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. |
b) 23 × 34×423×32 |
| Answer» | |
| 2. |
what is the percentage of 120 / 110 |
|
Answer» 91.666 is the write answer Percentage will be120/110 × 100= 109.09 % |
|
| 3. |
LO. A186 represent its solution on the number line.20Solve the inequation 12 + 1 x 55+ 3x, x ⏠R. Represent the solution in a muumline. |
| Answer» | |
| 4. |
5. In an A.P., if ph term isand qh term is , prove that the sum of first pq9 |
| Answer» | |
| 5. |
20. For what value of x will2x +land 3Xrand3x-2 x+ 8represent the same fraction? Also find the fraction. |
|
Answer» for x both have same fractionso (2x+1)/(3x-2)=(3x-4)/(x+8)(2x+1)(x+8)=(3x-4)(3x-2)2x^2+17x+8=9x^2-18x+87x^2-35x=0x^×-5x=0x(x-5)=0so x=0 or x=5 |
|
| 6. |
(9*(5/6))/((5/6)) |
| Answer» | |
| 7. |
Findif x = a cos' and y = a sine at 0 = |
| Answer» | |
| 8. |
ब(009 (हिलत७4/८मद... |
| Answer» | |
| 9. |
It sine - CosoI then find thevalieSino tease |
| Answer» | |
| 10. |
In Fig. 10.11, if TP and TQ are the two tangentsto a circle with centre O so thatthen L PTQ is equal to(A) 60°(C) 80°2.110°POQ 110B) 70(D) 90Fig. 10.11 |
| Answer» | |
| 11. |
\begin{array} { l } { \text { (a) } 8526 \div 2 } \\ { \text { (e) } 5355 \div 7 } \\ { \text { (i) } 5705 \div 7 } \end{array} |
| Answer» | |
| 12. |
200-(120-{110-(80+70)}-60) |
|
Answer» 200-(120-(110-(80+70)-60-200-(120-(110-150)-60-200-(120+40-60)-200-(100)-100 200 - 100 = 100 is the correct answer |
|
| 13. |
The value of 52 900980The value of 7 90 110Ram990023432B)A)C) 201527 |
| Answer» | |
| 14. |
Rc1850607080902004006008001000 |
|
Answer» total = 50+60+70+80+90=350sum= 50x200 + 60x400 + 70x600+80x800+90x1000= 230000mean = sum/total = 657.14285714 |
|
| 15. |
If the perimeter of a semi-circular park is 72 m, find its area and radius. |
| Answer» | |
| 16. |
The total number of real tangents that canbe drawn to the ellipse 3x2 +5y 32and 25x2 +9y(3,5) is1)0 2)2450 passing through3) 34) 422 |
|
Answer» thank you sir |
|
| 17. |
6. If the perimeter of a semi-circular park is 72 m, find its area and radius. |
| Answer» | |
| 18. |
9 เค�เคน &. 009 * |
|
Answer» રકમ આખી આપો આકૃતિ સહિત |
|
| 19. |
( 7 ) ^ 8 \div ( 7 ) ^ 13 |
| Answer» | |
| 20. |
7^{5} \div 7^{2} |
| Answer» | |
| 21. |
1. If sinefind the values of cost and tan0.25 ' |
| Answer» | |
| 22. |
009 (2)ëśě ons |
|
Answer» The answer will be 600 because the time 60 mins , 45 mins are multiples of it . |
|
| 23. |
(149 ()जी Hbbo| ¢ il [’ E |
| Answer» | |
| 24. |
की R e e: ८ 24 धर 2. 2 2. vl दि W L . |
| Answer» | |
| 25. |
पापकल्पाय-हो, तो x का मान है*-5x+42-9x+149(a) 2! ! |
| Answer» | |
| 26. |
Median of the givien data is : 144, 145, 147,148, 149, 151,152.154,1 55,160, |
|
Answer» number of term= evenmedian = (Nth + (N+1)th term)/2 =(149+151)/2=150 |
|
| 27. |
1 दि दे questions _ हननतले व) प्टणदीा ब्लेड 00 pach) O whicha guadtaTof dhe poitis ७:८5 ५, (3, रि.9मी |
|
Answer» (-2,4)in second quadrant(3,1)in first quadrent(-4,0)- x axis2,3)first quadrant |
|
| 28. |
DIMAAG LAGAO?QUIZI –A bag has 30 Red, 25 Blue, 25 Black, 20Green toffees and without seeing you cantake out as many toffees at Rs.22 per tof-fee. If your color comes you will get money.Which color will you select to make money?Red_BlueBlueBlackGreen |
|
Answer» Red is the write answer.. option (a) red is the right answer of following questions (red) is the correct answer of the correct Red is the right answer of the following a is the correct answer |
|
| 29. |
shape of a garden is rectangular in the middle and semi circularat the ends as shown in the diagram. Find the area and the perimeterTof this garden [Length of rectangle isThe7 m 20 (3.5 + 3.5) metres20 m |
| Answer» | |
| 30. |
| → निम्नलिखित प्रश्नों के उत्तर गणना करके दीजिए।$ APQR में 2 P = 70° और 2 R = 30° है तो त्रिभुज की किस भुजा का माप सबसे ।बड़ा होगा ? क्यों ?(p 10 |
|
Answer» given, P+Q+R =180° 70°+Q+30°=180° Q=180°-100° Q=80°hence,Q is the largest angle in a triangle. P+Q+R=180°70+Q+30=180°Q=80°Therefore Q is largest angle in ∆PQR.Therefore PR is the largest side |
|
| 31. |
COMPLEX NUMBERLEVEL-I (H.W)INTEGRAL POWERS OF iThe value of i+P+P + ......2n terms =D-1 21 30 4i(* +r")=1)i 2)-13)-450ame |
|
Answer» the answer of this question is 1 ( I) 2) 1is a correct answer 2) is a correct answer |
|
| 32. |
Ifa : b = 5 : 6 and b : c = 18:23,finda:b : c. |
|
Answer» a:b = 5:6 = 15:18 So a:b:c = 15:18:23 |
|
| 33. |
Find the ratio ofA : B : C, ifA :B-5:4and B : С_6:7. |
| Answer» | |
| 34. |
\frac { 6 } { 13 } \div 7 |
|
Answer» 0.065 |
|
| 35. |
If sine 25. find the values of cos0 and tano.25 ' |
| Answer» | |
| 36. |
99+999 |
|
Answer» 1098 is correct answer 1098 is right answer for this question 1098 is the correct answer 99 + 999 = 1098 ANSWER.. 1098 is the correct answer. 1098 is the right answer 1098is the best answer 1098 is correct answer ragini gupta hii can u do my help 1098 is the correct answer 1098 is the correct answer |
|
| 37. |
Ifa- 6+5 and b-then what will be the value of a + b |
|
Answer» a = √6+5 b = 1/a = 1*(√6-5)/(√6+5)*(√6-5) = √6-5 so, a²+b² = (√6+5)² + (√6-5)² = (√6)²+(5)²+10√6 +(√6)²+(5)²-10√6 = 6+25+6+25 = 62 a = √6+5 b = 1/a = 1*(√6-5)/(√6+5)*(√6-5) = √6-5 so, a²+b² = (√6+5)² + (√6-5)² = (√6)²+(5)²+10√6 +(√6)²+(5)²-10√6 = 6+25+6+25 = 62 can you explain 1/a |
|
| 38. |
The sum of deviations of 'n' observations from 25 is 25 and sum of deviations of then'observations from 35 is - 25. Find the mean of the observations |
| Answer» | |
| 39. |
rtalse and justify your answer:is aABC) 24 cmllelogram and X is the mid point of AB. if arlAXCD)=24 cm, thenAREA 9F TRANGLES AND QUADRILATERAL201 |
|
Answer» only give answers thanks |
|
| 40. |
Which term of the series 96, 48, 24,... W |
| Answer» | |
| 41. |
149 mm into lem |
|
Answer» 0.000049 is the right answer 0.000049 is the right answer .000049 is the correct answer of the given question 0.000049 is the correct answer 0. 000049 is answer of this statement |
|
| 42. |
13.54 x 9915.47 x 999 |
| Answer» | |
| 43. |
99*999 |
| Answer» | |
| 44. |
1 If P = ab and q=a3b, then find thisLHCF and LEM, |
|
Answer» Factors of p and q p = a*b*bq = a*a*a*b HCF = a*b = ab LCM = a*a*a*b*b = a^2b^2 p=a*b*bq=a*a*a*bL.C.M=a^3b^2 H.C.F=a*b=ab |
|
| 45. |
ofindLEM (306, 1316) itHCF (6306,1314) - 18 |
|
Answer» correct answer is LCM(306, 1314) = 22,338 please like and accept as best answer lcm= 306/2=153/3=51/3=17; lcm=1314/2=657 product of zeroes = hcf×lcm306×1314=18×x402084/18=x22338=x |
|
| 46. |
Find the Lem of |
| Answer» | |
| 47. |
If two adjacent angles of parallelogram are 3x - 5 and 6x + 5 then find the value of |
|
Answer» Weknowthatthesumofadjacentanglesofaparallelogramis180°.⇒2x+25+3x−5=180⇒5x+20=180⇒5x=180−20⇒5x=160⇒x=1605⇒x=32Therefore,thevalueofxis32. In Parallelogram sum of two adjacent angles is 180° by the property of Co-interior angles Hence,(3x - 5) + (6x + 5) = 1809x = 180x = 180/9x = 20 Therefore,Value of x = 20 |
|
| 48. |
10 p q , 20 q r , 30 r p |
|
Answer» first take common and solve like: 10pq=2*5*p*q and so on. |
|
| 49. |
HCE (9520, 6600) = 120 and Lem (252.0, Clo) 25Find the value of K. |
|
Answer» We know,If a and b are two numbers Then,a*b = HCF(a, b) * LCM(a, b)2520*6600 = 120*LCMLCM = 2520*6600/120LCM = 2520*55 = 138600 Given,LCM = 252KTherefore, k = 0*55 |
|
| 50. |
13. Find the mean of the following frequency distribution115125135145155165254020EDIAN |
|
Answer» Mean = Summation of f*x / Summation of f= (115*5 + 125*25+ 135*4 + 145*6 + 155*40+ 165*60 ) / ( 5+25+4+6+40+60)= 21210 / 140= 151.5 PLEASE HIT THE LIKE BUTTON |
|