Number Play Class 7 Long Question Answer

Number Play Class 7 Long Question Answer

Number Play Class 7 Long Question Answer

Question 1.

Construct a magic square using nine non-consecutive numbers with magic sum 18.

Solution:

Note : Answer is not unique.

Question 2.

Solve the puzzle :

Solution:

Product can be 111, 222, 333, 444, 555, etc.

BBB must be divisible by 6.

Ill is not divisible by 6. So, 111 is rejected.

Similarly, we reject 222 and 333.

Now, 444 ÷ 6 = 74.

So, we can take BBB as 444. That is B. = 4 with B = 4, we have :

This gives A = 7.

So, we have :

Question 3.

Two consecutive numbers in the Fibonacci sequence are 34 and 55. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?

Solution:

To find the next two numbers in the Fibonacci sequence after 34 and 55, we follow the rule that each number is the sum of the two preceding numbers.

(a) Next number after 55 : 34 + 55 = 89

(b) Next number after 89 : 55 + 89 = 144

So, the next two numbers are 89 and 144. Now, let’s find the previous two numbers in the sequence before 34 and 55.

(a) Previous number before 34 : 55 - 34 = 21

(b) Previous number before 21 : 34 - 21 = 13

Thus, the previous two numbers are 21 and 13.

Question 4.

Two consecutive numbers in the Fibonacci sequence are 233 and 377. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?

Solution:

To find the next two numbers in the Fibonacci sequence after 233 and 377, we again apply the same rule.

(a) Next number after 377 : 233 + 377 = 610

(b) Next number after 610 : 377 + 610 = 987

So, the next two numbers are 610 and 987. Now, let’s find the previous two numbers in the sequence before 233 and 377.

(a) Previous number before 233 :

377 - 233 = 144 (Since 144 + 233 = 377)

(b) Previous number before 144 :

233 - 144 = 89 (Since 89 + 144 = 233)

Thus, the previous two numbers are 144 and 89.

Question 5.

Solve the cryptarithm :

Solution:

(a) The two-digit number XY can be expressed as 10X + Y.

1. The two-digit number YX can be expressed as 10Y + X.
2. Number (ZZO) can be expressed as (110Z).

(b) Setting up the equation :

The equation from the cryptarithm can be set up as follows :

(10X + Y) + (10Y + X) = 110Z

Simplify gives : 11X + 11Y = 110Z

Rearranging leads to : X + Y = 10Z

(c) Finding possible values :

1. Since X and Y are digits (0-9), the maximum value for X + Y is 18 (when both are 9).
2. This means 10Z can only equal 10, but since Z must be a digit, Z can only be 1.

(d) Testing values :

1. If Z = l, then: X + Y = 10
2. Possible pairs (X, Y) that sum to 10 are : (1, 9), (9, 1)
3. Each pair can be checked :
4. For (X = 1, Y = 9) : 91 + 91 = 110 (Z = 1)