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

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 = $48/7$

Answer: (d)

Let the third number be x,

∴ First number = 3x

∴ Second number = ${3x}/4$

According to the question,

3x + ${3x}/4$ + x = 3 × 95

⇒ ${12x+3x+4x}/4$ = 285

⇒ 19x = 285 × 4

⇒ x = ${285×4}/19$ = 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 = 4, b = $3/4$, x = 95

Third Number = $3/\text"1+b+ab"$ ×x

= $3/{1+{3/4}+4×{3/4}}$×95

= ${3×4}/{4+3+12}$×95 = 60

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 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: (a)

Let the first number be x,

then, x = ${240 - x}/4$

⇒ 4x = 240 – x

⇒ 5x = 240

⇒ x = $240/5$ = 48