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) 2
(b) 4
(c) 4.5
(d) 5
The correct answers to the above question in:
Answer: (a)
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
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Read more twice thrice one third Based Quantitative Aptitude Questions and Answers
Question : 1
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) 48
b) 60
c) 15
d) 45
Answer »Answer: (a)
Let the first number be x,
then, x = ${240 - x}/4$
⇒ 4x = 240 – x
⇒ 5x = 240
⇒ x = $240/5$ = 48
Question : 2
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) 24
b) 48
c) 12
d) 20
Answer »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
Question : 3
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) 77
b) 132
c) 33
d) 66
Answer »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
Question : 4
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) 54
b) 60
c) 24
d) 42
Answer »Answer: (d)
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
Question : 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) 21
b) 19
c) 20
d) 18
Answer »Answer: (a)
$\text"3a+4b"/2$ > 50
⇒ 3a + 4b > 100
⇒ 3a + ${4a}/2$ > 100 [Since a = 2b]
⇒ 3a + 2a > 100
⇒ 5a > 100
⇒ a > 20
∴ Minimum value of a = 21
Question : 6
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
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
-
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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