**Class 9 Maths - Important Identities**

(a+b)² = a² + 2ab+ b²

(a–b)² = a² –2ab + b²

a² – b² = (a+b)(a–b)

Showing posts with label **Olympiad-maths**. Show all posts

Showing posts with label **Olympiad-maths**. Show all posts

Answer:

This magic square looks similar to any other square.

Sum of any rows is 139

(a) 5

(b) -5

(c) 1/5

(d) -1/5

1. A number is called a **perfect square** if it is expressed as the square of a number.

2. E.g. 1, 4, 9, 16, 25, ... are called perfect squares (1x1 = 1, 2x2 = 4, 3 x 3 = 9...)

3. In square numbers, the digits at the unit’s place are always 0, 1, 4, 5, 6 or 9.

4. The numbers having 2, 3, 7 or 8 at their units' place are not perfect square numbers.

a) 622

b) 393

c) 5778

d) 625

5. If a number ends with the odd number of zeros then it is not a perfect square.

a) 81000

b) 8100

c) 900

d) 6250000

6. The square of an even number is an even number while the square of an odd number is an odd number.

7. If n is a positive whole number then

or 2n numbers in between the squares of the numbers n and (n + 1)

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

(best viewed in Chrome or Firefox).

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

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

See Maths Challenges (Questions)...

Answer1: When two distinct lines cut each other, they always meet at single point. These two lines can cut the circle maximum at four different points. Therefore, maximum number of intersection points is 5 i.e (d). Following figure explains this.

Answer 2:

Total possible outcomes are 16 i.e.

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT

The catch here is

$\Rightarrow$ the probability is $\frac{11}{16}$

Answer 3:

Here product is a prime number. Since we know prime number is divisible by self and number 1 only. So the only choice for m is 1.

(a) 2 (b) 3 (c) 4 (d) 5

(a) 5/6 (b) 3/8 (c) 11/16 (d) 5/8

smaller than n, find the value of m.

key. If the 9 keys have been mixed up, find the maximum number of attempts Jane must make

before she can open all the boxes. (Singapore Asia-Pacific Mathematical Olympiad Competition)

9797797779777797777797777779......?

Follow the link to see the answers Maths Challenges (Answers)

Subscribe to:
Posts (Atom)