NCERT Solutions For Class 8 Maths Chapter 6 Ex 6.1
ncert solutions for class 8 maths chapter 6, Exercise 6.1 involve complete answers for each question in the exercise 6.1. The solutions provide students a strategic methods to prepare for their exam. Class 8 Maths Chapter 8 Squares and Square Roots Exercise 6.1 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 8 Maths Chapter 8 Squares and Square Roots Exercise 6.1 prepared by our subject matter experts in very delicate, easy and creative way.
Question 1:
What will be the unit digit of the squares of the following
numbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555
Answer:
We know that if a number has its unit’s place digit as a, then its
square will end with the unit digit of the multiplication a × a.
(i) 81
Since the given number has its unit’s place digit as 1, its square
will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.
(ii) 272
Since the given number has its unit’s place digit as 2, its square
will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.
(iii) 799
Since the given number has its unit’s place digit as 9, its square
will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.
(iv) 3853
Since the given number has its unit’s place digit as 3, its square
will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.
(v) 1234
Since the given number has its unit’s place digit as 4, its square
will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.
(vi) 26387
Since the given number has its unit’s place digit as 7, its square
will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.
(vii) 52698
Since the given number has its unit’s place digit as 8, its square
will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.
(viii) 99880
Since the given number has its unit’s place digit as 0, its square
will have two zeroes at the end. Therefore, the unit digit of the square of the
given number is 0.
(xi) 12796
Since the given number has its unit’s place digit as 6, its square
will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.
(x) 55555
Since the given number has its unit’s place digit as 5, its square
will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.
Question 2:
The following numbers are obviously not perfect squares. Give
reason.
(i) 1057 (ii) 23453
(iii) 7928 (iv) 222222
(v) 64000 (vi) 89722
(vii) 222000 (viii) 505050
Answer:
The square of numbers may end with any one of the digits 0, 1, 5,
6, or 9. Also, a
perfect square has even number of zeroes at the end of it.
(i) 1057 has its unit place digit as 7. Therefore, it cannot be a
perfect square.
(ii) 23453 has its unit place digit as 3. Therefore, it cannot be
a perfect square.
(iii) 7928 has its unit place digit as 8. Therefore, it cannot be
a perfect square.
(iv) 222222 has its unit place digit as 2. Therefore, it cannot be
a perfect square.
(v) 64000 has three zeros at the end of it. However, since a
perfect square cannot end with odd number of zeroes, it is not a perfect
square.
(vi) 89722 has its unit place digit as 2. Therefore, it cannot be
a perfect square.
(vii) 222000 has three zeroes at the end of it. However, since a
perfect square cannot end with odd number of zeroes, it is not a perfect
square.
(viii) 505050 has one zero at the end of it. However, since a
perfect square cannot end with odd number of zeroes, it is not a perfect
square.
Question 3:
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
Answer:
(i) 243 = 3 × 3 × 3 × 3 × 3
Here, two 3s are left which are not in a triplet. To make 243 a
cube, one more 3 is required.
In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect
cube.
Hence, the smallest natural number by which 243 should be
multiplied to make it a perfect cube is 3.
(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, two 2s are left which are not in a triplet. To make 256 a
cube, one more 2 is required.
Then, we obtain
256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect
cube.
Hence, the smallest natural number by which 256 should be
multiplied to make it a perfect cube is 2.
(iii) 72 = 2 × 2 × 2 × 3 × 3
Here, two 3s are left which are not in a triplet. To make 72 a
cube, one more 3 is required.
Then, we obtain
72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube.
Hence, the smallest natural number by which 72 should be
multiplied to make it a perfect cube is 3.
(iv) 675 = 3 × 3 × 3 × 5 × 5
Here, two 5s are left which are not in a triplet. To make 675 a
cube, one more 5 is required.
Then, we obtain
675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube.
Hence, the smallest natural number by which 675 should be
multiplied to make it a perfect cube is 5.
(v) 100 = 2 × 2 × 5 × 5
Here, two 2s and two 5s are left which are not in a triplet. To
make 100 a cube, we require one more 2 and one more 5.
Then, we obtain
100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube
Hence, the smallest natural number by which 100 should be
multiplied to make it a perfect cube is 2 × 5 = 10.
Question 4:
Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1…2…1
100000012 = …
Answer:
In the given pattern, it can be observed that the squares of the
given numbers have
the same number of zeroes before and after the digit 2 as it was
in the original number. Therefore,
1000012 = 10000200001
100000012 = 100000020000001
Question 5:
Observe the following pattern and supply the missing number.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = …
…2 =
10203040504030201
Answer:
By following the given pattern, we obtain
10101012 = 1020304030201
1010101012 = 10203040504030201
Question 6:
Using the given pattern, find the missing numbers.
12 +
22 +
22 =
32
22 +
32 +
62 =
72
32 +
42 +
122 =
132
42 +
52 +
_2 =
212
52 +
_2 +
302 =
312
62 + 72 + _2 = __2
Answer:
From the given pattern, it can be observed that,
(i) The third number is the product of the first two numbers.
(ii) The fourth number can be obtained by adding 1 to the third
number.
Thus, the missing numbers in the pattern will be as follows.
42+ 52 + = 212
52 +
+ 302 =
312
62 + 72 + =
Question 7:
Without adding find the sum
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Answer:
We know that the sum of first n odd natural numbers is n2.
(i) Here, we have to find the sum of first five odd natural
numbers.
Therefore, 1 + 3 + 5 + 7 + 9 = (5)2= 25
(ii) Here, we have to find the sum of first ten odd natural
numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2=
100
(iii) Here, we have to find the sum of first twelve odd natural
numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 =
(12)2 =
144
Question 8:
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer:
We know that the sum of first n odd natural numbers is n2.
(i) 49 = (7)2
Therefore, 49 is the sum of first 7 odd natural numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11)2
Therefore, 121 is the sum of first 11 odd natural numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Question 9:
How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Answer:
We know that there will be 2n numbers in between the squares of
the numbers n
and (n + 1).
(i) Between 122 and 132, there will be 2 × 12 = 24 numbers
(ii) Between 252 and 262, there will be 2 × 25 = 50 numbers
(iii) Between 992 and 1002, there will be 2 × 99 = 198 numbers
