10. |
Iso uorkers were engagadto Pinish a job in g certainno o das ywoorke dr opped out on. Ind daymore workes droppes conthethind day and soHnisFInd the nccompleted.daye in sohich thul ark atwerk aas |
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| 11. |
3. A computer manufacture company earn a profit of 100 on every laptop and loss of 7 40 onevery desktop computer.a. The company sells 5000 pieces of laptop and 4000 pieces of desktop computers in a month.What is its profit or loss?b. Find the number of laptops it must sell to have neither profit nor loss, if the number of solddesktop computers is 3000.ost. |
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| 12. |
17. चित्र में PQ || RS | TU तथा TPL PQ ZQTU=4501हो तो x,y,z के मान ज्ञात करो। PSTERबरUN |
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Answer» x=45. y=135. z=45 <PTR=90<QTR=45<OTP=90-45=45°<PQT=180-(45+90)=45°<TQR=90-45=45°Y=180-(90+45)=45° X=45. y=135. z=45 is the right answer <PTR=90<QTR=45<OTP=90-45=45°<PQT=180-(45+90)=45°<TQR=90-45=45°Y=180-(90+45)=45° x 35 y 135 z 45 answer |
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| 13. |
dy+ v cot x = cosec xах |
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| 14. |
sz Wt9 ८ el 5e b L (e दर 128 २ है. कपूर, पिंड 012 |
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| 15. |
\sqrt { \frac { \operatorname { cosec } x - 1 } { \operatorname { cosec } x + 1 } } = \frac { 1 } { \operatorname { sec } x + \operatorname { tan } x } |
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| 16. |
6 1-coxx + 1-tan x = (sec x cosec x + 1) |
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Answer» tanx/1−cotx+cotx/1−tanx=tan^2x/tanx−1+1/tanx(1−tanx =tan^3x−1/tanx(tanx−1) =tan^2x+tanx+1/tanx= =1+sec^2x⋅cosx/sinx= =1+secx⋅cscx= |
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| 17. |
\left. \begin{array} { l } { \text { If } \operatorname { tan } \frac { \theta } { 2 } = \sqrt { \frac { 1 - e } { 1 + e } } \operatorname { tan } \frac { \phi } { 2 } \text { then } } \\ { \text { ove that } \quad \operatorname { cos } \phi = \frac { \operatorname { cos } \theta - e } { 1 - e \operatorname { cos } \theta } } \end{array} \right. |
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| 18. |
(tan(A) %2B tan(B))/(cot(A) %2B cot(B))=tan(A) |
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Answer» sinA/ cosA + sinA/ cosA/ cosB/ SinB + cosB/ sin B= sinacosa+sinacosa/ cosasina x cosBsinb + cosbsinb/ sinBsina=1/ cosa-sin a x SinB sinB== sina/ cosa x sinb/cos b= Tan a tanb |
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| 19. |
\tan ^ - 1 \frac 2 5 %2B \tan ^ - 1 \frac 1 3 %2B \tan ^ - 1 \frac 1 12 = \frac \pi 4 |
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| 20. |
(tan(theta) %2B sec(theta) - 1)/(tan(theta) %2B sec(theta) - 1)=tan(theta) %2B sec(theta) |
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| 21. |
(tan(A) %2B 2)*(2*tan(A) %2B 1)=5*tan(A) %2B 2*sec(A)^2 |
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| 22. |
EXERCISE 4.1The cost of a notebook is twice the cost of a pen. Write a linear equation in twoSvariables to represent this statement.t.thot of a nen to hev) |
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| 23. |
(sin(theta) %2B tan(theta))/(-sin(theta) %2B tan(theta))=(theta*(S*(c*e)) %2B 1)/(sec(theta) - 1) |
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| 24. |
(ii) (1 + cot x -- cosec x)(1 + tan x + sec x) = 2 |
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| 25. |
5. A company sold 4595 bikes in January and 4957 bikes in February.How many bikes did the company sell in these two months ?Lin a city, there are 3957 m |
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Answer» January=4595 bikesFebruary=4957 bikesTotal= 9552 bikes number of bikes sold by company in January = 4595number of bikes sold by company in February = 4957total numbers of bikes sold in two months= 4595+4957= 9552 bikes |
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| 26. |
24 workes can build a wall in 15 days how many days will 9 workes take to build simlar wall, |
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Answer» thanks |
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| 27. |
Find the HCF of 405 and 2520. |
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Answer» 405 = 3 x 3 x 3 x 3 x 5 2520 = 3 × 3 × 3 × 3 × 5 × 5 HCF = 405 |
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| 28. |
Find the HCF of 405 and 2520 |
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Answer» 405/5=81/3=27/3=9/3=3; 2520/5=504/2=252/2=126/2=63/3=21/3=7 the HCF of this number s is45 |
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| 29. |
SZ=. |
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Answer» Multiply the numerator and denominator with 4 to make the calculation easier. |
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| 30. |
) (xg-9)(xE-9) (®)__.‘%_ ifl.i‘—lflfi % ZXG -sz LG et |
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Answer» 25 - 9x^2 = (5)^2 - (3x)^2Using identitya^2 - b^2 = (a + b)(a - b) = (5 + 3x)(5 - 3x) |
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| 31. |
\begin { equation } \begin{array}{l}{\text { Prove that: }} \\ {2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=0}\end{array} \end { equation } |
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| 32. |
(18/12)*((4/2)*((3/9)*x)) |
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Answer» 4 1/2 * 3/9 * 18/12 = 9/2 * 3/9 * 18/12 = 3/2 * 18/12 = 9/4 41/2*3/9*18/129/2*3/9*18/123/2*18/129/4 |
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| 33. |
117.In Δ ABC, < A-100° and 3 < BSZ C. Find < B and <C |
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Answer» A=100B=5/3CA+B+C=180100+5/3C+C=1808/3C=80so.C=30so B=50 |
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| 34. |
: निम्नलिखित को सिद्ध कीजिए-sin’ 8- sin’ x cos’ 6 &e‘ cos” 1) tan + cotf = secl x cosec |
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| 35. |
11. Counting by twos write numerals from 393 to 405. |
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Answer» 393,395,397,399,401,403,405, |
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| 36. |
315^circ*cot(-405^circ) %2B tan(-585^circ)*cot(495^circ) |
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Answer» the question answer is 2 |
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| 37. |
the height of the lighthouse is 90 metres, then fiProve :1 + sin A_ 1 + sin A +cos Acos A 1 + cos A - sin A19)20)Find the possible values of sin x if 8 sin x-cosProve :cosecx -1 _ 1V cosec x +1 sec X + tan xA kite is flying at a height of 60m oh21.to the |
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Answer» 1 + sina/ cosa × 1- sina/ 1- sina = (1+ sina)( cosa)/ ( 1 + sina)(1- sina) = ( 1+ sina)(cosa)/1- sina^2 = 1+ sina(cosa)/ cosA^2 = 1 + sina/ cosa |
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| 38. |
7. How many terms are there in the AP 6, 10, 14, 18,1742 |
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Answer» thanks bro for helping me |
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| 39. |
Find the value of the middle term of the followingAP:-6,-2,2_____128 |
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| 40. |
How many terms of the AP6.4.-5.... are needed to give the sum-25? |
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| 41. |
405 % 4 = |
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Answer» 1 is the answer of the following 1 Is a. right. answer the correct answer is 1 1 is the correct answer 1 is right ans . |
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| 42. |
512729isi) which term of the GP, 18,-12, 8, |
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Answer» it is the right answer thanks man |
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| 43. |
TE . B are two different values of x lying between 0 and 2 which satisfy the equation6 cos x + 8 sin x=9, find the value of sin (a + b). |
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| 44. |
0) 210.-130(11) 405-675(2) Express the followin0)I112Sz(vi |
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| 45. |
Find the80405 |
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Answer» a) 80/405 = 16/81 |
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| 46. |
30) The sum of the digit of a two digit number is 12. The number obtained by interchanging its digit exceedsthe given number by 18. Find the number. |
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Answer» Given:x + y = 12....(1)10y + x = (10x + y) + 18....(2) 10y - y = 10x - x + 189y = 9x + 18y = x + 2 ...(3)Substituting y = x + 2 in (1) x + (x + 2) = 122x + 2 = 122x = 12 - 22x = 10x = 10/2x = 5substituting in (3) y = 5 + 2y = 7 Thus the number is 57 |
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| 47. |
. In a two digit number, the unit's digit exceeds its ten's digit by 2. The productgiven number and the sum of its digits is equal to 144. Find the number. |
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| 48. |
in a two digit number, the digit at theunit place is double the digit in itstense place. The no. exceeds the sumof its digit by 18 . Find the number12:38 pm |
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| 49. |
18, 12 जनवरी, 1979 को कौन दिन था?Which day was on 12" January, 1979? |
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Answer» As 12th January 1979 = 1900 years + 78 years + period from 1st Jan. to 12 Jan. 1979 So No. of odd days in 1600 years = 0 No. of odd days in 300 years (1900 " 1600) = (5 * 3) ÷ 7 = 2 weeks + 1 odd day No. of odd days in 78 years = 19 leap year + 59 ordinary years = {(19 * 2) + (59 *1)} ÷ 7= (38 + 59) ÷ 7 = 97 ÷ 7 =13 weeks + 6 odd days No. of odd days in period from 1st Jan. to 12 Jan. 1979 = 12 days = 12 ÷ 7 = 1 week + 5 odd days Total odd days = 1 + 6 + 5 = 12 days = 12 ÷ 7 = 1 week + 5 odd days So 12th January 1979 is Friday as Sunday is 0; Monday is 1 and so on. 12 odd days+ up to 1978 ::1600 has 0 odd days.300 has 1 odd day.so total 13 days+ in 78::19 leap years has 38 and 59 days for normal years. so totally59+38+13=110%7 (per week)=5 that is friday |
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| 50. |
s. The sum of the digits of a two-digit number is 8. The number obtained by interchanging the twodigits exceeds the given number by 36. Find the number |
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