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. |
(a) 400 cm2 (b) 200 cm2 (c) 300 cm2 (d) 2-v100cm.27.If theand breadth of the plot isand breadth or the eo the length and the perimeter of a rectangular plot is 1 :3, then the ratio betwen the length(a) 1:2 (b) 2:1 (c) 3:2 (d) 2:3 |
| Answer» | |
| 2. |
if the sum of first n ,2n,3n terms of an AP be S1,S2 and S3 respectively then proved that S3=(S2-S1). |
| Answer» | |
| 3. |
A 5 m wide cloth is used to make a conicaltent of buse diameter 14 mand height 24 m. Find the cost of clothused at the rate of 25 per metre square |
| Answer» | |
| 4. |
5. Find Lc.m. (105, 91) using g.c.d. (a, b) Lc.m. (a, b) ab |
|
Answer» thank you so much bro |
|
| 5. |
If A= 4L3 2and B =show thatAB # BA |
| Answer» | |
| 6. |
2, b(i) (ab+ba)-1If a3, find the values of |
|
Answer» (2^3+3^2)^-11/8+91/17 (2^2+3^3)^-11/4+91/13 |
|
| 7. |
5. The length and the breadth of a rectangular field are in the ratio 7: 4. The cost of fencingtbe.. field atて25 per metre is 3300. Find the dimensions of the field. |
| Answer» | |
| 8. |
The length and the breadth of a rectangular field are in the ratio 7: 4. The cost of fencing thefield at.25 per metre isて3300. Find the dimensions of the field. |
| Answer» | |
| 9. |
Findthe20. A 5m wide cloth is used to make a conical tent of base diameter 14m and height 24m.cost of cloth used at the rate of Rs. 25 per metre. |
|
Answer» 1267938337+639399373639939 |
|
| 10. |
22. A wide cloth is used to make a conical tent of base diameter 14 m and becost of cloth used at the rate of 25 per metre.and height 2Am. Width |
| Answer» | |
| 11. |
sarimade a square from a wire of length 100 cm. What is its area?ăŞ100 cm 2(2) 400 cm 2(3) 625 cm2(4) 10,000 cm2 |
| Answer» | |
| 12. |
9. Let V5 + 13 be rational equal toThen, |
| Answer» | |
| 13. |
91+2 |
|
Answer» 91+2=93 is the answer 91+2=93(ans.)........ |
|
| 14. |
(91) (S1) |
|
Answer» hit like if you find it useful |
|
| 15. |
54-72x=91 |
| Answer» | |
| 16. |
9191/91*2 |
|
Answer» 101×2202 |
|
| 17. |
4+2+5+-91 |
|
Answer» 102 is the right answer of the following |
|
| 18. |
21. यदि किसी ऐनक के लैंस का पावर +2 डायप्टर हो तो उसकीफोकस दूरी होगी-(A) 200 cm(B) 100 cm(C) 50 cm(D) 2 cm |
|
Answer» uska focus bhi 2 cm hoga bhai uska focas length 2 hoga 2 cm will be the focus of the lence |
|
| 19. |
If $A=\left[ \begin{array}{cc}{1} & {2} \\ {3} & {-4}\end{array}\right] B=\left[ \begin{array}{ll}{4} & {3} \\ {2} & {1}\end{array}\right]$ then verify that $(A B)^{\prime}=B^{\prime} A^{\prime}$ |
| Answer» | |
| 20. |
8. The length of the diagonal of a square is 20 cmem. Its area isc) 300 mfield at(a) 400 cm(b) 200 cm() 100-2d a souare2 25 per metre is 7 2000 |
|
Answer» A=a^2d=√2aA=1/2d^2=1/2 (20)^2=200cm^2 |
|
| 21. |
n of the diagonal of a square is 20 cm. Its area is(a) 400 cm2(b) 200 cm2(c) 300 cm2(d) 100/2 cm |
| Answer» | |
| 22. |
3 +13. Find the value ofif V5 2.2363 V5OrArrange 32, 43, $/5 in descending order. |
|
Answer» √3+2.236/3-2.236=√5.226/0.744=√5.97=2.44335 |
|
| 23. |
rdthe value of,ifs-2236.3+V513. Find the value of2236 |
|
Answer» Rationalize,√[(3+√5)/3+√5)/(3-√5)(3+√5)] 9+5+6√5/414+6√5/47+3√5/2 √(7+3√5/2)Put √5=2.236=√6.854=2.618 wrong answer |
|
| 24. |
Which term of the AP 24, 21, 18, 15,is the first negative term? |
| Answer» | |
| 25. |
AP: 24, 21, 18 ..... के कितने पद लिए जाएँ, ताकि उनका योग 78 हो ? |
|
Answer» first term,a = 24 common difference,d = 21-24 = -3 let the number of terms to get sum 78 is n. Sn = n/2 (2a +( n-1) d) 78 = n/2(48+ (-3)n + 3) 78 = n/2(51 + (-3)n) 78*2=n (51+ (-3)n) 156= 51n + -3n^2 3n^2-51n+156=0 n^2 - 17n +52 =0 n^2-13n-4n +52 =0 n(n-13)-4(n-13)=0 (n-4)(n-13) =0 n = 4,13 Solving the quadratic equation, we get n=4 and n=13. So you can take either 4 terms or 13 terms to get the sum 78. |
|
| 26. |
HEMATIVAP : 24, 21, 18 ..... के कितने पद लिए जाएँ, ताकि उनका योग 78 हो ? |
|
Answer» first term,a = 24 common difference,d = 21-24 = -3 let the number of terms to get sum 78 is n. Sn = n/2 (2a +( n-1) d) 78 = n/2(48+ (-3)n + 3) 78 = n/2(51 + (-3)n) 78*2=n (51+ (-3)n) 156= 51n + -3n^2 3n^2-51n+156=0 n^2 - 17n +52 =0 n^2-13n-4n +52 =0 n(n-13)-4(n-13)=0 (n-4)(n-13) =0 n = 4,13 Solving the quadratic equation, we get n=4 and n=13. So you can take either 4 terms or 13 terms to get the sum 78. |
|
| 27. |
Which term of the AP 24, 21, 18, 15,. is the first negative term? |
| Answer» | |
| 28. |
2) There are 37 terms in an Đ.Đ., the sum of three terms placed exactly at the middle is 225 and thesum of last three terms is 429. Write the A.IP. |
|
Answer» Tysm |
|
| 29. |
1 Find the HCF of 81 and 237 |
|
Answer» HCF will be81 = 3⁴237 = 3× 79so HCF is 3 h c f=3 |
|
| 30. |
चŕĽ.237 + 153 = |
|
Answer» 237 + 153 = the answer was 390 |
|
| 31. |
Name1. Convert and write an equation with an exponent. Use your metersta. 2 meters to centimeters 2m - 200 cmb. 108 centimeters to meters108 cmmC.2.49 meters to centimetersd. 50 centimeters to meterscm =e.6.3 meters to centimetersf.7 centimeters to meterscm = __mg. In the space below, list the letters of the problems where |
|
Answer» b) 1.08 mc) 249 cmd) 0.5 me) 630 mf) 0.07 m |
|
| 32. |
10.The area of a rectangular field is 0:5 ha. If one side is 12.5m,find theother side in meters. |
| Answer» | |
| 33. |
24. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sumof the last three terms is 429. Find the A.P |
|
Answer» Let the first term and the common difference of the A.P areaanddrespectively. Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19thterm. Thus, three middle most terms of this A.P.are 18th, 19thand 20thterms. Givena18+a19+a20= 225 ⇒ (a+ 17d) + (a+ 18d) + (a+ 19d) = 225 ⇒ 3(a+ 18d) = 225 ⇒a+ 18d= 75 ⇒a= 75 – 18d… (1) According to given information a35+a36+a37= 429 ⇒ (a+ 34d) + (a+ 35d) + (a+ 36d) = 429 ⇒ 3(a+ 35d) = 429 ⇒ (75 – 18d) + 35d= 143 ⇒ 17d= 143 – 75 = 68 ⇒d= 4 Substituting the value ofdin equation (1), it is obtained a= 75 – 18 × 4 = 3 Thus, the A.P. is 3, 7, 11, 15 … |
|
| 34. |
n A P consists of 37 terms. The sum of the three middle terms is 225 and thesum of last three terms is 429. Find the AP23. A |
|
Answer» Let the first term and the common difference of the A.P areaanddrespectively. Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19thterm. Thus, three middle most terms of this A.P.are 18th, 19thand 20thterms. Givena18+a19+a20= 225 ⇒ (a+ 17d) + (a+ 18d) + (a+ 19d) = 225 ⇒ 3(a+ 18d) = 225 ⇒a+ 18d= 75 ⇒a= 75 – 18d… (1) According to given information a35+a36+a37= 429 ⇒ (a+ 34d) + (a+ 35d) + (a+ 36d) = 429 ⇒ 3(a+ 35d) = 429 ⇒ (75 – 18d) + 35d= 143 ⇒ 17d= 143 – 75 = 68 ⇒d= 4 Substituting the value ofdin equation (1), it is obtained a= 75 – 18 × 4 = 3 Thus, the A.P. is 3, 7, 11, 15 … |
|
| 35. |
to find the value of n, if a= 23x3b=2×3×5, 385 and 2cm tabe)=23x3²x52 |
|
Answer» 2^3×3^2×5=8×9×5=360; 2^3×3=27; 2x3x5=30; |
|
| 36. |
MPLE 2İfx=asin θ and y=btan θ, prove that(1..2x1 |
|
Answer» x = a sin theta : y = b tan theta ....(Given)L.H.S. = a²/x² - b²/y² = a²/(a sin theta)² - b²/(b tan theta)²a² and b² are cancelled and then we get1/sin² theta - 1/sin² theta/cos² theta ...[∴ tan theta = sin theta/cos theta]1/sin² theta - cos² theta/sin² theta= (1-cos² theta)/sin² theta ....[∴ sin² theta + cos² theta = 1]sin² theta/sin² theta = 1= R.H.S⇒ a²/x² - b²/y²Hence proved. |
|
| 37. |
16. An A.P consists of 37 terms. The sum of the three middle mostterms is 225. And the sum of the last three terms is 429. Find the A.P |
| Answer» | |
| 38. |
1)10200, 1.8. 20th term is 200.Somple 13: How many terms of the AP: 24, 21, 18,sum is 78?must be taken so that theirSolutiun Here |
| Answer» | |
| 39. |
\begin{array}{r}{\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}+\frac{\csc ^{2} \theta}{\sec ^{2} \theta-\csc ^{2} \theta}} \\ {=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}}\end{array} |
| Answer» | |
| 40. |
n AP consists of 37 terms. The sum of thethree middle most terms is 225 and thesum of the last three terms is 429. Find theAP24:"X |
|
Answer» Like if you find it useful |
|
| 41. |
nd the sum of last three terms is 429. Write the11产|ffirst term ofan AP.is a, second term is b and last term is c, then show that suof all terms is (a+cb+c-2a)2(b-a) |
| Answer» | |
| 42. |
e 48. An AP consists of 37 terms. The sulast three is 429. Find the APm of the three middle most terms is 225 and the sum ofthe |
| Answer» | |
| 43. |
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29thterm. |
| Answer» | |
| 44. |
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29thterm.8. |
| Answer» | |
| 45. |
24. An AP consists of 50 terms of which 3"d tern is 12 and the last term is 106. Find the 29term. |
|
Answer» n=50 It is an AP. So, tn=a+(n-1)d t3=a+(3-1)d 12 =a+2d (1) Last term is 106. t50=a+(50-1)d 106=a+49d (2) From (1) and (2), 106=12-2d+49d 94=47d d=94/47 =2 a= 12-2d =12-2(2) =8 a=8 So t29=a+(n-1)d =8+(29-1)2 =8+28(2) =8+56 =64. |
|
| 46. |
7.Find the HCF of 81 and 237 and express it as a linear combination of 81 and 229 |
|
Answer» Since, 237 > 81 On applying Euclid’s division algorithm, we get 237 = 81 × 2 + 75 ...(i) 81 = 75 × 1 + 6 ...(ii) 75 = 6 × 12 + 3 ...(iii) 6 = 3 × 2 + 0 ...(iv) Hence, and HCF (81, 237) = 3. 1 Write 3 in the form of 81x + 237y, move backwards 3 = 75 – 6 × 12 [From (iii)]= 75 – (81 – 75 × 1) × 12 [Replace 6 from (ii)] = 75 – (81 × 12 – 75 × 1 × 12)= 75 – 81 × 12 + 75 × 12= 75 + 75 × 12 – 81 × 12= 75 ( 1 + 12) – 81 × 12= 75 × 13 – 81 × 12 = 13(237 – 81 × 2) – 81 × 12 [Replace 75 from (i)] = 13 × 237 – 81 × 2 × 13 – 81 × 12= 237 × 13 – 81 (26 + 12)= 237 × 13 – 81 × 38= 81 × (– 38) + 237 × (13) = 81x + 237y Hence, x = – 38 and y = 13 |
|
| 47. |
ORFind HCF of 81 and 237 and express it as a linear combination of 81 and 237 ie., HCF (81,237) = 81x +237y for some x and y. |
|
Answer» By Euclid's Division Algorithm, 237=81(2)+(75) 81=75(1) + (6) 75=6(12)+(3) 6=3(2)+(0) Hcf =3 Expressing it in the form of 237x+81y=HCF 3=75-6(12) { From 2nd last step} 3=75-(81-75)(12) {Substituting} 3=75-(81*12-75*12) 3=75-81*12+75*12 3=75(13)-81(12) 3=(237-81*2)(13)-81(12) 3=237(13)-81(38) 3=237(13)+81(-38) {we need an expression in the form 237x + 81y } Therefore, x =13 , y =- 38 Like my answer if you find it useful! |
|
| 48. |
16), Find i ICF of 81 and 237 and express it us a linear combination of 81 and 237, le HCF (81,237)-81x+237yfor some x and y |
|
Answer» By Euclid's Division Algorithm, 237=81(2)+(75) 81=75(1) + (6) 75=6(12)+(3) 6=3(2)+(0) Hcf =3 Expressing it in the form of 237x+81y=HCF 3=75-6(12) { From 2nd last step} 3=75-(81-75)(12) {Substituting} 3=75-(81*12-75*12) 3=75-81*12+75*12 3=75(13)-81(12) 3=(237-81*2)(13)-81(12) 3=237(13)-81(38) 3=237(13)+81(-38) {we need an expression in the form 237x + 81y } Therefore, x =13 , y =- 38 |
|
| 49. |
नकद 3= T e(zz) Lim 573 —X —>0 X —Qa |
|
Answer» thank u so much we know krke h vo frmula h Yes that's the formula.you could use it. thanx mam |
|
| 50. |
210. Find the costThe cost of 5 meters of a particular quality of cloth is(iii) 10 (iv) 13 meters of cloth of the same quality |
|
Answer» rong |
|