model 3 twice, thrice, one third etc. of numbers Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Let the numbers be 2x, x and 4x respectively

∴ Average = ${2x+x+4x}/3$

⇒ ${7x}/3$= 56

⇒ x = ${3 × 56}/7$ = 24

∴ First number = 2x = 2 × 24 = 48

Third number = 4x = 4 × 24 = 96

∴ Required difference = 96 – 48 = 48

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 = $1/4$, x = 56

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

= ${3×2×{1/4}}/{1+{1/4}+2×{1/4}$×56

= ${{3/2}×4}/{4+1+2}$×56 = 48

Third number = $3/\text"1+b+ab"$ ×x

= $3/{1+{1/4}+2×{1/4}$×56

= ${3×4}/{4+4+2}$×56 = 96

Required difference = 96–48 = 48

Answer: (b)

Let the third number = x

∴ Second number = 2x

First number = 4x

Now, x + 2x + 4x = 3 × 77

⇒ 7x = 3 × 77

⇒ x = ${3 × 77}/7$ = 33

∴ First number = 33 × 4 = 132

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 = 2, x = 77

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

= ${3×2×2}/{1+2+2×2}$×77

= $12/7$×77

= 12 ×11 = 132

Answer: (b)

Let the third number be x.

∴ Second number = 3x

First number = 6x

∴ ${6x+3x+x}/3$ = 40

⇒ 10x = 120 ⇒ x = 12

∴ Required difference = 6x – x = 5x = 5 × 12 = 60

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 = 40

Largest Number = First Number

= $\text"3ab"/ \text"1+b+ab"$ ×x

= ${3×2×3}/{1+3+2×3}$×40

= $18/10$×40 = 72

Smallest Number = Third Number

= $3/\text"1+b+ab"$ ×x

= $3/{1+3+2×3}$×40

= $3/10$ ×40 = 12

Difference = 72 – 12 = 60

Answer: (d)

Let the first number be x.

∴ Second number = 2x

and third number = ${2x}/3$

According to the question,

x + 2x + ${2x}/3$ = 33 × 3

⇒ 3x + ${2x}/3$ = 99

⇒ ${9x +2x}/3$ = 99

⇒ 11x = 99 × 3

⇒ x = ${99×3}/11$= 27

∴ Largest number = 2x = 2 × 27 = 54

Answer: (d)

$\text"a+b+c"/3$ =2d

⇒ a + b + c = 6d ...(i)

Also, $\text"a+b+c+d"/4$ = 12

⇒ a + b + c + d = 48

⇒ 6d + d = 48

⇒ 7d = 48

⇒ d = $48/7$