Algebra Play Class 8 Extra Questions Maths Part 2 Chapter 6 - #NCSOLVE 📚

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Class 8 Algebra Play Extra Questions

Class 8 Maths Chapter 6 Algebra Play Extra Questions

Algebra Play Extra Questions Class 8

Question 1.
The sum of the digits of a 2-digit number is 12. If the digits are reversed, the new number is 36 more than the original number. Find the original number.
Solution:
Let the ones digit be x.
Then, the tens digit = 12 – x
Original number = 10(12 – x) + x
= 120 – 10x + x
= 120 – 9x
New number (reversed) = 10x + (12 – x) = 9x + 12
According to the question,
New number = Original number + 36
⇒ 9x + 12 = (120 – 9x) + 36
⇒ 9x + 12 = 156 – 9x
⇒ 18x = 144
⇒ x = 8
Tens digit = 12 – 8 = 4
Thus, the required number is 48.

Question 2.
Half of a herd of deer is grazing in the field, and three-fourths of the remaining are playing nearby. The remaining 9 are drinking water from the pond. Find the number of deer in the herd.
Solution:
Let the total number of deer in the herd be x.
Based on the information given, we can categorize the herd as follows:
Deer grazing in the field: Half of the herd is grazing, which is represented as \(\frac {x}{2}\).
The remaining deer in the herd = x – \(\frac {x}{2}\) = \(\frac {x}{2}\)
Deer playing nearby: Three-fourths of the remaining deer are playing.
Number of deer playing = \(\frac {3}{4}\) of \(\frac {x}{2}\)
= \(\frac{3}{4} \times \frac{x}{2}\)
= \(\frac {3x}{8}\)
Deer drinking water: The number of deer remaining to drink water is given as 9.
The total number of deer is the sum of those grazing, those playing, and those drinking water.
Therefore, \(\frac{x}{2}+\frac{3 x}{8}+9\) = x
Solving the equation, we have
\(\frac{4 x+3 x+72}{8}\) = x
⇒ \(\frac{7 x+72}{8}\) = x
⇒ 7x + 72 = 8x
⇒ x = 72
Hence, the total number of deer in the herd is 72.

Question 3.
Solve the following number pyramids:

Solution:
(i) 8 + p = 27
⇒ p = 27 – 8
⇒ p = 19
p + 15 = r
⇒ r = 19 + 15
⇒ r = 34
15 + q = 19
⇒ q = 19 – 15
⇒ q = 4
Now, s = 27 + r
= 27 + 34
= 61
t = r + 19
= 34 + 19
= 53
u = s + t
= 61 + 53
= 114

(ii) a + 11 = 16
⇒ a = 5
b + 13 = 30
⇒ b = 17
11 + b = c
⇒ c = 11 + 17 = 28
Now, d = 16 + c
= 16 + 28
= 44
e = c + 30
= 28 + 30
= 58
f = d + e
= 44 + 58
= 102

Question 4.
Decode the puzzle and find the final value.

Solution:

Question 5.
Create a Number Trick to make the final answer:
(a) 3 and (b) 7
Solution:

Now we have the required number 7.

Question 6.
Look at ‘Birthday date trick’.
Think of your birthday date → Multiply the month by 5 → add 4 in it → multiply it by 4 → add 20 more → again multiply by 5 → Add the date you thought of.
Find the date of (a) 1082, (b) 1405.
Solution:
The algebraic equation 100M + 180 + D.
(a) 1082
The constant number is 180.
Subtract it from the given number
1082 – 180 = 902
9 → Month
02 → Date
So the date is 02/9.

(b) 1405
Subtract the constant number from the given number.
1405 – 180 = 1225
Month → 12
Date → 25
So the date is 25/12.

Question 7.
Fill in the following pyramids.

Solution:

(b) Ist row (Bottom row):
a = 11 – 2 = 9
IInd row:
b = 9 + 7 = 16 and c = 7 + 8 = 15
IIIrd row:
d = 16 + 15 = 31
IVth row (top):
27 + 31 = 58

Question 8.
Recall the Virahahka-Fibonacci number sequence. Write the last bottom row with four continuous Virahanka-Fibonacci numbers and find the top of it. What is your observation of this pyramid?
Solution:
Virahanka-Fibonacci number sequence = 1, 2, 3, 5, 8, 13, 21,….

In this pyramid, we only get the number present in the Virahanka-Fibonacci numbers sequence.

Question 9.
Find the numbers of a 3 × 3 grid of a calendar if the sum of the numbers is 162.
Solution:
Let us say the middle number of the grid = x

The required grid will look like this after applying the pattern of the calendar.
Given: Sum of the grid is 162.
x – 8 + x – 7 + x – 6 + x – 1 + x + x + 1 + x + 6 + x + 7 + x + 8 = 9x
Now, 9x = 162
⇒ x = 18
After putting the value of x in the grid, we get the required numbers

Question 10.
Create algebraic equations to find the sums of different grid shapes within the 2 times table up to 100.

Solution:
(a) Let 26 = x. Then

Because the grid is in multiples of two, there are 10 columns and 5 rows.
So when we go one row up and down, we need to add or subtract 20, respectively.
And if we move right or left horizontally, we need to add or subtract 2 from the number, respectively.
For diagonally, we need to put both rules together to get the required number.
Therefore, the required equation is x + x – 2 + x + 2 + x – 20 = 4x – 20

(b) From above,
If we take 34 = y
y + y + 2 + y + 4 + y + 6 = 4y + 12

(c) Similarly, if we take 66 = p
p + p + 2 + p + 20 + p + 18 = 4p + 40

(d) Similarly, if we take 92 = w
w + w – 20 + w + 2 + w + 4 + w + 6 = 5w – 8

Question 11.
Find the values of and from the given grid.

Solution:

The post Algebra Play Class 8 Extra Questions Maths Part 2 Chapter 6 appeared first on Learn CBSE.

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