28th ARYABHATTA INTER-SCHOOL MATHEMATICS COMPETITION – 2011 CLASS - VIII

  

                                                  SECTION –A

Q1. Find the missing number in the series: 2, 4, 5, 15, 17, 68, 71,

355, 359, ____________.

Solution. The series is moving as

2 × 2 = 4

5 ×3 = 15

17 × 4 = 68

71 × 5 = 355

359 × 6 = 2154

 

Solution. 4 = 2^2, 25 = 5^2, 9 = 3^2, 81 = 9^2

270 = 2 × 5 × 3 × 9

Similarly,

2 × 8 × 9 × 6 = 844

  

Solution. The pattern follows as the opposite pair of numbers and their respective cubes. Thus, 5^3 = 125.

 

Q4. If K = 11 and KAMAL = 38, then PRIZE = __________.

Solution. K = 11 represents the alphabet number starting from A = 1, B = 2, C = 3... K = 11... Z = 26.

Therefore, KAMAL = 11 + 1 + 13 + 1 + 12 = 38.

Similarly, PRIZE = 16 + 18 + 9 + 26 + 5 = 74.

5. What is the angle between the minute hand and the hour hand of a clock at 15 minutes past 7?

Solution. Angle formed between hour and minute hand @ 7 = 7 × 30° = 210°

Angle formed by minute hand in next 15 minutes = 15 × 6° = 90°

Angle formed by hour hand in next 15 minutes = 15 × 1/2° = 7(1/2)°

Total angle = 210° - {90° - 7(1/2)°} = 127(1/2)° 

6. If MACHINE is coded as 197914152011, how will you code DANGER?

Solution. The coding follows as

MACHINE = 13  1  3  8  9  14  5 (as per their respective position in English alphabets)

Now, look at the coding given, each position is increased by 6.

Therefore, DANGER = 4  1  14  7  5  18

Now, increase their positions by 6.

DANGER = (4 + 6) (1 + 6) (14 + 6) (7 + 6) (5 + 6) (18 + 6)

Therefore, DANGER = 10  7  20  13  11  24 

 

Solution. The pattern follows as

16 = 3 × 5 + 1

81 = 16 × 5 + 1

406 = 81 × 5 + 1

Next term = 406 × 5 + 1 = 2031

  

8. A, B, C, D, E, F, G and H are the family members. B is the sister of G and G is the brother of C. E is the wife of A, whose father is H. D is the husband of B and F is the son of G. A is the father of B. How ‘F’ is related to ‘E’?

Solution. . F is the grandson of E. 

Solution. New equation after interchanging the symbols would look like as

40 ÷ 20 + 10 x 2 – 14 = 40 × 20 ÷ 10 - 2 + 14

= 40 × 2 - 2 + 14 = 80 + 12 = 92

10. Five poles are standing in a row. M is on the left of N, O is on the right of P which is on the right of N. If L is on the left of M, which pole is in the centre?

Solution. The correct arrangment would be

L M N P O

Therefore, N is standing at the centre.

  

                                                            SECTION - B

11. Ankit goes straight 6 km eastwards, then turns right and goes straight 2 km and turns right again and goes straight 8 km. In which direction is he from the starting point?

Solution. South-west 

Solution. 5^(a + b) = 5^(c - d)

As the bases are same, hence, the powers will be equal.

a + b = c - d

d = (c - a - b)

13. If the perimeter of a semicircular protractor is 36cm, its diameter is :

Solution. Perimeter of a semi-circular protractor = 2r + pie × radius

36 = 2r + 22r/7

36 = 36r/7

r = 7 cm

Therefore, diameter = 2 × 7 = 14 cm

14. A purchased 4 chairs and 3 tables for Rs. 1650. B purchased 3 chairs and 2 tables for Rs. 1150. The cost per chair is :

Solution. Let the price of a chair and a table be Rs. c and Rs. t respectively.

4c + 3t = 1650 ... (i)

3c + 2t = 1150 ... (ii)

Solving (i) & (ii), we get

t = 350 and c = 150

15. If the height of a cone is halved and its radius is doubled, then its volume is increased by :

Solution. Initial volume of a cone = 1/3 × 22/7 × r^2 × h

New dimensions,

radius = 2r

height = h/2

New volume of the cone = 1/3 × 22/7 × (2r)^2 × (h/2) = 2/3 × 22/7 × r^2 × h

Change in volume = 2/3 × 22/7 × r^2 × h - 1/3 × 22/7 × r^2 × h = 1/3 × 22/7 × r^2 × h

Therefore, a change of 100%.

16. An equilateral triangle is formed on a diagonal of a square of side ‘a’cm. The area of the triangle is :

Solution. Side of square = a cm

diagonal of the square = aSqrt (2) cm

side of equilateral triangle = diagonal of the square = a sqrt (2) cm

Area of equilateral triangle = sqrt (3)/4 × (side)^2

= sqrt (3)/4 × {a sqrt (2)}^2 = Sqrt (3)/4 × a^2 square cm

  

Solution. USing equality we can say that both, a & b are positive but less than 1.

I. a is less than b which implies (a - b) is negative. Hence, TRUE.

II. a × b will be positive which when divides 1 will result in positive value. Hence, TRUE.

III. (a - b) is negative and ab is positive. Thus, when a negative value is divided by a positive value will always result in negative qunatity. Hence, FALSE.

  

Solution. 110°

Draw the appropriate figure and mark the given dimensions accordingly. Use the theorem for angles formed by a transversal on parallel lines

19. A triangle and a trapezium are equal in area. They also have the same altitudes. If the base of the triangle is 36 cm, the mean of the parallel sides of the trapezium is :

Solution. AS it is given that there areas are equal

½ × 36 × altitude = ½ × (Sum of parallel sides) × altitude

Þ Sum of the parallel sides = 36

Þ Mean of the parallel sides = 36 ÷ 2 = 18 cm

20. The maximum number of points of intersection of 8 lines is:

Solution. Two straight lines can intersct atmost at 1 point.

Three straight lines can intersect atmost at 1 + 2 points.

Four straight lines can intersect atmost at 1 + 2 + 3 points.

Similarly, and so on

We can generalize this result by the formula

Þ n(n – 1)/2 where ‘n’ is the number of lines.

Here, for n = 8,

Maximum points of intersection of 8 straight lines = 8 × 7 ÷ 2 = 28

Solution. 50°

22. The minute hand of a wall clock is of the length 10.5cm. The area covered by it in 1 hour is :

Solution. In 1 hour, the hand of the clock will swap the entire 360°.

Area covered = pr² = 346.5 cm²

23. The unit digit in (7²⁷ – 3¹⁴ ) is :

Solution. These type of questions are asked in previous years. I have explained this type in detail. Kindly refer to it. If there is any problem, then revert back.

24. When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4?

Solution. Using division algorithm,

Dividend = Divisor × Quotient + Remainder

N = 4x + 3

2(n) = 2(4x + 3) = 8x + 6

2n = 8x + 6 = {(8x + 4) + 2} = {4(2x + 1) + 2}

Now, 4( 2x + 1) is divided by 4 and 2 is left as remainder.

  

Solution. 1/a + 1/b = 1/c and ab = c

(a + b)/ab = 1/c

(a + b)/c = 1/c

(a + b) = 1

Average of a and b = (a + b)/2 = ½

Option (c)

26. 9 men visited a hotel, 8 of them spent Rs.4 each over their meal and the 9th spent Rs. 2 more than the average of all the nine. The total money spent by them on the meal is :

Solution. Let the average money spent by 9 men be Rs. x

Total money spent by 8 of them = Rs. 4 × 8 = Rs. 32

Money spent by 9^th men = Rs. (x + 2)

Total money spent by 9 men = Rs. {(x + 2) + 32}

Average money spent by 9 men = {(x + 2) + 32}/9

x = {(x + 2) + 32}/9

On solving, we get

x = 4.25

Total money spent by 9 men = 4.25 + 2 + 32 = Rs. 38.25

Option (a)

27. Of the three numbers, the first is twice the second and is half the third. If the average of three numbers is 56 , the three numbers in order are :

Solution. Let the first number x

Second number = x/2

Third number = 2x

Sum of three numbers = x + x/2 + 2x = 7x/2

Average of three numbers = 56

(7x/2) ÷ 3 = 56

On solving, we get

X = 48

Therefore, the numbers are 48, 24, 96

NOTE: The question can be solved much more easily by checking the options

Option (a)

28. The circumradius of a triangle whose sides are 6cm, 8 cm and 10 cm is :

Solution. We know that angle subtended in a semi-circle is a right angle. Thus, hypotenuse of the triangle will become the diameter of the circle or 10 cm ÷ 2 = 5 cm

 

29. If a, b, c, d and e are five consecutive odd integers , then their mean is :

Solution. Mean = Sum of all observations ÷ Total number of observations

Mean = (a + b + c + d + e) ÷ 5

But this is not among the given options.

Lets try substituting some random values on this statement

1, 3, 5, 7, 9 and 37, 39, 41, 43, 45

Mean for 1^st set = (1+ 3+ 5+7+9) ÷ 5 = 5 or 1(First number) + 4 = 5 (Mean of given set of values)

Mean of 2^nd set = (37 + 39 + 41 + 43 + 45) = 41 or 37 (First number) + 4 = 41 (Mean of given set of values)

 Thus, we can conclude that the average/mean of five consecutive odd integers is always 3^rd consecutive integer or 4 more than the 1^st consecutive integer. 

                                                                    Section C

Solution. Let x = 0.34676767…

10x = 3.4676767…

100x = 34.676767…

100x – x = (34.676767…) – (0.34676767…)

99x = 34.33

x = 3433/9900

Similarly, let y = 0.1333333…

10y = 1.33333…

10y – y = (1.33333…) – (0.1333333…)

9y = 1.2

y = 12/90 = 2/15

Þ x + y = 3433/9900 + 2/15 = 4753/9900

32. The sum of a two digit number and the number obtained by interchanging the digits of the number is 121. If the digits of the number differ by 5 , then find the number.

Solution. Let the 2-digit number be 10x + y.

Number obtained by interchanging the digits = 10y + x

x – y = 5 or y – x = 5                          … (1)

Þ (10x + y) + (10y + x) = 121

Þ x + y = 11                          … (2)

Solving equation (1) & (2), we get

x = 8 or 3  

y = 3 or 8

Therefore, the number is 38 or 83.

33. P and Q can do a piece of work in 12 days. Q and R in 15 days, R and P in 20 days. In how many days R alone can do the same work?

Solution. 1/P + 1/Q = 1/12,     1/Q + 1/R = 1/15,   1/R + 1/P = 1/20

Adding the above three equations we get

Þ 2{1/P + 1/Q + 1/R} = 1/12 + 1/15 + 1/20 = 1/5

Þ 1/P + 1/Q + 1/R = 1/10

Þ In 1 day, R alone can do it in = {1/P + 1/Q + 1/R } – {1/P + 1/Q }

= 1/10 – 1/12 = 1/60

Therefore, R alone can complete the whole work in 60 days. 

36. Two cylinders of same volume have their heights in the ratio 1 : 3. Find the ratio of their radii.

Solution. V₁ = p × r₁² × h₁         & V₂ = p × r₂² × h₂

Þ V₁ ÷ V₂ = {r₁²  ÷  r₂² }×{h₁ ÷ h₂}

V₁ = V₂

Þ {h₂ ÷ h₁} = {r₁ ÷ r₂}²

Þ r₁/r₂ = √ h₂ /h₁ = √3 : 1

37. Find the number of spherical bullets, each 0.6cm in diameter be made out of a rectangular solid 9cm x 11 cm x 12 cm.

Solution. Volume of solid block = n × Volume of each bullet

Where ‘n’ is number of bullets

Þ 9 × 11 × 12 = n × 4p/3 × (0.3)³

Þ n = 35 × 3 × 100 = 10500 

 Q38. The largest sphere is cut off from a cube of diagonal 5 cm. Find the volume of the sphere.

Solution. Largest sphere implies that the sphere would be fitted exactly into the cube.

Þ Diagonal of sphere = Length of the edge of cube

Þ    Diagonal of cube = side√3

Þ    Side of the cube = 5 cm

Largest sphere = 4p/3 × (5)³ = 500p/3 

 

Solution. From the given figure

b = c and 3a = (a + 2b)

b = c and a = b

Þ a = b = c

3a + c = 180°

4c = 180°

c = 45° 

40. A cylindrical vessel 60cm in diameter is partially filled with water. A sphere 60cm in diameter is gently dropped into the vessel. To what height will water rise in the cylinder?

Solution. Due to the sphere the volume of water in vessel will increase.

Volume of sphere = p × 30 × 30 × h

Þ h = 4/3 × 30 = 40 cm

41. If I drive at a speed of 24 kmph, I reach school 5 minutes late and if I drive at 30kmph , I reach 4 minutes too soon. Find the distance of the school from my residence.

Solution. Let the distance from the residence to school be d km and usual time to reach be t hr.

Þ d = 24 × {t + 5/60}           … (1)

and d = 30 × {t – 4/60}           … (2)

Solving equation (1) and (2), we get

Þ 24 × {t + 5/60} = 30 × {t – 4/60}

Þ t = 1/3 + 1/3 = 2/3 hr or 40 minutes 

42. A chord of a circle is 12cm in length and its distance from the centre is 8cm. Find the length of the chord of the same circle whose distance from the centre is 6cm.

Solution. Refer to the given figure 

In ∆OAE,

Þ OA² = 8² + 6² = 100

OA = 10 cm

In ∆ODF,

Þ DF² = 10² – 6² = 64

DF = 8 cm

  

 

 

44. How many integers are there between 100 and 1000 all of whose digits are odd?

Solution. All the numbers between 100 to 1000 will be 3-digit numbers.

111      113      115      117      119

131      133      135      137      139     

151      153      155      157      159

171      173      175      177      179

191      193      195      197      199

Total numbers starting with ‘1’ = 5 × 5 = 25

So, in the same manner there would be 25 numbers each starting with 3, 5, 7 and 9.

Therefore, total such numbers = 5 × 25 = 125

  

Solution. a^m × aⁿ = a^{m +  n}

Þ a^{m +  n} = a^{m n}

Þ m + n = mn

Þ m (n – 2) + n (m – 2) = mn – 2m + mn – 2n = 2mn – 2mn = 0 

46. What is the area of a circle that is inscribed in a square of area 4 sq cm? (Write the answer in terms of π ).

Solution. Area of square = 4

Þ Side of square = √4 = 2 cm

Side of square = Diagonal of circle

Þ Radius of circle = 1 cm

Area of circle = pr²  = p sq cm 

 

Solution. x⁶ – ax⁵ + x⁴ – ax³ + 3x – a + 2

Þ x⁵(x – a) + x³(x – a) + 3(x – a) + 2(a + 1)

Now,

Þ (x – a) {x⁵ + x³ + 3} + 2(a + 1)

If (x – a) is factor of above expression, then the remainder left would be 0.

Þ 2(a + 1) = 0

Þ a = –1

48. Some problem with the question itself

49. The following observations have been arranged in the ascending order. If the median of the data 29, 32, 48, 50, x, x+2, 72, 78, 84 and 95 is the mean of 75, 64, 136, 25, 15, then find the value of x.

Solution. Mean = (75 + 64 + 136 + 25 + 15) ÷ 5 = 63

Arranging the given numbers in ascending order

29, 32, 48, 50, 61(x), 63(x + 2), 72, 78, 84, 95

Number of terms = 10 (even)

Median = (n/2 + 1)^th term

= 6^th term 

 

Solution. (√3/7) = (√21/49) = (√21)/7 » 4.58/7 = 0.65

                                                                

                                                                  Section D

52. A hemispherical bowl is made of steel of 0.25cm thickness. The inner radius of the bowl is 5cm. Find the volume of the steel used.

Solution. Outer radius = Inner radius + Thickness = 5 + 0.25 = 5.25 cm

Volume of hemi-spherical bowl = 2p/3 × {r₂² – r₁²}

= 2p × {(5.25)² – (5)²} = 5.125p cc

  

55. Three numbers are in the ratio 2 : 3 : 4 and the sum of their cubes is 0.334125. Find the largest number.

Solution. Let the numbers be 2x, 3x, 4x respectively.

(2x)³+ (3x)³ + (4x)³ = 0.334125

99x³ =0.334125

On solving, we get

x = 0.15  

57. The compound interest on a certain sum of money for 2 years at 5% per annum is Rs102.50. Find the compound interest on the same sum for the same period at 4% per annum.

Solution. Let the amount of Rs. x.

Þ  (x + 102.5) = x(1 + 5/100)²

Þ  (x + 102.5) = x(105/100)²

Þ  (x + 102.5) = x(11025/10000)²

10000x + 1025000 = 11025x

1025x = 1025000

x = 1000

58. A train travelling with a constant speed crosses a 96m long platform in 12 seconds and another 141 m long platform in 15 seconds. Find the length of the train.

Solution. Let the length of the train be l m.

Þ (96 + l)/12 = (141 + l)/15

Þ 480 + 5l = 564 + 4l

l = 84 m  

60. A shopkeeper makes a profit of 20% even after giving a discount of 10% on the advertised price of a printer. If he makes a profit of Rs. 450, find the advertised price.

Solution. Let the advertised price be Rs. 100.

SP after discount = 100 – 100 × 10% = Rs. 90

CP = 90/1.2 = Rs. 75

Profit = SP – CP = 90 – 75 = Rs. 15

15 = Rs. 450

1 = Rs. 450/15 = Rs. 30

100 = 30 × 100 = Rs. 3000