Question

SHARSHAVAS
Class: 8th
MATHEMATICS HOLIDAY ASSIGNMENT
1. Fill in the blanks:
(i) The ones digit in the square of 77 is
(ii) The number of non-square numbers between 242 and 252 is
(iii) If a number ends with 5, its square ends with
(iv) A square number will not end with numbers
(v) The number of perfect square numbers between 300 and 500 is
(vi) The ones digits in the cube of 73 is
(vii) The maximum number of digits in the cube of a two digit number is
(viii) The cube root of 540x50 is
(ix) The cube root of 0.000004913 is
(x) The smallest number to be added to 3333 to make it a perfect cube is​

Answer 1

Answer:

(i)9

(ii)10

(iii)5

(iv)3,7,8

(v)5

(vi)9

(vii)6

(viii)30

(ix)0.017

(x)42

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Answer 2

i) 9

ii) Nine

iii) 5

iv) 2, 3 & 8

v) five

vi) 7

vii) six digits

viii) 30

ix) 0.17

x) 42 is the smallest no. that when added to 3333 makes it a perfect cube.

Step-by-step explanation:

i)

Concerned about ones digits we operate only on ones digits with the multiplication:

$...7\times ...7=.....49$

we get 9 at ones place.

ii)

Nine, because we have the square nearest to the smallest given no. as 15 which results 225 then the next no. 16 has its square as 256 so we count all the numbers lying between the given numbers.

iii)

Squaring a number ending with 5 also involves the concept of working on ones digits.

$....5\times ...5=.....25$

iv)

2, 3 and 8 Since these digits are never there in the unit place when all the digits are checked after squaring.

v)

Five, ($18^2,19^2,20^2,21^2,22^2$)

vi)

Operating with ones digits:

$...3\times ....3\times ....3=.....27$

we get 7 at ones place.

vii)

We check it after cubing the largest two digit number:

6 digits can be there.

viii)

$\sqrt[3]{540\times 50}$

$=\sqrt[3]{27000}$

$=30$

ix)

$\sqrt[3]{0.000004913} =\sqrt[3]{\frac{4913}{1000000000}}$

$=\frac{17}{1000}$

$=0.017$

x)

On factorizing we have:

$11\times 3\times 101=3333$

Now we observe for the nearest perfect cube:

$3^3=27$ for 33

and

$4^3=64$ for 101

On taking the product:

$3\times 4=12$ the no. with its cube nearest and just greater will be larger than this now we check step by step for 13,14, 15

we get :

$15^3=3375$

so,

$3375-3333=42$

42 is the smallest no. that when added to 3333 makes it a perfect cube.

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TOPIC: arithmetic

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