Aryabhatta Exam Solutions: Ramanujan Society of Born Mathematicians (JR), Inter School mathematics Competition (2005)

Solution: Largest 6 digit number = 999990

Largest 5 digit number = 99999

Quotient = 999990 ÷ 99999 = 10

Option C

Solution. 50 can be written as the product of two numbers as

(50 × 1), (25 × 2), (10 × 5) {As each of these are given in the options}

If smaller number is 1, then the larger number will be 50. But 1 is not given as any of the options.

If smaller number is 2, then the larger number will be 25. But 2 is not the factor of 25.

If smaller number is 5, then the larger number will be 10. In this case, 5 is the factor of 10. Hence, correct answer.

Option C

Solution. Total number of parts formed after each chocolate is divided in 3 parts

= 3 × 9 = 27

If 2 pieces of chocolates are left, then it implies that he has distributed 25 pieces.

Now, each friend got a piece of chocolate, then 25 pieces will be distributed to 25 friends.

Option C

Solution. First 3 prime numbers = 2, 3 and 5

Sum of first 3 prime numbers = 2 + 3 + 5 = 10

Twice the sum of first 3 prime numbers = 2 × 10 = 20

As per the given condition,

¼ of number is 20.

Then the number will be 4 × 20 = 80

Option D

Solution: In this question, we have to just check whether after applying the given time constraints the time in two clocks become same.

Case 1:

Assume if clock I is 15 min fast, then the actual time will be 10:35. Checking for the clock which is 10 min slow. Clock II and Clock III will show 10:45 and 10:20 respectively.

Case 2:

Assume if clock I is 30 min fast (half an hour), then the correct time will be 10:20. Checking for the clock which is 10 min slow. Clock II and Clock III will show 10:45 and 10:20 respectively.

Here, the correct time shown by two clocks (I & III) is 10:20. Now, only one condition is left and a clock is left. Lets apply the remaining condition on the Clock II, 15 min fast.

The correct time shown by Clock II will be 10:20.

Thus, the correct time is 10:20

Option B

Solution:

Solution. As children of Class 5 are not exposed to simple equations as per NCF

Therefore, I am not providing the solution to the given question

Solution. Remember the whole circle constitutes an angle of 360° at the centre.

Arjun’s share is half of the circle

Bhanu’s share is one-fourth of the circle

Remaining one-fourth of the circle is divided equally between Chandini and Devansh.

Share of Devansh = ½ of one-fourth = ½ × ¼ = 1/8 or 2/16

Option B

Solution. This is a simple problem based on the application of HCF.

In this case, the milkman has to calculate the HCF of the capacities of three cans containing milk.

HCF of (39, 52, 26) = 13

Now, you can see that using a container of same size he can transfer milk 3 times from 39 L can, 4 times from 52 L can and 2 times from 26 L can.

Option B

.

Solution.

Option B

Solution. We know a central angle formed at the centre of a circle measures 360°.

If we have to place 6 spokes equally at the centre of a wheel, then the measure of the angle between them =

Option B

Solution. Now, Madhav’s weight can neither be 2.345 kg nor 234.5 kg nor 2345 kg. Thus, the only option we are left with is 23.45 kg.

Therefore, we can say that Madhav’s weight may be 23.45 kg.

Option B

Solution.

Option A

MDLLLXXXVIII = 1000 + 500 + 50 + 50 + 50 + 10 + 10 + 10 + 8 = 1688

Option B

MDCCCLXXXVIII = 1000 + 500 + 100 + 100 + 100 + 50 + 10 + 10 + 10 + 8 = 1888

Option C

CMDLLXXVIII = (1000 – 100) + 500 + 50 + 50 + 10 + 10 + 8 = 1528

One has to remember that whenever a smaller value precedes a larger value, the smaller value is subtracted from the larger value

Option D

MDCCCLXXXIVIII = 1000 + 500 + 100 + 100 + 100 + 50 + 10 + 10 + 10 + (5 – 1) + 3 = 1887

Therefore, it is clearly evident that MDCCCLXXXVIII is the largest among the given numerals.

Option B

Solution. 1 million = 10 lakhs = 10,00,000

Option A

Solution. Average of ‘n’ observations can be calculated by dividing the sum of ‘n’ observations with the total number of observations i.e. n

Now, there are 10 children in the party. So, ‘n’ = 10

Sum of 10 observations or we can say sum of the ages of the 10 children = 10 × 18 = 180 years

But Renu’s brother also joined the party, which means we now have 10 + 1 = 11 children

So, we have to calculate the total age of 11 children = 11 × 17 = 197 years

Therefore, age of Renu’s brother can be calculated by subtracting the age of Renu’s brother from the total age of 11 children

Age of Renu’s brother = 197 – 180 = 17 years

Option C

Solution. Sum of (Difference between 18 and 4) and (Product of 9 and 3)

Difference between 18 and 4 = 14

Product of 9 and 3 = 27

Sum of (14) and (27) = 14 + 27 = 41

Option D

Solution. As children of Class 5 are not exposed to simple equations as per NCF

Therefore, I am not providing the solution to the given question

Solution. This is a simple question based on the conversion of units

Total distance covered = 5 mm + (1/50) m + (3/4) cm + 2.5 cm

Changing all the units to mm

Total distance covered = 5 mm + 20 mm + 7.5 mm + 25 mm = 57.5 mm or 5.75 cm

Option B

Solution. Divide 30 by ½ =

Add 10 to 60 = 70

Option D

Solution. Jai earns 2 times of as much as Dinesh earns

Similarly, Jai earns 3 times as much as Binu earns

Now, if Binu earns Rs. 1200

It implies that Jai earns earns 3 times of Rs. 1200 = Rs. 3600

But Jai earns 2 times as much as Dinesh

It implies that Dinesh earns Rs. 1800

To find the average earning, we must find the toatla earning of the three.

Total earning of the three = Jai + Dinesh + Binu = 3600 + 1800 + 1200 = Rs. 6600

Option B

Solution.

It is clearly evident that red colored layer will have a total of 4 × 4 = 16 cubes

We can see that there is 1 cube less in blue colored layer. Similarly, there are 4 pink colored cubes less in pink layer and 9 green colored cubes less in green layer.

Total number of cubes less from all layers = 1 + 4 + 9 = 14 cubes

If all these cubes were placed in their respective layers, then the total number of cubes = 4 × 16 = 64 cubes

Now, if we remove the number of cubes that are less in each layer, it will give us the number of cubes that are visible from the front view

Number of cubes visible = 64 – 14 = 50

Option C

Solution. In this question, the thing that we have to learn that is

Freezing point of water in Fahrenheit = 212°F (100°C)

Boiling point of water in Fahrenheit = 32°F (0°C)

Now, we have to find out in what time temperature rises to 212°F from 32°F

For that we need to calculate the difference between the freezing point and boiling point

Difference between freezing point and boiling point = 212 – 32 = 180°F

It is given that temperature rises by 5°F every minute

By applying unitary method, we can find that in what time temperature difference of 180°F can be covered

For 5°F temperature rise, time taken = 1 minute

For 1°F temperature rise, time taken = 1/5minute

For 180°F temperature rise, time taken = 180/5= 36 minutes

Option D

Solution. So, we put together all the pieces of wire then it be equal to the total length of the wire i.e. 800 m or 8000 cm

Length of each piece of wire = 200 cm

Number of pieces cut from 8000 cm of wire =

Option B

Solution. Carry out the calculation, step-wise

5.2 × 10 = 52

52 + 0.5 = 52.5

52.5 × 100 = 5250

Option D

Solution. Banta spends ¼ of 32 = Rs. 8

Money left = 32 – 8 = Rs. 24

Further, he spends ½ of what was left = ½ of 24 = Rs. 12

Total money spend = 12 + 8 = Rs. 20

Similarly, Santa spends ½ os 32 = Rs. 16

Money left = 32 – 16 = Rs. 16

Further, he spends ¼ of what was left = ¼ of 16 = Rs. 4

Total money spend = 16 + 4 = Rs. 20

Option C

Solution. The schedule of the commercial breaks during the show time can be represented as:

Solution. ½ of 2 + 2 = 1 + 2 = 3

Option C

Solution. As children of Class 5 are not exposed to simple equations as per NCF

Therefore, I am not providing the solution to the given question

Solution. As percentage is not taught in Class 5 as per NCF, I am not providing solution to this question.

Solution. Bina is 2 years and 9 months older to Mini.

Adding 2 years to Mini’s date of birth, we get

1^st June 1999 + 2 years = 1^st June 2001

Now, 9 months are left to add, so adding 9 months

1^st June 2001 + 9 months = 1^st March 2002

Option D

Solution. A cube given below has 6 surfaces.

If we put a cube on top of it, the figure will look like as following

Now, we can see that the new figure will also have 6 plane surfaces.

Option A

Solution. As per NCF, these type of questions are not taught to a Class 5. But still there is a way to solve.

As the answer will be the smallest number which when increased by 21, becomes exactly divisible by 3, 8, 9, 12, 16 and 18

You can add 21 to each of the option and check in which option the result obtained after adding 21 is exactly divisible by 3, 8, 9, 12, 16 and 18

Option B

Solution. Total quantity of orange squash after adding water to it = 1200 ml + 60 ml = 1260 ml

Quantity of orange squash given to Tinku = 1/3 of 1260 ml = 420 ml

Quantity of orange squash left with her = 1260 – 420 = 840 ml or 0.840 l

Option C

Solution. Cost of a dozen chocolates = Rs. 132

Cost of half a dozen chocolates = 132 ÷ 2 = Rs. 66

Cost of a chocolate = 66 ÷ 6 = Rs. 11

Cost of a chocolate after rounding off to nearest ten = Rs. 10

Cost of half a dozen chocolates after rounding off to nearest ten = 6 × 10 = Rs. 60

Cost of half a dozen chocolates after rounding off = Rs. 70

Difference between the cost of half a dozen chocolates = Rs. 70 – Rs. 60 = Rs. 10

Option D

Solution.

Option A, is wrong as its is reciprocal of 2/5.

Option B, is wrong as there is percentage sign included.

Option C, when reduced to simplest form will not be equal to 2/5

Thus, we are left with only option D which is correct.

Option D

Solution. Total time taken including stoppages to reach Jaipur = 5 hours

Total time taken excluding stoppages to reach Jaipur = 5 – 1 = 4 hours

It means that the bus traveled for 4 hours.

In 1 hour, bus can travel a distance of 70 km

In 4 hours, bus can travel a distance of 4 × 70 = 280 km

Option A

Solution. 0.5 ÷ 0.05 = 10

Option D

Solution. Check each of the option.

Option A, 9 – 35 = –26 and 53 – 9 = 44

Option B, 18 – 35 = –17 and 53 – 18 = 35

Option C, 44 – 35 = 9 and 53 – 44 = 9

Option D, 71 – 35 = 36 and 53 – 71 = –18

The given condition holds true for option C only i.e. 44

Option C

Solution. Odd primes from 1 to 10 = 2, 3, 5 and 7

HCF of odd primes from 1 to 10 = 1

Product of even primes = 2

Difference = 2 – 1 = 1

Option B

Solution. Remember

1 score = 20 and a gross = 12 dozen or 12 × 12 = 144

¼ of 200 = 50

2 score = 2 × 20 = 40

A gross = 144

Þ 50 + 40 + ? = 144

Þ 90 + ? = 144

? = 144 – 90 = 54

Option C

Solution. As per NCF, a student is not exposed to angles and types of angles. So, I am not providing solution to the question.

Solution. Overtime wages = 1.5 × 8.30 = Rs. 12.45

If Arun worked 44 hours last week, that means he worked 40 hours regular and 4 hours as overtime.

Wages paid for regular working hours = 8.30 × 40 = Rs. 332

Wages paid for overtime working hours = 12.45 × 4 = Rs. 49.80

Total wages paid = Rs. 381.80

Option C

Solution. Greatest even number of 5-digits = 99998

Smallest number of 6-digits = 100000

Difference = 100000 – 99998 = 2

Option B

Solution. Check by placing each option in place of (?).

Option D

Solution. The boy reads 20 pages per day and completes it in 32 days.

It implies that in 32 days, he has read 20 × 32 = 640 pages

Half of 640 = 320 pages

Remaining pages = 640 – 320 = 320 pages

Now, he can read 20 pages in 1 day.

So, to complete 320 pages he needs 320 ÷ 20 = 16 days.

Remaining 320 pages can be read in 320 ÷ 40 = 8 days

Total number of days = 16 + 8 = 24 days

Option A

Solution. Total cost of the two = 6 + 10 = Rs. 16

Now, machine only accepts half rupee coins and there are two half rupee coins in a rupee.

Thus, to buy both items, Arjun must put 16 × 2 = 32 coins in the machine.

Option C

Solution. Product of 1, 0.1 and 0.01 = 0.001

Number to be added = 1 – 0.001 = 0.999

Option D

Solution. Assume that the required rectangle is given below

Now, you have to draw lines in such a manner that lines would cut the length and breadth of the rectangle to become the sides of square. As we know that all the sides of a square are equal. This means, we have to divide 48 cm and 84 cm in parts in such a way that they form sides of square. To divide it in such a way, we can use HCF

HCF of (48, 84) = 12 cm

So, we have to divide the length and the breadth of the rectangle is parts of 12 cm.

The following figure represents the process