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. |
10. Two poles of equal heights are standing opposite each other on either side of the road,which is 80 m wide. From a point between them on the road, the angles of elevation ofthe top of the poles are 60° and 30°, respectively. Find the height of the poles and thedistances of the point from the poles. |
|
Answer» Thanks |
|
| 2. |
Examine that sin | χ l is a continuous function. |
| Answer» | |
| 3. |
Show that |cosx| is continuous function. |
| Answer» | |
| 4. |
Show that cos |x|is a continuous function. |
| Answer» | |
| 5. |
Show that sec x is a continuous function. |
|
Answer» -cosx is a continuous functionfor everyx∈R , so thatsecxas the quotient of twocontinuous functionsiscontinuousfor everyxexcept where the denominator is zero. -sec X=1/cos X -Sosecxin undefined wherecosx=0, and that happens at odd multiples ofπ/2,like,-π/2 to π/2 -sec x is undefined at -π/2 and π/2 It-iscontinuous on the open interval (-π/2,π/2) |
|
| 6. |
tne building.10 Two poles of equal heights are standing opposite each other on either side of the roadwhich is 80 m wide. From a point between them on the road, the angles of elevation dfthe top of thé poles are 60 and 30P, respectively. Find the height of the poles and tedistances of the point from the poles. |
| Answer» | |
| 7. |
= 4 x 5 x 6 - 4x 5 = 100Example 12 Find the value of n such thate p. objects are of anif any, are of differen"P= 42 "P, n>4mn>4Solution (1) Given that |
| Answer» | |
| 8. |
Example 12Find the value of n such that(i)nP,-42"R, n > 4 |
|
Answer» nPk = n!/(n-k)! nP5 = 42 × nP3 n!/(n-5)! = 42 × n!/(n-3)! (n-3)!/(n-5)! = 42 (n-3)(n-4)(n-5)! / (n-5)! = 42 (n-3)(n-4) = 6×7 = (10-3)(10-4) n = 10 If you find this answer helpful then like it. |
|
| 9. |
15. Two poles are of height 28 m and 36mand distance between them is 1tlhe distance between the top of the poles.Mathematics7 |
|
Answer» Let the distance between height of the towers is zTheheight diffeence of towers is 36-28 = 8m.(x)The distance between two towersis 15m.(y) by pythogaras theorem x²+y²=z² 8²+15²=z² 64+225=z² 289=z² 17=z²the distance between towers is 17m |
|
| 10. |
12. There are a certain number of trees in a garden. The number of trees in each row iseighteen times the number of rows. Find the number of rows in the garden and thenumber of trees in each row if the total number of trees is 2668050. |
| Answer» | |
| 11. |
3.The height of Mount Everest is 8848 m. Write it in standard form.The speed of light is 300000000m/sec. Express it in standard formThe distance from the earth to the sun is 149600000000 m. Write it in standard |
| Answer» | |
| 12. |
Writein standard form. |
|
Answer» -35/30-7/6 is the right answer -35/30-7/6 is the correct answer -7/6 is a correct answer -7/6 is the right answer of following questions The answer is = -7/6 -7/6 is the correct answer -7.6 is the right answer -7/6 is the correct answer |
|
| 13. |
1) The height of Mount Everest is 8848 m. Write it in standard form.form |
| Answer» | |
| 14. |
0.000001 in standard form |
|
Answer» Standard form1 × 10 power ( -6) |
|
| 15. |
31470000000 in standard form |
|
Answer» 3.147 * 10 ^10 is the standard form i guess it is right |
|
| 16. |
write 456928 in standard form |
| Answer» | |
| 17. |
write 39087.8 in standard form? |
| Answer» | |
| 18. |
1. f(x) cos(x) is a continuous function because(a) a composition of continuous function is a cofunction(b) product of continuous function is a continuotion(c) cosine is an even function(d) sum of continuous functions is continuousThe value of b for which the function2. |
|
Answer» 1 2 |
|
| 19. |
111. Every absolutely continuous function is(a) continuous(b) constant(c) absolutely discontinuous(d) of the above1osined |
|
Answer» Every absolutely continuous function is uniformly continuousand, therefore,continuous.EveryLipschitz-continuous functionisabsolutely continuous. ... If f: [a,b] → R isabsolutely continuous, then it has the Luzin N property (that is, for any such that , it holds that , where stands for the Lebesgue measure on R). |
|
| 20. |
Solution C is the correct answI.Example 29 The function f(x) r+fr - 1 is(A)(B)(C)(D)continuous at x 0 as well as at x 1.continuous at x = 1 but not at x = 0discontinuous at x = 0 as well as at x = 1continuous at x = 0 but not at x = 1. |
|
Answer» option A is correct... see the graph below.. |
|
| 21. |
16. Show that the square of an odd positive integer can be of the form 6q tor 6q3 for some+ 1integer q. |
| Answer» | |
| 22. |
Prove that the function f defined byifx#0f(x)-| x"sin0ifx=0continuous and derivable at x = 0 but its derivative is not continuous at x = 0. |
| Answer» | |
| 23. |
(569-479 B.C.)m of the squares of the other two sides.e. We are given a right triangle A DG6.8 Ln a right triangle, the square of the hypotenuse is equal to theÇInis equal to the |
|
Answer» Given: A right angled ∆ABC, right angled at B To Prove- AC²=AB²+BC² Construction: draw perpendicular BD onto the side AC . Proof: We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other. We have △ADB∼△ABC. (by AA similarity) Therefore, AD/ AB=AB/AC (In similar Triangles corresponding sides are proportional) AB²=AD×AC……..(1) Also, △BDC∼△ABC Therefore, CD/BC=BC/AC (in similar Triangles corresponding sides are proportional) Or, BC²=CD×AC……..(2) Adding the equations (1) and (2) we get, AB²+BC²=AD×AC+CD×AC AB²+BC²=AC(AD+CD) ( From the figure AD + CD = AC) AB²+BC²=AC . AC Therefore, AC²=AB²+BC² This theroem is known as Pythagoras theroem. |
|
| 24. |
) Mr. Kasam runs a small business of makingearthen pots. He makes certain number of potson daily basis. Production cost of each pot isR 40 more than 10 times total number of pots, hemakes in one day. If production cost of all potsper day is 600, find production cost of one potand number of pots he makes per day.(4 marks) |
|
Answer» Let the total No. of pots per day = x Production cost per pot = 10x + 40 Production cost of all pots per day; = (10x + 40)x = 600 10x² + 40x - 600 = 0 x² + 4 - 60 = 0 x² + 10x - 6x - 60 = 0 x(x + 10) - 6(x + 10) = 0 (x - 6)(x + 10) = 0 Either x - 6 = 0 x = 6 or x + 10 = 0 x = -10 Since the total No. of pots can not be negative, we take x = 6 Production cost per pot = 10x + 40 = 10(6) + 40 = Rs. 100 Total No. of pots per day = x = 6 pots |
|
| 25. |
6. Mr. Kasam runs a small business of making earthen pots. He makes certain numberof pots on daily basis. Production cost of each pot is 40 more than 10 times totalnumber of pots, he makes in one day. If production cost of all pots per day is ?600, find production cost of one pot and number of pots he makes per day. |
| Answer» | |
| 26. |
Sameer bought 1600 bananas at 3.75 a dozen. He sold 900 of them at 2 for ₹1 and the remaining at 5 for 2. Find his gain or loss per cent. |
| Answer» | |
| 27. |
Find the value of p, for which one root of the quadratic equation px2- 14x + 80 is 6 times the other. 2 marks |
|
Answer» Like if you find it useful |
|
| 28. |
12.Find the value of p, for which one root of thequadratic equation px2-14x +8 0 is 6 timesthe other.(AI 2017) |
| Answer» | |
| 29. |
Find the value of p, for which one root of the quadratic equationpx2-14x + 8 = 0 is 6 times the other. |
| Answer» | |
| 30. |
bought 1600 bananas atăŚ3.75 a dozen. He sold 900 of them at 2 for: 2. Sameer:at 5 for t2. Find his gain or loss per cent.1 and the remaining |
| Answer» | |
| 31. |
1. IfA- 0,6,TT2and f: A → B isa surjection defined by f(x)-cos x then find-B |
|
Answer» cos0=1cospie/6=root(3)/2cospie/4=1/root(2)cospie/3=1/2cospie/2=0so B={1,root(3)/2,1/root(2),1/2,0} |
|
| 32. |
bought bananas at the rate of 10 for 45 and sold at the rate oas for51. Find his gain or loss per cent. |
|
Answer» Cost of 10 bananas = Rs. 45 Cost of 1 banana = 45/10 = Rs. 4.5 Selling price of 12 bananas = Rs. 51 Selling price of 1 banana = 51/12 = Rs. 4.25 Loss = Cost price of 1 banana- Selling price of 1 banana Loss = 4.5 - 4.25 Loss = Rs. 0.25 Loss percent = (Loss*100)/Cost price ⇒ (0.25*100)/4.5 ⇒ 25/4.5 ⇒ 5.55% |
|
| 33. |
2. PQR is a triangle right angled at P and M isapoint on QR such that PM L QR. Show thatPM2 QM. MR. |
| Answer» | |
| 34. |
and respeedMuzaffarpur to Jammutawi.our dhwaj Express starts from Muzaffarpur on Sunday at 07:00aches in Jammutavi on Monday at 11: 30. If the averageof the train is 50 km/h then find the distance between |
|
Answer» Speed=50km/hrTotal time=28.5 hoursDistance=Speed*time=50*28.5=1425Km From which book have you taken this question? Please tell us so that we can provide you faster answer. Ganit bharati |
|
| 35. |
9. If x, y, z: are in proportion, find x if y - 6 and z -12 |
| Answer» | |
| 36. |
man bought two cycles for 1040. He sold one at a loss of 15% and other at a pront ofthen he found that each cycle was sold for the same price. Find the C.P. of each cycle. |
|
Answer» CP of one bicycle = 1040 ÷ 2 = Rs.520 SP of first cycle = 520 x 85/100 = Rs.442 SP of second cycle = 520 x 136/100 = Rs.707.20 If there is no profit then CP of one bicycle = Rs.442 CP of second bicycle = Rs.707.20 707.20 is the correct answer of the given question |
|
| 37. |
one knogram, r nict ins gan F for 45000 . He sold one at a loss of 5% andA man bought two computers 450If the selling price of each set is same, find the cost price of each set.the other at a profit of 5%. |
| Answer» | |
| 38. |
The marked price of a watchwas 720/-. A man bought thesame for 550.80, after gettingtwo successive discounts, thefirst 10%. What was the seconddiscount rate ? |
|
Answer» Marked price of watch MP = 720 Let second discount be x% Then,550.80 = 720(1 - 10/100)(1 - x/100)550.80 = 720(9/10)(100-x/100)55080/72*9 = 100-x85 = 100 - xx = 100 - 85x = 15 Therefore, second discount rate is 15% |
|
| 39. |
Find the value of p, for which one root of the quadratic equation px2 14x + 80 is 6 times the other.and 3, 10, |
|
Answer» px^2-14x+8=0let a and b be the roots of the equationb=6asum of roots = -b/aproduct of roots = c/ahere a=p , b=-14 , c=8a+b=14/pab=8/p a+6a=14/p7a=14/pa=2/p a*6a=8/p6a^2=8/p3a^2=4/p 3*(2/p)^2=4/p 3*4/p^2=4/p p=3 |
|
| 40. |
Find the value of k, for which one root of the quadratic equationkx2-14x + 12 = 0is six times the other. |
|
Answer» suppose roots are a and 6aso a+6a=14/kso 7a=14/kso a=2/k so k=2/a6a*a=12/ka^2=2/k=2a/2=aso a(a-1)=0if a=0 k=infinite which in not possiblea=1 so k=2 |
|
| 41. |
1. A man boughiEXERCISE 9 (D)en bought an article for 25 and sold it for 40. Another man bought an article Towar 365. What rate of profit is greater and by what per cent ?40. Another man bought an article for 50 and soldnSameer bought 1600 bananas atmanas at 3.75 a dozen. He sold 900 of them at 2 for 1 and the remaining atSfor 72. Find his gain or loss per cent.A woman bought so |
|
Answer» 1.ans=CP1 = Rs. 25SP1 = Rs. 40CP2 = Rs. 50SP2 = Rs. 65 Profit = SP1 - CP1= Rs. (40-25) = Rs. 15Profit%1 = 15/25×100 = 60% Profit = SP2 - CP2= Rs. (65-50) = Rs. 15Profit%2 = 15/50×100 = 30% Answer. Profit%1 is greater by 30%. CP1 = Rs. 25SP1 = Rs. 40CP2 = Rs. 50SP2 = Rs. 65 Profit = SP1 - CP1= Rs. (40-25) = Rs. 15Profit%1 = 15/25×100 = 60% Profit = SP2 - CP2= Rs. (65-50) = Rs. 15Profit%2 = 15/50×100 = 30% Answer. Profit%1 is greater by 30%. |
|
| 42. |
= 4—‘ BC. fag sifeg 1 AD® =10 धर'ख) सिद्ध कीजिए किन A+l J—-—l sin A\ [ +3 / 2 A.1-sin A sin A+1 . |
| Answer» | |
| 43. |
4 0 I- isa solution of the equation 3x2 + 2kox -3 - 0, find the value of k |
|
Answer» x=-1/2 is the solutionso 3/4-k-3=0so k-9/4 |
|
| 44. |
I and m are two parallel lines intersectedanother pair of parallel lines p and(see Fig. 7.19). Show that Δ ABC Δ CDA. |
| Answer» | |
| 45. |
19 An aero plane flying horizontally 1000-,3 m above the ground isobserved at an angle ofon the ground after 10 seconds, the angle of elevation at the point ofobservation changes to 30° Find the speed of the plane In m |
| Answer» | |
| 46. |
4. The length of the shadow of a tower onthe plane ground is 3 times the heightof the tower. The angle of elevation of sun |
| Answer» | |
| 47. |
91 and m are two parallel lines intersected byanother pair of parallel lines p and a(see Fig. 7.19). Show that A ABC=ACDA. |
|
Answer» land mareare parallel and they are equal and congruent l am liked Your answer like my answer |
|
| 48. |
The length of shadow of a tower on plane ground is root 3 times of the tower find the angle of elevation of the sun |
|
Answer» it doesn't showing answer it doesn't showing answer love you |
|
| 49. |
the correct alternatives in each of the following1. The rational number which is neither apositive nor a negative is00 1(4) none of tZ in standard form is-81(c)그-81s)If 6-3, then the value of 1 is(4) 8(4) 12± in its simplest form is written as7211(b)(c)之185. Absolute value ofis(a)(b) 24) none of thee112 n standard form is |
|
Answer» 3.x/6 = 4/(-3)x = 6x4/(-3)x=-4 x 2x = -8 |
|
| 50. |
Two poles of heights 6 m and 11 m stand on a plane ground. If the distatheir feet is 12 m, find the distance between their tops. |
| Answer» | |