what is the answer for this question
1.Find the least number which must be added to 1750 so as to get a perfect
square. Also, find the square root of the obtained number.
2. Find the smallest whole number by which 2800 should be divided so as to
get a perfect square.
Answers
Step-by-step explanation:
1. Therefore 41^2 < 1750. Next perfect sqiare number 42^2 =1764.
2. Hence, 2800 needs to be divided by 7 to become a perfect square. Thus, the required smallest whole number by which it should be divided so as to get a perfect square number is 7 and the square root is √400= 20
Step-by-step explanation:
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
Sol: (i) ∵ 1 × 1 = I
∴ The unit’s digit of (81)2 will be 1.
(ii) 2 × 2 = 4
The unit’s digits of (272)2 will be 4.
(iii) Since, 9 × 9 = 81
The unit’s digit of (799)2 will be 1.
(iv) Since, 3 × 3 = 9
The unit’s digit of (3853)2 will be 9.
(v) Since, 4 × 4 = 16
The unit’s digit of (1234)2 will be 6.
(vi) Since 7 × 7 = 49
The unit’s digit of (26387)2 will be 9.
(v) Since, 8 × 8 = 64
The unit’s digit of (52698)2 will be 4.
(vi) Since 0 × 0 = 0
The unit’s digit of (99880)2 will be O.
(vii) Since 6 × 6 = 36
The unit’s digit of (12796)2 will be 6.
(x) Since, 5 × 5 = 25
The unit’s digit of (55555)2 will be 5.
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
Sol: (i) 1057
Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)
∴1057 is not a perfect square.
(ii) 23453
Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9).
∴23453 is not a perfect square.
(iii) 7928
Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).
∴7928 is not a perfect square.
(iv) 222222
Since, the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
∴222222 is not a perfect square.
(v) 64000
Since, the number of zeros is odd.
∴64000 is not a perfect square.
(vi) 89722
Since, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
∴89722 is not a perfect square.
(viii) 222000
Since, the number of zeros is odd.
∴222000 is not a perfect square.
(viii) 505050
The unit’s digit is odd zero.
∴505050 can not be a perfect square.
3. The squares of which of the following would be odd numbers?
(i) 431(ii) 2826(iii) 7779(iv) 82004
Sol: Since the square of an odd natural number is odd and that of an even number is an even number.
(i) The square of 431 is an odd number.
[∵ 431 is an odd number.]
(ii) The square of 2826 is an even number.
[∵ 2826 is an even number.]
(iii) The square of 7779 is an odd number.
[∵ 7779 is an odd number.]
(v) The square of 82004 is an even number.
[∵ 82004 is an even number.]
4. Observe the following pattern and find the missing digits.
112 =
121
1012 =
10201
10012 =
1002001
1000012 =
1.............2 ............. 1
100000012 =
............
Sol: Observing the above pattern, we have
(i) (100001)2 - 10000200001
(ii) (10000001)2 = 100000020000001
5. Observe the ,following pattern and supply the missing number.
112 = 121
1012 = 10201
101012 = 102030201
10101012 =
..............2 = 10203040504030201
Sol: Observing the above, we have
(i) (1010101)2 = 1020304030201
(ii) 10203040504030201 = (101010101)2
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
Note: To find pattern:
Third number is related to first and second number. How?
Fourth number is related to third number. How?
Sol: The missing numbers are
(i) 42 + 52 + 202 = 212
(ii) 52 + 22 + 302 = 312
(iii) 62 + 72 + 422 = 432
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
Sol: (i) The sum of first 5 odd = 52
= 25
(ii) The sum of first 10 odd numbers = 102
= 100
(iii) The sum of first 12 odd numbers = 122
= 144
8. (i) Lxpress 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Sol: (i) 49 = 72 = Sum of first 7 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = 112 = Sum of first 11 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
(i) 12 and 13(ii) 25 and 26(iii) 99 and 100
Sol: Since between n2 and (n + 1)2, there are 2n non-square numbers.
∴ (i) Between 122 and 132, there are 2 × 12, i.e. 24 numbers
(ii) Between 252 and 262, there are 2 × 25, i.e. 50 numbers
(iii) Between 992 and 1002, there are 2 × 99, i.e. 198 numbers