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Tuesday 16 October 2012

28th Aryabhatta Inter School Mathematics Competition 2011



Q1. The largest natural number, by which the product of three consecutive even numbers is always divisible, is _____.
Solution. First of all list down different set of three consecutive even numbers
(2, 4, 6) (4, 6, 8) (6, 8, 10) (8, 10, 12) etc
Product of numbers in first set = 2 × 4 × 6
Product of numbers in second set = 4 × 6 × 8
Product of numbers in third set = 6 × 8 × 10
Product of numbers in fourth set = 8 × 10 × 12

NOTE: The product of three consecutive natural numbers is always divisible by 6.
Now, the largest natural number which divides the product is 12.
As each set can written as
2 × 4 × 6 = 2 × {1 × 2 × 3}
4 × 6 × 8 = 2 × {2 × 3 × 4}
The product of numbers given in brackets will always be divisible by 6.
Therefore, when 2 is multiplied by 6, the product will be divided by 12.
(Hell of a question for a Class V student, no matter how much brilliant he/she is!)

Q2. To the product of smallest twin prime numbers, add the HCF of two co-primes. The difference of the face value and the place value of the digit in Tens place of the sum, is _________.

Solution. Twin prime numbers: A pair of prime numbers which differ by 2. The smallest twin prime numbers are (3, 5) etc.
CO-prime: Two integers a and b are said to be coprime if the only positive integer that evenly divides both of them is 1.

Product of smallest twin prime numbers = 3 × 5 = 15
HCF of two co-prime numbers = 1
After sum, the result = 15 + 1 = 16
Place value of digit in Tens place of sum = 10
Face value of digit in Tens place of sum = 1
Difference = 10 ~ 1 = 9

Q3. 0.2% of a number is 0.3, the number is ________.
Solution. 0.2% of a number = 0.3
(Observe carefully there is only single after decimal on both sides)
Þ   2% of a number = 3
Þ  100% of a number = 150 (multiply both sides by 50)
(Remember the rule that every quantity is 100% in itself) 


Q4. In a school, there are four boys to every three girls. If there are 304 boys, the number of girls in the school is ________.
Solution. 4 Boys for every 3 girls
 Let’s understand this as 


 





The above figure represents a group of 4 boys for every 3 girls.
Now, to count 304 boys how many group of such 4 boys will be required = 304 ÷ 4 = 76
 
In each of these 76 groups, there will be 3 girls.
So, total number of girls = 76 × 3 = 228 


Q5. A snack bar sells five items with an average price of Rs. 0.60. Which two items from the following items can be added to the menu without changing the average price?
Solution. Average price changes with number of items. If we do not want to change it, then we should try keeping the average price constant.
So, the average price of each item is Rs. 0.60. Therefore, for two items it will become 0.60 + 0.60 = Rs. 1.20.
Now, check for options only the price of Cake slice and Chewing Gum together makes Rs. 1.20 possible.

Q6. If 40 nails are used in making one shoe, the number of nails needed to make 20 pairs of shoes are _______
Solution. Total number of shoes in 20 pairs = 2 × 20 = 40 shoes
One shoe requires = 40 nails
40 shoes will require = 40 × 40 = 1600 nails

Q7. Students of a class took a maths test, 1/3 of the class got B grade, ¼ got B+ grade, 1/6 got C grade and 1/8 failed. The remaining students got A grade. The number of students who got an A grade is ________. (Number of students in the class is less than 30)
Solution. Total number of students in class = Students who got A grade + students who got B grade + students who got (B+) grade + Students who got C grade + student who failed
= Students who got A grade + 1/3 + ¼ + 1/6 + 1/8
= student who got A grade + 7/8
Students who got A grade will be 1/8.

Q8. Manya is 5 year 8 months old. Her sister Sanya is three-quarters her age. When Manya is 9 year 8 months old, the age of her sister would be _______.
Solution. Age of Manya = 5 yrs 8 months or 68 months
Age of her sister Sanya = ¾ × 68 = 51 months
Now, Manya will be 9 yrs 8 months or 116 months old.
Manya’s age increased by = 116 months – 68 months = 48 months
Therefore, Manya’s sister will also become older by 48 months
Sanya age = 51 + 48 = 109 months or 9 yrs 1 month.
 

Q9. Half of a pie is divided into 3 equal pieces. Each piece is _______of the whole pie. 





 









Q10. In a division sum, the divisor is 12 times the quotient and 5 times the remainder. If the remainder is 48, then the dividend is _________.
Solution. Divisor is 5 times the remainder.
Þ Divisor = 48 × 5 = 240
Divisor is 12 times the quotient.
Þ Quotient = 240 ÷ 12 = 20
Dividend = Divisor × Quotient + Remainder
= 240 × 20 + 48 = 4800 + 48 = 4848

Q11. Bananas cost Rs. 36 per dozen, and an apple costs Rs. 7 each. A plate of fruit salad consists of 3 bananas and 1/4th of an apple. Find the cost of 12 such plates.
Solution. Cost of each banana = 36 ÷ 12 = Rs. 3
Cost of each apple = Rs. 7
Cost of a plate consisting of 3 banana and 1/4th apple
= 3 × 3 + 7 × ¼ = 9 + 7/4 = 43/4
Cost of 12 such plates = 12 × 43/4 = 3 × 43 = 129

Q12. 4 boxes contain a total of 96 sweets. If each sweet costs 45p, the cost of each box of sweet is ________.
Solution. Number of sweets per box = 96 ÷ 4 = 24
So, each box contains 24 sweets.
Cost of each box = 24 × 0.45 = Rs. 10.80 


Q13.   









Solution. The pattern to find middle number is:
58 = 9 × 8 – 10 – 4
Similarly, 15 × 8 – 10 – 9 = 101
(In such type of questions generally arrange given digits using four basic mathematical operators i.e. addition, subtraction, multiplication and subtraction)


Q14. The least number which when divided by 5, 6, 7 and 8 leaves remainder 3, but when divided by 9 leaves no remainder is _______.
Solution. Least number divided by 5, 6, 7 and 8 = LCM of (5, 6, 7 and 8)
= 840
But they leave a remainder of 3. Thus, actual number which when divided by 5, 6, 7 and 8 leaving remainder 3 in each case will be 840 + 3 = 843.
843 when divided 9 leave a remainder of 6.
Series of such numbers will be (1 × 843), (2 × 843), (3 × 843) etc
Out of these numbers, 3 × 843 is exactly divisible by 9 and leaves a remainder of 3 with 5, 6, 7, and 8.
 







Solution. Size of each paper = ¾ m
Total length of the paper = 17(1/2) m or 35/2 m.
Number of pieces cut = Total length of the paper ÷ Size of each paper
= 35/2 ÷ ¾ = 70/3 or 23(1/3)
This shows that exactly 23 pieces can be cut and some paper would be left.
Total length paper used in 23 pieces = 23 × ¾ = 69/4 m
Paper left = 35/2 – 69/4 = ¼ m

Q16. Puneet cut 85 cm of Brass from one end of a piece of Brass rod 2.3 m long. What decimal fraction of 5 m is the length of the remaining piece?
Solution. Length of the remaining piece of rod = 2.3 m – 0.85 m = 1.45 m
Required decimal fraction = 1.45/ 5 = 0.29

Q17. Jay bought 3 packets of pins, 2 packets of tattoos, and 1 packet of glitter for Rs. 19.10. Each packet of pin costs Rs. 1.90 and the packet of glitter costs twice as much as a packet of tattoos. The packet of tattoos costs him _________.
Solution. Total cost = 3 packets of pins + 2 packets of tattoos + 1 packet of glitter
19.10 = 3 × 1.90 + 2 packets of tattoos + 2 packets of tattoos
(Since, a packet of glitter costs twice the packet of tattoos)
Cost of 4 packets of tattoos = 19.10 – 5.70 = Rs. 13.40
Cost of a packet of tattoo = 13.40 ÷ 4 = Rs. 3.35

Q18. Using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 each exactly once, write any three 3-digit numbers so that the second number is twice the first and third number is thrice the first number.
1) __________         2) __________         3) ___________
Solution. Highly time consuming question. Even not at par with the knowledge level of a class V student.

Q19. In a party, 52% of the guests liked the decorations and 40% liked the food, 27% liked both. The percentage of guests who did not like the food as well as the decoration is __________.
Solution. Answer to this question will be (52 + 40 – 27 = 35%). But to teach this concept to a class V student is a tedious task. So, I am leaving the solution part of the question.

Q20. A super fast running at the speed of 94 km/hr develops a snag and stops. It had covered (1/17)th of the total distance in the 4 hors it had been running. The distance yet to be covered is ___________.
Solution. In 4 hours the train has traveled some distance. Let’s calculate it first.
In 1 hour, train covers = 94 km
In 4 hours, train covers = 4 × 94 = 376 km
Now, this 376 km is (1/17)th part of the total distance to be covered.
Therefore, the total distance will be = 376 × 17 =6392 km 








CDI = (500 – 100) + 1 = 401
XLIV = (50 – 10) + (5 – 1) = 44
XVIII CMXII = 10 + 8 + (1000 – 100) + 10 + 2 = 930
CXCVII = 100 + (100 – 10) + 7 = 197
Þ CDI + XLIV × XVIII CMXII ÷ CXCVII
= 401 + 44 × 930 ÷ 197

Q22. The average weight of 6 students increase by 3 kg when one of the students whose weight is 58 kg is replaced by a new student. The weight of the new student is
Solution. As Class V students are not exposed to ‘Average’, answering this question to explain them will be bit difficult.

Q23. Mohit had 390 marbles. He shared it with 8 boys and 13 girls so that each boy got 2 marbles more than each girl. The number of marbles left with him is _______.
Solution. Each boy got 2 marbles more than each girl. As there are 8 boys, then in all 8 × 2 = 16 extra marbles are distributed between boys and girls beside the number of equally divided marbles.

Number of marbles left to be divided equally between boys and girls = 390 – 16 = 374

Now, these 374 marbles are divided between boys and girls in such a way that they get equal number of marbles.

Total number of person = 8 boys + 13 girls = 21
Number of marbles each of them got = 374 ÷ 21 = 17 quotient and 17 remainder.

This implies that each boy and girl got 17 marbles.
Total number of marbles distributed = 17 × 21 = 357

Number of marbles left to be distributed = 374 – 357 = 17


CDI = (500 – 100) + 1 = 401
XLIV = (50 – 10) + (5 – 1) = 44
XVIII CMXII = 10 + 8 + (1000 – 100) + 10 + 2 = 930
CXCVII = 100 + (100 – 10) + 7 = 197
Þ CDI + XLIV × XVIII CMXII ÷ CXCVII
= 401 + 44 × 930 ÷ 197

Q22. The average weight of 6 students increase by 3 kg when one of the students whose weight is 58 kg is replaced by a new student. The weight of the new student is
Solution. As Class V students are not exposed to ‘Average’, answering this question to explain them will be bit difficult.

Q23. Mohit had 390 marbles. He shared it with 8 boys and 13 girls so that each boy got 2 marbles more than each girl. The number of marbles left with him is _______.
Solution. Each boy got 2 marbles more than each girl. As there are 8 boys, then in all 8 × 2 = 16 extra marbles are distributed between boys and girls beside the number of equally divided marbles.

Number of marbles left to be divided equally between boys and girls = 390 – 16 = 374

Now, these 374 marbles are divided between boys and girls in such a way that they get equal number of marbles.

Total number of person = 8 boys + 13 girls = 21
Number of marbles each of them got = 374 ÷ 21 = 17 quotient and 17 remainder.

This implies that each boy and girl got 17 marbles.
Total number of marbles distributed = 17 × 21 = 357

Number of marbles left to be distributed = 374 – 357 = 17




















Solution. i) To find the bus which takes least time, we have to find the time taken by bus after departing from Noodle Pond till it arrives at Lake Cola.

Time taken by Bus 1 = 02:35 – 20:10 = 6:25 hours
Time taken by Bus 2 = 16:35 – 11:25 = 5:10 hours
Time taken by Bus 3 = 13:10 – 08:10 = 5:00 hours
Thus, we can conclude that Bus 3 takes the least time to reach Lake Cola from Noodle Pond.

ii) In the same manner, we will find which bus takes least time (to be fast) to arrive Salad Town after departing from Burger Junction.

Time taken by Bus 1 = 01:05 – 21:55 = 03:10 hours
Time taken by Bus 2 = 15:30 – 12:20 = 02:50 hours
Time taken by Bus 3 = 11:15 – 09:05 = 02:10 hours

Thus, we can conclude that Bus 3 takes least time. Hence, it is fastest.

iii) As the person has missed Bus 2 at Chocopur, he has to wait till the next Bus arrives. It implies that we have found the time difference between the departures from Chocopur to arrival at Salad Town for different buses.
Time difference for Bus 1 = 01:05 – 23:45 = 01:20 hours
Time difference for Bus 2 = 15:30 – 14:05 = 01:25 hours
Time difference for Bus 3 = 11:15 – 10:55 = 20 minutes

iv) TO find which is the fastest Bus from Burger Junction to Chocopur, we have to find which Bus takes the least time.

Time difference from Bus 1 = 23:20 – 21:55 = 01:25 hours
Time difference from Bus 2 = 13:50 – 12:20 = 01:30 hours
Time difference from Bus 3 = 10:40 – 09:05 = 01:35 hours

Thus, it is clear that Bus 1 takes the least time and hence, the fastest.
 

Thursday 4 October 2012

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
1st June 1999 + 2 years = 1st June 2001
Now, 9 months are left to add, so adding 9 months
1st June 2001 + 9 months = 1st 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 
Option B













Solution. Refer to the following figure
 
Distance from starting point = 2 + 6 = 8 km
Option B





















Solution. Eaten part of first pizza = 8/9

Remaining part of first pizza = 1/9
 
Eaten part of second pizza = 2/3
 
Remaining part of second pizza = 1/3

Eaten part of third pizza = 7/9
 
Remaining part of third pizza = 2/9
 
Remaining part of all the pizzas left = 1/9 + 1/3 + 2/9 = 6/9 = 2/3
Option A


  



















Solution. There are more than 24 triangles in the given figure.