Saved Bookmarks
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. |
5.Find the marked price of a washing machine which is sold at * 8,400 afterallowing a discount of 16%. |
|
Answer» let marked price =x Then selling price =x16/100X=0.84xaccording to the question 0.84x=8400x=10000 |
|
| 2. |
22. Prove that from a given point, one and only one perpendicular can be drawn to a given line. |
|
Answer» LetAbe the point andLbe the line. Suppose thatBandCare two different points on lineL withABandACboth perpendicular toL. Then triangleABChas right angles atBandC, plus an angle atAwhich add up to more than180°. That does not work for a triangle in the plane whose angles have to add up to exactly180°. |
|
| 3. |
The number of lines that can passthrough a given point is |
|
Answer» Infinite number of lines can pass through one point. Very Good 👌👌 |
|
| 4. |
x + y + z = 0 \text { then } \frac { x ^ { 2 } } { y z } + \frac { y ^ { 2 } } { z x } + \frac { z ^ { 2 } } { x y } |
| Answer» | |
| 5. |
Calculate the probability of picking up a blue ball randomlyfrom a bag in which 5 red, 4 blue and one black bails arepresent.6. |
|
Answer» total balls = 10blue balls= 4so probability of getting one blue ball = 4/10 |
|
| 6. |
14. What is the probability of picking the letter'N' from the word 'RANDOM'? |
|
Answer» Total letters =6Number of times N come = 1Probability of picking N is 1/6 |
|
| 7. |
\left(a^{\frac{1}{x-y}}\right)^{\frac{1}{x-z}} \times\left(a^{\frac{1}{y-z}}\right)^{\frac{1}{y-x}} \times\left(a^{\frac{1}{z-x}}\right)^{\frac{1}{z-y}} |
| Answer» | |
| 8. |
11. Find the probability of picking up a card having 10 written on it from a deck of playing cards |
|
Answer» There is no card written 10 on it. So, probability of finding a 10 written card is 0.Please hit the like button if this helped you. |
|
| 9. |
\frac { z } { 1 } z + \frac { b } { \varepsilon } z + \frac { 8 } { s } \varepsilon |
| Answer» | |
| 10. |
Prove that AA and A'A are symmetric for the[1 2]matrix A= |
| Answer» | |
| 11. |
16. If a- 2 find the value of +a+find the value of a + aa+1 |
| Answer» | |
| 12. |
14. A film show lasted for 3hours. Out of this time I hours was spent on advertisememWhat was the actual duration of the film? |
|
Answer» Total duration of film = 3 2/3 hours Duration of advertisement = 1 1/2 hours Therefore,Actual duration of film= 3 2/3 - 1 1/2= 11/3 - 3/2= (22 - 9)/6= 13/6= 2 1/6 hours |
|
| 13. |
3427. A film show lasted for 31 hours. Out of this time, l-hours was spent on advertiseme3What was the actual duration of the film? |
| Answer» | |
| 14. |
(5) i) in a calendar, a square of four numbers is marked.The sum of the numbers is 80. What are thenumbers? |
|
Answer» let the 1st no =x 2nd no=x+1 3rd no=x+7 4th no.=x+8 4x+16=80 4x+=64 x=16 It have some more |
|
| 15. |
5.Find the marked price of a washing machine wallowing a discount of 16%.taxehich is sold at8,400 afteral |
| Answer» | |
| 16. |
If a line touches a circle at only one point, then it is known as |
|
Answer» If a line touches a circle at only one point then it is known as tangent |
|
| 17. |
Construction : Construct a tangent to a circle at a given point when the centre of the circle isknown. |
| Answer» | |
| 18. |
Manju deposited18,000 at 6% pcr annum for 2 years in a bank. If the interest is calculatedhalf-yearly, what amount will she get after 2 years? |
|
Answer» A=18000(1+6/200)(1+6/200)(1+6/200)(1+6/200) =18000(1.03)(1.03)(1.03)(1.03) =20,259.15858 |
|
| 19. |
10 shident 00... (008: 04 नै0 [00- केश 90569 Yb0१) 4 ७छतण फैथए |
|
Answer» Please like the solutionand check previous Answer as well This is related to simple equations , Let the number of boys = x Number of girls = 4 more than the number of boys = x + 4According to the problem given , Total students participated in quiz = 10 Number of boys + Number of girls = 10 x + x + 4 = 10 2x + 4 = 10 2x = 10 - 4 2x = 6 x = 6 / 2 x = 3 Therefore Number of boys = x = 3 Number of girls = x + 4 = 3 + 4 = 7 I hope this helps you. |
|
| 20. |
In Fig. 8.13, find tan P-cotR. |
| Answer» | |
| 21. |
1300 hours is the same asI) 13 aclock(ii) 1:00 a.m.(iii) 1:00 pm(iv) I hour before noon |
| Answer» | |
| 22. |
36. A piece of wood is of length 127 m. If it is cut into two pieces in such a way that the length of onepice is i, elhnr isthe lanh o tpice? |
|
Answer» convert from mixed fraction to fraction12 3/4 =51/4 5 1/4 = 21/4wood left 51/4 - 21/4 = 30/4 = 15/2 the length of other piece is 15/2 m |
|
| 23. |
Water is lowing at the rate of 15 kni per hour through a pipe of diameter 14cm into a3rectangular tank which is 50 m long and 44 m wide. Find the time in which the level ofwater in the tank will rise by 21 cm21. Water is flowing at the rate of 15 km per hour through a pipe of diameter 121. |
| Answer» | |
| 24. |
0.Iiingpiceofthegrainperkg.Ravisells his motorcycle for36,000 and incurs a loss of 10%. Find the cost price and theloss amount. |
| Answer» | |
| 25. |
balls UCI UW027. 7 times the reciprocal of a number added tois 3, find the number11 |
|
Answer» 3 is the correct answer of the given question 3 is the correct answer of the given question |
|
| 26. |
हू कल 2) +5=0 (VI |
|
Answer» Ans :- These two lines are parallel to each other. step it is graph |
|
| 27. |
A milk shop opens at 6: 30 a.m. and closes at 9:00 p.m. How long does the shop remainkind Open?thrtr at 10 15 a.m. and closes at 7: 15 p.m. Find the time duration of |
|
Answer» Duration for which shop opens=2hours and 30 minutes. |
|
| 28. |
of the same work in one hour. Who finished,4uisni3Manu finished 2 of his work in an hour whereas Pritifinished3.the work earlier?o Wasim and kent the remaining for |
|
Answer» Mahesh finished his work in 5 upon 9 hours Kapil finished the same work in 3 upon 4 hour hu talk by what fraction was it longer |
|
| 29. |
in 8.13, find tan p - cot R |
| Answer» | |
| 30. |
. In Fig. 8.13, find tan P-cot R |
| Answer» | |
| 31. |
In Fig. 8.13, find tan P-cot R. |
|
Answer» using Pythagoras law we got QR² = PR² - PQ² => QR = √ PR² - PQ² => QP = √ ( 13 )² - ( 12 )² => QP = √ 169 - 144 => QP = √25 => QP = 5 .°. tan P = QR/PQ tan P = 5/12 and Cot R = QR/PQ Cot R = 5/12 now tan P - Cot R = 5/12 - 5/12 = 0 Read more on Brainly.in - https://brainly.in/question/1345179#readmore |
|
| 32. |
= p then Prove that tancot - |
|
Answer» cot pi/16. cot 2pi/16. cot3pi/16. cot4pi/16. cot5pi/16. cot6pi/16. cot7pi/16= cot(pi/2-7pi/16)cot(pi/2-5pi/16)cot(5pi/16)cot(5pi/16)cot(7pi/16)=tan(7pi/16) tan(5pi/16)cot(5pi/16)cot(5pi/16)=( (1/cot(7pi/16)(1/ cot5pu/16)cot5pi/16)=cot(5pi/16)cot(7pi/16)= cot(5pi/16) answer |
|
| 33. |
12. In AABC, if cos A sin B-cos C, then showthat it is a right angled triangle. |
| Answer» | |
| 34. |
12. In AABC, if cos A- sin B-cos C, then showthat it is a right angled triangle. |
| Answer» | |
| 35. |
In the given figure, AABC is a right-angled trianglein which Z A is 90°. Semicircles are drawn on ABAC and BC as diameters. Find the area of the shadedregion.7 |
|
Answer» BC²=AC²+AB² =8²+6² =64+36 =100 BC=10cm Area of biggest semicircle= 1/2 * 3.14 * r² =0.5*3.14*5*5 =0.5*78.5 =39.25 cm² Area of 2nd Largest Semicircle = 0.5*3.14*r² =0.5*3.14*3² =0.5*3.14*9 =14.13 cm² Area of smallest Semicircle = 0.5*3.14*r² =0.5*3.14*4² =0.5*3.14*16 =25.12cm² Area of Triangle = 1/2 *b*h = 0.5*6*8 =24 cm² Area of shaded region = (24+25.12+14.13) - 39.25 = 63.25-39.25 =24 cm² |
|
| 36. |
32. AABC is a right angled triangle, in whichA(0, 2) and B(2, 0) are given. Then, findthe coordinates of C.CBSE 2013 |
|
Answer» B is the best answer me please and thank u for the t you have a nice day and I have no idea what to tell you about it and (y - 0)/(x - 2) × (y - 2)/(x - 0) = -1 or, y(y - 2)/x(x - 2) = -1 or, y² - 2y + x² - 2x = 0 it is only when x = 0 or 2 also y = 0 or 2 so, C (0,0) and (2,2) |
|
| 37. |
31. There are four candles which can burnfor 5 hours, 4 hours, 3 hours and 2 hoursrespectively. All of them are lighted at thesame time. They are kept lighted till the time3 candles are finished. If the cost of lightingeach candle is 1-50 per hour, what will bethe total expenditure ? |
|
Answer» after 4 hrs 3 of them will burn so total cost is (2h+3h+4h+4h)×1.5=13×1.5=19.5Rs |
|
| 38. |
b ^ { 2 } \operatorname { sin } 2 C + c ^ { 2 } \operatorname { sin } 2 B = 2 b c \operatorname { sin } A |
| Answer» | |
| 39. |
\operatorname { sin } ^ { 2 } 0 + \operatorname { sin } ^ { 2 } ( \frac { \pi } { 6 } ) ^ { c } + \operatorname { sin } ^ { 2 } ( \frac { \pi } { 3 } ) ^ { c } + \operatorname { sin } ^ { 2 } ( \frac { \pi } { 2 } ) ^ { c } |
| Answer» | |
| 40. |
\left. \begin{array} { c } { \text { If } \angle A + \angle B + \angle C = 180 ^ { \circ } \text { then, prove that: } } \\ { \operatorname { sin } A + \operatorname { sin } B - \operatorname { sin } C = 4 \operatorname { sin } \frac { A } { 2 } \operatorname { sin } \frac { B } { 2 } \operatorname { cos } \frac { C } { 2 } } \end{array} \right. |
| Answer» | |
| 41. |
1)18 inutes to 1 hour (vi)Thinutes to 1 hour (vi0%)2. Expresseach of the following ratios in the simplest form:(i) 36: 90(iv) 480: 384(ii) 324 144(v) 186 403 |
| Answer» | |
| 42. |
\lim _{h \rightarrow 0} \frac{\tan (x+h)-\tan x}{h} |
|
Answer» put in limit it forms 0/0 formhence using l hospital ruledifferentiating both numbers for and denominator separately wrt to hsec^2(X+h)/1now put in limitshencesec^2x |
|
| 43. |
If the angle of elevation of a cloud from a point h meeters above a lake is aangle of depression of its reflection in the lake is β, prove that the height of thecloud ish (tan α+ tan β )tanß-tan α |
|
Answer» Let a be a point h metres above the lake AF and B be the position of the cloud. Draw a line parallel to EF from A on BD at C. But, BF = DF Let, BC = m so, BF = (m + h) ⇒ BF = DF = (m + h) metres Consider ΔBAC, AB = m cosec α ---------- (1)and, AC = m cot α Consider ΔACD, AC = (2h + m) cot β Therefore, m cot α = (2h + m) cot β ⇒ m = 2h cot β / (cot α - cot β) Substituting the value of m in (1) we get, AB = cosec α [2h cot β / (cot α - cot β)] = 2h sec α / (tan β - tan α) = h (tan α + tan β)/(tan β - tan α) Hence proved. Like my answer if you find it useful! |
|
| 44. |
From the top of a lighthouse, the angles of depression of tuo shipsonthe opposite sides of it are observed to be oz and ß. If the height of helighthouse be h metres and the line joining the ships passes throuekthe foot of the lighthouse, show that the distance between the shipsish(tan α + tan β)metres.tan α tan βHOTS |
| Answer» | |
| 45. |
The top of a hill observed from the top and bottom of a building of height h is at angles of elevation p andq respectively. The height of the hill ish cot pcot p- cot qhtan ptan p- tan qh cot pcot p+ cot qh cot ptan p+ tan qb)c)d) |
|
Answer» Thank u very much |
|
| 46. |
If \frac{\sin (\theta+A)}{\sin (\theta+B)}=\sqrt{\frac{\sin 2 A}{\sin 2 B}} then tan^2 theta=?a)tan A b)tanB c)tanA+tanB d)tanAtanB |
| Answer» | |
| 47. |
tan (A-B)tan A-tan Bam A tanFind the valueton 15ăam 15I+tanA tanB |
|
Answer» Let tan(15°) = tan(45°-30°) We know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B) tan(45°-30°)= (tan45°- tan30°)/ (1+tan45°tan30°) = {1- (1/√3)}/{1+(1/√3)} ∴ tan15° = (√3 - 1) / (√3 + 1) |
|
| 48. |
if A+B+C=π then prove that tanA+tanB+tanC=tanA tanB tanC |
| Answer» | |
| 49. |
tanA+tanB=cosA+cosB |
| Answer» | |
| 50. |
In a right angled triangle AABC, A and B are acute angles. Iftan2A + tan2B = 14then tanA + tanB = |
|
Answer» x^2+x^2=14; 2x^2=14; x^2=14/2=7; Tan(V7)+tan(V7)=tan(V7)^2=7 Jo Bhi Mera answer Dekhe usko teri maa ki kasam please mujhe like kar main bahut Gareeb Hoon please like 7 is the correct answer 7 is the correct answer . |
|