Numbers Exercise 2

⁸⁄₁₁ = 0.727, ⁵⁄₇ = 0.714, ⅗ = 0.6, ⁴⁄₉ = 0.44, ⁵⁄₁₂ = 0.416, ²⁄₇ = 0.285, ⅛ = 0.125

Descending order :

⁸⁄₁₁, ⁵⁄₇, ⅗, ⁴⁄₉, ⁵⁄₁₂, ²⁄₇, ⅛

So, ⅗ is the third.

Decimal equivalent of given fractions:

⅖ = 0.4; ¾ = 0.75; ⅘ = 0.8; ⁵⁄₇ = 0.714; ⁶⁄₁₁ = 0.545

Clearly, 0.4 < 0.545 < 0.714 < 0.75 < 0.8

∴ ⅖ < ⁶⁄₁₁ < ⁵⁄₇ < ¾ < ⅘

Divisor = [Sum of remainders]

– [ Remainder when sum is divided]

= 11 + 21 – 4 = 28

Let number be N.

Then, N = Divisor × Q₁ + 23

2N = Divisor × Q₂ + 11, where Q₁ and Q₂ are quotients respectively.

Let the number be 5q + 3, where q is quotient

Now (5q + 3)² = 25q² + 30q + 9

= 25q² + 30q + 5 + 4

= 5[5q² + 6q + 1] + 4

Hence, remainder is 4

56 = d₁ × d₂

∴ required remainder = d₁r₂ + r₁

where d₁ = 7 and r₁ = 3 and r₂ = 5

Let the quotient be q when divided by 8. n= 3{5(8q+4)+2}+1 =3(40q+22)+1 =120q+66+1 =120q+67
Now, if it is divided by 8. We have n=120q+67=8(15q+8)+3, remainder 3. And if 15q+8 is divided by 5, we get remainder 3. And if 3q+1 is divided by 3, we get remainder 1. Hence, this all gives us the remainders as 3, 3 and 1.
Trick: Complete remainder = d₁d₂r₃ + d₁r₂ + r₁ = 3 × 5 × 4 + 3 × 2 + 1 = 67 Divided 67 by 8, 5 and 3, the remainders are 3, 3, 1.

By actual division, we find that 999999 is exactly divisible by 13. The quotient 76923 is the required number.

By division Algorithm,

49471 = 246 × D + 25

⇒ D = 201

Let the number be z. Now

385 = 5 × 7 ×11

x = 11 × 102 + 10 = 1132

y = 7x + 6 = 7 × 1132 + 6 = 7930

z = 5y + 4 = 5 × 7930 + 4 = 39654

Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $.

Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.

Hence, digits in place of * and $ are 4 and 0 respectively.

On dividing 803642 by 11, we get remainder = 4.

∴ Required number to be added = (11 – 4) = 7.

z = 6 × 1 + 4 = 10

y = 5 × 10 + 3 = 53

x = 4 × 53 + 2 = 214

On dividing 6709 by 9, we get remainder = 4.

∴ Required number to be subtracted = 4.

On dividing 427398 by 15, we get remainder = 3.

∴ Required number to be subtracted = 3.

Number = (31 × Q) + 29.

Given data is inadequate.

The number is divisible by 18 i.e., it has to be divisible by 2 and 9.

∴ B may be 0, 2, 4, 6, 8.

A + 4 + 5 + 7 + 1 + 2 + 0 + 3 + B

= A + B + 22.

A + B could be 5, 14 (as the sum can’t exceed 18, since A and B are each less than 10).

So, A and B can take the values of 6, 8.

Number is of the form

= 7n + 3; n = 1 to 13

So,

a, a + 2, a + 4 are prime numbers.

Put value of ‘a’ starting from 3, we will have 3, 5 and 7 as the only set of prime numbers satisfying the given relationships.

987 = 3 × 7 ×47

So, required number must be divisible by each one of 3, 7, 47.

None of the numbers in (a) and (b) are divisible by 3, while (d) is not divisible by 7.

∴ Correct answer is (c).

Since 111111 is divisible by each one of 7, 11 and 13, so each one of given type of numbers is divisible by each one of 7, 11, and 13. as we may write, 222222 = 2 × 111111, 333333 = 3 × 111111, etc.

= (7³)²⁸/(7³ - 1)

= {(7³)²⁸ - 1 + 1}/(7³ - 1)

= {(7³)²⁸ - 1}/(7³ - 1) + 1/(7³ - 1)

((7³)²⁸ – 1) / (7³ – 1) is always divisible as it is in the form of (xⁿ – yⁿ) / (x – y), hence the remainder is 1.

The required numbers are 307, 317, 327, 337, 347, 357, 367, 370, 371, 372, 373, 374, 375, 376, 378, 379, 387, 397.

Hence there are 18 numbers.

Here, number of integers next higher and next lower are same (=4).

Now, since 81 is divisible by 9, therefore, the sum is divisible by 9

Let the original number is x. Then

∴

A number is divisible by 9 if the sum of its digits is divisible by 9.

Here 5 + 4 + 3 + 2 + * + 7 = 21 + *

So, the digit in place of * is 6

Decimal equivalents of given fractions:

½ = 0.5; ⅔ = 0.67; ⁵⁄₉ = 0.56; ⁶⁄₁₃ = 0.46; ⁷⁄₉ = 0.78

∴ 0.46 < 0.5 < 0.56 < 0.67 < 0.78

⁶⁄₁₃ < ½ < ⁵⁄₉ < ⅔ < ⁷⁄₉

∴ Fourth fraction = ⅔

Decimal equivalents of fractions

⅞ = 0.875, ⅘= 0.8, ⁸⁄₁₄ = 0.57, ⅗= 0.6, ⅚ = 0.83

∴ 0.875 > 0.83 > 0.8 > 0.6 > 0.57

∴ ⅞ > ⅚ > ⅘ > ⅗ > ⁸⁄₁₄