Monday, November 7, 2011

NTSE Maths Challenges

Following are the typical problems asked in NTSE, Oympiad Maths,  and other Mathematical based aptitude tests
(best viewed in Chrome or Firefox).

Q1: Ramesh walks from his home to Railway station at the rate of 5 kmph, he misses the train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance of the station from his home.

Answer: Let 'd' be the distance between Ramesh's house to Station.
We know that distance = speed x time.

If he walks at speed 5 kmph, let time taken be t1
⇒ t1 = d/5

Let the time t2 is taken when he walks with a speed of 6 kmph.

⇒  t2 = d/6

Since the time gap between two cases is 12 mins = 1/5 hrs
⇒ t1 - t2 = d/5 - d/6 = 1/5 hrs
⇒  6d - 5d = 6
⇒ d = 6km

Q2: Roma travels from place A to B and then from B to C.
From A to B Distance is 55 km which she covers with an average speed of 22 km per hour
It takes her 1 hour 30 minutes moving from B to C with an average speed 30 km per hour
Calculate her average speed over the whole of her journey from A to C.


Answer: We know that $ Time = Distance ÷ Speed

∴ Time taken from A to B = 55/22 = 2.5 hours  = 2 Hrs and 30 mins

Distance from A to B = Speed x Time = 20 x 1.5  = 45 km.  (note time taken from B to C is given as 1 Hrs 30 minutes i.e. 1.5 Hrs)

Average Speed = Total Distance  ÷ Total Time
∴  = (55+45) ÷ (1.5+2.5) = 100  ÷ 4 = 25 kmph



Q3: What is the value of


Answer:




Q4: What is the difference between the sum of the first 2008 even numbers and the sum of the
first 2008 odd numbers?
              (Singapore Maths Olympiad for Primary)
Answer:

Let us try to identify the pattern an try for small set.
Sum of first 5 even numbers
        N(Even) = 2 + 4 + 6+ 8 + 10
Similarly, sum of first 5 odd numbers will be:
        N(Odd)  =  1 + 3 + 5 + 7 + 9

If we subtract Odd list from Even list, i.e.
        N(Even) - N (Odd) = (2+4+8+10+12) - (1+3+5+7+9)
 ⇒  = (2-1) + (4-3) + (6-5) + (8-7) + (10-9)
⇒    = 1 + 1 + 1 + 1 + 1 = 5  (or 5 ones)

Similarly, the answer for the first 2008 numbers would be? 2008

Q5: The pages of a book are numbered consecutively, beginning with 1. The digit 7 is printed 25 times in numbering the pages. What is the largest number of pages the book can have?
Answer:
Count number of 7 appears how many times between 1 and 100 pages numbers = 20
(e.g. 7, 17, 27, 37, 47, 57, 67, 77, 87, 97, 70, 71, 72, 73, 74, 75, 76, 78, 79 = 20, 77 has two 7 digits)
 Count next five time 7 digit appears from page 101 onwards i.e. 107, 117, 127, 137, 147.

Thus the largest number of pages the book will have: 157-1 = 156

Q6: A student multiplies the month and the day in which he was born by 31 and 12 respectively. The sum of the two resulting products is 170. Find the month and the date in which he was born. (SMOP 2008)

Answer: Let d be the day and m represents month number.
It is obvious, both d and m are natural numbers and 1 ≤ d ≤ 31.
Similarly m lies in range 1 ≤ m ≤ 12

The sum of two digits is 31m + 12d = 170
Since 170 is an even number, 12d product is also an even number. 31m must be an even number.
∴ m should be an even number. Possible values of m be 2, 4, 6, 8, 10, 12.
By hit and trial we can find the m cannot be greater than 6, because (31x6 = 186 > 170).
Possible values of m are 2 and 4.
Let m = 2, then d = (170 - 62)/12 = 9
Let m = 4, then d = (170 - 124)/12 = 3.8334 (m can't be in decimal.)
∴ Possible answer is m = 2 and d = 9 i.e. Feb 9 ... (answer)

Q7: 291 digits have been used for printing the page numbers of a book. How many pages does the book have?

Answer:
Let us count how many digits appear page-wise.
From Page 1 to Page 9, no. of digits are: 9
From Page 10 to Page 99, number of digits are: 90 x 2 = 180

It implies digits remaining will be part of 3-digit pages,
 i.e. 291 - (180 + 9) = 102 digits
Number of Pages having 3-digits = 102 ÷ 3 = 34

∴ Total pages are: 9 + 90 + 34 = 133 ...(answer)

Q8: 9 locks and 13 keys are mixed up. If each lock can be opened by only one of the keys and no two locks can be opened by the same key, what is the smallest number of tries required to ensure that the correct key for every lock is found?

Answer:  Minimum tries to unlock the first lock is: 12  (keys left = 12)
Now to unlock  the second lock, no. of tries = 11 (keys left = 11)

Lock to open 1st 2nd 3rd 4th 5th  6th 7th 8th 9th
No. of tries 12 11 10 9 8 7 6 5 4

Total tries = 12 + 11 + 10 + 10 + 9 + 7 +6 + 5 + 4 = 72

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