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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. |
4 Twg cylinders have same lateral surface area Their radi are in theratio 32, Find the ratio of their heights |
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Answer» As their radii are in the ratio 3:2 Let them be 3x and 2x Let Height of first cylinder=h And Height of second cylinder=H ATQ 2πrh=2πRH 3xh=2xH h/H=2x/3x h/H= 2/3 Hence, the answer is 2/3… |
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| 2. |
\begin{array}{c}{1 \Rightarrow(5+\sqrt{5})(5-\sqrt{5})} \\ {99 \%=F}\end{array} |
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Answer» 5(5-root(5))+root(5)(5-root(5))=25-5root(5)+5root(5)-5=20 Ans 20 |
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| 3. |
\operatorname{lt}_{x \rightarrow 5} \frac{1-\sqrt{x-4}}{x-5} |
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| 4. |
\lim _{x \rightarrow 5} \frac{x^{2}-9 x+20}{x^{2}-6 x+5} |
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| 5. |
The perimeter of the floor of a square room is 18 m and its height is 3m. What is tie ares of t4 walls of the roomat is the area of the1 121 |
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| 6. |
: 5 \operatorname { tan } \theta = 4 \Rightarrow \operatorname { tan } \theta = \frac { 4 } { 5 } |
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Answer» 5tanx=45/4tanx=1so tanx=4/5 |
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| 7. |
A football team won 10matches out of the total number of matches they played. Iftheir win percentage was 40, then how many matches did they play in all?IfChameli had 600 left after sne nding 750, ofher4.5. |
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| 8. |
4y A football team won 10 matches out of the total number of matches they played. Iftheir win percentage was 40, then how many matches did they play in all? |
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| 9. |
72% of25 students are good in mathematics. How many are not good in mathematics?A football team won 10 matches out of the total number of matches they played. Iftheir win percentage was 40, then how many matches did they play in all? |
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| 10. |
<-उ SR ला.ol el oon कह o ! ('x »a)[<x' + '/a) है |
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| 11. |
T e oL e e है वLl el R o |
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| 12. |
4Twe cylinders havesamurfacn area Their radi are in theo32 ind the ratte of their heights |
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Answer» As their radii are in the ratio 3:2 Let them be 3x and 2x Let Height of 1st cylinder=h And Height of 2nd cylinder=H Acc. to 1uestion 2πrh=2πRH 3xh=2xH h/H=2x/3x h/H= 2/3 Hence, the answer is 2/3 or 2:3 |
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| 13. |
>सिक कल छा है रन kg ol किसी ale aou Wik| el vy 8 0 |
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| 14. |
The area of a square is 144 sq. cm. Find its perimeter. |
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Answer» area of a square = 144 a×a = 144 a = 12Perimeter = 4a = 4×12 = 48 area of square=144 a*a=144a =12 perimeter of square= 4a = 4*12 48 ans area of square = side × sideAs all the sides of square are equal , let side be x.Area of square = 144 cm^2 144 = x × x144 = x^2√144 = x12 = x Perimeter of square = 4 × side = 4 × 12 = 48 cm Perimeter of square is, 48 cm |
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| 15. |
What is the side of asquare room whose area is 144 sq. m? |
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Answer» who 24 |
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| 16. |
The area of the circle is 25 π sq. cms. Find thelength of its arc subtending an angle of 144° at thecentre. Also find the area of the correspondingsector |
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| 17. |
13.If the area of a rectangular plot is 144 sq. m and its length is 16 m. Find the breadthof the plot and the cost of fencing it at the rate Rs 6 per metre. |
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| 18. |
D) 8,1030.The sum of integers from 1 to 100 that are divisible by 2 (or) 5 is[IITJEE-IC) 5030D) 7030B) 3070A) 3050 |
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Answer» 1 The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100. This forms an A.P. with both the first term and common difference equal to 2. ⇒100 = 2 + (n–1) 2 ⇒n= 50 2 The integers from 1 to 100, which are divisible by 5, are 5, 10… 100. This forms an A.P. with both the first term and common difference equal to 5. ∴100 = 5 + (n–1) 5 ⇒ 5n= 100 ⇒n= 20 3 The integers, which are divisible by both 2 and 5, are 10, 20, … 100. This also forms an A.P. with both the first term and common difference equal to 10. ∴100 = 10 + (n–1) (10) ⇒ 100 = 10n ⇒n= 10 4 ∴Required sum = 2550 + 1050 – 550 = 3050 Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050. |
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| 19. |
For tue fhilowing cistribation, write the median ciass:MarhBelow 0Number of students27Below 40Belw 50Всилу 6()75S0 |
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Answer» median class is 30-40 |
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| 20. |
T के.- उन eT M s et ) e g51%] 1) -ol g o . . . el Sl sl AN |
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Answer» (1 - i) - ( -1 + i6) 1 - i + 1 -6i 2 - 7i |
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| 21. |
// व ऊ::- उड़ उरऊगा: "नी el—ffifw\t\w ol o of otl =ATL स्| VI o ad LE 3 |
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Answer» Answer: Step-by-step explanation: Sum of natural numbers n ( n+1) / 2 Sum of natural numbers between 1 to 1000 2 + 3+ .......+ 999 [999 x ( 999+1) / 2 ] - 1 Since it starts from 2 499500 - 1 = 499499 Divisible by 2 2+ 4+ ...........+ 998 2 ( 1+2+......+ 499) 2 x [ 499 x ( 499+ 1)] / 2 499 x 500 249500 Div by 5 5+10+.........+ 995 5 ( 1+2+.......+ 199) 5 [ 199 x ( 199 + 1)] / 2 5 x 100 x 5 99500 Neither div by 2 nor 5 499499 - ( 99500 + 249500 ) 150499 |
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| 22. |
Megha bought 10 note books for40 and sold them at4.75 per notebook find her gain percent. |
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Answer» CP=40total SP=4.75×10=47.5gain percentage=(47.5-40)/40*100 =7.5*100/40=18.75% |
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| 23. |
Exercise 1.1Besert gemmes and rewrite in words occording to the Indian syst21201-212.801hond hundred on |
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Answer» please send clear pic |
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| 24. |
or6q+5,whereqis2. Show that any positive odd integer is of the form 6q +1, or 6+3,some integerhond o32 memhers in |
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| 25. |
An ornament is bought for240 and sold foră210. The loss % is |
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Answer» Cost price=240RsSelling price=Rs210Loss=240-210=30 Rs Loss%= Loss*100/CP=30*100/240=100/8=12.5% |
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| 26. |
Fiha the perfneter ol20736 sq. m.the field at the rate5. One side of a square field is 92 m. Find the cost of raising a lawn onsq. m. |
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| 27. |
The base of a triangular field is three times its altitude. If the cost of levelling the field atmetre is 76588, find its base and height.3050 per sq |
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Answer» Let the altitude of the triangular field be h. then base of the triangle will be 3h. Area of triangle =1/2×base×height. =1/2×3h×h =3/2h^2 cost of levelling =6588 =3/2×h^2×30.50=6588 =h^2=6588/45.76 h^2=143.96 h=11.99 so altitude of triangle=11. 99m base =11. 99×3=35.97m |
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| 28. |
The area of a square field is 35144 Sq. m. Find its side. |
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| 29. |
what if i bought 10 phones for 20 RS and I got them for 25 RS then how much profit |
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Answer» writing is not good I asked you the wrong question but still you answered it😂😂 |
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| 30. |
Aman wanted to exchange Rs 1000 in two types of notes of Rs 5 and Rs 10denominations. If he got 180 notes in all, ind the number of notes of each kind |
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Answer» Please like the solution 👍 ✔️ send me video solution thanks |
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| 31. |
an got 10% increase in his salary. If his new salary is Rs.1 ,54,000, find hisy. |
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| 32. |
d)1.()10160. Tarun got 30% concession on the labelled price of an article and sold it for Rs. 8750with 25% profit on the price he bought. What was the labelled price ?(a) Rs. 10,000(b) Rs. 12,000(c) Rs. 16,000 of these |
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| 33. |
The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168metres, find its dimensions.2. |
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| 34. |
y,5, Vikram got an increase ofher aggregate percentage.10% in his salary. If the salary after increawas Rs 36355, find his salary before increase. |
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Answer» Let salary before increase be x10% increasex+x*10/100=36355x+x/10=3635510x+x/10=3635511x=363550x=33050 So, salary before increase is 33050Rs |
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| 35. |
The base of a triangular field is three times its height. If the cost of cultvati1080 per hectare is? 14580, find its base and height. |
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Answer» thanks |
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| 36. |
The base of a triangular field is three times its altitude. If the cost of leveling the field at3050sq. metre is 7350, find its base and height. |
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| 37. |
hase of a triangular field is three times its altitude. If the cost ofsowing the field at t 58 per hectare is 783, find its base and height |
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| 38. |
Find each of the following ratios in the simplest form:(i) 24 to 56(ii) 84 paise to 3(ii) 4 kg to 750 g |
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Answer» thank you |
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| 39. |
24. यदि L.C.M. (96, 168) = 672 है, तो H.C.F. (96, 168) कामान ज्ञात करें।(a) 13(b) 17(c) 24(d) 28 |
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Answer» H.C.F of 98,168=24 is answer Answer:c)24 Explanation: The answer of this question is 24 HCF of 98,196 is 24 .this the answer H.C.F. = (98 × 168)÷672 =24(ans) |
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| 40. |
Set of even prime numbers |
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Answer» The onlyeven prime numberis 2. All othereven numberscan be divided by 2. |
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| 41. |
2) Which of the following sets are empty sets?A) set of intersecting points of parallel lines B) set of even prime numbers.C) Month of an english calendar having less than 30 days.DP = {xlx E I, -1<x<1} |
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Answer» Answer. option A)set of intersecting points of parallel lines option A is the correct answer |
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| 42. |
A pair of prime numbers with difference 2 is called a pair of(A) composite numbers(C) even primes(B) twin primes(D) co-primes |
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Answer» Two consecutiveprime numberswhose differenceis2 are calledtwin primes. |
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| 43. |
17 6 125 1 5 7| 4 6 ?168 90 168(A) 4(B) 6(C)3(D)5 |
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Answer» 4 is the correct answer the correct answer is A 4 is the right answer 4 is the correct answer 4 is the right answer 3 is right answer.... (a) is the right answer 4 is the right answer the option (a) is the right answer of following questions option a. is the correct answer 4 is the right answer 4 is the right answer.. A is the correct answer option a is the correct answer |
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| 44. |
The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168metres, find its dimensions |
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| 45. |
Hef of 168 and Rois |
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Answer» the hcf of 168 and 126 is 42 Answer is 42 The factors of 126 is 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, and 126The factors of 168 is 1, 2, 3, 4, 5, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56 according to prime factor168=2^3×7×3126=2×3^2×7HCF= 2×3×7= 42as per Euclid lemmaa=bq+r q > r less or equal to 0when r=0, HCF=q168>126126)168(1 126 -------- 42)126(3 126 ---------168=126×1+42126= 3×42+0so HCF=42 |
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| 46. |
the parameter of a sector of a circle whose radius is 5.6 is 27.2 cm. find the area of the sector. |
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Answer» show me diagram,... |
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| 47. |
24.-sat 7, 13, 19,, 241 |
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| 48. |
The angles of elevation of the top of a lighthouse from 3 boats A, B and C in a straightlineof same side of the light house are a, 2a, 3arespectively. Ifthe distance between the boatsA and Band the boats B and C arex and y respectively find the height of the light house? |
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Answer» here, <Q = 90, PQ = h , AB = xSo, IN ∆ BPQ, tan2a = PQ/QB QB = h/tan2a----------( 1 ) now, IN ∆PQC, tan3a = PQ/QC QC = h/tan3a-----------( 2 ) similarly, IN ∆APQtana = PQ/(QA) tana = h/(QC + BC + AB) [ as QA = QC + BC + AB] (QC + BC + x) = h/tana [ AB = x]----------( 3 ) We may write [BC = QB - QC] we get, [ QC + QB - QC + X] = h/tana From-------( 1 ) , & -------( 3 ). [h/tan 2a + x] = h/tana[(h + xtan2a)/tan2a] = h/tana tana[h + xtan2a] = htan2a htana + xtana.tan2a = htan2a h(tan2a - tana) = xtana.tan2a |
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| 49. |
LEPatI8Byvolumeofaceticacidtowaterinthissample.23. The salaries of A, B and C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% areallowed respectively in their salaries, then what will be the new ratio of their salaries?Hint: Let the salaries of A, Band C (initially) be 2x, 3x and 5x respectivelyThen,new, ratio (115% of 2x) : ( 1 10% of 3x) : ( 120% of 5x)] |
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| 50. |
84kg 800 g into g |
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Answer» 1kg =1000g54kg= 54*1000g54kg=54000g 54kg 800g=54000+800g54kg 800g=54800g |
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