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Thursday 28 February 2013



ARYABHATTA INTER-SCHOOL MATHS COMPETITION 2012
CLASS V
PART I: ARITHMETIC
 
Q1. The product of (1 – ½) (1 – 1/3) (1 – ¼) (1 – 1/5) … (1 – 1/100)
Solution. In this question, solve each bracket by taking LCM. 
½ × 2/3 × ¾ × 4/5 … 99/100
Now, as you can see that the denominator and numerator of the consecutive terms can be cancelled.
You need not to carry cancellation of each and every term. Just assume that the cancellation goes on till the last term. So, after canceling all the terms, the term we will be left with is numerator of first term and denominator of last term i.e. 1/100.
Q3. In a certain year, January had exactly four Tuesday and five Saturday. The day on which January 1 falls on that year is
Solution. I am leaving the solution of this question as from my perspective it is too much to ask from a Class V student.

Q4. A tank containing 19.62 L of water can be emptied in 10 min. Amount of water that can be emptied in 1 sec ___ L.
Solution. This question is simply based on unitary method.
10 min = 10 × 60 sec or 600 sec
In 600sec, the quantity of water emptied is 19.62 L
In 1 sec, the quantity of water emptied is 19.62 ÷ 600 = 0.0327 L or 32.7 ml

Q5. The fraction halfway between 1/5 and 1/3 is ___ .
Solution. In order to find the halfway between the given two numbers, first we have to convert them in to like fraction.
1/5 = 3/15 and 1/3 = 5/15
Now, it is easy to find the number halfway between the given two numbers.

Q6. The number of three-digit numbers divisible by 13 is
Solution. If we find all the numbers that are divisible by 13, from 1 – 999, then subtract all the numbers that are divisible by 13 , from 1 – 99.
The number that will be left will lie between 100 – 999.
Total three-digit numbers between 1 – 99 is 99.
Numbers lying in this range divisible by 13 = 7 (Simply 13’s table with numbers less than 100)
Total three-digit numbers between 1 – 999 is 999.
Numbers lying in this range divisible by 13 = 76

Þ Total three-digit numbers divisible by 13 = 76 – 7 = 69

Q7. Greeting cards are sold in the pack of 6, 8 and 24 cards. The minimum number of packs needed to buy exactly 110 cards is
Solution. In order to buy exactly 110 cards, 19 packs of 6 cards each, 14 packs of 8 cards each and 5 packs of 24 cards each are needed. 











Q9. Date and Time 2012 minutes after the beginning of 19th January, 2012 was
Solution. To answer the question, we must calculate 2012 minutes is equivalent to how many days.
2012 minutes = 2012/60 hr = 2012/(60 × 24) day or 1(143/360) day
This implies that 2012 minutes after 19th January, 2012 will be 1(143/360) day.
So, date will be 20th January, 2012.

To calculate time, we need to convert (143/360)th part of a day in to the time.
(143/360)th part of a day = (143/360) × 24 hr = 143/15 hr
143/15 hr = 9(8/15) hr or 9 hr 32 min

Q10. The difference between the first 1591 even numbers and 1591 odd numbers is
Solution. Now, this question involves too much calculation if you go by general method of finding sum for even & odd numbers, then subtracting them. For quicker method, we can generalize it as

Consider first five consecutive even numbers and odd numbers.
Þ (2 + 4 + 6 + 8 + 10) & (1 + 3 + 5 + 7 + 9)
Þ 30 & 25
So, the difference is 5.

Consider another case of first ten consecutive even numbers and odd numbers.
Þ (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20) & (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)
Þ 110 & 100
So, the difference is 10.

Similarly, here, if we consider first 1591 even numbers and odd numbers.
Þ (2 + 4 + 6 + …) & (1 + 3 + 5 + …)
Here, the difference will be 1591. 



Q12. A tennis singles tournament had six players. Each player played every other player once with no ties. If the first player won 4 games, second player won 3 games, third player won 2 games, fourth player won 2 games, fifth player won 2 games, then the sixth player won ___ games.
Solution. In this question, we need to find out how many games were played between six players with no ties.
Here, the following pairs provide the information for matches played between any two players, given they played with other once. Assume there are six players with numbers1, 2, 3, 4, 5, 6.
(1, 6) (1, 5) (1, 4) (1, 3) (1, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6) (4, 5) (4, 6) (5, 6)
So, total number of matches played between six players = 15

Total number of games won by given five players = 4 + 3 + 2 + 2 + 2 = 13

Number of games won by sixth player = 15 – 13 = 2

Q13. A factory packs 48 chocolates in each box. They have 243 boxes to be filled. They have already made 10,000 chocolates. The number of chocolates they have to make more to fill the boxes is
Solution. In this question, we have to find out how many chocolates need to pack in 243 boxes.
Number of chocolates need to pack = 243 × 48 = 11664
Now, they have already made 10000 chocolates.
Number of chocolates needed = 11664 – 10000 = 1664











































Q15. Neha and Priya were once the same height. Since then Priya has grown 30% while Neha has grown half as many centimeters as Priya. Priya is now 65 cm tall. Present height to Neha is
Solution. Assume the height of both Neha and Priya be 100 cm.
According to question, Priya has grown 30%.
Increased height of Priya = 30% of 100 = 30 cm
New height of Priya = 100 + 30 = 130 cm
Also, ‘Neha has grown half as many centimeters as Priya’.
Increased height of Neha = 30 ÷ 2 = 15 cm
New height of Neha = 100 + 15 = 115 cm

Present of height Priya is given as 65 cm. We can now calculate the present height of Neha by using Unitary method as given below.
When Priya’s assumed height is 130 cm, then her present height is 65 cm
When Priya’s assumed height is 1 cm, then her present height is 65/130 cm
Þ If Neha’s assumed height is 115 cm, then her present height is 65/130 × 115 = 57.5 cm
Þ Present height of Neha is 57.5 cm



Q17. For a party, the chef is making 30 kg of fruit salad using 25% guavas, 30% apples and 45% bananas. In anticipation of more guests, he adds 5 kg of apples more to his salad. The percentage of apples in the fruit salad now is
Solution. Quantity of guavas in salad = 25% of 30 kg = 7.5 kg
Quantity of apples in salad = 30% of 30 kg = 9 kg
Quantity of bananas in salad = 45% of 30 kg = 13.5 kg
Quantity of apples added in salad = 5 kg
Total quantity of salad = 30 + 5 = 35 kg
Total quantity of apples in salad = 9 + 5 = 14 kg
Þ Percentage of apples in salad = 14/35 × 100 = 40%

Q18. A store normally sells T-shirt at Rs. 100 each. This week the store is offering one T-shirt free for each purchase of four. Samaira needs 7 T-shirts and Naisha needs 8 T-shirts. Money they will save if they purchase together is
Solution. Together they need 7 + 8 = 15 T-shirts
If each T-shirt costs Rs. 100, then total cost of 15 T-shirts = 15 × 100 = Rs. 1500

But with the discount scheme 5 T-shirts (4 + 1 free) for Rs. 400.
Þ With same scheme 15 T-shirts will cost = 400 × 3 = Rs. 1200 


Q20. Suppose 1/5 of Class V students of a school participate in a maths competition and 0.95 of these students get maths teacher of their choice next year. Only 50% of the Class V students who do not participate in the maths competition get the maths teacher of their choice next year. The percentage of students who get the maths teacher of their choice the next year is
Solution. Fraction of students participated in maths competition = 1/5
Percentage of students participated in maths competition = 1/5 × 100 = 20%
Percentage of these 20% students, who get teacher of their choice = 0.95 of 20% = 19%
Now, percentage of students who do not participated in maths competition = 100 – 20 = 80%
Percentage of these 80% students, who get teacher of their choice = 80% ÷ 2 = 40%
Þ Total number of students who get the teacher of their choice = 19 + 40 = 59%

Q21. A hare is running at a rate of 1 m every min, while a tortoise is crawling at a rate of 1 cm every second. In 1 hr, the difference between the distance covered by the hare and the tortoise would be
Solution. TO calculate the difference between them in 1 hr or 60 min, we should calculate the distance covered by them in that time.
Distance covered by hare in 60 min
In 1 min, the hare covers 1 m
Þ In 60 min, the hare covers 60 m

Distance covered by tortoise in 60 min
In 1 second, the tortoise covers 1 cm
In 60 second or 1 min, the tortoise covers 60 cm
Þ In 60 min, the tortoise covers 60 × 60 = 3600 cm or 36 m

Þ Difference in distance covered in 1 min = 60 – 36 = 24 m

Q22. At a seminar 2/5 of the audience were children, 3/10 were ladies. The rest were men. If the number of children was 28 more than men, then the total number of people in the audience was
Solution. As the difference between the number of children and men in given, we must calculate their respective fractional values.
Fraction of children = 2/5
Fraction of ladies = 3/10
Þ Fraction of men = 1 – (2/5 + 3/10) = 3/10
Þ Difference between the fraction of men and children = 2/5 – 3/10 = 1/10
Now, it is given that this 1/10 is equal to 28
Þ Total number of people in the audience was 28 × 10 = 280

Q23. Look at the given time table and answer the following questions.


  1. The fastest train going to Pluto from Venus is
  2. Train that take shortest time from Mars to Pluto is
  3. The fastest train between Mars and Jupiter is
  4. The train that takes the longest time fom Venus to Jupiter is
Solution. The solution to this type of questions is already given on this blog in some previous year papers. Kindly refer to them. In case of any query, comment. 

PART 2: GEOMETRY
 








































Q3. The measures of six angles of a heptagon are 126, 109, 168, 132, 189 and 113. Find measure of the seventh angle of the triangle.
Solution. The solution to this type of question is already given on this blog. Refer it form there. In case of any query, comment.

Q4. Number of isosceles triangles that can be formed having a perimeter 23 cm is
NOTE: Sides have lengths in whole numbers
Solution. I am leaving the solution of this question as from my perspective it is too much to ask from a Class V student.

Q5. The measure of the supplement of the smaller angle formed by the hands of a clock that displays a time of four o’clock is
Solution. Angle between the hands of the clock at 4 O’clock = 4 × 30° = 120°
Supplement of this angle = 180° - 120° = 60°



Q7. In order to walk 1.5 km in a rectangular park, Rohan has to walk the length 30 times or walk the perimeter 10 times. The area of this rectangular park is
Solution. To cover a distance of 1.5 km, one needs to walk 30 times the length or 10 times the perimeter of the rectangular park.
Length of the park = 1.5 km/30 = 50 m
Perimeter of the park = 1.5 km/10 = 150 m

To find the area of the park, we require its length and breadth. We know its length, the breath can be calculated as
Þ Perimeter = 2(length + breadth)
Þ Breadth = 75 – 50 = 25 m
Þ Area of the park = 50 × 25 = 750 sq m

Q8. There are 36 students in a class. The class teacher wants to make a pie graph to show the club preference of the students in her class. 11 students are in mathematics club, 9 students are in skating club, and 4 students are in gymnastics club. Out of the remaining, ¼ are in painting club. Rest of them is in the dance club. The number of degrees the teacher will use to show the students in the dance club is
Solution. To find the number of degrees for dance club, we must know the number of students in the club.
Number of students in dance club = 8
Now, 36 students are represented by 360°
Þ 1 student will be represented by 10°
Þ 8 students will be represented by 80°



Q10. A 2 cm by 3 cm rectangle and 3 cm by 4 cm rectangle are combined within a square without overlapping and the sides of the square are parallel to the sides of the given two rectangles. The smallest possible area of this square is
Solution. Try it on your own. In case of any query, comment.