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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]