ICOME
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Unit 1

Unit-1:


(a) VBODMAS (b) Divisibility (c) Cube and cube roots (d) Square and square roots

(e)Surds, indices and rationalisation


Q1.If √12 + √27 = 8.66,so find the value of 3√3 + √147 .

Solution.
= √12 + √27
2√3 + 3√3
5√3 = 8.66
√3 =
8.66/5

√3 = 1.73
= 3√3 + 7√3
= 10√3
= 10×1.73
= 17.3
Solution

Q2. One-fifth of the cube root of a number is 10.find the number.

Solution.
1/5
(x)1/3 = 10
{x = is a numbers}
x1/3 = 50
(x1/3)3 = (50)3
x = 125000.
Solution

Q3. If 2√80 = 17.89, find the value of 10√5.

Solution.
2√80
2 × 4√5
8√5 = 17.89
√5 = 2.24
10√5 = 10 × 2.24 = 22.4

Solution

Q4. If √5 = 2.23, find the value of √80 -
1/2
√20 - √20 .

Solution.
√16×5 -
1/2
√20 - √20 .
4√5 -
1/2
2√5 - 2√5
= 4√5 - 3√5
= √5 = 2.23
Solution

Q5. If one-tenth of square-root of x is 1. find x ?

Solution.
Let number is x
1/10
√x = 1
√x = 10
x = 100
Solution

Q6. In
17/8
, nfind the number of decimal places after which the decimal expansion terminates.

Solution.
x =
15/8
=
15/23 x 50


x =
P/2m x 5n
,

x terminates after k places of decimals, where k is the larger of m and n .
Solution

Q7. x = 6n, ends with the digit zero, if n is .

Solution.
x = 6n = (3 x 2)n,

→ The only prime factor is 2 , 5 does not occur.
Solution

Q8. Find the remainder when the square of any prime number greater than 3 is divided by 3 is .

Solution.
1
Solution

Q9. Find the greatest number of 5 digits which exactly divisible by 24, 15, 36 .

Solution.
Required number = 99999 - Remainder

when 99999 is divided by the LCM of 24, 15 and 36 .

= 99999 - 279

= 99720
Solution

Q10. For which value of n, 2n x 5n ends in 6 .

Solution.
No value .
Solution