model 3 twice, thrice, one third etc. of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Let second number be x

∴ The first number = 3x

and the third number = 15x

Now, x + 3x + 15x = 3 × 57

⇒ 19x = 3 × 57

⇒ x= ${3×57}/19$= 9

∴ Required difference

= 15x – x = 14x

= 14 × 9 = 126

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

a = 3, b = 5, x = 57

First number = $\text"3ab"/ \text"1+b+ab"$x

= ${3×3×5}/{1+3+15}$× 57

= $45/19$ × 57 = 135

Second number = $\text"3ab"/ \text"1+b+ab"$x

= ${3×5}/{19}$× 57

= $15/19$ × 57 = 45

Third number = $3/ \text"1+b+ab"$x

= $3/19$×57 = 9

Required result = 135 – 9 = 126

Answer: (a)

$\text"3a+4b"/2$ > 50

⇒ 3a + 4b > 100

⇒ 3a + ${4a}/2$ > 100 [Since a = 2b]

⇒ 3a + 2a > 100

⇒ 5a > 100

⇒ a > 20

∴ Minimum value of a = 21

Answer: (d)

Let the third number be x.

∴ Second number = 3x

and first number = 6x

∴ 6x + 3x + x = 3 × 20

⇒ 10x = 60 ⇒ x = 6

∴ Required sum = 6x + x = 7x = 7 × 6 = 42

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

Here, a = 2, b = 3, x = 20

Largest Number= $\text"3ab"/ \text"1+b+ab"$ x

= ${3×2×3}/{1+3+2×3}$×20

=$18/10$×20 = 36

Smallest Number = $3/\text"1+b+ab"$ ×x

= $3/{1+3+2×3}$×20

= $3/10$×20 = 6

Sum = 36 + 6 = 42

Answer: (c)

Let the second number be x.

∴ First number = 2x

∴ Third number =${2x}/3$

∴ 2x+x+${2x}/3$ = 49.5×3

⇒ 6x + 3x + 2x =49.5×9 = 445.5

⇒ 11x = 445.5

⇒ x = ${445.5}/11$ = 40.5

∴ Required difference

= 2x - ${2x}/3$ = ${4x}/3$

= ${4×40.5}/3$ = 54

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

Here, a = 2, b = $3/2$, x = 49.5

First Number = $\text"3ab"/ \text"1+b+ab"$ 5x

= ${3×2×{3/2}}/{1+{3/2}+2}×{3/2}$×49.5

= ${18/2}/{11/2}$×49.5

= ${18×49.5}/11$ = 18×4.5

= ${18×45}/10$ = 81

Third Number = $3/\text"1+b+ab"$ ×x

= $3/{1+{3/2}+2×{3/2}}$×49.5

= $3/{11/2}$ ×49 5.

= 6 ×4.5 = 27

Difference = 81 – 27 = 54

Answer: (d)

x + y + z = 180

x =$1/4$(y+z)

⇒ 4x = y + z

⇒ 5x = 180,

∴ x = 36

Answer: (a)

Let the third number be x.

∴ Second number = 3x

First number = 6x

Now, $\text"x + 3x + 6x"/3$ = 10

⇒ 10x = 30 ⇒ x = 3

∴ The largest number = 6x = 6 × 3 = 18.

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

a = 2, b = 3, x = 10

Largest number = $\text"3ab"/ \text"1+b+ab"$x

= ${3×2×3}/{1+3+2×3}$×10

= $18/10$ ×10 = 18