type 7 finding sum difference product based ratio & proportion problems Section-Wise Topic Notes With Detailed Explanation And Example Questions
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The following question based on ratio & proportion topic of quantitative aptitude
(a) 8 : 7
(b) 4 : 3
(c) 10 : 7
(d) 12 : 7
The correct answers to the above question in:
Answer: (b)
Using Rule 21,
Initially number of boys
= $8/{8 + 5} × 286 = 8/13 × 286$ = 176
Number of girls
= $5/13 × 286$ = 110
22 more girls get admitted.
∴ Required ratio
= $176/{110 + 22} = 176/132 = 4/3$ = 4 : 3
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Read more finding sum difference product problems Based Quantitative Aptitude Questions and Answers
Question : 1
Two numbers are in the ratio 5 : 7. If 9 is subtracted from each of them, their ratio becomes 7 : 11. The difference of the numbers is
a) 15
b) 18
c) 12
d) 6
Answer »Answer: (c)
Let the numbers be 5x and 7x.
Now, ${5x - 9}/{7x - 9} = 7/11$
11 (5x - 9) = 7 (7x - 9)
55x - 99 = 49x - 63
55x - 49x = 99 - 63
6x = 36 ⇒ x = 6
Required difference
= 7x - 5x = 2x = 2 × 6 = 12
Using Rule 35Two numbers are in the ratio a:b and if x is subtracted from each number the ratio becomes c:d. The two numbers will be= ${xa(d-c)}/{ad-bc}$ and ${xb(d-c)}/{ad-bc}$
Here, a = 5, b = 7, x = 9, c = 7, d = 11
1st Number = ${xa(d-c)}/{ad-bc}$
= ${9 ×5(11 - 7)}/{5 ×11 - 7 × 7}$
= ${45 × 4}/{55 - 49}$
= ${45 × 4}/6$= 30
2nd Number = ${xa(d-c)}/{ad-bc}$
= ${9 ×7(11 - 7)}/{5 ×11 - 7 × 7}$
= ${63 × 4}/{55 - 49}$
= ${63 × 4}/6$= 42
Their difference = 42 - 30 = 12
Question : 2
The average of 11 numbers is 36, whereas average of 9 of them is 34. If the remaining two numbers are in the ratio of 2 : 3, find the value of the smaller number (between remaining two numbers).
a) 54
b) 36
c) 48
d) 45
Answer »Answer: (b)
According to the question,
Sum of remaining two numbers
= 11 × 36 - 9 × 34
= 396 - 306 = 90
Ratio of the remaining two numbers = 2 : 3
Smaller number
= $2/5$ × 90 = 36
Question : 3
The total number of students in a school was 660. The ratio between boys and girls was 13 : 9. After some days, 30 girls joined the school and some boys left the school and new ratio between boys and girls became 6 : 5. The number of boys who left the school is :
a) 30
b) 60
c) 40
d) 50
Answer »Answer: (a)
In the first case,
Boys = $660 × 13/22$ = 390
Girls = $660 × 9/22$ = 270
If x boys leave the school, then
${390 - x}/{270 + 30} = 6/5$
390 - x = 360
x = 390 - 360 = 30
Question : 4
Two numbers are such that the ratio between them is 4 : 7. If each is increased by 4, the ratio becomes 3 : 5. The larger number is
a) 56
b) 64
c) 48
d) 36
Answer »Answer: (a)
Let the numbers be 4x and 7x.
${4x + 4}/{7x + 4} = 3/5$
21x + 12 = 20x + 20
21x - 20x = 20 - 12
x = 8
Larger number
= 7x = 7 × 8 = 56
Using Rule 34,
a = 4, b = 7, c = 3,
d = 5, x = 4
Larger number = ${xb(c-d)}/{ad-bc}$
= ${4 ×7(3 -5)}/{4 × 5 - 3 × 7}$
= ${4 × 7 × (-2)}/{20 - 21}$= 56
Question : 5
Two numbers are in the ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. The smaller number is
a) 66
b) 77
c) 49
d) 39
Answer »Answer: (c)
Let the numbers be 7x and 11x respectively.
${7x + 7}/{11x + 7} = 2/3$
22x + 14 = 21x + 21
x = 7
Smaller number
= 7x = 7 × 7 = 49
Using Rule 34Two numbers are in the ratio a:b and if each number is increased by x, the ratio becomes c:d. Then the two numbers will be${xa(c-d)}/{ad-bc}$ and ${xb(c-d)}/{ad-bc}$
Here, a = 7, b = 11, x = 7, c= 2, d = 3
1st Number = ${xa(c-d)}/{ad-bc}$
= ${7 × 7(2 - 3)}/{7 × 3 - 11 × 2}$
= ${49 × -1}/{21 - 22}$ = 49
2nd Number = ${xb(c-d)}/{ad-bc}$
= ${7 × 11(2 - 3)}/{7 × 3 - 11 × 2}$
= ${77 × -1}/{21 - 22}$ = 77
Smallest number = 49
Question : 6
Three numbers are in the ratio 1 : 2 : 3 and the sum of their cubes is 4500 . The smallest number is
a) 6
b) 10
c) 5
d) 4
Answer »Answer: (c)
Let the numbers be x, 2x and 3x.
According to the question,
$x^3 + (2x)^3 + (3x)^3$ = 4500
$x^3 + 8x^3 + 27x^3$ = 4500
$36x^3$ = 4500
$x^3 = 4500/36$ = 125
$x = √^3{125}$= 5 = smallest number
GET ratio & proportion PRACTICE TEST EXERCISES
type 1 basic concepts of ratio & proportion
type 2 age based ratio & proportion problems
type 3 addition subtraction product on ratio & proportion
type 4 income & expenditure based ratio & proportion problems
type 5 shares & partnership based ratio & proportion problems
type 6 fractions based ratio & proportion problems
type 7 finding sum difference product based ratio & proportion problems
type 8 alligation & mixtures based ratio & proportion problems
type 9 coins & rupees based ratio & proportion problems
ratio & proportion Shortcuts and Techniques with Examples
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