QUANT TEST PAPER 3

**Explanation
to Paper III
**

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)cm^{2}

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

r^{3}*8 = 512

r^{3}= 512/8 = 64

r = 4

diameter = 8

Hence[4]

16. sc = 2*22/7*r*h = 2*22/7*5*3 = 30*22/7 cm^{2}

volume = 22/7*r*r*h = 75*22/7 cm^{2}

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*10^{5}m^{3}

weight of water = 2*10^{5}*1000kgs

= 2*10^{5 }tons

Hence[3]

19. Length of water column = 6000/60 = 100m/min

volume of water = 100*10*200 = 2*10 ^{5}m^{3}

Weight of water = 2*10^{5}*1000kgs

= 2*10^{5 }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 m^{3}

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 = w_{1}g

Weight of liquid = w_{2}-w

Weight of liquid(half filled with liquid) = w_{2}-w_{1}

Weight of liquid filled = 2(w_{2}-w_{1})

Weight of jar + weight of liquid = w_{1}+2(w_{2}-w_{1})

= 2w_{2}-w_{1}

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

x^{2}= 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*4cm^{3}

Volume of core with paper = 22/7*d/4*d/4*4cm^{3}

Volume of paper alone = 22/7[d^{2}/16-10^{2}/16]*4cm^{2}

= 22/7*[d^{2}-100/4]cm^{3}

Also, volume of paper = 22/7*100*100*4*0.1/10cm^{3}

= 100*22/7*400*0.1/10cm^{3}

= 400cm^{3}

400*22/7 = 22/7(d^{2}-100)/4

d^{2} = sqrt(1700) = 41.2(approx)

Hence[1]

44. Total are = 4 area(triangleOAB)

= 4*1/2*10*1.5 = 30cm^{2}

Hence[1]

45. floor space required = 100m^{2}

Air space required = 100*5.5m^{3} = 550m^{3}

Height = 550/100 = 5.5m

Hence[1]

46. 64+64x^{2} = 64(1+x^{2})

25+25x^{2} = 25(1+x^{2})

sqrt(64+64x^{2})-sqrt(25+25x^{2}) = 8sqrt(1+x^{2})-5sqrt(1+x^{2})

= 3sqrt(1.x^{2})

Hence [1]

47. Measure of arc XYZ = XOY

= 2(XDY)

=2(XYP)

=2*50^{o} = 100^{o}

Hence [1]

48. Let the two no.s be x and y.

(x^{2}/y^{3})/(x^{3}/y^{2}) = 1/20

x^{2}/y^{3}*y^{2}/x^{3} = 1/20

1/xy = 1/20

We cannot find x:y

Hence[4]

49. 1^{st} investment = An investment of Rs.150 fetches a dividend of Rs.5.50

Rate = 5.50/150*100 = 36.67%(approx)

2^{nd} 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 a^{3}+b^{3}+c^{3} = 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]