8th Grade > Mathematics
SQUARES AND SQUARE ROOTS MCQs
:
A
Numbers ending with 2, 3, 7, or 8 at unit's place are never perfect squares.
Estimation without actually finding roots:
484 can be the perfect square of a number ending with 2 or 8.
576 can be the perfect square of a number ending with 4 or 6.
529 can be the perfect square of a number ending with 3 or 7.
:
D
Between the squares of any two consecutive numbers n and (n+1), lie 2n non-perfect squares.
Therefore between squares of 75 and 76 (n = 75 and n + 1 = 76), lie (2 x 75) = 150 non-perfect squares.
:
C
Since the square sheet needs to be divided equally into 4, 5, and 6 equal parts respectively, its area must be a perfect square divisible by 4, 5, and 6; or the area of the square sheet should be a multiple of 4, 5, and 6.
L. C. M. of 4, 5, and 6 = 60. But 60 is not a perfect square.
60=2×2×3×5=22×3×5.
If we multiply the LCM by 3 and 5 both, it will become a perfect square. Therefore the minimum area of the square sheet =60×3×5=900 sq. units.
Now 900 is a minimum perfect square, which is divisible by 4, 5, and 6. Therefore the side of the square should be the square root of the area of the square (Area of a square = side2 sq. units).
Hence, the minimum length of the side of the square = √(900)=√(22×32×52) = 30 units.
:
A
182+242=324+576=900=302
Since the squares of 18 and 24 add up to the square of 30, the numbers 18, 24 and 30 form a Pythagorean triplet, and hence, can be used to represent the three sides of a right- angled triangle.
:
A
Given number: 3364
By grouping the numbers from the left, we get ¯¯¯¯¯¯33 ¯¯¯¯¯¯64.
Now, perform the long division.
5 85¯¯¯¯¯¯33 ¯¯¯¯¯¯6425 ↓108 864 864 0
Thus, the square root of 3364 is 58.
:
B
Number of trees = 1024. Since number of rows = number of columns, this means that the number of trees is a perfect square. Therefore number of rows/columns = square root of 1024 (number of trees).
√1024= √2×2×2×2×2×2×2×2×2×2=32
Therefore, the initial number of rows/columns = 32. After cutting 1 row and one column, number of rows/columns = 32-1 = 31.
The remaining number of trees = 312.
Therefore, number of trees that were cut = 322−312 = 1024 - 961 = 63.
:
Sum of n consecutive odd integers starting from 1 is equal to n2, where 'n' is the number of terms involved in the series.
Therefore,
i. 1+3+5+7+9+11+13+15+17+19+21 = sum of first 11 odd integers = 112 = 121.
:
A
1024=2×2×2×2×2×2×2×2×2×2
=22×22×22×22×22⇒ √(1024)=√(22×22×22×22×22)=2×2×2×2×2=32
2025=3×3×3×3×5×5
=32×32×52⇒ √(2025)=√(32×32×52)=3×3×5=45
:
The square root of a number can be found by repeatedly subtracting odd integers from the number starting from 1.
So in this case,
256- 1-3-5-7-9-11-13-15-17-19-21-23-25-27-29-31 = 0.
(16 steps, hence square root of 256 is 16.)
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C and D
Given: 'p' is the square of 'q'.
⇒ 'q' is the square root of 'p'
p>400
The square root of 400 is 20.
So if "p">400, its square root 'q' can be any number greater than 20.