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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. |
Tp denotes the number of permatations of a + 3) things taken all at a time, a denotes the number ofpermutations of things taken ll at a time and r denotes the number of permutations of (0-10)things taken all at a time such that pa182 qr, then the value of ris3) 124) 102) 11 |
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Answer» 2)11 is correct answer 2) 11 is the right answer of the following 2)11 is the correct answer |
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| 2. |
12.Find the real and imaginary part of (2-31) |
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Answer» 2 is the real part and -3 is the imaginary part |
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| 3. |
A number is first increased by 15%and then reduced by 15%. By whatpercentage does the number change? |
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Answer» Let the number be 100increased no.= 115decrease= 115 ×15/100= 17.25decreased no. = 115-17.25 = 97.75change= 100-97.75= 2.25percent change= 2.25/100×100= 2.25% Plz like my answer and accepted as best plz Let the no. be x15% + x = 15/100 + x= 3/20 + x3/20 + x - 15%= 3/20 + x - 3/20so both 3/20 was subtracted leaving nothing.=xso by 0% or none percentage the no. change. Hope it helps!!! the answer is 2.25%to get more answeraccept me best let no be 10015% increase no = 115again 15% reduced no = 115×85/100=9755/100= 97.55so total no changed = 100-97.55= 2 45% let no be 10015% increase no become= 11515% reduce no become 115×85/100= 9775/100= 97.75no reduced= 100-97.75= 2.25% 2.35℅ percentage the number change....becuase first its increase then it decrease. Let the number be 100increased no.= 115decrease= 115 ×15/100= 17.25decreased no. = 115-17.25 = 97.75change= 100-97.75= 2.25percent change= 2.25/100×100= 2.25% plz like my answer 2.35% percentage the number change.... because first it's increase then it decrease |
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| 4. |
Find the ysof e faypormiknd CoRfFicient |
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| 5. |
A number is first increased by 15% athen reduced by 15%. By what percentdoes the number change? |
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Answer» Let the number be 100increased no.= 115decrease= 115 ×15/100= 17.25decreased no. = 115-17.25 = 97.75change= 100-97.75= 2.25percent change= 2.25/100×100= 2.25% |
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| 6. |
9. Show that each of the relation R in the set A = {x € Z:0<x< 12), given by(1) R = {(a, b) : la -- bl is a multiple of 4}(ü) R = {(a,b): a = b)is an equivalence relation. Find the set of all elements related to 1 in each case1.3Thefun |
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| 7. |
The price of a shirt first increased by 20% and was then reduced by 25%. Now itsprice isE)450. Find its original price. |
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| 8. |
2. People who attended a fair on the first day were 3000. If the number increased to 3600 the second day, findthe rate of increase.ntities 165 |
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| 9. |
Find ten rational numbers between and2 |
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Answer» LCM of 5 and 2 is 10.(-4/5)=(-8/10), (-1/2)=(-5/10). Rational numbers between these two numbers = -7/10, -6/10, -13/20, -61/100,-62/100,-63/100,-64/100,-66/100,-67/100,68/100. Please hit the like button if this helped you out.. |
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| 10. |
. Why did the narrator feel that he would never be able to discoversomething about the girl's looks? |
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Answer» this is not a question of mathematics.... it is drama I think |
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| 11. |
Mike sold two horses for18,000 each. On one he got a profit of 20% and onthe other he lost 20%. Find his total gain or loss. |
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| 12. |
18.If 13,1 = 1(3,1) and ,1then show that the real part of s, is zero.(NCER |
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| 13. |
Q. 21. If a, b, c are non-zero real Hulloelt fucQ. 22. Ifx -y 2 and y 15, find the value of -y3. |
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| 14. |
1. Amm buys aof agricuhral and fr Rs. 300000, He sells one-third at a loss of 20% andtwoรกths at a gain of 25%. At what price must he sell the remaining land so as to make anovenll prat of 10%. |
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| 15. |
Salaryof a person is first increased by20%, then it is decreased by 20%ercentagechange in his salary is |
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| 16. |
A number is first increased by 10% and then decreased by 10%, and then thenumber isa) 15%5) 10%c) 20%d) |
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Answer» b is the correct answer b is the correct answer option b is the correct answer of the given option b is the answer of the question |
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| 17. |
3+1521. Ifpa nd qfind the value of p?+q. |
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Answer» it's wrong |
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| 18. |
If a+1/p= p and a-1/a= q , find relation between p and q. |
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| 19. |
Mr Joshi buys an estate for 334 lakh. He buildsa boundary wall around it at a cost of 2 lakh.He sells two-fifths of it at a loss of 20% andone-third of it at a gain of 25%. At whatgain per cent must he sell the remainderso as to make a profit of 5% on thewhole?loss of 20%. |
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Answer» Thanks Thank you so much |
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| 20. |
an buys a pit of land at 3,60,000. H sells one third of the plot at a loss of 20% Again he sellsof the plot left at a profit of 25%. At what price should he sell the remaining plot in order tog prices of the commodity in both the cascget a profit of 10% on the whole?. A shopkeener houuhtce looks for34 and sold them at the rate of 12 locks for:57. |
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Answer» Cost = Rs 3, 60, 000. cost of 1/3 of the plot = Rs 1,20,000.loss due to its sale = 20% = Rs 24,000. cost of the second 1/3 of the plot = Rs 1,20,000Profit due to its sale = 25% = Rs 30,000. Net profit to be gained on the overall sales = 10% = Rs 36, 000. Profit to be gained from the sale of third part = Rs 36,000 - 30, 000 + 24,000 = Rs 30,000 So Sale price for the last 1/3 of the plot = Rs 1,20, 000 + Rs 30, 000= Rs 1, 50, 000 |
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| 21. |
Ans. 17 %97,An article was bought for? 300, and sold at 6% profit. Find its Selling Price.f 10% Find |
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| 22. |
(f) Mike sold two horses for18,000 each. On one he got a profit of 20% and onthe other he lost 20%. Find his total gain or loss.(g) Aryan bought a cupboard for1,520 and sold it at a profit of 12%. Find theselling price of the cupboard. |
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Answer» 1.S.p=18000 gain 20%C.P 15000Sp 18000 loss 20%Cp 22500Total cp 37500Total sp 36000Loss 1500Loss% = 4%2 .selling pricecp=1520profit%= 12 1/2= 25/2 %=25/200=0.125amount gained=1520×0.125=190selling price=c.p+ profit amount= 1520+190=1710 |
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| 23. |
JUNE megelm.14. Under what condition the polynomiaat condition the polynomial ax+bx+c does not have any real zero. |
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Answer» Roots of Polynomial ax2+ bx+ c = 0Can be calculated by using quadratic equation x=−b±b2−4ac√2aSo we can find out nature of roots by calculate the value of b2- 4ac , As :1 )If b2- 4ac > 0 the quadratic has two real distinct roots2 )If b2-4ac = 0 the quadratic has real and equal roots3 )If b2-4ac < 0 the quadratic has no real roots: they are complexLets have a examplex2+x+ 1Here a = 1 , b = 1 And c = 1Soroots will bex=−1±1−4√2 x=−1±i3√2 ( As we know−1−−−√= i , where i is any imaginary number )So we can see that both roots are not real .SoWe can calculate b2- 4ac = 1 - 4 = -3 < 0So,If the polynomial ax2+bx +c does not have any real zero , than b2- 4ac< 0 . c<0 is the correct answer of the given question c <0 is the correct answer |
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| 24. |
(July 2015 M) |
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Answer» sin 45= cos 45= 1/√2so theeta= 45°please like the solution 👍 ✔️ |
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| 25. |
The price of an article is first decreased by 20% and then increased by 30%. If the resulting priceis Rs.416, the original price of the article is |
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Answer» suppose price is xso x*(0.8)*(1.3)=416so x=416/((0.8)*(1.3))=400 Thanxx sir |
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| 26. |
If p =-3-153+53+15and q=-=, find the value of p'+q.3-75 |
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| 27. |
What is the condition for (a -by-+ ve ? |
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Answer» it is possible only if a>b |
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| 28. |
A man buys a plot of agricultural land for 300000. He sells one third ata loss of 20% and two fifth at a gain of 25%. At what price must he sellthe remaining land so as to make an overall profit of 10%.F |
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Answer» CP of the total land = Rs 300000 Sold 1/3 of land at loss of 20% CP of 1/3 land = 1/3* 300000 = 1,00,000 SP of 1/3 land = CP * ( 1- Loss%/100) = 100000 * ( 1- 20/100) = 100000* 80/100 =Rs. 80,000/- Similarly sold 2/5 at a gain of 25% CP of 2/5 land = 2/5*300000= 1,20,000/- SP of 2/5 land = 120000 * ( 1+25/100) = 120000* 125/100 = Rs. 1,50,000/- Total gain % reqd = 10% Total SP reqd = Total CP * ( 1+ Gain%/100) = 300000 * (1+10/100) = 300000* 110/100 = Rs. 3,30,000/- SP of the balance land = 3,30,000- 80,000 1,50,000 = 1,00,000/- Therefore he has to sell the balance land at Rs 1,00,000/- (Answer) |
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| 29. |
3YIn the figure, if O is the centre of the circle and LAOC-140, find x.2140° |
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| 30. |
B 2 मा .बए ... 0७४० ७ का, 1, 2. elकि. |
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Answer» a=8,d=10-8=2 60th term=a+(60-1)d=8+(59)2=8+118=126 AP:8,10,12.....120,122,124,126 To find sum of last ten terms, reverse the order of AP 126,124,122,120,.....,12,10,8 Now we have to find sum of first ten terms=(n/2)[2a+(n-1)d]=5[2*126+(10-1)(-2)]=5[252-18]=5(134) =670 |
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| 31. |
2 el* J/3+/2 " J/a+/3 |
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| 32. |
Dsubject = mathssumit is as old as Anita |Euve year ago his agewas three times Anito'sage. Find their presentadeTheconse |
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Answer» let the age of sumit be xand the age of anita be y1st conditiom sumit age = anita age2nd condition 5 years agox-5=3(y-5)x-5=3y-15x-3y=-10 y-3y=-10 :(X=Y)-2y=-10y=5years oldso anita is 5 years old.and sumiy age is also 5 years old. ANSWER 5 is the correct answer of the given question is Anita is 5 year old and sumiy age is also 5 year old |
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| 33. |
An article was purchased for?!239 including GST of 18%. Find the price ofthearticle before GST was added?11. |
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| 34. |
rind the value of the polynomial ys_5y +6 at y 0.In which quadrant, the points P (2,-3) and Q (-3, 2) lie?In the given figure, PQI IRS and ZACS = 127", find <BA |
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| 35. |
66) A dealer sells an article for 75 and gains as much percent as the cost price of the article.Findthe cost price of the article. |
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| 36. |
The selling price of an article is – times its cost price. Find the gain percent of thearticle. |
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Answer» 20% is the answer of the following 11it is right answers |
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| 37. |
The list price of an article is 800 and isavailable at a discount of 15 percent. Find:(i) the selling price of the article;(i) the cost price of the article, if a profit f% is made on selling it.313 |
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Answer» MP or LP = Rs 800Discount = 15% Discount amount = L P* 15/100 = Rs 800 * 15/100 = Rs120Selling price = LP - discount amount = Rs800 - Rs 120 = Rs 680Let Cost Price be CPProfit made = 40/3 %Profit amount = CP * 40/(3*100) = 0.4CP/3 Selling Price = SP = CP + profit = CP + 0.4 CP/3 = CP ( 1 + 0.4/3) = CP * (3+ 0.4)/3 Hence 680 = CP * 3.4/3 CP = 3 * 680 / 3.4 = Rs 600 |
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| 38. |
- 12 < 4 - \frac { 3 x } { - 5 } \leq 2 |
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| 39. |
2. (cos2 67° -sin2 23) 5What is the value of (cos2 67° - sin2 239) ? |
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Answer» As we know thatcos 67= sin(90-67)= sin23 so sin²23- sin²23= 0 |
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| 40. |
MATHEMA11.58in θ + sin2 6-1, then cos2 θ + cosfg-(b) 1420, Ifs(c) 0(d) noneofte(a) -12 |
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Answer» another method plz |
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| 41. |
6. (i) From the formula cos(A - B) cos A cos B + sin A sin B deduce that cos2 A+ sin2 A 1. |
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| 42. |
(cos2 67° - sin2 23What is the value of (cos 67 -sin2 239)O |
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Answer» cos²67 - sin²23 = sin²23 - sin²23 (∵ cos(90 - θ) = sinθ) = 0 |
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| 43. |
L: 2 el 140 STUIS + ,SOLWS |
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Answer» Please hit the like button |
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| 44. |
r. To raise awareness about hazards of smoking, a school decides to start 'NO SMOKING' campaigiStudents were asked to prepare banners in the shape of triangle as shown in the Fig. 4. Design ofthe banner is such that AD is the median and E is the mid-point ofAD. Show that ar(ABED) = 1ar(AABC)Smoking isProhibitedFig. 4 |
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Answer» Thnx |
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| 45. |
irS |
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| 46. |
1. In a trapezium PQRS, PQ | IRS; and P-70 and ZQ-80°. Calculatethe measure of S and R. |
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| 47. |
\frac{2^{3} \cdot 2^{-2} \cdot 2^{4}}{2^{-1} \cdot 2^{0} \cdot 2^{-3}} |
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| 48. |
A man sold an article for594 and gained 10% on it, Find the cost price of the article. |
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Answer» sp is 594 Selling Price,SP= Rs 594 Let the cost price be CP. CP + 10%of CP= 594 (1+0.1)CP= 594 CP=Rs 540 Hence, the cost price of article was Rs 540. |
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| 49. |
\left. \begin{array} { l } { x \leq 2 \pi \text { is } } \\ { \frac { \pi } { 3 } } \\ { \frac { \pi } { 3 } , \frac { 5 \pi } { 3 } , \operatorname { cos } ^ { - 1 } ( - \frac { 3 } { 2 } ) ( \text { d } ) \text { } } \end{array} \right. |
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Answer» 6cosx - 3 + 4cos^2x - 2cosx = 04cos^2x + 4cosx - 3 = 0D = 16 + 48 = 64root D = 8cosx = (-4 + 8)/8 or (-4-8)/8= 1/2 or -3/2so, cosx = 1/2and x = 2npi +_ (pi/3) |
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| 50. |
\frac { 4 ^ { 2 n } \cdot 2 ^ { n + 1 } } { 2 ^ { n - 3 } \cdot 4 ^ { 2 n + 1 } } |
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Answer» 4^(2n)×2^(n+1)----------------------2^(n-3) × 4^(2n+1) 2^(4n) × 2^(n+1)= ------------------------ 2^(n-3) × 2^(4n+2) = 2^ { 4n+n+1-(n-3+4n+2)} = 2^ { 5n+1-5n +1}= 2^2 = 4 |
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