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. |
Duu.. ITULIFind the value of (71)" by the short cut method. |
|
Answer» 71³ (70 + 1)³ 70³ + 1³ + 210 ( 71) 343000 + 1 + 14,700357,911 |
|
| 2. |
Is= 20 -EXERCISE - 10.2Find the area of the rectangles with the give50 cm and 20 cm65 m and 45 m25 cm and 16 cm(iv) 7 km and 19 kmFind the area of squares with the giver |
|
Answer» 1000cm,2925m,300cm,133km 1000cm,2925m,300cm,133km 1000cm,2925m,300cm,133km |
|
| 3. |
A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing with a speedof 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water isdesired?INCERT EXEMPLARI |
|
Answer» Width 300cm=3mdepth 120cm=1.2mspeed of canal 20km/h20×1000m/hvol. of water flow in 1 hour = width of canal ×depth of canal ×speef of canal water 3×1.2×20×1000=72000m3in 20 in vol. of water flows = 72000×20÷60=24000m3area irrigated in 20 min 2400÷0.08=300000m230 hectare |
|
| 4. |
ute the value of x, in the following figure.120°112° |
|
Answer» 180-120=60(=angle BAC)Now,60+x=112(Exterior angle property)x=112-60=52° correct answer |
|
| 5. |
ute (x+y).(x + y) (x + y) (x + y)? |
| Answer» | |
| 6. |
. Find an acute angle θ, when\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}} |
| Answer» | |
| 7. |
If the arcs of the same lengths in two circlessubtend angles 65° and 110° at the centre. Findthe ratio of their radii. |
|
Answer» thanks |
|
| 8. |
11. If the arcs of the same lengths in twocircles subtend angles 65° and 110° at thecentre, find the ratio of their radii. |
| Answer» | |
| 9. |
the scale on snap is 1mm:4m.Find the distance on the map for an actual distance for 84m |
|
Answer» If you find this solution helpful, Please give it a 👍 thanku |
|
| 10. |
10 Anita has to drive from village A to village B. She measures a distance of 3.5 cmbetween these villages on the map. What is the actual distance between the villagesif the map scale is 1 cm 20 km? |
|
Answer» 1cm = 20km on ground. So, 3.5 cm on map = 20*3.5= 70km. Please hit the like button if this helped you |
|
| 11. |
5.Explain why a rectangle is a convex quadrilateral. |
|
Answer» Both diagonals lie in its interior, so it is a convex quadrilateral. |
|
| 12. |
4 Amap is given with a scale of 2 cm - 1000 km.What is the actual distancebetween the two places in kms, if the distance in the mapis 2.5 cm? |
| Answer» | |
| 13. |
In Fig. 6.39, sides QP and RQ of A POR are produced to points S and T respectively.If< SPR = 135° and < PQT-110°, find < PROˇ一620/ XYZ = 54°. If YO and ZO are the bisectors of < XYZ and |
| Answer» | |
| 14. |
8. Prove that the sum of angles of any convexquadrilateral is 360 |
| Answer» | |
| 15. |
72. त्रिभुज POR इस प्रकार है कि PQ = 9 सेमी., 6 सेमी., PR =7.5 सेमी. और त्रिभुज PQR त्रिभुज XYZ के समरूप है। यदिXY = 18 सेमी है, तो YZ का मान ज्ञात कीजिए।(a) 15 सेमी.(b) 18 सेमी(c) 12 सेमी(d) 9 सेमी |
|
Answer» a is the correct answer option a is the right answer A is the right answer option (a) is the right answer of following questions 15 cm is correct answer a is the correct answer a is the Right Answer option a is the right answer a is correct answer option a is the right answer a is correct answer |
|
| 16. |
X Y2, Draw an acute angled Δ POR. Draw all of its altitudes. Name the point ofconcurrence as O |
| Answer» | |
| 17. |
segmen2. Drow an acute angled Aconcurrence as 'o'.ute angled A POR. Draw all of its altitudes. Name the point oNon its medians and show the centroid.to the orthocentre |
|
Answer» Solution:- Draw the line QR of any length.At Q draw an acute angle (i.e. greater then 0 degree and less then 90 degree),with the help of protector and scale.DRAW A Line from alphabetic character with the given angle ,mark the purpose P.join P and alphabetic character. Δ PQR is created.too draw altitudes , draw lines from all points i.e. PQR, be a part of them with broken lines.where all lines meet mark that time as O. Here is your acute angle I am sending photo |
|
| 18. |
11. The graph of y = f(x) is given below. Find the number of zeroes of f(x).CAST Sale Dona |
|
Answer» 2 is the correct answer of the given question No. of zeroes is 2 as the graph touches the x- axis at two points. The number of zerose of f(x) os 2 |
|
| 19. |
Short Rnswe16. From a paint on the ground 40 m away from the foover, the angle of clevation of the top of theof a toer is B0. The angle of elevation of the top of awater tank (on the top of the tower) is() hcight of the tower (l) depth of the tank.townud ita subtend angles of |
| Answer» | |
| 20. |
laly pehcils are there in the packet now?a. 75+ 42c. 1,278+3,564 d. 35,276+28,357Frame simple word problems for the following addition statements.b. 135 +578 |
|
Answer» a. 75 is added to 42b. 135 is added to 578c. 1278 is added to 3564d. 35276 is added to 28357 |
|
| 21. |
Let A(5,6]; how many binary operations can be defined on this set ?(A)8(B) 10(C) 16D)20 |
| Answer» | |
| 22. |
Find the value ofx, which satisfies the inequation:1 2x5Graph the solution on the number line. |
|
Answer» hit like if you find it useful |
|
| 23. |
2000x3x(3-5)3x3-2x5 |
| Answer» | |
| 24. |
xample 1 State with reason which of the following are surds and which0 V45 (1) 120 x 145 (1) 24 x 254 (10) 3/12 + 600 |
|
Answer» the answer was ³√45 is yes √45=3√5 it's surds √20×√45=2√5×3√5=30 it's not surds(2)^1/3×3(√6)1/3. it's call be surds 3√12÷6√27=6√3÷18√3=1/3 it's not surds |
|
| 25. |
25 Given that x e R, solve the following inequation and graph the solution on theline :-1 < 3 + 4x < 23. |
| Answer» | |
| 26. |
xample 1: Find the value of(110)^2. |
|
Answer» (10+100)^2 |
|
| 27. |
A contractor undertakes to dig a canal,6 kilometre long, in 35 days and employed90 men. He finds that after 20 days only 2 kmof canal have been dug. How many more menmust he employ to finish the work in time? |
|
Answer» 90 men can dig a canal of length 2000 m in 20 days.⇒ Work capacity of 1 man =2000 m/ (20 days × 90 men)= 10/9 meter per day per manRemaining length of the canal = 6000 - 2000 = 4000 mNumber of days remaining = 35 - 20 = 15 daysWork done by one man in 15 days = 10/9 × 15= 50/3metersNumber of men required to complete the work in 15 days= (Remaining length of the canal) / (Work capacity of 1man) =4000 / (50/3) = 240menAddition men required to complete the work = 240 – 90 = 150 men. |
|
| 28. |
xample 1:Find the equation of the circle with centre at (-1, 2) which passes through the point (3.1). |
|
Answer» , find the equation of the circle with center at (-3, -2) and radius 6 |
|
| 29. |
(c) Which word from the passage means the same as 'to renew"?(i) rejuvenate(ii) enhance(iii) release(iv) relieve |
|
Answer» release is answer of this question answer is release for the question which you asked iii) release is correct answer release is the right answer The correct answer is rejuvenate. Release means to let out. the the correct answer is rejuvenate . release means to let out rejuvenate is your answer |
|
| 30. |
Solve the inequation: 6x - 5 < 3x + 4, x € I. |
|
Answer» 6x-5<3x+46x-3x-5<3x+4-3x3x-5<43x-5+5<4+53x<93x/3<9/3x<3 |
|
| 31. |
12 List the soruFind the values of x, which satisfy the inequation:1 2x5Graph the solution set on the number line.3 2x -1 |
|
Answer» how did the sign change |
|
| 32. |
Draw the graph of the inequation x +0. |
|
Answer» priti ji app question samjhi nahi phir dekhiye |
|
| 33. |
8+x=%1-x£Z (¥)Bl s 28 MATEMAT ATES RATOE “जा |
| Answer» | |
| 34. |
estion 1Solve the inequation1 2x52 3 63 < |
| Answer» | |
| 35. |
=4 is one of the roots of theequation 3x2 + kx -2-0, find the value of k| 4. If xGEST : STANDARD X |
|
Answer» Given2x2+ kx + 3 = 0 has equal rootsHence b2– 4ac = 0That is (k)2– 4(2)(3) = 0k2– 24 = 0k2= 24∴k =± 2√6 X=43x4x4+k x 4 - 2 =048+4k-2=04k=-48+24k=-46K=-46\4K=11.5 |
|
| 36. |
Let, f. is a function defined as f(x),xERA,f : R â R . Determine the range off |
|
Answer» f(x) = x²/(x²+1) Observe the f that x² < x² + 1 So, x²/(x²+1) < 1 As square the f is always positive and at x = 0, f is 0 So, Range = [0, 1) |
|
| 37. |
Iwill the median class and modal classof grouped data always be different ?Justip 1 доия олад ел . |
|
Answer» No, themedianand modalclass of grouped data canalso be same in some cases. The cumulative frequency just greater than N/2 is 26, then themedian classis 120 - 140. ... Hence themedian classand modalclassare the same |
|
| 38. |
xample 2: Find the radius of curvature of the cr\sqrt{x}+\sqrt{y}=1 \text { at }(1,1) |
| Answer» | |
| 39. |
5(b). Find the derivative ofusing rules of differentiation.sin x |
| Answer» | |
| 40. |
6 8 73 64424 48 ?Given options are• A 13. B. 42. C. 18• D. 3010. |
|
Answer» 6x2 = 12 12x3 =36 36-12= 248x2= 16 16x4= 64 64-16= 48In same manner :7*2= 14 14*4= 56 56*14 = 42 |
|
| 41. |
Solve the inequation 12+13x, xeR. |
|
Answer» hit like if you find it useful |
|
| 42. |
f:R→Rgivenas f(x)3x+7,forallxeR= |
|
Answer» let x and y belongs R f(x) = f(y) 3x + 7 = 3y + 7 x = y So, one - one f(x) = 3x + 7 = y x =( y - 7)/3 f inverse (x) = (x-7)/3 So, onto |
|
| 43. |
i) fRR given as fx) 3x+7, for all xeR |
|
Answer» let x and y belongs R f(x) = f(y) 3x + 7 = 3y + 7 x = y So, one - one f(x) = 3x + 7 = y x =( y - 7)/3 f inverse (x) = (x-7)/3 So, onto |
|
| 44. |
18. 16(3x - 5) - 1 0(ा- 8) > 40 |
|
Answer» 16 ( 3x - 5 ) - 10(4x - 8) = 40=> 48x - 80 - 40x + 80 = 40=> 8x = 40=> x = 5. PLEASE HIT THE LIKE BUTTON |
|
| 45. |
the cost price of 9 articles is equal to the selling price of 11 articles.find the loss percentage |
| Answer» | |
| 46. |
4. Find the domain of f(x) x2 +2, xeR:(d) 12, ) |
|
Answer» If you like the solution, Please give it a 👍 |
|
| 47. |
भाजक भागफल का 25 गुना हैं और शेषफल का 5 गुना है। यदिभाषफल 16 हो, तो भाज्य है-(1) 6400(2) 6480(3) 40(4) 480 |
|
Answer» 2 because it has the highest number |
|
| 48. |
ose the correct answer from the given four options (1 to 9):1In the adjoining figure, the length of BC is(a) 2/3 cm(b) 3 3 cm(c) 4/3 cm(d) 3 cm2 In the adjoining figure, if the angle of elevation is 60and the distance AB 10/3 m, then the height of thetower isTower(a) 20 V3 cm(c) 30 m(b) 10 m(d) 30/3 mHeights and Distances 557 |
|
Answer» 1. sin (30°) = BC/AC1/2 = BC/6BC = 3option (d) is correct |
|
| 49. |
5. tan®60° + 3०6१30* का मान ज्ञात कीजिए।g 5 LT LS |
| Answer» | |
| 50. |
16 + 40 y + 25 y ^ { 2 } \text { by } 5 y + 4 |
| Answer» | |