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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. |
Plz answer this is urgent |
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Answer» Answer: 3 Step-by-step EXPLANATION: FIRST MARK me BRAINLIEST and FOLLOW me |
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| 2. |
11.Find the volume of rectangular box with length breadth and height respectively 2p,4q,8r(1 Point)32qr64pqr648pqr |
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Answer» 64pqr is the answer. Step-by-step EXPLANATION: LENGTH = 2p Breadth = 4q Height = 8r Volume = L×B×H => 2p×4q×8r => 64pqr |
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| 3. |
Pls help me with this one .don't spam |
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Answer» Step-by-step explanation: PLEASE LET ME KNOW ... it helps you in COMMENT section ..
THANK YOU |
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| 5. |
Change the voice (voice change) ? |
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Answer» Step-by-step EXPLANATION: it is CORRECT ✅✅✅ |
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| 6. |
Antilogarithm of -1.5082 |
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Answer» Step-by-step EXPLANATION: this is a answer please FOLLOW me |
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| 7. |
Which term of the A.P. 35, 28, 21, ... is zero? |
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Answer» Answer: N = 6 Step-by-step explanation: 0 = a+ (n-1)d 0 = 35 + (n-1)-7 0 = 35 + (-7n+7) 0 = 35-7n+7 7n = 35+7 7n = 42 n = 42/7 n = 6 |
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| 8. |
A loud sound can be heard at a large distance but a feeble or soft sound cannot be heard at a large distance. Explain, why? |
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Answer» A loud |
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| 9. |
The largest number of the 3 consecutive numbers is x+1, then the smallest number is |
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| 10. |
82 × 35 by expansion method |
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Answer» Answer: 8*10+2*1+3*10+5*1=2870 |
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| 11. |
For number 4 and 6 hcf is 12 |
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| 12. |
In 5⁴,4is called the_______ |
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Answer» exponent Step-by-step EXPLANATION: it may HELP youTHANK my answer |
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| 13. |
The pairs of line represented by the equation 3x-6y=6 and 6x +KY =10 will be parallel if k is |
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Answer» Step-by-step explanation: 3x- 6y=6 , 6x+KY = 10 3x-6y =6 (i) 6x+ky=10 (ii) Multiplying EQUATION i & ii 6x-12 y = 12 (iii) SUBTRACTING equation (ii) by (iii) 6x - 12y - 12 =0 6x+ ky - 10 =10 ________________ 12ky-22 12y -22=k y=10 putting value of Y in equation (iii) 6x+12× 10 = 12 6x-120 =12 6x+ky = 10 6x- 120 -12=0 6x+ky -10 =0 _____________ -120- 22=k
K=98. so, ANS is k= 98.. |
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| 14. |
Divide a^4-a^2+5 by a^2-3 |
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Answer» Answer: 243 Step-by-step EXPLANATION: |
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| 15. |
Find the value 5/6÷9/2 |
Answer» ANSWER... |
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| 16. |
The daily sales at a 24-hours minimart produce a distribution that is approximately normal with a mean of 1210 and a standard deviation of 143.Find the probability that the sales on a given day at this store are between RM 1200 and RM 1300. |
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| 17. |
50 + 4 + 0/10 + 5/100 is equal to _________. |
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Answer» Your answer is 54.05 Step-by-step explanation: Hope it will be HELPFUL...☺☺☺♥️♥️ |
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| 18. |
In the following figure O is the centre of the circle chord PQ and RS intersect inside the circle at M . Prove that: Angle POR + Angle QOS = 2Angle PMR. |
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Answer» Answer: O is the centre of the circle. chord PQ≅ chord RS (Given) ⇒ ARC PQ≅ arc RS (Correspondidng arcs of CONGRUENT chords of a circle are congruent) ⇒m(arcPQ)=m(arcRS) ⇒m(arcPQ)=80 o [m(arcRS)=80 o ] (1) m(arcPR)=∠POR=70 o (Measure of a minor arc is the measure of its central angle) (2) m(arcPR)+m(arcPQ)+m(arcQS)+m(arcRS)=360 o
⇒70 o +80 o +m(arcQS)+80 o =360 0
⇒m(arcQS)=360 o −230 o =130 o
(3) m(arcQSR)=m(arcQS)+m(arcRS)=130 o +80 o =210 o |
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| 19. |
A line which intersects two or more loner in different points is known as |
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Answer» RESPIRATION through SKIN is CALLED as |
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| 20. |
It's urgent answer fast.. find the roots of the quadratic equation:x^2 -3x-10 =0 |
Answer» x²-3x-10x²-5x+2x-10 x(x-5)+2(x-5) (x-5)(x+2) –» x=5,-2 |
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| 21. |
Solve for :(sin 3 0 -1) (2cos -1) =0 |
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Answer» I don't UNDERSTAND the QUESTION PLEASE mark me as brainlest answer |
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| 22. |
Monts: 8 9 10 11 12 13 14 15No of students ! 4 5 3 8 4 10 5 1 |
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Answer» SORRY But UNABLE To UNDERSTAND The Question ♀️♀️ |
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| 23. |
Determine the nature of the roots for the following quadratic equation1) 15 xsquare +11x+2=0 |
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Answer» Answer: 15x² + 11x +2=0solution: a=15 b=11 c=2 we KNOW that, d=b²-4ac = (11)²-4×15×2 =121-120 = 1 The value of d is GREATER than ZERO,hence this equation has real and unequal roots.... hope it HELPED you... |
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| 24. |
In figure the side QR of A PQR is produced to a point S. If the bisectors of ZPQR and angle PRS meet at point T, then prove that angle QTR= 1 2 angle QPR . |
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Answer» Step-by-step EXPLANATION: stanza SAID I to my heart here's a lesson for me that MAN's but a PICTURE of what I might be but thanks to my FRIEND for their care in my breeding ho |
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| 25. |
(a^2-b^2)^2 plz answer |
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Answer» a⁴ + b⁴ - 2a²b² Step-by-step EXPLANATION: ( a² - b² )² = [ ( a + B ) ( a - b ) ]² = [ ( a + b )² ] [ ( a - b )² ] = [ a² + b² + 2AB ] [ a² + b² - 2ab ] = a² [ a² + b² - 2ab ] + b² [ a² + b² - 2ab ] + 2ab [ a² + b² - 2ab ] = a⁴ + a²b² - 2a³b + a²b² + b⁴- 2ab³ + 2a³b + 2ab³ - 4a²b² = a⁴ + b⁴ + a²b² + a²b² - 4a²b² - 2a³b + 2a³b - 2ab² + 2ab² = a⁴ + b⁴ + 2a²b² - 4a²b² - 0 - 0 = a⁴ + b⁴ - 2a²b² |
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| 26. |
Find the perimeter of a sector of angle 45° of a circle with radius 7 cmplease gimme the ans fast |
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Answer» hope this will help you FRIEND |
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| 27. |
23. The circumference of acircle of diameter d is |
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Answer» CIRCUMFERENCE = π × DIAMETER = 2π × RADIUS |
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| 28. |
1. Simplify: (2 + V5) () |
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Answer» Answer: 7+2root10 Step-by-step EXPLANATION: PLEASE MARK ME AS THE BRAINLIEST!!! |
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| 29. |
SECTION-A1. Class mark of 22.5-32.5 isa) 25 b) 27.5 c) 26d) 26.5 |
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Answer» Step-by-step EXPLANATION: nxxhccddodieskdkddijxixxixxixidjs |
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| 30. |
Can I get an answer for this |
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Answer» njjdjdjdddd nnnjdjdj Step-by-step EXPLANATION: hhhddeeeeeodidjd dkdidjsjwjw |
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| 31. |
If the sum of the zeroes of the quadratic polynomial x2 + 2x + 3k isequal to the product of its zeroes, then k =? |
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Answer» Answer: k = -2/3 Step-by-step explanation: LET ax^2 +bx + C be quadratic equation SUM of ROOTS(zeroes) = -b/a = -2/1 = -2 prodcts of roots(zeroes) = c/a = 3k/1 = 3k A/c to QUESTION, sum of zeroes = product of zeroes => -2 = 3k => k = -2/3 Hope it helps Please mark BRAINLIEST |
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| 33. |
एक समद्विबाहु त्रिभुज की रचना करो जिसमें 30 सेंटीमीटर और ए बी प्लस एसी 10 सेंटीमीटर एंगल 90 डिग्री का है |
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Answer» such if GC I Iedtuytrrrrtt678yui8uhgyuhfgji8 r |
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| 34. |
Divide 3450among a b c in the ratio 3:5:7: |
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Answer» Step-by-step EXPLANATION: SUM of RATIOS = 3 + 5 + 7 = 15 Share of A =( 3 × 3450)÷ 15 =690 Share of B = (5×3450)÷15 =1150 SHARE OF C = (7×3450) ÷ 15 =1610 |
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| 35. |
In the adjoining figuretheadjoining figure PQ, RS bisect eachother at 'O' then prove that TRIANGLE POR = TRIANGLE ROSotherPs |
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Answer» Step-by-step EXPLANATION: Maths - Simple Equations So, the EQUATION is not SATISFIED. (ii) x + 3 = ... LHS = 4 * 1 - 3 = 4 - 3 = 1 ≠ RHS. So, p = 1 is not the solution of given equation. (e) 4p – 3 ... a = 35/15 [35 and 15 are DIVIDING by 5]. => a = 7/ ... |
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| 36. |
Standers form of 5600000 is |
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Answer» Answer: 5.6 X 10^6 Explanation: 5600000 = 5.6 X 10^6 |
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| 37. |
Adrian has to take 2 Spoons ofmultivilamin syrup each day. If thecapacity of the spoon is 15 ml. Whatquantity of syrup will be taken by himin the month of december?please explain |
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Answer» Step-by-step EXPLANATION: oqkkqkqlq wjjq saw always did does e I'd definitely s s ausb a Amanda Asa.sbjskw sisnmsmq qshsbs |
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| 38. |
lalitha bought vegetables weighing 10kg ,out of this 4kg onions,3kg 750g tomatoes and the rest is potatoes .what is the weight of potatoes |
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Answer» Answer: 3.85kg since, WT. of POTATOES= wt. of total vegetables- (wt. of TOMATOES + wt. of onions) Total weight =10kg Weight of onions = 3.5kg Weight of tomatoes = 2.075kg Let weight of potato be x ⇒3.5+2.075+x=10 ⇒5.575+x=10 ⇒x=10−5.575 ⇒x=4.425 ∴x=4kg425g Therefore weight of potatoes =4kg 425G |
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| 39. |
Find the area of region bounded by x^2+y^2=4 and y=√3x and x axis in the first quadrant |
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Answer» jsjsjsjjsjsjsjjsjsjjsjsjsks |
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| 40. |
21. Area of a rectangle of lengthI and breadth bisxb |
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Answer» Answer: Step-by-step EXPLANATION: Area of rectangle is =l×b |
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| 41. |
The third angle of an isosceles triangle is 120 less than the sum of two equalangles.Find all three angles of the trianglePLZ ANSWER QUICKLY!!!! |
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Answer» Answer: o'clock obligation VIVID office office JELLYFISH TRAFFIC xo idiot |
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| 42. |
24. In given fig. ABCD is Rhombus. Find the values of x,y and z. write proper reasons forfinding x, y and z.A300B ВС C |
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Answer» Step-by-step explanation: about the wonderful thing that had happened. Haninah's wife had something to cook and to bake. And most of them were happy. Now Ben Dosah's wife had one extremely nosy neighbour. When she saw the smoke coming out of the chimney, she THOUGHT: "The Rabbi's wife has nothing in her oven. What is all that smoke about? I must go in to see what she is burning in her oven." So the nosy neighbor threw her shawl over her shoulders and walked across to the Rabbi's house. She knocked at the door and waited a while. The Rabbi's wife, however, was ashamed to MEET her. Therefore, instead of answering the knock, she ran into the next room. Again the nosy neighbour knocked at the door. "O dear!" she thought. "I guess she is ashamed to meet me. Well, I'll go in anyway." So without waiting for any answer to her knock, she quickly opened the door and WENT into the house. What do you think she saw? There on the stove all kinds of latkes and festive dishes were being prepared. And she could smell cherry pies and honey cookies in the stove. It even seemed to her that she SMELLED the cookies burning. When she saw all this she called to the Rabbi's wife, "Hurry up. Your cookies are burning." "What was that? What did you say?" the Rabbi's wife called from the other room. She was sure that she hadn't heard right, because she knew that there was nothing in the stove. "Is it possible that she is making fun of me?" the Rabbi's wife thought. "Well, what are you waiting for? Hurry before they are all burnt," the neighbour's voice was heard again. The Rabbi's wife, full of joy, ran into the kitchen. She was blushing. But it didn't MATTER. She looked as if she were hot because of the hard work. Going over to the stove, she touched the cookies to make sure that they were real, and said: "'I had just run out to get my wooden spoon when you knocked on the door." |
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| 43. |
Express 98 as aproduct of its primes |
Answer» So continue with the next prime factor, 7. ... So, the prime factors of 98 are written as 2 X 7 x 7 or 2 x 72, where 2, and 7 are the prime numbers. |
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| 44. |
If HCF (143, 440) = 11, then what is the LCM (143, 440) |
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| 45. |
Find using Distributive property: 163 X105 |
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Answer» 163*(100+5) 163*100+163*5 16300+815 17115 Hope it HELPS you |
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| 46. |
what sum of money lent at 12 1/2% per annum will produce the same interest in 4 years as ₹8560 produces in 5 years at 12% per annum? |
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Answer» <P> the interest =(p*4*12.5)/100 =RS . p/2 PART II given SUM = rs 8560, time =5 years and rate of interest = 12% p.a then interest=Rs . (8560*5*12)/100 . RS .5136 PART III by the given condition interest on rs point p in four years at 12.5%p.a=interest on rs .8560 in 5 year at 12% p.a then P/2=5136 Or, P=2*5136 or. p=10272 |
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| 47. |
How much thrust will be required to exert a pressure of 20,000 pa on an areaof 1 cm2 |
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Answer» PRESSURE = 20,000 Pa
FORCE = Pressure × area =20,000×10^-4 =2 NEWTON |
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| 48. |
₹90000 borrowed at 5.5% p.a. for 3years.find the amount to be paid at the end of third year |
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Answer» Answer: Find the simple interest on: Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also. Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.Solution: Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.Solution:P = $ 900, Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.Solution:P = $ 900,R = 5% p.a. Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.Solution:P = $ 900,R = 5% p.a.T = 3 years 4 months = 40/12 years = 10/3 years Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.Solution:P = $ 900,R = 5% p.a.T = 3 years 4 months = 40/12 years = 10/3 yearsTherefore, S.I = (P × R × T)/100 = (900 × 5 × 10)/(100 × 3) = $ 150 Find the simple interest on:(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.Solution:P = $ 900,R = 5% p.a.T = 3 years 4 months = 40/12 years = 10/3 yearsTherefore, S.I = (P × R × T)/100 = (900 × 5 × 10)/(100 × 3) = $ 150Amount = P + S.I = $ 900 + $ 150 = $ 1050 Step-by-step EXPLANATION: $ 1000 for 6 months at 4% per annum. Find the amount also. $ 1000 for 6 months at 4% per annum. Find the amount also.Solution: $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000, $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a. $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 years $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20 $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20Therefore, A = P + I = $( 1000 + 20) = $ 1020 $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20Therefore, A = P + I = $( 1000 + 20) = $ 1020(c) $ 5000 for 146 days at 15¹/₂% per annum. $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20Therefore, A = P + I = $( 1000 + 20) = $ 1020(c) $ 5000 for 146 days at 15¹/₂% per annum.Solution: $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20Therefore, A = P + I = $( 1000 + 20) = $ 1020(c) $ 5000 for 146 days at 15¹/₂% per annum.Solution:P = $ 5000, R = 151/2% p.a. T = 146 days $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20Therefore, A = P + I = $( 1000 + 20) = $ 1020(c) $ 5000 for 146 days at 15¹/₂% per annum.Solution:P = $ 5000, R = 151/2% p.a. T = 146 daysS.I = ( 5000 × 31 × 146)/(100 × 2 × 365) $ 1000 for 6 months at 4% per annum. Find the amount also.Solution:P = $ 1000,R = 4% p.a.T = 6 months = 6/12 yearsS.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20Therefore, A = P + I = $( 1000 + 20) = $ 1020(c) $ 5000 for 146 days at 15¹/₂% per annum.Solution:P = $ 5000, R = 151/2% p.a. T = 146 daysS.I = ( 5000 × 31 × 146)/(100 × 2 × 365)= $ 10 × 31 = $ 310 |
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| 49. |
व्हाट इज रिसिप्रोकल ऑफ - 5 |
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Answer» RECIPROCAL of -5 is -1/5.
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| 50. |
8 bunches of spinach cost $20 how many bunches of spinach can you get for $5? |
Answer» Answer :Cost of 8 bunches = $20 Cost of 1 bunch = 20/8 = $2.5 Let x be the no. of bunches then, 2.5×x = $5 (SINCE cost of one bunch is 2.5 so we multiply with the number of bunches to get the total cost) x = 5/2.5 x = 2 Alternate method8 bunches - - - - - - - > $20 x bunches - - - - - - - -> $ 5 Cross multiplying, 8×5 = 20×x 40 = 20x x = 40/20 x = 2 |
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