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

Answer: (c)

Let the second number be x.

Then first number = $x/2$

and third number = 2x

According to the question,

$x/2$+x+2x=28×3

⇒ ${x+2x+4x}/2$=28×3

⇒ 7x = 28 × 3 × 2

⇒ x = $168/7$ = 24

∴ Third number = 2 × 24 = 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 = $1/2$, b = $1/2$, x = 28

Third Number= $3/\text"1+b+ab"$ ×x

= $3/{1+{1/2}+{1/2}×{1/2}}$×28

= $3/{{4+2+1}/4}$×28

= ${3×4×28}/7$= 48

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

Answer: (d)

x + y + z = 180

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

⇒ 4x = y + z

⇒ 5x = 180,

∴ x = 36

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: (a)

Let the third number be x.

∴ Second number = 3x

First number = 6x

∴ (x + 3x + 6x) = 100 × 3

⇒ 10x = 300

⇒ x = 30

∴ The largest number = 6x = 6 × 30 = 180

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 = 100

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

= ${3×2×3}/{1+3+2×3}$×100

= ${18 × 100}/10$ = 180