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) 3 : 7
(b) 3 :5
(c) 5 : 6
(d) 6 : 5
The correct answers to the above question in:
Answer: (d)
Tricky Approach
Required ratio
= 15 × 22 : 11 × 25 = 6 : 5
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Read more finding sum difference product problems Based Quantitative Aptitude Questions and Answers
Question : 1
A and B have money in the ratio Rs.2 : 1. If A gives 2 to B, the money will be in the ratio 1 : 1. What were the initial amounts they had?
a) Rs.8 and Rs.4
b) Rs.6 and Rs.3
c) Rs.16 and Rs.8
d) Rs.12 and Rs.6
Answer »Answer: (a)
Let A and B have Rs.2x and Rs.x initially.
2x - 2 = x + 2
x = 4
Initial amount with A = Rs.8
Initial amount with B = Rs.4.
Question : 2
Two numbers are in the ratio of 3 : 5. If 9 is subtracted from each then they are in the ratio 12 : 23. The smaller number is
a) 28
b) 36
c) 33
d) 55
Answer »Answer: (c)
Numbers = 3x and 5x (let)
According to question,
${3x - 9}/{5x - 9} = 12/23$
69x - 207 = 60x - 108
69x - 60x = 207 - 108
9x = 99 ⇒ $x = 99/9$ = 11
Smaller number = 3x = 3 × 11 = 33
Question : 3
Two numbers are in the ratio of 3 : 5. If 9 be subtracted from each, then they are in the ratio of 12 : 23. Find the numbers.
a) 33, 55
b) 60, 69
c) 36, 115
d) 15, 28
Answer »Answer: (a)
According to the question,
${3x - 9}/{5x - 9} = 12/23$
(Numbers = 3x and 5x )
69x - 207 = 60x - 108
9x = 207 - 108 = 99
x = 11
Required numbers
3 × 11 = 33 and 5 × 11 = 55
Using Rule 35,
Here, a = 3, b = 5, c = 12,
d = 23, x = 9
1st Number = ${xa(d-c)}/{ad-bc}$
= ${9 × 3(23 - 12)}/{3 × 23 - 5 × 12}$
= ${27 × 11}/{69 - 60}$
= ${27 × 11}/9$= 33
2nd Number= ${xb(d-c)}/{ad-bc}$
= ${9 × 5(23 - 12)}/{3 × 23 - 5 × 12}$
= ${45 × 11}/{69 - 60}$
= ${45 × 11}/9$ = 55
Numbers are 33, 55
Question : 4
The ratio of the number of boys and girls of a school with 504 students is 13 : 11. What will be the new ratio if 12 more girls are admitted?
a) 9 : 10
b) 10 : 9
c) 81 : 91
d) 91 : 81
Answer »Answer: (d)
Total numbers of girls in the school
= $504 ×11/{13 + 11}$
= $504 × 11/24$ = 231
Total numbers of boys in the school
= $504 × 13/{13 + 11}$
= $504 × 13/29 = 273$
Now, total no. of girls when 12 more girls are admitted
= 231 + 12 = 243
Required ratio
= 273 : 243 = 91 : 81
Question : 5
What must be added to each term of the ratio 7 : 11, so as to make it equal to 3 : 4 ?
a) 6.5
b) 5
c) 7.5
d) 8
Answer »Answer: (b)
Let the required number be x.
${7 + x}/{11 + x} = 3/4$
28 + 4x = 33 + 3x
x = 33 - 28 = 5
Question : 6
The students in three classes are in the ratio 2 : 3 : 5. If 40 students are increased in each class, the ratio changes to 4 : 5 : 7. Originally, the total number of students was :
a) 200
b) 400
c) 180
d) 100
Answer »Answer: (a)
Let the number of students in three classes be 2x, 3x and 5x respectively.
Due to increase of 40 students in each class,
we have ${2x + 40}/{3x + 40} = 4/5$
10x + 200 = 12x + 160
2x = 200 - 160
2x = 40 ⇒ x = 20
Original strength
= 10x = 10 × 20 = 200
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
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