model 3 twice, thrice, one third etc. of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 9 EXERCISES
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The following question based on average topic of quantitative aptitude
(a) 42
(b) 45
(c) 30
(d) 36
The correct answers to the above question in:
Answer: (d)
x + y + z = 180
x =$1/4$(y+z)
⇒ 4x = y + z
⇒ 5x = 180,
∴ x = 36
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Read more twice thrice one third Based Quantitative Aptitude Questions and Answers
Question : 1
Among three numbers, the first is twice the second and thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is
a) 39.5
b) 41.5
c) 54
d) 28
Answer »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
Question : 2
Of the three numbers, the first is twice the second and the second is 3 times the third. If their average is 100, the largest of the three numbers is :
a) 180
b) 300
c) 120
d) 150
Answer »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
Question : 3
Of the three numbers, the first is 3 times the second and the third is 5 times the first. If the average of the three numbers is 57, the difference between the largest and the smallest number is
a) 126
b) 135
c) 9
d) 18
Answer »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
Question : 4
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) 18
b) 30
c) 12
d) 15
Answer »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
Question : 5
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) 24
b) 18
c) 48
d) 36
Answer »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
GET average PRACTICE TEST EXERCISES
model 1 basic average questions
model 2 average of consecutive numbers
model 3 twice, thrice, one third etc. of numbers
model 4 find nth average from 1st & last number
model 5 find new average from error
model 6 find average of excluded number
model 7 average on ages/weight
model 8 find monthly income
model 9 average on cricket/exam
average Shortcuts and Techniques with Examples
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model 1 basic average questions
Defination & Shortcuts … -
model 2 average of consecutive numbers
Defination & Shortcuts … -
model 3 twice, thrice, one third etc. of numbers
Defination & Shortcuts … -
model 4 find nth average from 1st & last number
Defination & Shortcuts … -
model 5 find new average from error
Defination & Shortcuts … -
model 6 find average of excluded number
Defination & Shortcuts … -
model 7 average on ages/weight
Defination & Shortcuts … -
model 8 find monthly income
Defination & Shortcuts … -
model 9 average on cricket/exam
Defination & Shortcuts …
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