Practice Twice thrice one third - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   The average of first three numbers is thrice the fourth number. If the average of all the four numbers is 5, then find the fourth number.

(a)

(b)

(c)

(d)

Explanation:

Let the numbers be a, b, c,

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

⇒ a + b + c = 9d

Again, ${\text"a + b + c + d"}/4$ = 5

⇒ a + b + c + d = 20

⇒ 9d + d = 20

⇒ 10d = 20 ⇒ d = 2


Q-2)   The average of three numbers is 28, the first number is half of the second, the third number is twice the second, then the third number is

(a)

(b)

(c)

(d)

Explanation:

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


Q-3)   Of the three numbers, the first is twice the second and the second is thrice the third. If the average of the three numbers is 10, the largest number is :

(a)

(b)

(c)

(d)

Explanation:

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


Q-4)   Of the three numbers whose average is 60, the first is one fourth of the sum of the others. The first number is :

(a)

(b)

(c)

(d)

Explanation:

x + y + z = 180

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

⇒ 4x = y + z

⇒ 5x = 180,

∴ x = 36


Q-5)   If the arithmetic mean of 3a and 4b is greater than 50, and a is twice b, then the smallest possible integer value of a is

(a)

(b)

(c)

(d)

Explanation:

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

⇒ 3a + 4b > 100

⇒ 3a + ${4a}/2$ > 100 [Since a = 2b]

⇒ 3a + 2a > 100

⇒ 5a > 100

⇒ a > 20

∴ Minimum value of a = 21


Q-6)   Of the three numbers, the first number is twice of the second and the second is thrice of the third number. If the average of these 3 numbers is 20, then the sum of the largest and smallest numbers is

(a)

(b)

(c)

(d)

Explanation:

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


Q-7)   Out of 4 numbers, whose average is 60, the first one is one fourth of the sum of the last three. The first number is

(a)

(b)

(c)

(d)

Explanation:

Let the first number be x,

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

⇒ 4x = 240 – x

⇒ 5x = 240

⇒ x = $240/5$ = 48


Q-8)   Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, then difference of first and third number is

(a)

(b)

(c)

(d)

Explanation:

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


Q-9)   The average of three numbers is 77. The first number is twice the second and the second number is twice the third. The first number is :

(a)

(b)

(c)

(d)

Explanation:

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


Q-10)   The average of three numbers is 40. The first number is twice the second and the second one is thrice the third number. The difference between the largest and the smallest numbers is

(a)

(b)

(c)

(d)

Explanation:

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