# Ratio and Proportion Questions and Answers updated daily – Aptitude

Ratio and Proportion Questions: Solved 256 Ratio and Proportion Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Ratio and Proportion Questions

1. A got twice as many marks in English as in Science. His total marks in English, Science and Mathematics is 180. If the ratio of his marks in English and Mathematics is 2 : 3, what is his marks in Science?

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Correct Ans:30

Explanation:

English : Mathematics = 2:3

English : Science = 2:1

English : Mathematics : Science = 2:3:1

English + Mathematics + Science = 180

Marks in Science = (1/6)*180 =

English : Science = 2:1

English : Mathematics : Science = 2:3:1

English + Mathematics + Science = 180

Marks in Science = (1/6)*180 =

**30**
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2. In a book store house, the ratio of English to hindi books is 7:2. If there are 1512 English books and due to increase in demand of English books, few English books are added by the shopkeeper and the said ratio become 15:4. The number of English books added is:

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Correct Ans:108

Explanation:

---> The ratio of English books/Hindi books

---> 7 : 2 = 1512 : Hindi books

---> Hindi books = 432

---> Let x books are added:

---> 15 : 4 = (1512 + x) : 432

--->

---> 7 : 2 = 1512 : Hindi books

---> Hindi books = 432

---> Let x books are added:

---> 15 : 4 = (1512 + x) : 432

--->

**on solving we get x = 108**
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3. By mistake, instead of dividing Rs. 117 among A, B and C in the ratio 1/2,1/3,1/4 it was divided in the ratio of 2:3:4. Who gains the most and by how much?

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Correct Ans:C, Rs. 25

Explanation:

----> Original ratio of A, B, and C = 1/2 : 1/3 : 1/4 = 6: 4: 3

----> Share of A = 6/13 * 117 = Rs. 54

----> Share of B = 4/13 * 117 = Rs. 36

----> Share of C = 3/13 * 117 = Rs. 27

----> The ratio of A, B, and C by mistake = 2 : 3 : 4

----> Share of A = 2/9 *117 = Rs.26

----> Share of B = 3/9 *117 = Rs.39

----> Share of C = 13 * 4 = Rs.52

----> Therefore, it is clear from the above calculation that C gains maximum i.e.

----> Share of A = 6/13 * 117 = Rs. 54

----> Share of B = 4/13 * 117 = Rs. 36

----> Share of C = 3/13 * 117 = Rs. 27

----> The ratio of A, B, and C by mistake = 2 : 3 : 4

----> Share of A = 2/9 *117 = Rs.26

----> Share of B = 3/9 *117 = Rs.39

----> Share of C = 13 * 4 = Rs.52

----> Therefore, it is clear from the above calculation that C gains maximum i.e.

**52 - 27 = Rs. 25**
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4. Dhiraj have coins are in the ratio of Rs.1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the numbers of Rs.1, 50p, and 25p coins are reversed, find the new total value of coins in the bag of Dhiraj?

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Correct Ans:Rs. 82

Explanation:

Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order; the only possibility for the ratio is 3:5:7

---> Let the number of Rs.1, 50p and 25p coins be 3x, 5x and 7x respectively.

---> Hence, total value of coins in paise

---> = 100*3x+50*5x+25*7x = 725x

---> Given 725x = 5800

---> therefore x = 8

---> If the number of coins of Rs. 1,50p and 25p is reversed, the total value of coins in the Bag (in paise)

---> = 100*7x+50*5x+25*3x = 1025x

---> Substitute x = 8

--->

---> Let the number of Rs.1, 50p and 25p coins be 3x, 5x and 7x respectively.

---> Hence, total value of coins in paise

---> = 100*3x+50*5x+25*7x = 725x

---> Given 725x = 5800

---> therefore x = 8

---> If the number of coins of Rs. 1,50p and 25p is reversed, the total value of coins in the Bag (in paise)

---> = 100*7x+50*5x+25*3x = 1025x

---> Substitute x = 8

--->

**New total = RS. 82**
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5. Karan, Hari and Kowshik play cricket. The runs got by Karan to Hari and Hari to Kowshik are in the ratio of 5:3. They get altogether 588 runs. How many runs did Karan get?

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Correct Ans:300 runs

Explanation:

Karan, Hari and Kowshik play cricket. The runs got by Karan to Hari and Hari to Kowshik are in the ratio of 5:3. They get altogether 588 runs. How many runs did Karan get:

Reference :

---> Karan : Hari = (5:3)

---> Hari : Kowshik = (5:3)

---> Karan : Hari : Kowshik = (25:15:9)

---> Therefore, the runs made by Karan,

---> = (25/49) * 588

---> = 25 * 12 =

Reference :

---> Karan : Hari = (5:3)

---> Hari : Kowshik = (5:3)

---> Karan : Hari : Kowshik = (25:15:9)

---> Therefore, the runs made by Karan,

---> = (25/49) * 588

---> = 25 * 12 =

**300 runs**
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6. An outgoing batch of students wants to gift books worth Rs 4,200 to their teachers. If the boys, offer to pay 50% more than the girls and an external sponsors gives three times the boy’s contribution, then how much should the boys donate?

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Correct Ans:Rs 900

Explanation:

----> The ratio of the share girls: boys: sponsors = (1:1.5:4.5)

----> So the proportion to boy’s share = (1.5/7)

----> Hence, the boys would donate = (1.5/7) * 4200 =

----> So the proportion to boy’s share = (1.5/7)

----> Hence, the boys would donate = (1.5/7) * 4200 =

**Rs 900**
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7. Chiku, Tipu and Pinku have some candies with each. Five times the number of candies with Pinku equals seven times the number of candies with Chiku while five times the number of candies with Chiku equals seven times the number of candies with Tipu. What is the minimum number of candies that can be there with all three of them put together?

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Correct Ans:109

Explanation:

---> Let the candies with Chiku, Pinku and Tipu be a,b and c respectively.

---> Given, 5b = 7a, and 5a = 7c

---> 25b = 35a & 35a = 49c

---> 25b = 35a = 49c => (b/49) = (a/35) = (c/25)

---> The least possible integral values for a,b & c will be a = 35 , b = 49 & c = 25

--->

---> Given, 5b = 7a, and 5a = 7c

---> 25b = 35a & 35a = 49c

---> 25b = 35a = 49c => (b/49) = (a/35) = (c/25)

---> The least possible integral values for a,b & c will be a = 35 , b = 49 & c = 25

--->

**Total = 35+49+25 = 109**
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8. The ratio of the number of UG students and PG students in a college is 5:4. If 40% of the UG students and 50 % of the PG students are scholarship holders. What percentage of the students does not get the scholarship?

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Correct Ans:55.50%

Explanation:

---> Let UG students be = 5x and

---> PG students be = 4x respectively

---> No. of those who do not get scholarship,

---> = 60% of 5x + 50% of 4x

---> = (60/100*5x) + (50/100*4x)

---> = 300x/100+ 200x /100

---> = 3x + 2x

---> = 5x

---> Required percentage = (5x/9x) *100

--->

---> PG students be = 4x respectively

---> No. of those who do not get scholarship,

---> = 60% of 5x + 50% of 4x

---> = (60/100*5x) + (50/100*4x)

---> = 300x/100+ 200x /100

---> = 3x + 2x

---> = 5x

---> Required percentage = (5x/9x) *100

--->

**= 55.5%**
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9. The wages of Naveena and Menaka are in the ratio ratio 6:5. If the wages of each is increased by Rs.6000, the new ratio becomes 38:35. What is Menaka"™s present salary?

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Correct Ans:4500

Explanation:

---> Let the original salary of Naveena and Menaka be 6x and 5x respectively

---> Then, (6x+6000) / (5x+6000) = 38/35

---> 35(6x+6000) = 38(5x +6000)

---> 210x +210000 = 190x + 228000

---> 210x â€“ 190x = 228000 â€“ 210000

---> 20x = 18000

---> X = 900

---> Menakaâ€™s present salary = 5x

---> = 5*900 =

---> Then, (6x+6000) / (5x+6000) = 38/35

---> 35(6x+6000) = 38(5x +6000)

---> 210x +210000 = 190x + 228000

---> 210x â€“ 190x = 228000 â€“ 210000

---> 20x = 18000

---> X = 900

---> Menakaâ€™s present salary = 5x

---> = 5*900 =

**Rs. 4500**
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10. One year ago the ratio between Lalitha's and Ganga's salary was 3 : 4. The ratios of their individual salaries between last year's and this year's salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Lalitha, now is

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Correct Ans:Rs. 1600

Explanation:

Let the salaries of Lalitha and Ganga one year before be Lâ‚, Gâ‚ respectively and present salary be Lâ‚‚, Gâ‚‚ respectively.

Given ratio:

Lâ‚/Gâ‚ = 3/4

Lâ‚/Lâ‚‚ = 4/5

Gâ‚/Gâ‚‚ = 2/3

At present the total of their salary is Rs. 4160.

Lâ‚‚ + Gâ‚‚ = 4160

Substitute Gâ‚‚ value in above equation,

Lâ‚‚ + (3Gâ‚/2) = 4160

Now, substitute Gâ‚ value,

Lâ‚‚ + [3(4Lâ‚/3)/2] = 4160

Lâ‚‚ + 2Lâ‚ = 4160

Then, substitute Lâ‚ value,

Lâ‚‚ + 2(4Lâ‚‚/5) = 4160

5Lâ‚‚ + 8Lâ‚‚ = 20800

13Lâ‚‚ = 20800

Lâ‚‚ = Rs. 1600.

Hence, present salary of Lalitha is Rs. 1600

Given ratio:

Lâ‚/Gâ‚ = 3/4

Lâ‚/Lâ‚‚ = 4/5

Gâ‚/Gâ‚‚ = 2/3

At present the total of their salary is Rs. 4160.

Lâ‚‚ + Gâ‚‚ = 4160

Substitute Gâ‚‚ value in above equation,

Lâ‚‚ + (3Gâ‚/2) = 4160

Now, substitute Gâ‚ value,

Lâ‚‚ + [3(4Lâ‚/3)/2] = 4160

Lâ‚‚ + 2Lâ‚ = 4160

Then, substitute Lâ‚ value,

Lâ‚‚ + 2(4Lâ‚‚/5) = 4160

5Lâ‚‚ + 8Lâ‚‚ = 20800

13Lâ‚‚ = 20800

Lâ‚‚ = Rs. 1600.

Hence, present salary of Lalitha is Rs. 1600

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11. Ratio of present age of Veer, Sameer, Divyaraj, Ayush and Sumit is 14 : 15 : 13 : 12 : 16 and sum of age of Veer, Divyaraj & Sumit four years hence will be 44 years more than sum of present age of Sameer & Ayush. Find the ratio of age of Veer, Sameer, Divyaraj, Ayush and Sumit after 10 years?

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Correct Ans:(19 : 20 : 18 : 17 : 21)

Explanation:

Let the present age of Veer, Sameer, Divyaraj, Ayush and Sumit be 14x, 15x, 13x, 12x, 16x respectively.

Sum of age of Veer, Divyaraj & Sumit four years hence = 44 years more than sum of present age of Sameer & Ayush

(14x + 13x + 16x + 4*3) = 44 + 15x + 12x

43x + 12 = 27x + 44

43x - 27x = 44 - 12

16x = 32

x = 2

Present age of Veer = 14*2 = 28 yrs

Present age of Sameer = 15*2 = 30 yrs

Present age of Divyaraj = 13*2 = 26 yrs

Present age of Ayush = 12*2 = 24 yrs

Present age of Sumit = 16*2 = 32 yrs

Ratio of age of Veer, Sameer, Divyaraj, Ayush and Sumit after 10 yrs = (28 + 10) : (30 + 10) : (26 + 10) : (24 + 10) : (32 + 10)

= 38 : 40 : 36 : 34 : 42

= 19 : 20 : 18 : 17 : 21.

Sum of age of Veer, Divyaraj & Sumit four years hence = 44 years more than sum of present age of Sameer & Ayush

(14x + 13x + 16x + 4*3) = 44 + 15x + 12x

43x + 12 = 27x + 44

43x - 27x = 44 - 12

16x = 32

x = 2

Present age of Veer = 14*2 = 28 yrs

Present age of Sameer = 15*2 = 30 yrs

Present age of Divyaraj = 13*2 = 26 yrs

Present age of Ayush = 12*2 = 24 yrs

Present age of Sumit = 16*2 = 32 yrs

Ratio of age of Veer, Sameer, Divyaraj, Ayush and Sumit after 10 yrs = (28 + 10) : (30 + 10) : (26 + 10) : (24 + 10) : (32 + 10)

= 38 : 40 : 36 : 34 : 42

= 19 : 20 : 18 : 17 : 21.

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12. Distance travelled by a train is 1830 km and the speed of train is one more than twice the time taken to travel the distance. Find the ratio of the speed of train and time taken by the train.

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Correct Ans:61 : 30

Explanation:

Let the time taken by train be 't' hr.

As per question, speed = 1 + 2t

WKT, Speed = Distance /Time

1 + 2t = 1830/t

2t

2t

2t(t - 30) + 61(t - 30) = 0

(2t + 61) (t - 30) = 0

t = 30; -61/2

So, t = 30 hrs.

Hence, speed = 1830/30 = 61 km/hr

Required ratio = 61 : 30

As per question, speed = 1 + 2t

WKT, Speed = Distance /Time

1 + 2t = 1830/t

2t

^{2}+ t - 1830 = 02t

^{2}- 60t + 61t - 1830 = 02t(t - 30) + 61(t - 30) = 0

(2t + 61) (t - 30) = 0

t = 30; -61/2

So, t = 30 hrs.

Hence, speed = 1830/30 = 61 km/hr

Required ratio = 61 : 30

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13. If the digits of the age of Mr. Suman are reversed then the new age so obtained is the age of his wife. 1/11 of the sum of their ages is equal to the difference between their ages. If Mr. Suman is elder than his wife then find the ratio of their ages.

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Correct Ans:6 : 5

Explanation:

Let the present age of Mr. Suman = (10x + y) yrs.

Present age of his wife = (10y + x) yrs

According to the question,

(1/11)Sum of their ages = Difference of their ages

(1/11)(10x + y + 10y + x) = (10x + y) - (10y + x)

(1/11)(11x + 11y) = 9x - 9y

x + y = 9x - 9y

-8x = -10y

x/y = 10/8

x/y = 5/4

x : y = 5 : 4

Age of Mr. Suman = 10(5) + 4 = 54 yrs

Age of his wife = 10(4) + 5 = 45 yrs

Required ratio = 54/45 = 6 : 5.

Present age of his wife = (10y + x) yrs

According to the question,

(1/11)Sum of their ages = Difference of their ages

(1/11)(10x + y + 10y + x) = (10x + y) - (10y + x)

(1/11)(11x + 11y) = 9x - 9y

x + y = 9x - 9y

-8x = -10y

x/y = 10/8

x/y = 5/4

x : y = 5 : 4

Age of Mr. Suman = 10(5) + 4 = 54 yrs

Age of his wife = 10(4) + 5 = 45 yrs

Required ratio = 54/45 = 6 : 5.

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14. The ratio of the density of 3 kinds of petrol A1, A2 and A3 is 13 : 17 : 19. The density of A1 is 39 gm/cc and A1, A2 and A3 are mixed in the ratio of 3 : 5 : 7 by weight. If a litre of A3 cost Rs. 38 then find the cost of A3 in 1050 kg of mixture of A1, A2 and A3.

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Correct Ans:Rs. 326.67

Explanation:

Given:

Ratio of weight in mixture = 3 : 5 : 7

Ratio of density of 3 petrol = 13 : 17 : 19

Total weight of mixture = 1050 kg

Density of A1 = 39 gm/cc

Density of A2 = (17/13)*39 = 51 gm/cc

Density of A3 = (19/13)*39 = 57 gm/cc

WKT, Volume = Weight/Density

Now, the weight of A3 in 1050 kg mixture = (1050 x 7)/15 = 490 kg

Volume of A3 = 490/57 liters

Cost of A3 = (490/57)*38 = Rs. 326.67.

Ratio of weight in mixture = 3 : 5 : 7

Ratio of density of 3 petrol = 13 : 17 : 19

Total weight of mixture = 1050 kg

Density of A1 = 39 gm/cc

Density of A2 = (17/13)*39 = 51 gm/cc

Density of A3 = (19/13)*39 = 57 gm/cc

WKT, Volume = Weight/Density

Now, the weight of A3 in 1050 kg mixture = (1050 x 7)/15 = 490 kg

Volume of A3 = 490/57 liters

Cost of A3 = (490/57)*38 = Rs. 326.67.

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15. Sachin and Anurag have monthly incomes in the ratio 6 : 7 and Their monthly expenditures in the ratio 5 : 4. If they save Rs. 700 and Rs. 2100 respectively, find the monthly expenditure of Anurag.

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Correct Ans:2800

Explanation:

Given:

Ratio of income = 6 : 7

Ratio of expenditure = 5 : 4

Let the ratio of income of Sachin and Anurag be 6x and 7x respectively.

Let the ratio of expenditure of Sachin and Anurag be 5y and 4y respectively.

As per the question,

6x - 5y = 700 ...(1)

7x - 4y = 2100 ...(2)

On Solving (1) and (2),

we get x = 700, y = 700

Monthly Expenditure of Anurag = 4y = 4*700 = Rs.2800.

Ratio of income = 6 : 7

Ratio of expenditure = 5 : 4

Let the ratio of income of Sachin and Anurag be 6x and 7x respectively.

Let the ratio of expenditure of Sachin and Anurag be 5y and 4y respectively.

As per the question,

6x - 5y = 700 ...(1)

7x - 4y = 2100 ...(2)

On Solving (1) and (2),

we get x = 700, y = 700

Monthly Expenditure of Anurag = 4y = 4*700 = Rs.2800.

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16. The ratio of salary of two persons X and Y is 5:8. If the salary of X increases by 60% and that of Y decreases by 35% then the new ratio of their salaries become 40:27. What is X's salary?

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Correct Ans:data inadequate

Explanation:

Ratio of salary of X and Y is 5:8

Let the original salary of X and Y be Rs.5k and Rs.8k respectively.

After increasing 60%, new salary of X = 160% of 5k = 160x5k/100 = 80k/10 ...(1)

After decreasing 35%, new salary of Y = (100-35)% of 8k = 65% of 8k = 52k/10 ...(2)

Given that, new ratio is 40:27

That is, 80k/10 : 52k/10 = 40/27

This does not give the value of k;

So that we cannot find X's exact salary.

Hence the answer is data inadequate.

Let the original salary of X and Y be Rs.5k and Rs.8k respectively.

After increasing 60%, new salary of X = 160% of 5k = 160x5k/100 = 80k/10 ...(1)

After decreasing 35%, new salary of Y = (100-35)% of 8k = 65% of 8k = 52k/10 ...(2)

Given that, new ratio is 40:27

That is, 80k/10 : 52k/10 = 40/27

This does not give the value of k;

So that we cannot find X's exact salary.

Hence the answer is data inadequate.

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17. Daniel had 10 paise, 25 paise and 50 paise coins in the ratio of 5 : 4: 9 respectively. After giving Rs. 30 his mother he has Rs. 30. How many 50 paise coins did he have?

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Correct Ans:90

Explanation:

Total money = 30 + 30 = 60

In this question the total money is given so we need to multiply the value of the coins in the ratio.

5 * 0.10 x +4* 0.25x + 9 * 0.50x = 60

.50x + x + 4.5x = 60

6x = 60

x = 10

Number of 50 paise coins = 9 × 10= 90

Hence,

In this question the total money is given so we need to multiply the value of the coins in the ratio.

**coins * ratio of coins = Total money****x = No Of Coins**5 * 0.10 x +4* 0.25x + 9 * 0.50x = 60

.50x + x + 4.5x = 60

6x = 60

x = 10

Number of 50 paise coins = 9 × 10= 90

Hence,

**the answer is 90**
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18. Two numbers are in the ratio of 1(1/2) : 2(2/3). When each of these is increased by 15, the ratio changes to 1(2/3) : 2(1/2). The larger of the numbers is,

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Correct Ans:48

Explanation:

Two numbers are in the ratio of,

1(1/2) : 2(2/3) => 3/2 : 8/3 = > 9 : 16 (9x, 16x)

1(2/3) : 1(1/2) => 5/3 : 5/2 => 2 : 3

(9x + 15)/(16x + 15) = 2/3

27x + 45 = 32x + 30

5x = 15

x = 3

The larger of the numbers is, 16x = 16*3 = 48

1(1/2) : 2(2/3) => 3/2 : 8/3 = > 9 : 16 (9x, 16x)

1(2/3) : 1(1/2) => 5/3 : 5/2 => 2 : 3

(9x + 15)/(16x + 15) = 2/3

27x + 45 = 32x + 30

5x = 15

x = 3

The larger of the numbers is, 16x = 16*3 = 48

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19. Out of three positive numbers, the ratio of the first and the second numbers is 4 : 3 that of the second and the third numbers is 6 : 5 if the product of the second and the third numbers is 4320. What is the sum of three numbers?

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Correct Ans:228

Explanation:

Ratio of 1st and 2nd numbers =4 : 3

Ratio of 2nd and 3rd numbers = 6 : 5

Let the 2nd number = 6x, third number = 5x

Product of 2nd and 3rd numbers = 4320

5x × 6x = 4320

x

x = 12

2nd number = 72, 3rd number = 60 ,

1st number = (72/3)× 4 = 96

Sum of three numbers = 60 + 72 + 96 = 228

Ratio of 2nd and 3rd numbers = 6 : 5

Let the 2nd number = 6x, third number = 5x

Product of 2nd and 3rd numbers = 4320

5x × 6x = 4320

x

^{2}= 144x = 12

2nd number = 72, 3rd number = 60 ,

1st number = (72/3)× 4 = 96

Sum of three numbers = 60 + 72 + 96 = 228

**Hence, the answer is 228.**
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20. Ratio of the salary of Balu& Peelu is 5:8. If the salary of Balu increase by 40% and those of Peelu decrease by 15%, the new ratio of their salary becomes 10:9. What is Balu 's salary?

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Correct Ans:Data inadequate

Explanation:

Let the original salary of Balu& Peelu be 5x & 8x respectively

New salary of Balu = 40% of 5x

= (140/100) * 5x

= (700x /100)

New salary of Peelu =

= (185 /100 )

= (74x/5 )

Therefore 7x : (74x/5) = (10:9)

7x*(5/74x) = (10:9)

New salary of Balu = 40% of 5x

**+ 5x**= (140/100) * 5x

= (700x /100)

New salary of Peelu =

**8x +**85% of 8x= (185 /100 )

= (74x/5 )

Therefore 7x : (74x/5) = (10:9)

7x*(5/74x) = (10:9)

**We can't find x, so the given data is inadequate**
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