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Nairaland Forum / Nairaland / General / Education / Snepco Past Questions (4119 Views)
Nnpc/snepco Past Questions / Nnpc/snepco Past Questions / NNPC/SNEPCO Scholarship Past Questions 2000  2019 PDF (2) (3) (4)
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Re: Snepco Past Questions by femi4: 9:03am On Oct 14 
NisforNicky:x + y/2 = 3.2 x + y/4 = 2.3 Solve simultaneously y/4 = 0.9 y = 3.6kg Mass of empty bottle x = 3.2 1.8 = 1.4 kg 1 Like 
Re: Snepco Past Questions by Xandre: 9:09am On Oct 14 
ThankGod001:I doubt o A call was made to dragnet and it was stated that if Calcs were needed,it would be made available in the system. 
Re: Snepco Past Questions by NisforNicky: 9:13am On Oct 14 
femi4:wow dat was fast thanks bro pls don't forget the rest 
Re: Snepco Past Questions by femi4: 9:37am On Oct 14 
Second moving average of order 3 (50+10+70)/3 = 130/3 = 43.3 
Re: Snepco Past Questions by NisforNicky: 9:37am On Oct 14 
cc femi4 now, this one below is from the recently concluded SPDC mete3

Re: Snepco Past Questions by NisforNicky: 9:48am On Oct 14 
femi4: does this mean that we go to the second value on the data log, and take three values and then find their average dats the only thing I understand about dis 
Re: Snepco Past Questions by mete3: 9:52am On Oct 14 
Are this question also for Snepco
NisforNicky: 
Re: Snepco Past Questions by NisforNicky: 9:53am On Oct 14 
mete3: spdc but they ask similar questions all the ones above are SNEPCo tho 
Re: Snepco Past Questions by femi4: 10:05am On Oct 14 
Re: Snepco Past Questions by femi4: 10:06am On Oct 14 
NisforNicky:Exactly 1 Like 
Re: Snepco Past Questions by NisforNicky: 10:06am On Oct 14 
femi4:no 12 in options 
Re: Snepco Past Questions by NisforNicky: 10:07am On Oct 14 
Re: Snepco Past Questions by femi4: 10:08am On Oct 14 
NisforNicky:Well , I was surprised too 
Re: Snepco Past Questions by mete3: 10:08am On Oct 14 
NisforNicky:Okay.. Thanks Buh pls, Did u hav them in pdf or what. 
Re: Snepco Past Questions by NisforNicky: 10:10am On Oct 14 
mete3: yh but too big to upload on Nairaland ill try posting them as screenshots later when i buy data using free mb now I hv posted virtually all the math sef jxt varbal left 
Re: Snepco Past Questions by femi4: 10:13am On Oct 14 
The set doesn't have a mode ( no repetition of numbers) 1 Like 
Re: Snepco Past Questions by mete3: 10:13am On Oct 14 
NisforNicky:Can i pm u Boss, I need it pls. 
Re: Snepco Past Questions by NisforNicky: 10:16am On Oct 14 
femi4: exactly u think these are negative questions? 
Re: Snepco Past Questions by NisforNicky: 10:18am On Oct 14 
mete3:no need pls il post it here later for everyone 
Re: Snepco Past Questions by mete3: 10:25am On Oct 14 
NisforNicky: I already did but no p. the thing is, am nt using android for nw nd was unable to expand the size of qtns been posted. reason why i ask u to help me snd them in pd format. 
Re: Snepco Past Questions by NisforNicky: 10:25am On Oct 14 
femi4:I don't think they re working at d sam rate 
Re: Snepco Past Questions by femi4: 10:25am On Oct 14 
NisforNicky:May be 
Re: Snepco Past Questions by ivumar: 10:57am On Oct 14 
NisforNicky: In a game of cards, there are 4 kings. Therefore, prob that the cards are kings will be 4/52 x 3/51 = 1/221 As long as it is not stated otherwise in the question, drawing should always be without replacement. 2 Likes 
Re: Snepco Past Questions by ivumar: 10:59am On Oct 14 
femi4: Lol...Thank God I'm not the only one.. I started doubting the little knowledge I have when I could not find 12 in the options. 1 Like 
Re: Snepco Past Questions by NisforNicky: 11:08am On Oct 14 
ivumar: where did you get 3/51 pls 
Re: Snepco Past Questions by ivumar: 11:16am On Oct 14 
NisforNicky: Based on the without replacement I emphasized on. If initially, there were 4 kings. How many kings will remain after you have drawn one out? Also the initial number of cards was 52. How many will remain after one card is drawn out? You get right? 1 Like 
Re: Snepco Past Questions by NisforNicky: 11:30am On Oct 14 
ivumar: no. weren't the cards drawn simultaneously? 1 Like 
Re: Snepco Past Questions by ivumar: 11:57am On Oct 14 
NisforNicky: simultaneously was not specified.. Even if it was, it does not matter. You still have to treat the cards individually. Reason being that, if you are to dig through layers of abstraction, then two objects picked 'simultaneously' will still differ in the time picked even if the difference in time is usually so small(in the order of micro seconds). Therefore, there will always be a first card picked and a second(although they were picked simultaneously). That's why we account for the first card picked, after which we account for the second, bearing in mind, that a card has been picked already. 2 Likes 1 Share 
Re: Snepco Past Questions by NisforNicky: 12:07pm On Oct 14 
ivumar: oh thanks I finally get it you are a good tutor bro pls help attempt the rest 1 Like 
Re: Snepco Past Questions by ivumar: 12:41pm On Oct 14 
NisforNicky:From the sequence 3, 4, 1, 3, 4, 1....., it can be observed that the nth term is gotten from subtracting the (n2)th term from the (n1)th term. In other words, nth term = (n1)th term  (n2)th term. Okay lets validate.. According to the sequence, the 3rd term is 1. Lets check if it tallies with the emboldened. 3rd term = 2nd term  1st term. i.e, 3rd term = 4  3 = 1. This shows correspondence. Now you can check for 4th, 5th and 6th term for yourself and see that they agree with the emboldened. Solving for the 7th term, we have 7th = 1  (4) = 1 + 4 = 3 Therefore, the 7th term is 3. Now we solve for the difference between btw the 7th and 6th term, which is 3  (1) = 3 + 1 = 4. Answer not in the options I guess they mistakenly did '3  1' instead 1 Like 
Re: Snepco Past Questions by ivumar: 12:55pm On Oct 14 
NisforNicky: The question on angle of elevation is, I think incorrect. Or what do you guys think... because I don't get how angle of elevation can be greater than 90 degrees. 1 Like 
Re: Snepco Past Questions by ivumar: 1:27pm On Oct 14 
NisforNicky: The graph of y=x and y=x² intersects when, at the same value of x, both graphs have the same value of y. Observing the options, there are only two possible values... 0 and 1. So lets use them to test. for the graph, y = x, when x = 0, y = 0. Will it be same for graph of y = x²? lets check. when x = 0, y = 0², therefore, y = 0. Therefore, for both graphs, when x=0, y will also = 0. Therefore, they intersect at (0,0) Next, test for when x = 1, you will also discover that for both graphs, when x = 1, y will also = 1. Therefore, they also intersect at (1,1). Alternatively, you can solve using the equation of both graphs. y1 = x y2 = x². Recall that at the point of intersection, y1 must = y2 So at what value of x can y1 = y2 We solve this by equating y1 to y2, i.e, y1 = y2, i.e x = x². Lets solve x = x² x²  x = 0 Factorize x(x  1) = 0 Therefore, x = 0 or x = 1 Input this values of x in each of the equations to get y. When x is 0, you will get 0 as y for both graphs And when x is 1, you will get 1 for both graphs. Therefore point of intersection is (0,0) and (1,1) 1 Like 
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