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Tuesday 8 May 2012

Solutions: 29th Aryabhatta Exam Class VIII, 2012




29th ARYABHATTA INTER-SCHOOL MATHS COMPETITION – 2012
CLASS - VIII


Section A
1. Find the next number in the series:
14, 28, 20, 40, 32, 64, . . .
a) 52
b) 56
c) 96
d) 128
Solution. The series is moving as
14 × 2 = 28
20 × 2 = 40
32 × 2 = 64 and so on
Also, the difference between 28 and 20, 40 and 32 is 8. Similarly, the difference between 64 and 56 will also be 8.

2. Choose the alternative which closely resembles the mirror image of the given combination.
a) 1
b) 2
c) 3
d) 4
Solution. First of all understand the concept of mirror image.
Hmmm…see in the given word last letter is 2. So, 2 will be the initial letter of the mirror image which is given only in option (2).

3. Choose the set of figures which follows the rule:
 Any figure can be traced by a single unbroken line without retracting.

a) 1                  b) 2                 c) 3                 d) 4
Solution. No short-cut can be provided here. It’s purely based on your reasoning capabilities.

4. What number comes inside the circle?

a) 5
b) 6
c) 8
d) 12
Solution.

5. Peter went 20m to the east then he turned left and after walking 15m turned right and went 25m, and then turned right and went 15m. How far Peter was from the starting point?
a) 40m
b) 30m
c) 45m
d) 50m
Solution. 


6. Which number completes the puzzle?



a) 21
b) 30
c) 19
d) 23
Solution. The numbers in the puzzle constitute the list of consecutive prime numbers 5 onwards.

7. The figures given below are divided into certain parts. Each part bears a number and one part is blank. Numbers follow a certain pattern of rule. You are required to analyse the given figures and then fill in the blank.

a) 33
b) 88
c) 14
d) 24
Solution. The pattern is
6 + 7 = 13 and 6 × 7 = 42,
4 + 9 = 13 and 4 × 9 = 36 and
8 + 3 = 11 and 8 × 3 = 24.

8. Five books are lying in a pile. E is lying on A and C is lying under B. A is lying above B and D is lying under C. Which book is lying at the bottom?
a) A
b) B
c) C
d) D
Solution. Using first two conditions we can say

As there are only five books, so A is lying above B implies that
Using last conditions, we get the final arrangement as

9. Which number replaces the question mark?

a) 8
b) 11
c) 7
d) 9
Solution. The pattern is
7 + 9 – 4 = 12,
1 + 4 – 1 = 4 and
6 + 5 – 2 = 9.

10. The figure given below is divided into certain parts. Each part bears a number and one part is blank. Numbers follow a certain pattern of rule. You are required to analyse the given figures and find the missing number.

a) 60
b) 62
c) 70
d) 77
Solution. The pattern is
90 – 85 = 40 – 35 = 5,
36 – 31 = 49 – 44 = 5,
78 – 73 = 80 – 75 = 5 and
72 – 67 = 65 – 60 = 5.

SECTION - B
Write the correct option in the Answer Sheet.

11. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
a) 276
b) 299
c) 322
d) 345
Solution. Let the numbers be 23m and 23n respectively.
LCM of (23m, 23n) = 23 × m × n, where m & n are natural numbers and co-prime to each other.
So, ‘m’ and ‘n’ are the factors. There value may be either 13 or 14.
Therefore, numbers are 23 × 13 = 299 and 23 × 14 = 322.

12. Two circles both of radii 6 cm each have exactly one point in common. If A is a point on one circle and B is a point on the other circle, what is the maximum possible length (in cm) for the line segment AB?
a) 12
b) 18
c) 24
d) 9
Solution. Two circles have exactly one point in common means that they touch each other. The maximum length of a line segment in a circle is of diameter.
Therefore, the maximum length of line segment AB = 12 + 12 = 24 cm.


13. It was Sunday on Jan 1, 2006. What was the day of the week on Jan 1, 2010?
a) Sunday
b) Saturday
c) Friday
d) Wednesday
Solution. This is the problem based on finding odd number of days between any two given dates.
365 days can be written as 52 weeks + 1 odd day
This 1 odd day is taken in to consideration for counting days.

No. of days between Jan 1, 2006 – Jan 1, 2007 = 365 or 1 odd day
= one day after Sunday = Monday
No. of days between Jan 1, 2007 – Jan 1, 2008 = 365 or 1 odd day
= one day after Monday = Tuesday
No. of days between Jan 1, 2008 – Jan 1, 2009 = 366 or 2 odd day
= two day after Tuesday = Thursday
No. of days between Jan 1, 2009 – Jan 1, 2010 = 365 or 1 odd day
= one day after Thursday = Friday

14. The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
a) 3 : 3 : 10
b) 10 : 11 : 20
c) 23 : 33 : 60
d) Cannot be determined
Solution. Let their salaries be 2x, 3x and 5x respectively.
New salaries
A = 2x × 1.15 (115/100) = 23x
B = 3x × 1.10 = 33x
C = 5x × 1.20 = 60x
New ratio = 23x : 33x : 60x = 23:33:60

15. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
a) 9
b) 10
c) 12
d) 20
Solution. Speed of bus excluding stoppages is 54 kmph.
If stoppages are included then Bus stops every hour for that time in which it can travel 9 km (54 – 45).
Therefore, stoppage per hour is time taken by bus to travel 9 km =  9 × 60/54
= 10 min

16. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
a) 40
b) 80
c) 120
d) 200
Solution. Let the numbers be 3x, 4x and 5x respectively.
LCM of (3x, 4x and 5x) = 60 × x
60 × x = 2400
x = 40

17. If there are two examination rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
a) 20
b) 80
c) 100
d) 200
Solution. Let the number of students in room A and B be a and b respectively.
(a – 10) = (b + 10) and
(a + 20) = 2(b – 20)
Solving these two equations we get,
a = 100 and b = 80

18. A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
a) 30 birds
b) 60 birds
c) 72 birds
d) 90 birds
Solution. Let the total number of shots be x.
Shots fired by A = 5x/8
Shots fired by B = 3x/8
Killing shots by A = (1/3) × (5x/8) = 5x/24
Killing shots by B = (½) × (3x/8) = 3x/16
Shots missed by B = (1/2) × (3x/8) = 3x/16
Therefore, 3x/16 = 27
x = 144
Killings by A = 5 × 144/24 = 30

19. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
a) 6.25
b) 6.5
c) 6.75
d) 7
Solution. Total runs scored in 10 overs = 3.2 × 10 = 32
Remaining runs to be scored in 40 overs to reach the target = 282 – 32 = 250
Required run rate = 250/40 = 25/4 = 6.25

20. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
a) 4 years
b) 8 years
c) 10 years
d) 12 years
Solution. Let the ages of the children be x yrs, (x + 3) yrs, (x + 6) yrs, (x + 9) yrs and (x + 12) yrs.
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x + 30 = 50
5x = 20
x = 4 yrs

21. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. The ratio of its diameter to its height is
a) 3 : 7
b) 7 : 3
c) 6 : 7
d) 7 : 6
Solution. Curved surface area of cylindrical pillar = 2prh
Volume of cylindrical pillar = pr2h
2prh : pr2h = 264 : 924
Solving we get,
r = 7 cm and h = 6 cm
Therefore, diameter : height = 14 : 6 = 7 : 3

22. 461+462+463+464+465 is divisible by
a) 5
b) 11
c) 17
d) None of the above
Solution. 461+462+463+464+465 = 461{1 + 41 + 4+ 43 + 44} = 461 × 341
But 341 can be factorized as 11 × 31. So, divisible by both 11 and 31.

23. The last digit of the expansion 212n64n is
a) 2
b) 4
c) 0
d) None of the above
Solution. 212n64n = 4096n – 1296n
The power of ‘n’ is same in both which will result in 0.

24. 999 999 × 222 222 + 333 333 × 333 334 =
a) 333 333 000 000
b) 444 444 333 000
c) 444 444 000 000
d) 333 333 444 000
Solution. 999 999 × 222 222 + 333 333 × 333 334
999 999 × 222 222 = 3 × 333 333 × 222 222
3 × 333 333 × 222 222 + 333 333 × 333 334
333 333{3 × 222 222 + 333 334}
333 333{666 666 + 333 334}
333 333 000 000

25. M being the mean of x1, x2, x3, x4, x5, x6, find the value of :
(x1-M) +( x2-M) +(x3-M)+ (x4-M)+ (x5-M)+ (x6-M)
a) 1
b) 0
c) 6M
d) M/6
Solution. Mean = Sum of all the observations/Number of observations
M = x1 + x2 + x3 + x4 + x5 + x6/6
6M = x1 + x2 + x3 + x4 + x5 + x6
(x1-M) +( x2-M) +(x3-M)+ (x4-M)+ (x5-M)+ (x6-M)
= x1 + x2 + x3 + x4 + x5 + x6 – 6M
= 6M – 6M
= 0

26. A basket ball is just packed in a cube of side 20cm, then the surface area of the ball is :
a) 120 cm2
b) 2400 cm2
c) 400π cm2
d) 1256 cm2
Solution. The basketball is packed in a cube implies that the basketball touches the surfaces of the cube. This in turn implies that the diameter of the basketball is equal to the side of the cube.
Diameter = 20 cm
Radius = 10 cm
Surface area of basketball = 4pr2 = 400p cm2

27. The class marks of a frequency distribution are given as follows: 15, 20, 25,…. The class corresponding to the class mark 20 is :
a) 12.5-17.5
b) 17.5-22.5
c) 18.5-22.5
d)19.5-20.5
Solution. The sum of the upper mark and lower mark should be 2 × 20 in case of class mark is 20. Just check the options in which the sum of the intervals is 40.

28. Diagonal of a cube is 6 cm . Then its lateral surface area is
a) 6 cm2
b) 36 cm2
c) 12 cm2
d) 8 cm2
Solution. Diagonal of a cube = side√3
Side of the cube = 6/√3 = 2√3
Lateral surface area of a cube = 4 (side)2 = 4 × (2√3)2 = 4 × 12 = 48 cm2

29. Let m be the mid-point and l be the upper limit of a class in a continuous frequency distribution. The lower limit of the class is :
a) 2m+l
b) 2m- l
c) m-l
d) m+2l
Solution. Mid point = (upper limit + lower limit)/2
Lower limit = 2m – l

30. What decimal of an hour is a second?
a) 0.00029
b) 0.000228
c) 0.00027
d) 0.00026
Solution. Second /hour = 1/60 × 60 = 1/3600 = 0.00027


SECTION – C
Write only the answers of the following questions in the Answer
Sheet.
31. Evaluate:





Solution. 1+ xb – a + xc – a can be written as:
1 + xb/ xa + xc/ xa = (xa + xb + xc)/ xa
Similarly, other denominators can be written.
xa/ (xa + xb + xc) + xb/ (xa + xb + xc) + xc/ (xa + xb + xc)
= (xa + xb + xc)/(xa + xb + xc)
= 1

32.







Solution. Area of sector = θ/360° × pr2
Angle = arc/radius = 4p/radius
Radius = 4p/40° = p/10°
Area of sector = θ/360° × pr2 = 40°/360° × p × p/10° × p/10°
= p3/900

33. How many diagonals do a 63 sided convex polygon have?
Solution. For a convex n-sided polygon, there are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n (n -3). However, this would mean that each diagonal would be drawn twice, (to and from each vertex), so the expression must be divided by 2.
The number of diagonals in a polygon = n (n -3)/2, where n is the number of polygon sides.
Number of diagonals = 63 × 60/2 = 63 × 30 = 1890

34. What is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9, 40 and 41?
Solution. first check whether the given sides make a Pythagorean triplet.
412 = 402 + 92
1681 = 1681
The sides form a Pythagorean triplet. Hence, the triangle is right angled.

As we know that right angle is formed in a semi-circle, this, implies 41 cm is the hypotenuse and lies on the diameter of the circle.
Therefore, radius of the circle = 41/2 = 20.5 cm





Solution. √5/2 – 10/√5 + 5√5
= √5/2 - 2√5 + 5√5 = √5/2 + 3√5 = 7√5/2

36. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. Find the percentage of decrease in area.
Solution. Let the initial length and breadth of the towel be x mt and y mt respectively.
Initial area of towel = xy sq. units
New length = 0.08x mt
New breadth = 0.09y mt
New area = 0.08x × 0.09y = 0.0072xy sq units
Change in area = xy – 0.0072xy = 0.28xy
Percentage change = 0.28xy/xy × 100 = 28%

37. A large cube is formed from the material obtained by melting three smaller cubes of sides 3, 4 and 5 cm. What is the ratio of the total surface areas of the smaller cubes to that of the large cube?
Solution. To compare first we need to find out the side of the bigger cube.
Total volume of the bigger cube = sum of the volumes of the smaller cubes
Total volume of the bigger cube = (3)3 + (4) + (5) = 27 + 64 +125 = 216
Side of the bigger cube = 6 cm
Surface area of the bigger cube = 6 (side)2 = 6 × 36 sq units
Total surface area of the smaller cubes =  6 (3)2 + 6 (4)2  + 6 (5)2
= 6 × {9 + 16 +25} = 6 × 50 sq units
Required ratio = 36 : 50 = 18 : 25

38. In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 per kg?
Solution. let the respective qunqtities be m kg @ Rs. 15 and n kg @ Rs. 20.
Quantity of mixture = (m + n) kg
15 × m + 20 × n = 16.5 × (m + n)
15m + 20n = 16.5m + 16.5n
1.5m = 3.5n
m/n = 3.5/1.5
m:n = 7:3

39. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?
Solution. 3 pumps can empty a tank in 2 days working = 8 hours
3 pumps can empty a tank in 1 day working = 2 × 8 hours
1 pump can empty a tank in 1 day working = 2 × 8 × 3 hours
4 pumps can empty a tank in 1 day working = 48/4 = 12 hours

40. Find the least perfect square which is divisible by 21, 36 and 66.
Solution. LCM of (21, 36, 66) = 36 × 7 × 11
36 is a perfect square and 7, 11 are co-prime. Thus, the least perfect square number will be 36 × 49 × 121.

41. A man takes 6 hrs 15 minutes to walk a certain distance and riding back. He could walk both ways in 7 hrs 45 minutes. In how much time can he ride both ways?
Solution. Let tw and tr be the time taken to walk and ride respectively.
tw + tr = 6(1/4)
tr + tr = 7(3/4)
tr = 31/4 × ½ = 31/8
tw = 25/4 – 31/8 = 50/8 – 31/8 = 19/8 or 2(3/8)
To ride both ways he will take = 19/4 = 4(3/4) minutes

42. A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. Find the speed of the second train.
Solution. Let the speed of the second train be m km/hr.
Relative speed = (50 + m) × 5/18 m/s
Time taken to cross second train = 6sec
 (108 + 112)/ (50 + m) × 5/18 = 6
m + 50 = 132
m = 82 km/hr

43. OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. If the radius of circle is 10 cm, find the area of rhombus.
Solution.

N is the point where the diagonals of the rhombus bisect each other.
OA = OB = OC = 10 cm
AC = 2 × NC
In ∆ONC,
OC2 = ON2 + OC2
NC2 = 100 – 25 = 75
NC = 5√3 cm
Area of rhombus = ½ × d1 × d2 = ½ × 10 × 5√3 = 25√3 sq units

44. At what time between 8 and 9 will the hands of the clock be in the same straight line?
Solution. In 12 hours, both the hands of a clock are in the same straight line for 11 times. (You can calculate!)







12 Hours / 11
= 1 + 1/11 hours
= 1 hour + 60/11 minutes
= 1 hour 5 5/11 minutes
= 1 hour 5 ½ minutes approximately

The next time after 6:00 that the
hands make a straight line
is approximately 7:05½.
After this, the hands make a straight line
is 1 hour 51/2 min after 7:05½
or 8:10 (10/11).

Short-cut: The formula for finding the
                   clock angle θ is

where H is the hour part of the time,
M is the time past the hour, in minutes,
such that 0 minutes ≤ M < 60 minutes.

45. A printer numbers the pages of the book starting from 1 and uses 3189 digits in all. How many pages does the book have?
Solution. No. of digits in 1-digit page nos. = 1 × 9 = 9.
No. of digits in 2-digit page nos. = 2 × 90 = 180.
No. of digits in 3-digit page nos. = 3 × 900 = 2700.
No. of digits in 4-digit page nos. = 3189 – (9 + 180 + 2700) = 3189 – 2889 = 300.
Therefore, No. of pages with 4-digit page nos. = (300/4) = 75.
Hence, total number of pages = (999 + 75) = 1074.







Solution. x + y = 2z
 x + y = z + z
x – z = z – y
x – z = –(y – z)
Replacing this value in given equation, we get,
x/(x- z) – z/(x – z)
= (x – z)/ (x – z)
= 1






Solution. 4a2 + 9b2 + 12ab – 8a – 12b
 (2a + 3b) 2 – 4(2a + 3b)
 (2a + 3b) {2a + 3b – 4}

48. The sum of the digits of a two digit number is multiplied by 8. The result is 8 more than the 2-digit number. Find the number.
Solution. Let the 2-digit number be 10x + y.
8 (x + y) – 8 = 10x + y
7y – 2x = 8
7y = 2x + 8 = 2(x + 4)
7y = 2(x + 4)
The above equation holds true for x= 3 & y = 2.
Therefore, required number = 10x + y = 32.

49. The mean of 1, 7, 5, 3, 4 and 4 is m. The observations 3, 2, 4, 2, 3, 3 and p have mean (m-1) and median q. Find p and q.
Solution. m = 24/6 = 4
Mean of 3, 2, 4, 2, 3, 3 and p = (m – 1) or 3
3 = (17 + p)/7
21= 17 + p
p = 4 & q = 4th term on arranging numbers in ascending order i.e. 3.



Solution. a + b = –c
 (a + b)2 = c2
a2 + b2 = c2 – 2ab
substituting this value in given expression,
= (2c2 – 2ab)/(c2 – ab)
= 2(c2 – ab)/(c2 – ab)
= 2

SECTION -D
Write only the answers of the following questions in the Answer Sheet.

51. A sector of a circle of radius 12 cm has the angle 120°. It is rolled up so that two bounded radii are joined together to form a cone. Find the volume of the cone.
Solution. Arc length = radius × angle
= 12 × 2p/3 = 8p cm                                                            {\120° = 2p/3  &   p = 180°}
The slant height of the cone will be equal to the radii of the circle. Similarly, the base of the cone is formed by arc length.
Circumference of cone = 8p
2pr = 8p
r = 4 cm
Height of the cone = √1442 – 42 = √128 = 16√8 cm
Volume of the cone = 1/3pr2h = 1/3 × p× 16 × 16√8 = 256√8/3 cubic cm

52. The sum of two numbers is 18 and the sum of their reciprocals is ¼.  Find the greater number.
Solution. Let the numbers be m and n.
m + n = 18 & 1/m + 1/n = ¼
 (m + n)/mn = ¼
18/mn = ¼
mn = 72
Therefore, the pair of numbers satisfying the above condition is (1, 72), (2, 36), (3, 24), (4, 18), (6, 12) and (8, 9). But out of these pairs only (6, 12) satisfies all the conditions.
 
 
 53. If the weight of a spherical shell is 7/8th of what it would be if it were a solid shell, find the ratio of the inner and outer radii of the shell.
Solution.

Volume of spherical shell = 4/3p{R3 – r3}
Volume of solid shell = 4/3pR3 (Assume R as radius of the solid shell because its size remains unchanged)
4/3p{R3 – r3} = 7/8 × 4/3pR3
R3 – r3 = 7/8 × R3
R3/8 = r3
r/R = (1/8)1/3
r/R = ½ or 1:2

54. A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
Solution. Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
Quantity of A left in mixture = 7x – 7/12 × 9 = (7x – 21/4) L
Quantity of B left in mixture = 5x – 5/12 × 9 = (5x – 15/4) L
The can is filled with B implies that after the mixture is drawn off, 9L of liquid B is poured in the can.
So, the total quantity of B in mixture =  (5x – 15/4) + 9L
Required ratio = (7x – 21/4)/ (5x – 15/4) + 9L
7/9 = (7x – 21/4)/ (5x – 15/4) + 9L

252x – 189 = 140x + 147
112x = 336
x = 3.
Therefore, quantity of A in can = 21 L

55. The area of a square field is 24200 sq m. How long will a lady take to cross the field diagonally at the rate of 6.6 km/hr?
Solution. Area of square field = (side)2
 Side of square field = 110√2 m
Diagonal of square field = side × √2 = 110√2 × √2 = 220 m
Time taken = 220 m × 18/6.6 × 5 = 120 seconds
  
56. What is the percentage discount that a merchant can offer on her Marked Price so that she ends up selling at no profit or loss, if she had initially marked her goods up by 50%?
Solution. The merchant had initially marked her goods up by 50%.
Let us assume that her cost price of the goods to be Rs. 100.
Therefore, a 50% mark up would have resulted in her marked price being Rs.100 + 50% of Rs. 100 = Rs100 + Rs50 = Rs150.
The question states that she finally sells the product at no profit or loss. This essentially, means that she sells the product at cost price, which in this case would be Rs100.
Therefore, she had offered a discount of Rs50 on her marked price of Rs150.
therefore, percentage discount offered by her = 50/150 × 100 = 33.33%.

57. The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. Find the simple interest on the same sum for double the time at half the rate percent per annum.
Solution. Let the principal be Rs. x.
Amount = Rs. (x + 525)
(x + 525) = x (1 + 10/100)2
(x + 525)/x = 121/100
x = 52500/121
Simple interest = PRT/100 = Rs 5250/121