This test has been downloaded free from www.freewebs.com/s3094
QUANTITATIVE APTITUDE TEST PAPER 3
QUANTITATIVE APTITUDE TEST PAPER 3 : EXPLANATORY ANSWERS
1.  A = 50pts., B = 40pts. 
  B = 50pts., C = 30pts.
l.c.m. of 50 and 40 is 200
  If B gets 200, A gets 250, C gets 150pts.
  If A gets 50, C gets 50*150/250 = 30pts.
  Hence[1]
  
  2. Work done by x and y in 5 min = 5(10+15/150) = 5/6
  Remaining work = 1/6
  for doing 1/15 work, q takes 1 min.
  for doing 1/6 work,?
  1/6*1*15 = 5/2 min
  Hence [4]
  
  3. Let time be t and speed be s.
  Distance travelled = st = 2/3 s*(t+2)
  t = 2/3 (t+2)
  t = 4hrs.
  Hence [2]
  
  4. Let the speed be x.
  When speed becomes 3/4th of the usual time, time becomes 4/3 i.e. 1/3rd more 
  of the usual
  time.
  1/3rd of usual time = 45min.
  1/3rd of usual time taken to travel by = 45-15
  1/3rd of usual time taken to travel AB = 10kms = 15min.
  Usual time taken to travel 10km = 45 min
  usual speed = 10*60/45 = 40/3 km/hr
  Usual time taken to travel Ay = 45*3 =135 min = 2 hrs.15min.
  Distance = 40/3*135/60 = 90/3 = 30 km
  Total distance = 30+30 = 60km
  Hence[1]
  
  5. 12km/hr = 12*100/60 = 200metres/min
  Distance of the cat from the rat = 50+200 = 250metres
  Since both are moving in same direction, the rat gains 15-12 = 3km/hr i.e. 50
metres/min
  Time required to overtake = 250/50 = 5 min
  Hence [1]
  
  6. Let time be t hrs after 5a.m.
  6.5t-5 = 30
  6.5t = 35
  t = 5 1/3 hrs.
  The resoirvoir will fill at 5a.m.+5 hrs20min = 10.20a.m.
  Hence[3]
  
  7. speed down the river = 5km/hr
  speed up the river = 5/(1 1/3) = 3.75 km/hr
  speed of the current = 1/2(5-3.75) = 0.625 km/hr
  Hence[1]
  
  8. 15m + 10b = 12m+20b
  3m = 10b
  money earned by 10 boys = Rs.100
  Rs.100 = money earned by 3 men.
  1 man should be paid 100/3 = Rs.33.33
  Hence[2]
  
  9.The cop gains 1/8-1/12 = 1/24 km/min = 41 2/3 m/min
  To gain 100m, time, 100/(41 2/3) = 2.4 min
  The thief has gone ahead by 2.4*1/12*1000 = 200m
  Hence [1]
  
  10. If the sides are 5x, 6x and 7x
  s = 18x/2 = 9x
  Area = sqrt[s(s-a)(s-b)(s-c)] = 800
  = sqrt[9x(9x-5x)(9x-6x)(9x-7x)] = 800
  x = sqrt[800/sqrt(216)]
  Hence [2]
  
  11. s = a+b+c+d/2 = 50/2 = 25
  Area of the quadrilateral = sqrt(10*13*15*12) = 30*sqrt(26)cm2
  Hence [1]
  
  12. Area of 4 walls = perimeter*height
  = 2(40+15)*h
  = 110*h
  Area = 7500/5 = 1500
  1500 = 110h
  h = 13.63
  Hence[2]
  
  13. Area = 1/2(sum of parallel sides * height)
  = 1/2 * sum of sides *20
  800 = 10*sum of parallel sides
  80 = sum of parallel sides
  x + (x+10) = 80
  x = 35
  other side = 45
  Hence[1]
  
  14. 450/200
  200r = 900
  r = 4.5 cm
  Hence[4]
  
  15 . density of steel is 1 i.e. iron is 8
  weight of iron ball = weight of steel
  volume of iron *8 = volume of steel *1
  r3*8 = 512
  r3= 512/8 = 64
  r = 4
  diameter = 8
  Hence[4]
  
  16.  sc = 2*22/7*r*h = 2*22/7*5*3 = 30*22/7 cm2
  volume = 22/7*r*r*h = 75*22/7 cm2
  Hence [4]
  
  17.  New volume/old volume = (22/7*49*r*r*h)/(22/7*r*r*25*h) = 49/25
  If old is 25, new is 49, then increase is 24
  increase% = 24/25*100 = 96%
  Hence[2]
  
  18.  Length of water column = 6000/60
  volume of water = 100*10*200 = 2*105m3
  weight of water = 2*105*1000kgs
  = 2*105 tons
  Hence[3]
  
  19.  Length of water column = 6000/60 = 100m/min
  volume of water = 100*10*200 = 2*10 5m3
  Weight of water = 2*105*1000kgs
  = 2*105 tons
  Hence[3]
  
  20. Total surface area of a cylinder = 2*22/7*r*h+2*22/7*r*r
  = 2*22/7*7*10+2*22/7*7*7
  = 440+308
  = 748
  Hence[1]
  
  21.  2*22/7*r*h
  = 2*22/7*2*20
  = 80*22/7
  Hence[2]
  
  22.  r+h = 40
  2*22/7*r*r+2*22/7*r*h = 2*22/7*r(r+h)
2*22/7*r(r+h) = 1760
  r = 7
  h = 33
  volume = 22/7*r*r*h = 22/7*7*7*33 = 5082 m3
  Hence[3]
  
  23. Let 3r,4r be the radii
  Let 2h, 3h be the height.
  Ratio of volumes = 1/2
  Hence [1]
  
  24.  Let heights = h,2h
  Let radii be r and R
  22/7*r*r*h = 2*22/7*R*R*h
  r*r = 2(R*R)
  r/R = sqrt(2)/1
  Hence[3]
  
  25. Let correct time be x.
  distance travelled in (x+10) min. at 20km/hr
  Distance travelled in (x+2) at 30km/hr
  (x+10)*20/60 = (x+2))*30/60
  x = 14 min.
  Hence [3]
  
  26. x/20-x/40 = 6
  2x-x/40 = 6x = 240km
  Hence[2]
  
  27. Let speed be x.
  360/x - 360/(x+20) = 3
  360(x+20)-360x = 3x(x+20)
  (x+60)(x-40)
  x = 40
  Hence [1]
  
  28. Suppose the first distance is covered in x hours and 2nd distance in y hours.
  4x+6y = 30
  6x+4y = 32
  Solving equations,
  x = 3.6 hrs.,y = 2.6 hrs.
  Total time = 6.2 hrs.
  Hence[2]
  
  29. Distance covered by thief in 1hr. is 20km.
  Now 10 km will be compensated in 1 hr.,
  20 km will be compensated in 2 hrs. 
  So, he overtakes the thief at 4p.m.
  Hence[3]
  
  30.  1/2:1/3 = 3:2
  If y takes 2 min., x takes 3 mins.
  If y takes 24 min, x takes 24*3/2 = 36 min.
  Hence[4]
  
  31.  -19,-17, -13,-11,-7
  
  32.  +1,-3
  
  33.  *2.5, *3.5, *4.5, *5.5, *6.5
  
  34.  +7,*7
  
  35.  Area of circle = 22/7*r*r = 22/7*21*21 = 1386
  Area of triangle = 1/2 b*h
  
  36.  no exp
  
  37.+88, -143
  
  38. 7695/855=9, 69255/7695=9, .....5609655/623295=9
  
  39. Weight of jar = w1g
  Weight of liquid = w2-w
  Weight of liquid(half filled with liquid) = w2-w1
  Weight of liquid filled = 2(w2-w1)
  Weight of jar + weight of liquid = w1+2(w2-w1)
  = 2w2-w1
  Hence [3]
  
  40.  In finding average speed, we use Harmonic mean.
H.M.= 1/{1/3[1/x+1/2x+1/3x]}= 18x/11 km/hr.
  Hence[4]
  
  41. Draw XE to YZ
  By pythagoras theorem,
(XE) 2= (XY) 2-(YE) 2= (15) 2-(9) 
  2 =144
  Let AB = CD = x.
AD/XE = AY/EY
  x = 71/5
  x2= 51.84
  Hence [3]
  
  42. Total money deposited = Rs.12*150 = Rs.1800
  Interest on Rs.150 for 12 months = 150*12/12*6*1/100 = 9
  Interest on Rs.150 for 11 months = 150*11/12*6*1/100 = 33/4
  Interest on Rs.150 for 1 month = 150*1/12*6*1/100 = 3/4
  Total interest = 9/1+33/4+3/4 = Rs.18
  Total amount due = 1800+18 = Rs.1818
  Hence [3]
  
  43. Volume of core = 22/7*10/4*10/4*4cm3
  Volume of core with paper = 22/7*d/4*d/4*4cm3 
  Volume of paper alone = 22/7[d2/16-102/16]*4cm2
  = 22/7*[d2-100/4]cm3
  Also, volume of paper = 22/7*100*100*4*0.1/10cm3
  = 100*22/7*400*0.1/10cm3
  = 400cm3
  400*22/7 = 22/7(d2-100)/4
  d2 = sqrt(1700) = 41.2(approx)
  Hence[1]
44. Total are = 4 area(triangleOAB)
= 4*1/2*10*1.5 = 30cm2
Hence[1]
45. floor space required = 100m2
Air space required = 100*5.5m3 = 550m3
Height = 550/100 = 5.5m
Hence[1]
46. 64+64x2 = 64(1+x2)
25+25x2 = 25(1+x2)
sqrt(64+64x2)-sqrt(25+25x2) = 8sqrt(1+x2)-5sqrt(1+x2)
= 3sqrt(1.x2)
Hence [1]
47. Measure of arc XYZ = XOY
= 2(XDY)
=2(XYP)
=2*50o = 100o
Hence [1]
48. Let the two no.s be x and y.
(x2/y3)/(x3/y2) = 1/20 
x2/y3*y2/x3 = 1/20
1/xy = 1/20
We cannot find x:y
Hence[4]
49. 1st investment = An investment of Rs.150 fetches a dividend of Rs.5.50
Rate = 5.50/150*100 = 36.67%(approx)
2nd investment = An investment of Rs.15 fetches a dividend of Rs.0.35
Rate = 0.35/15*100 = 86.37%(approx)
Hence[1]
50. If a+b+c = 0, then a3+b3+c3 = 3abc
Putting x-y = a,y-z = b, x-z = c
Then a+b+c = 0
The factors are 3(x-y)(y-z)(x-z)
Hence[1]