Edudel, Directorate of Education Govt. of NCT of Delhi Class 8 Mental Maths Question Bank Chapter 6 Square and Square Roots Questions Solution. In this chapter, there is total 50 math problems. We have given each questions solution step by step.
Edudel Class 8 Mental Maths Chapter 6 Square and Square Roots:
1.) How many perfect squares lie between 100 and 1000?
ANSWER:
There are 21 perfect squares lie between 100 and 1000.
2.) Write the smallest four digit perfect square number.
ANSWER:
1024 is the smallest four digit perfect square number.
3.) Find the value of x, if (3)2 + (4)2 + (12)2 = x2
ANSWER:
We know,
() 2 = 9
(4) 2 = 16
(12) 2 = 144
= 9 + 16 + 144
x2= 169
x = √169
x = 13
4.) Simplify: √12×8+12×4
ANSWER:
We have to find √12×8+12×4
√96+48
√144
√12×8+12×4= 12
5.) Simplify:√(0.01)+√0.0064
ANSWER:
We have to find√(0.01)+√0.0064
We know,
√0.0064
= 0.08
√0.01+0.08
√0.09
√ (0.01)+√0.0064= 0.3
6.) Simplify: √320+√9+√49
ANSWER:
We solve given square root from innermost square root.
We know,
√49= 7
√9+7
= √16 = 4
√320+4
√324= 18
√320+√9+√49 = 18.
7.) Simplify: √1000 X √4410/4.41
ANSWER:
We have to find √1000 X √4410/4.41
We know,
√100= 10
√= 210
√1000 X √4410/4.41
√1000 x √1000
√1000 X √4410/4.41 = 1000
8.) Find the value of 5022 – 4992
ANSWER:
We have to find 5022 – 4992
Here we put, a = 502 and b = 499
We form identity = a2 – b2 = (a + b) (a – b)
5022 – 4992 = (502 + 499) (502 – 499)
5022 – 4992 = 3003
9.) Simplify: 52 – 42 + 32 – 22 + 12
ANSWER:
We have to find 52 – 42 + 32 – 22 + 12
52 = 25, 42 = 16 32= 9, 22 = 4
25 – 16 + 9 – 4 + 1
52 – 42 + 32 – 22 + 12= 15
10.) Simplify: (√324 / 6)2
ANSWER:
We have to find (√324 / 6)2
We know,
√324= 18
(18 / 6)2
= 32
(√324 / 6)2= 9
11.) Simplify: √100 + √0.09
ANSWER:
We have to find √100 + √0.09
We know,
√100= 10
√0.09= 0.3
= 10 + 0.3
√100 + √0.09= 10.3
12.) Simplify: √5/24.2
ANSWER:
We have to find √5/24.2
√5/24.2=√5/2×12.2
By solving we get,
√5/24.2= 5 / 11
13.) How many 2’s are there in the prime factorization of 4000?
ANSWER:
There are 5 number of 2’s are there in the prime factorization of 4000.
14.) Find the least number that should be added to 221 to get a perfect square.
ANSWER:
Let, number added is x.
221 + X = perfect square
We know,
152 = 225
221 + X = 225
X = 225 – 221
X = 4
The number added is 4.
15.) If area of a square field is 1764 sq. m. find its perimeter.
ANSWER:
We know,
Area of square = (Side) 2
Given, area of a square field is 1764 sq. m
1764 sq. m = (Side) 2
(Side) = √1764
(Side) = 42 m.
We have to find Perimeter of square field.
Perimeter of square field = 4 x side
Perimeter of square field = 4 x 42
Perimeter of square field = 168 m.
16.) How many numbers lie between 102 and 152?
ANSWER:
We know,
102= 100
152= 225
Numbers lie between 102 and 152= (225 – 100) – 1
Numbers lie between 102 and 152= 124
17.) Find the value of (12 + 7 + 3 +2 + 1 + 0)2?
ANSWER:
We have to find (12 + 7 + 3 +2 + 1 + 0) 2
= (25)
(12 + 7 + 3 +2 + 1 + 0) 2 = 625
18.) Simplify: (√1024 − √961)
ANSWER:
We have to find (√1024 − √961)
We know,
√1024 = 32
Also,
√961= 31
(√1024 − √961) = 32 – 31
(√1024 − √961) = 1
19.) How many digits are there in the square root of 15625?
ANSWER:
We know,
√15625= 125
There are 3 digits in the square root of 15625.
20.) How many digits will be there in the square of 999?
ANSWER:
We have to find, (999)2
There are 6 digits in the square of 999.
21.) Find the value of√31+√21+√15+√1
ANSWER:
We solve given square root from innermost square root.
We know,
√16= 4
√21+4= √25= 5
√31
√31+5=√36 = 6
√31+√21+√15+√1= 6
22.) Find the value of√54-√21+√18-√4
ANSWER:
We solve given square root from innermost square root.
We know,
√4= 2
√18-2= √16 = 4
√25= 5
√54-5= √49 = 7
√54-√21+√18-√4= 7
23.) Find the value of 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
ANSWER:
We have to find 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
We are adding all numbers we get,
3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 143
24.) Find the value of x if, 62 + 72 + 422 = x2
ANSWER:
We have to find 62 + 72 + 422 = x2
We know,
62= 36
72= 49
422= 1764
36 + 49 + 1764 = x2
1749 = x2
x= 43
25.) Find the value of 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20
ANSWER:
We have to find 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20
Given number are even number.
We know,
Sum of first n natural even numbers= (n + 1)
Here, there are 10 even numbers. n = 10
Sum of first n natural even numbers = 10 x 11
Sum of first n natural even numbers = 110
26.) Find the value of x if, 42 + 52 + 202 = x2
ANSWER:
We have to find 42 + 52 + 202 = x2
We know,
42 = 16
52 = 25
202 = 400
16 + 25 + 400 = 441
x2= 441
x= 21
27.) Find the value of (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5)
ANSWER:
We have to find (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5)
2 + 6 + 12 + 20
= 20 + 20
(1 x 2) + (2 x 3) + (3 x 4) + (4 x 5) = 40
- Find the value of 312 – 292
ANSWER:
We have to find 312 – 292
Here we put, a = 31 and b = 29
We form identity = a2 – b2 = (a + b) (a – b)
312 – 292 = (31 + 29) (31 – 29)
312 – 292 = 120
29.) Find the value of √99 × √44
ANSWER:
We have to find √99 × √44
√99 = √11×9 = 3√11
√44=√11×4 = 2√11
3√11 x 2√11 = 6 x 11
√99 x √44 = 66
30.) What is the sum of first 14 odd natural numbers?
ANSWER:
We have to find the sum of first 14 odd natural numbers
We know,
Sum of first n natural odd numbers=n2
n = 14
The sum of first 14 odd natural numbers = 14 2
The sum of first 14 odd natural numbers = 196
31.) Find X and Y if (X) Y =441such that X>Y, where X and Y are positive integers.
ANSWER:
We have to find (X) Y = 441
We know,
√441= 21
441 = 212
(X) Y = 441= 212
By comparing
X = 21 and Y = 2
32.) If length of a rectangular park is 80 m and its breadth is 60 m. find the sum of length of its diagonals.
ANSWER:
Given that,
Length of a rectangular park is 80 m and its breadth is 60 m
We have to find the sum of length of its diagonals.
By using right angled Triangle theorem,
Diagonal of rectangle2 = 802 + 602
Diagonal of rectangle2 = 6400 + 3600
Diagonal of rectangle = 100 cm
The sum of length of its diagonals = 100 + 100 = 200 cm.
33.) Evaluate: (2 +√3)2 + (2 -√3)2
ANSWER:
We have to find (2 +√3)2 + (2 -√3) 2
(2 +√3)2= 4 + 3 = 7
(2 -√3) 2= 4 + 3 = 7
(2 +√3)2 + (2 -√3) 2= 7 + 7 = 14
34.) Find the least number that should be subtracted from 537 to get a perfect square.
ANSWER:
Let the number is x.
We form equation.
537 – x = perfect square.
We know, 232 = 529 which is perfect square.
537 – x = 529
After solving,
X = 8
35.) If the area of a square is 20.25 sq. m, find its perimeter.
ANSWER:
We know,
Area of square = (Side) 2
Given, area of a square is 20.25 sq. m
20.25 sq. m = (Side) 2
(Side) = √20.25
(Side) = 4.5 m.
We have to find Perimeter of square
Perimeter of square = 4 x side
Perimeter of square = 4 x 4.5
Perimeter of square = 18 m.
36.) Find the least number by which 288 must be multiplied so that it becomes a perfect square.
ANSWER:
Let the number is x.
We know, 576 is multiple of 288.
288 x X = 576
X = 576 / 288
X = 2
2 is the least number by which 288 must be multiplied so that it becomes a perfect square.
37.) Find the least number by which 147 must be divided so that it becomes a perfect square.
ANSWER:
Let the number is x.
We know, 441 is multiple of 147.
147 x X = 441
X = 441 / 147
X = 3
3 is the least number by which 147 must be multiplied so that it becomes a perfect square.
38.) The area of a square field is 30 (1/4) sq. m. Find the length of the side of the square.
ANSWER:
We know,
Area of square = (Side) 2
Given, area of a square field is30 (1/4)= 30.25 sq. m
30.25 sq. m = (Side) 2
(Side) = √30.25
(Side) = 5.5 m.
The length of the side of the square is 5.5 m.
39.) Find the length of a side of a square playground whose area is equal to the area of rectangular field of dimension 72m and 50m.
ANSWER:
Given that,
Square playground area = the area of rectangular field of dimension 72m and 50m.
We know,
Area of rectangular field = Length x Breadth
Area of rectangular field = 72m x 50m.
Area of rectangular field = 3600 m2
We know,
Area of square = (Side) 2
3600 m2= (Side) 2
(Side) = √3600
(Side) = 60 m.
The length of the side of the square field is 60 m.
40.) Find the value of√625 – √441 / √625 + √441
ANSWER:
We have to find √625 – √441 / √625 + √441
We know,
√625= 25
√441= 21
25 – 21 / 25 + 21
= 4 / 46
√625 – √441 / √625 + √441= 2 / 23
41.) Find: 9+11+13+15+17+19+21+23
ANSWER:
We have to find 9+11+13+15+17+19+21+23
After adding we get,
9+11+13+15+17+19+21+23 = 128
42.) How many perfect squares lie between 0 and 550?
ANSWER:
There are 23 perfect squares lie between 0 and 550
43.) Find the value of√√1296/2401
ANSWER:
We have to find √√1296/2401
We know,
√1296/2401= 36 / 49
Now again
√36/49= 6 / 7
√√1296/2401= 6 / 7
44.) Find the value of (44 + 3)2 – (44-3)2
ANSWER:
We have to find (44 + 3)2 – (44-3)2
(44 + 3)2= 472
(44 – 3)2= 412
472 – 412
Here we put, a = 47 and b = 41
We form identity = a2 – b2 = (a + b) (a – b)
472 – 412 = (47 + 41) (47 – 41)
472 – 412 = 528
45.) Find the value of (√441 – √196 +√121 -√64)
ANSWER:
We have to find (√441 – √196 +√121 -√64)
We know,
√441= 21
√196= 14
√121= 11
√64= 8
(√441 – √196 +√121 -√64)= 21 – 14 + 11 – 8
(√441 – √196 +√121 -√64)= 10
46.) Find the missing number: 175 x ____ = 352
ANSWER:
We have to find 175 x ____ = 352
We know,
352= 1225
The missing number = 1225 / 175
The missing number = 7.
47.) Find the Pythagorean triplet whose smallest number is 12.
ANSWER:
Pythagorean triplet whose smallest number is 12 is 12, 16, 20
48.) Find x if, 5x = 492 – 442
ANSWER:
Here we put, a = 49 and b = 44
We form identity = a2 – b2 = (a + b) (a – b)
5x = 492 – 442 = (49 + 44) (49 – 44)
5x = 492 – 442 = 93 x 5
x =93 x 5 / 5
x = 93
49.) Findx if, x = √0.01+0.03+0.07+0.11+0.14
ANSWER:
We have to find √0.01+0.03+0.07+0.11+0.14
We add all.
√0.01+0.03+0.07+0.11+0.14 = √0.36
√0.36= 0.6
√0.01+0.03+0.07+0.11+0.14 = 0.6
50.) Find xif, x = (9 + 4√5) (9 – 4√5)
ANSWER:
We have to find (9 + 4√5) (9 – 4√5)
Here we put, a = 9 and b = 4√5
We form identity = a2 – b2 = (a + b) (a – b)
92 –4√52 = (9 + 4√5) (9 – 4√5)
81 – (16 x 5)
(9 + 4√5) (9 – 4√5)= 81 – 80
(9 + 4√5) (9 – 4√5)= 1