\n Q-2 \u00e2\u20ac\u201c D<\/em>
\n Q-3 – D<\/em>
\n Q-4 – B<\/em>
\n Q-5 – D<\/em>
\n Q-6 – C<\/em>
\n Q-7 – B<\/em>
\n Q-8 – B<\/em>
\n Q-9 – D<\/em>
\n Q-10 – D<\/em>
\n Q-11 – D<\/em>
\n Q-12 – B<\/em>
\n Q-13 – C<\/em>
\n Q-14 – B<\/em>
\n Q-15 – C<\/em>
\n Q-16 – B<\/em>
\n Q-17 – C<\/em>
\n Q-18 – C<\/em>
\n Q-19 – C<\/em>
\n Q-20 – C<\/em>
\n Q-21 – D<\/em>
\n Q-22 – B<\/em>
\n Q-23 – D<\/em>
\n Q-24 – B<\/em>
\n Q-25 – C<\/em><\/td>\n
\n Q-27 – D<\/em>
\n Q-28 – D<\/em>
\n Q-29 – B<\/em>
\n Q-30 – C<\/em>
\n Q-31 – B<\/em>
\n Q-32 – C<\/em>
\n Q-33 – A<\/em>
\n Q-34 – A<\/em>
\n Q-35 – C<\/em>
\n Q-36 – C<\/em>
\n Q-37 – A<\/em>
\n Q-38 – C<\/em>
\n Q-39 – C<\/em>
\n Q-40 – A<\/em>
\n Q-41 – C<\/em>
\n Q-42 – –<\/em>
\n Q-43 – B<\/em>
\n Q-44 – D<\/em>
\n Q-45 – D<\/em>
\n Q-46 – D<\/em>
\n Q-47 – A<\/em>
\n Q-48 – D<\/em>
\n Q-49 – C<\/em>
\n Q-50 – B<\/em><\/td>\n
\n Q-52 – C<\/em>
\n Q-53 – C<\/em>
\n Q-54 – C<\/em>
\n Q-55 – C<\/em>
\n Q-56 – C<\/em>
\n Q-57 – C<\/em>
\n Q-58 – B<\/em>
\n Q-59 – B<\/em>
\n Q-60 – B<\/em>
\n Q-61 – B<\/em>
\n Q-62 – A<\/em>
\n Q-63 – B<\/em>
\n Q-64 – D<\/em>
\n Q-65 – B<\/em>
\n Q-66 – A<\/em>
\n Q-67 – D<\/em>
\n Q-68 – D<\/em>
\n Q-69 – A<\/em>
\n Q-70 – D<\/em>
\n Q-71 – C<\/em>
\n Q-72 – C<\/em>
\n Q-73 – D<\/em>
\n Q-74 – B<\/em>
\n Q-75 – C<\/em><\/td>\n
\n Q-77 – C<\/em>
\n Q-78 – D<\/em>
\n Q-79 – A<\/em>
\n Q-80 – A<\/em>
\n Q-81 – D<\/em>
\n Q-82 – A<\/em>
\n Q-83 – *<\/em>
\n Q-84 – C<\/em>
\n Q-85 – B<\/em>
\n Q-86 – B<\/em>
\n Q-87 – B<\/em>
\n Q-88 – A<\/em>
\n Q-89 – C<\/em>
\n Q-90 – C<\/em>
\n Q-91 – A<\/em>
\n Q-92 – B<\/em>
\n Q-93 – B<\/em>
\n Q-94 – C<\/em>
\n Q-95 – A<\/em>
\n Q-96 – C<\/em>
\n Q-97 – A<\/em>
\n Q-98 – A<\/em>
\n Q-99 – C<\/em>
\n Q-100 – C<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Solution-1<\/strong> Solution -2<\/strong> Solution -3<\/strong> Solution -4<\/strong> Solution -5<\/strong> Solution -6<\/strong> Solution -7<\/strong> Solution -8<\/strong> Solution -9<\/strong> Solution -10<\/strong> Solution -11<\/strong> Solution -12 – Question incorrect<\/strong> Solution -13<\/strong> Solution -14<\/strong> Solution -15<\/strong> Solution -16<\/strong> Solution -17<\/strong> Solution -18<\/strong> Solution -19<\/strong> Solution -20<\/strong> Solution -21<\/strong> Solution -22<\/strong> Solution -23<\/strong> Solution -24<\/strong> Solution-25<\/strong> Solution -26<\/strong> Solution -27<\/strong> Solution -28 (incorrect Option) (<\/strong>Option D is 4 instead of 24<\/em>)<\/strong> Solution -29<\/strong> Solution -30<\/strong> Solution -31<\/strong> Solution -32<\/strong> Solution -33<\/strong> Solution -34<\/strong> Solution -35<\/strong> Solution -36<\/strong> Solution -37<\/strong> Solution -38<\/strong> Solution -39<\/strong> Solution -40<\/strong> Solution -41<\/strong> Solution -42 –<\/strong><\/p>\n Solution -43<\/strong> Solution -44<\/strong> Question number 45 to 47<\/strong> Solution -45<\/strong> Solution -46<\/strong> Solution -47<\/strong> Solution -48<\/strong> Solution -49<\/strong> Solution -50<\/strong> PSPCL LDC \/ UDC (General) Practice Set – 1 https:\/\/www.punjabexamportal.com\/11052024\/2015\/12\/09\/pspcl-ldc-udcgeneral-practice-set\/ Mock Test Solution UDC LDC..<\/p>\n","protected":false},"author":1,"featured_media":1206,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,4,8,1,10],"tags":[],"class_list":["post-1198","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-jobs","category-mcq","category-pspcl","category-punjab-exam-portal","category-punjab-gk"],"_links":{"self":[{"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/posts\/1198","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/comments?post=1198"}],"version-history":[{"count":0,"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/posts\/1198\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/media\/1206"}],"wp:attachment":[{"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/media?parent=1198"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/categories?post=1198"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.punjabexamportal.com\/11052024\/wp-json\/wp\/v2\/tags?post=1198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n3*6*9 =162<\/p>\n
\na:b=2:3 b:c=5:8=(5*35:8*35)=3:245,
\na:b:c=2:3:245=10:15: 24
\nb=98*1549=30<\/p>\n
\nLet CP=100 & SP=175 also assume that he reduces x% amount to get no profit no loss.
\nX% of 175=100
\nX=400\/7
\nThat means he will offer=100-400\/7=300\/7=42.85%<\/p>\n
\nAverage after 11 innings should be 36
\nSo, Required score = (11 * 36) – (10 * 32)
\n= 396 – 320 = 76<\/p>\n
\nLet six years ago the age of Kunal and Sagar are 6x and 5x resp.
\nthen,(6x+6)+4(5x+6)+4=11\/10
\n10(6x+10)=11(5x+10)
\n5x=10
\nx=2
\nSo Sagar age is (5x+6) = 16<\/p>\n
\n‘m’ roots are 4 or -4<\/p>\n
\nDate on 27th February 1603 was Thursday<\/p>\n
\nLet 1 man’s 1 day work = x and 1 woman’s 1 days work = y.
\nThen, 4x + 6y = 1\/8
\nand 3x+7y = 1\/10
\nsolving, we get y = 1\/400 [means work done by a woman in 1 day]
\n10 women 1 day work = 10\/400 = 1\/40
\n10 women will finish the work in 40 days<\/p>\n
\nLet the total time be x mins.
\nPart filled in first half means in x\/2 = 1\/40
\nPart filled in second half means in x\/2 =
\n1\/60+1\/40=1\/24
\nTotal = x\/2*1\/40+x\/2*1\/24=1
\nx\/2(1\/40+1\/24)=1
\nx\/2*1\/15=1
\nx=30mins<\/p>\n
\nDifference between C.I and S.I for 2 years = 36.30
\nS.I. for one year = 330.
\nS.I. on Rs 330 for one year = 36.30
\nSo R% = 36.30\/330*100 = 11%<\/p>\n
\nSpeed=Distance\/Time = 200\/24
\nConvert km\/h multiply by 18\/5
\n200\/24 * 18\/5 = 30km\/h<\/p>\n
\nCorrect Question is – If the length of a rectangle is increased by 30% and the width of the rectangle is decreased by 20%,then the area of the rectangle ________<\/em>
\nAns-B) Increases by 4%<\/p>\n
\nLet Y\/128 = 162\/Y
\nThen x2 = 128 x 162
\n= 64 x 2 x 18 x 9
\n= 82 x 62 x 32
\n= 8 x 6 x 3
\n= 144.
\nx = 144 = 12<\/p>\n
\nThere are a total of 90 two digit numbers. Every third number will be divisible by ‘3’. Therefore, there are 30 of those numbers that are divisible by ‘3’.Of these 30 numbers, the numbers that are divisible by ‘5’ are those that are multiples of ’15’. i.e. numbers that are divisible by both ‘3’ and ‘5’. There are 6 such numbers — 15, 30, 45, 60, 75 and 90.
\nfind out numbers that are divisible by ‘3’ and not by ‘5’, which will be 30 – 6 = 24.
\n24 out of the 90 numbers are divisible by ‘3’ and not by ‘5’.
\nThe required probability is therefore, 24\/90 = 4\/15<\/p>\n
\nSpeed with current is 20,
\nspeed of the man + It is speed of the current
\nSpeed in still water = 20 – 3 = 17
\nNow speed against the current will be
\nspeed of the man – speed of the current
\n= 17 – 3 = 14 kmph<\/p>\n
\nP=(S.I.*100)\/(R*T)
\nSo by putting values from our question we can get the answer
\nP=(4016.25*100)\/(9*5)=8925<\/p>\n
\nGiven – 30 liters mixture in ratio of milk and water 7:3
\nmeans milk is 21 literss and water is 9 liters
\nbecause resultant water in mixture 40%, means milk is 60%
\nnew ratio is 60:40 = 3:2
\n21\/(9 + x)=3\/2
\nx=5<\/p>\n
\nAs Shyam is helped by Ram and Singhal every third day, Shyam works for 3 days while Ram and Singhal work for 1 day in every 3 days.
\nTherefore, the amount of work done in 3 days by Shyam, Ram and Singhal:
\n= 3\/20+1\/30+1\/60
\n=1\/5 of the job.
\nHence, it will take them 5 times the amount of time = 3\u00c3\u20145 = 15 days.<\/p>\n
\nThe total time of journey = 5 hours.
\nLet ‘x’ hours be the time that I traveled at 40 kmph
\nTherefore, 5 – x hours would be time that I traveled at 60 kmph.
\nHence, I would have covered x*40 + (5 – x)60 kms in the 5 hours = 240 kms.
\nSolving, for x in the equation 40x + (5 – x)*60 = 240, we get
\n40x + 300 – 60x = 240
\n20x = 60 or x = 3 hours.<\/p>\n
\nLet S be the selling price of 1 article.
\nTherefore, the selling price of 100 articles = 100 S.P. –(1)
\nThe profit earned by selling these 100 articles = selling price of 75 articles = 75 S.P — (2)
\nWe know that Selling Price (S.P.) = Cost Price (C.P) + Profit — (3)
\nSelling price of 100 articles = 100 S.P and Profit = 75 S.P from (1) and (2). Substituting this in eqn (3), we get
\n100 S.P = C.P + 75 S. Hence, C.P = 100 S.P – 75 S.P = 25 S.P
\nProfit % = (Profit\/C.P)*100 = (75\/25)*100 =300<\/p>\n
\nA:B:C = 35000:45000:55000
\n= 7:9:11
\nto get the share, first make total of above ratio.
\nthen get each share.
\nD) Rs. 10500, Rs. 13500, Rs. 16500<\/p>\n
\nLet the present age of Baskar be ‘b’ and that of Rajesh be ‘r’.
\nSo, r = b – 10 …. eqn (1).
\n10 years back Rajesh was (r – 10) years old. 10 years back Baskar was (b – 10) years old.
\nThe question states that 10 years back Rajesh was two thirds as old as Baskar was.
\ni.e., (r – 10) = (2\/3)*(b – 10) …. eqn (2).
\nCross multiplying, we get 3(r – 10) = 2(b – 10)
\nor 3r – 30 = 2b – 20 …. eqn (2)
\nFrom eqn (1) we can substitute r as (b – 10) in eqn (2)
\nSo, 3(b – 10) – 30 = 2b – 20
\nor 3b – 30 – 30 = 2b – 20
\nor b = 40.
\nThe present age of Baskar is 40 years.<\/p>\n
\n1\/4th of a number is 17
\nmeans actual number is = 17*4 = 68
\n45% of 68 = 68*45\/100 = 30.6<\/p>\n
\nDistance covered = 120+120 = 240 m
\nTime = 12 s
\nLet the speed of each train = x.
\nThen relative velocity = x+x = 2x
\n2x = distance\/time = 240\/12 = 20 m\/s
\nSpeed of each train = x = 20\/2 = 10 m\/s
\n= 10*18\/5 km\/hr = 36 km\/hr<\/p>\n
\nAll are prefect square except 20
\nSo ans is -20<\/p>\n
\nAns is A)\u00c2\u00a9$37 (In question paper option typed ‘\u00c2\u00a9S37’ by mistake)<\/p>\n
\nTOUR – 1234
\nCLEAR – 56784
\nSPARE – 90847
\nFrom above three given statement
\nwe can find code of SCULPTURE – 953601347
\n5th digit is – 0<\/p>\n
\nin first column , 13+7*2 = 27
\n2nd column = 54+45*2 = 144
\nlet missing number is x
\nNo 3rd column is x+32*2 =68
\nx = 4<\/p>\n
\n6*2 + 1 =13
\n13*2 + 2 = 28
\n28*2 + 3 = 59
\n59*2 + 4 = 122<\/p>\n
\nSouth-East becomes North – move on right side ‘South-East’->’East’->’Noth-East’->’North’
\nSouth becomes North-East – move on right side ‘South’->’South-East’->’East’->’Noth-East’
\nSame way
\nWest become ‘South-East’ – move on right side ‘West’->’South-West’->’South’->’South-East’<\/p>\n
\nFrom figure 1st 2nd and 4th we conclude that 6,4,1 and 2 dots appear adjacent to 3 dots.
\nClearly there will be five dots on the face opposite the face with 3 dots<\/p>\n
\nThe above three numbers are multiples of the numbwer at the bottom
\nClearly 45,18,27 are all multiple of 9, So missing number is 9<\/p>\n
\nThe middle letters are static, so concentrate on the first and third letters.
\nThe series involves an alphabetical order with a reversal of the letters.
\nThe first letters are in alphabetical order: F, G, H, I , J.
\nThe second and fourth segments are reversals of the first and third segments.
\nThe missing segment begins with a new letter.<\/p>\n
\nClearly, 1st, 8th, 15th, 22nd, and 29th October are Sundays.
\nSo, 31st October is Tuesday.
\nSo 1st November will be Wednesday.<\/p>\n
\nClearly, Number students behind the nitin in rank = (49 – 18) = 31
\nNitin is 32nd from the last<\/p>\n
\nHere Physics, Chemistry, Zoology deals with science but Geography is not.<\/p>\n
\nThe number of doctors who are neither artists nor players is 17.<\/p>\n
\nThe number of doctors who are both players and artists is 3.<\/p>\n
\nThe number of artists who are players is 22 + 3 = 25.<\/p>\n
\nThe number of players who are neither artists nor doctors is 25.<\/p>\n
\nThe number of artists who are neither players nor doctors is 30.<\/p>\n
\nMirror image of 8:35 is 3:25<\/p>\n
\nMirror image is 4th<\/p>\n
\n![\"Mock](\"https:\/\/www.punjabexamportal.com\/11052024\/wp-content\/uploads\/2015\/12\/cod.png\")
<\/p>\n
\nThe code for \u00e2\u20ac\u02dcmoney\u00e2\u20ac\u2122 is \u00e2\u20ac\u02dcla\u00e2\u20ac\u2122.<\/p>\n
\nThe code for \u00e2\u20ac\u02dcsupply\u00e2\u20ac\u2122 is either \u00e2\u20ac\u02dcmo\u00e2\u20ac\u2122 or \u00e2\u20ac\u02dcta\u00e2\u20ac\u2122<\/p>\n
\ndemand – \u00e2\u20ac\u02dcmo\u00e2\u20ac\u2122 or \u00e2\u20ac\u02dcta\u00e2\u20ac\u2122
\nonly – \u00e2\u20ac\u02dcne\u00e2\u20ac\u2122 or \u00e2\u20ac\u02dcki\u00e2\u20ac\u2122
\nThe code for \u00e2\u20ac\u02dcmore\u00e2\u20ac\u2122 may be \u00e2\u20ac\u02dcxi\u00e2\u20ac\u2122<\/p>\n
\nAnswer is ‘aacb’
\nNow complete series is [abc ab bca bc cab ca abc ab]<\/p>\n
\n3rd figure is right asnwer<\/p>\n
\nThe word ‘OPTICAL’ contains 7 different letters.
\nWhen the vowels OIA are always together, they can be supposed to form one letter.
\nThen, we have to arrange the letters PTCL (OIA).
\nNow, 5 letters can be arranged in 5! = 120 ways.
\nThe vowels (OIA) can be arranged among themselves in 3! = 6 ways.
\nRequired number of ways = (120 x 6) = 720.<\/p>\n","protected":false},"excerpt":{"rendered":"