Ratio and proportion questions

• Let the third proportional to 12 and 30 is 'a'.
• So, 12 : 30 :: 30 : a becomes in proportion.

• Therefore, products of extremes terms = products of means terms

  \(\Rightarrow\) a \(\times\)12 = 30 \(\times\) 30

  \(\Rightarrow\)12 a = 900

  \(\Rightarrow\) a = \(\frac{900}{12}\)

  \(\Rightarrow\) a = 75.

• Hence, the third proportional to 12 and 30 is 75. .

• Let \(x\) must be subtracted
  Then \(\frac{14-x}{17-x}= \frac{34-x}{42-x}\)
• \((14-x)(42-x)=(34-x)(17-x)\)
• \(x^2-56x+588=x^2-51x+578\) <br> \(5x=10 \implies x=2\)

• The mean proportional between the two terms of a ratio in a proportional is the square root of the product of these two.

• For example, in the proportion \( a:b :: c:d, \) we can define the mean proportional for the ratio a:b as the square root of the product of the two terms of the ratio or \( \sqrt{ab} \).

• Let the third proportional to 12 and 30 be \(x\)

• Then, \(12:30::30:x\)

  \(12x= 30 \times 30 \implies x=75\)

• Third proportional to 12 and 30 \( = 75 \).

• Mean proportional between 9 and 25
  = 15.

• Required ratio = \( 75 : 15 = 5 : 1 \).

According to the question,

A X 2/3 = B X 75/100 = C X 6/10

or, A X 2/3 = B X 3/4 = C X 3/5

therefore A : B : C = 9 : 8 : 10

• Let the two positive numbers be 4x and 5x

• According to the given condition,

  4x X 5x = 180

  x\( \pm \)3

• Only x = 3 is considered because the numbers are positive. So, the numbers are 12, 15.

• The difference between the two numbers is,

  d = 15-12 = 3

• Therefore, the difference between two numbers is 3.

• Volume of mixture = 30 liters
• Volume of milk in the mixture = 7/10 x 30 = 21 litres
• Volume of water in the mixture = 9 liters
• Let x liters water be added in the mixture.

• Now, 21/ 9+ x = 1/2

  42 = 9 + x

  X = 33 liters.

• No.of coins =\(1\times5x:2\times3x:4\times\;x \)

  \((5x+6x+4x)\rightarrow(465) \)

  15x = 465

  x=31.......(i)

• No.of coins continue =\(5\times31,6\times31,4\times31 \)

  =155,186,124

• If the third proportion of two numbers '\(a\)' and '\(b\)' is denoted to be that number 'c' such that,

  \(a:b = b:c\)

• Then, \(c\) = \(\frac{b^{2}}{a}\)

• Here \(a\) = 5 and \(b\) = 10

  \(\Rightarrow \frac{10^{2}}{5}\)

  = 20.

• The third proportion of two numbers '\(5\)' and '\(10\)' is 20.

• Male : female = 4x : 3x

• 42 more females joins the party = 3x + 42

  \(\Rightarrow\)4x /3x+42 = 3/4

  \(\Rightarrow\)16x = 9x+126

  \(\Rightarrow\)7x = 126

  \(\Rightarrow\)x = 18

• So total number of peoples in the party is,

  = 4x +3x = 7 x 18 = 126.

• Let the number of coins be in the ratio

  8x:9x:11x

• Then

  (₹1)8x+(₹0.5)9x+(₹0.25)11x=366

  8x+4.5x+ 2.75x=366

  15.25x=366

  x=366/15.25

  x=24

• That is,

  1rupee coin : 8x = 8x24= 192 coins

  50 paise coin : 9x = 9 x 24 = 216 coins

  25 paise coin : 11x = 11 x 24 = 264 coins

Calcultion:

• Let girls and boys in the class 4x and 5x respectively.

• Therefore, Total Students = 4x + 5x = 9x

• Girls students of mathematics = \(\frac{3}{4}\times 4x = 3x\)

• Girls students of Physics = \(\frac{1}{4} \times 4x = x\)

• % of girls studying in mathematics = \(\frac{3x}{9x} \times 100 = 33.3 \%\).

• Hence, option A is the correct answer.

• Given, the two ratios are 17 : 18 and 10 : 11.

• To compare two or more fractions, make the denominator same by taking their LCM and then compare (same can be done with numerator).

• Then, LCM (18, 11) = 198

  ⇒ 17/18 = 17/18 × 11/11 = 187/198

  ⇒ 10/11 = 10/11 × 18/18 = 180/198

  ∴ 17/18 is greater than 10/11.

• Given, ratio of angles of a triangle = 2 : 3 : 5.
• Sum of all three angles of a triangle is 180°.
• Ratio of angles of a triangle = 2x : 3x : 5x

• According to the question

  2x + 3x + 5x = 180°

  ⇒ 10x = 180°

  ⇒ x = 18°

• Smallest angle is = 2 × 18° = 36°

• Greatest angle is = 5 × 18° = 90°

• ∴ Required ratio = 90° : 36° = 5 : 2

A : B = 1 : 2, B : C = 3 : 4, C : D = 6 : 9, D : E = 12: 16

= A : B : C = 3 : 6 : 8

= A : B : C : D = 9 : 18 : 24 : 36

= A : B : C : D : E = 9 : 18 : 24 : 36 : 48

= 3 : 6 : 8 : 12 : 16

• Required ratio

  = 120% of Rs. 30 : 110% of Rs. 20

  = 36 : 22

  = 18 : 11

• Answer is 18 : 11.

Radii and perimeters are directly proportional to each other

• Let originally the salary of Mohan, Suresh and Nitin are 5x, 7x and 8x respectively.

• The increased salary will be 140% of 5x, 150% of 7x and 175% of 8x

  ( 140 X 5x)/100 : (150 X 7x)/100 : (175 X 8x)/100

• Therefore, the required ratio will be 7x : 21/2 x : 14x

  14x : 21x : 8x = 2:3:4.

• Ratio = \( \frac{1}{4}: \) \( \frac{1}{6}: \)\( \frac{1}{3} \)

  \( => \) \( \frac{1}{4} \times 12 : \) \( \frac{1}{6} \times 12 : \)\( \frac{1}{3} \times 12 \)

  \( => 3 : 2 : 4 \)

• Therefore, share of Ravi, Amit and Alok is \( 3x , 2x, 4x \).

• So, \( 3x + 2x + 4x = 630 \)

  \( x= 70 \).

• Alok Share = \( 4x \) = \( 4 \times 70 \) = \( 280 \) Rs.

• Let three numbers be A ,B and C . So mean proportional is \( A : B = B : C \) , which implies \( B^2 = A\times C \)

• so \( A\times C = 100 \) . \( A = \frac{100}{C} .\)

• And consider three numbers A, C and D , D is the third proportional ,

  \( A : C = C : 80 \) , \( \frac{100}{C} = \frac{C^2}{80} \) , \( C ^3 = 8000 \) , \( C = 20 \) .

• Therefore \(A = 5\) .

• Let the quatities of alloys mixed be x and y respectively.

• Now, \( \frac{7/29 x + 21/58y}{22/29x + 37/58y} \) = 25/62

  (14x + 21y)/(44x + 37y) = 25/62

  232x = 377y

  x/y = 377/232 = 13/8

Alternate method

• Quantity of gold in 1 kg of alloy A = \(\frac{7}{29}\)
• Quantity of gold in 1kg of alloy ‘B’=\(\frac{21}{58}\)
• Quantity of gold in 1 kg of alloy ‘C’ = \(\frac{25}{87}\)

• Required ratio=( \(\frac{21}{58}-\frac{25}{87}\)):(\(\frac{25}{87} -\frac{7}{29}\))

  = 13 : 8.

• Let weigth be 10 gm

• Price = 10\( ^{3} \) = 1000

• Weigth of their pices = 3 : 4 : 3

• Their values = 3gm, 4gm, 3gm

• Their prices = 3\( ^{3} \) + 4\( ^{3} \) + 3\( ^{3} \)

  =27+64+27=118

  Difference =1000-118=882

  % of difference = 882/1000 x 100 = 88.2%

• b is in mean proportion to a and c, so \( \frac{a}{b} = \frac{b}{c} \) = k

• a = bk and b = ck

  \( \frac{ (a-b)^{3} }{ (b-c)^{3} } \) = \( \frac{ (bk-b)^{3} }{ (ck-c)^{3} } \) = \( \frac{ b^{3} }{ c^{3} } \)

  But, \( \frac{a}{b} = \frac{b}{c} \)

• \( \frac{ (a-b)^{3} }{ (b-c)^{3} } \) = \( a^{3} \) : \( b^{3} \)

• Initially alcohol = \(\frac{2}{5} \times 25\) ltr and water is 15 ltr.

• To make a solution of 60% alcohol = \(\frac{10+x}{25+x} = \frac{60}{100}\).

  \(100\times(10+x) = 60\times(25+x)\)

  \(1000+100x = 1500 + 60x\)

  \(100x - 60x = 1500 - 1000\)

  \(40x = 500\)

  \(x = 12.5\)

• Suppose the values of Rs.1 coins, 50 paise coins and 25 paise coins are 2x, 3x and 5x respectively.

  \(\therefore\) The number of coins respectively are

  2x, 3x \(\times\) 2, 5x \(\times\) 4 = 2x , 6x and 20x

• According to the question,

  2x + 6x + 20x = 420

  28x = 420 \(\Rightarrow\) x = 15

• Number of Rs.1 coins = 2x = 15 \(\times\) 2 = 30.

• Let the angle of triangle be 3x, 5x, and 7x respectively

  3x + 5x + 7x = 180

  x= 12

• Angles will be = 36, 60 and 84 degrees.

• Required ratio of external angles = 144 : 120 : 96 = 6 : 5: 4.

Spirit : Water = {(6/5 + 12/10 + 20/15)} : {(9/5 + 28/10 + 55/15}

112 : 248 = 14:31

• Given, the numbers are 54, 71, 75, and 99.

• Assume that x be number subtracted from each of the numbers 54, 71, 75 and 99 so that their remainders will be a proportion.

  ∴ 54 - x , 71 - x , 75 - x , and 99 - x are in proportion.

  ∴ (54 − x)/(71 − x) = (75 − x)/(99 − x)

  ⇒ (54 − x)(99 − x) = (75 − x)/(71 − x)

  ⇒ x\(^{2}\)− 99x − 54x + 5346 = x\(^{2}\) − 75x − 71x + 5325

  ⇒ −153x + 146x = −5346 + 5325

  ⇒ −7x = −21

  ⇒ x = 3

• ∴ When x = 3 subtracted from each of the numbers so that the remainders will be a proportion.

• It is given that the number of employees are reduced in the ratio 8:5
• So, \(\frac{number\ of \ employees\ before}{number\ of\ employees\ after}\) = \(\frac{8}{5}\)
• It is also given that their wages(per employee) were increased in the ratio 7:9
• So, \(\frac{Initial\ wages}{New\ wages}\) =\(\frac{7}{9}\)
• Let the initial number of employees = 8x
• And the final number of employees = 5x
• Also, let the total initial wage = 7y
• And the new wage = 9y
• Remember that here, x and y are constants.
• The Overall Wage bill will depend on both, the number of employees and the wage of each employee. It will be equal to the product of the number of employees and the wage per person.
• Therefore, overall wages bill = number of employees × wage per person
• So, initial overall wages bill = 8x×7y=56xy

• Final overall wages bill =5x×9y=45xy

  ∴ Ratio = \(\frac{Initial\ overall\ wage\ bill}{Final\ overall\ wage\ bill}\)

  Ratio =\(\frac{56xy}{45xy}\) = \(\frac{56}{45}\)

  Ratio→ 56:45

  ∴ The overall wages bill is decreased in the ratio 56:45.

• Value of broken diamonds =(1+4+9+16)=30 unit
• Original diamond value (1+2+3+4)\(^2\)=100 unit

• Then (70 unit) \(\rightarrow70,000 \)

  100 unit \( \rightarrow\frac{70,000}{70}\times100 \)

  =100000

• Initial salary ratio = \(9 : 6 : 7\)

• Ratio of salary after increase = \(9\times\frac{168}{100} : 6\times\frac{189}{100} : 7\times\frac{144}{100}\)

  = \(12 : 9 : 8\)

Let ratio = k : 1

Let C as (x, 0)

Section Formula = (6k + 2)/ (k + 1) = x and (5k -3 )/(k+1) = 0

5k - 3 = 0

k = 3/5

Ratio = 3:5

• Let the amounts invested by A and B in the beginning be 2x and 3x.
• Let the amount by A after 6 months be y.

• A’s share by the end of the year will be proportional to

  \(2x \times 6 + y \times 6 = 12x + 6y\)

• B’s share by the end of the year will be proportional to

  \(3x \times 12 = 36x\)

• It has been given that

  \(\frac{12x+6y}{36x}=\frac{3}{2}\)

  \(y=7x\)

• Ratio of the amount invested by A initially to the amount invested after 6 months \(= 2x : 7x = 2:7\).

• Let the alcohol added be x liter

• ATQ

  \(\frac{9+x}{30+x}=\frac{2}{5}\)

  \(\Rightarrow\) \(45+5 x=60+2 x\)

  \(\Rightarrow\) 3 x = 15

  \(\Rightarrow\) x = 5 liter.

Calculation :

• When four number a, b, c and d are in proportion we can write

  ab =cd

• Now here we have to find the no. which when subtracted from numerator and denominator leaves the remainder in proportion.

• Thus let that no. Be “x”.

  So the equation will be :

  \(\frac{(36 - x)}{(48- x)} = \frac{(56 - x)}{(76 - x)}\)

  ⇒\(\frac{(36 - x)}{12}= \frac{(56 - x)}{20}\) (if a/b=c/d , then a/(b-a)=c/(d-c))

  Or

• \(720 - 20 x = 672 - 12 x\)

  \(8 x = 48\)

  \(x = 6\)

• Required number = \(6\)

• Alternate Solution:

  \(x = \frac{(ad-bc)}{(a+d)-(b+c)}\)

  = \(\frac{(36x\times 76 - 56 x\times 48 )} {(36+ 76) - (56 +48 )}\)

  = \(\frac{48}{8}= 6\) .

Hence, the answer is option B

For the initial solution,
Total volume = 60 litres
Volume of alcohol = \( \frac{5}{12} \ of \ 60 = 25 \ litres\)
Volume of water = \( \frac{7}{12} \ of \ 60 = 35 \ litres\)
Now we add water to change the proportion to 5:11
So, the volume of alcohol will not change.
Volume of alcohol in the new solution = 25 litres
Since water:alcohol = 11:5
=> Volume of new solution=\( 25 \times \frac{16}{5} = 80 \ litres\)
So, water added = 80-60 = 20 litres.
Volume of water in new solution = 55 litres.
Now 32 litres of the solution is removed. So, remaining volume of solution = 48 litres.
Volume of alcohol remaining = \( 25 - \frac{5}{16} \times 32 = 15 \ litres \)
Volume of water remaining = \( 55 - \frac{11}{16} \times 32 = 33 \ litres \)
To make ratio 2:3, amout of alcohol must be, \( \frac{2}{3} \times 33 = 22 \ litres \)
So, volume of alcohol added = 22-15 = 7 litres.

• Let the base be x and y

• Let their heights be 6h and 7h

\(\frac{(\frac{1}{2} \times x \times 6h)}{(\frac{1}{2} \times y \times 7h)} = \frac{2}{5}\) \(=> \frac{x}{y} = \frac{7}{15}\)

• Required ratio is 7:15 .

• A : B = 2 : 3 , B : C = 2 : 4 and C : D = 2 : 5

• \(\frac{A}{D} = \frac{A}{B} \times \frac{B}{C} \times\frac{C}{D}\)

  =\( \frac{2}{3} \times \frac{2}{4} \times\frac{2}{5}\)

  = 2 : 15

• Answer is \( 2:15 \).

Given that the selling prices of three articles, S.P1 =5X, S.P2 =6X, S.P3 =9X, And their cost prices are C.P1 =4Y, C.P 2 =5Y, C.P3 =8Y Given that , S.P1 –C.P1 =S.P2 –C.P2, 5X-4Y =9X-8Y, X =Y .

Their profit percentages are, p1 =\(\frac{5-4}{4}\)\(\times\)100 = 25% ,

p2 = \(\frac{6-5}{5}\)\(\times\) 100 = 20%,

p3 = \(\frac{9-8}{8}\)\(\times\)100 = \(\frac{121}{2}\) %

Ratio of the percentages is 25:20:\(\frac{121}{2}\) = 10:8:5

• Let boys and girls in the class \(5x\) and \(6x\) respectively.
• ∴ Total Students \(= 5x+6x=11x\)
• Boy students of mathematics \(= \frac{2}{3}\times 5x= \frac{10}{3} x\)
• Boy students of Physics \(= \frac{1}{3} \times 5x= \frac{5}{3}x\)
• Percentage of boy's studying in mathematics \(=\displaystyle \frac{{ \frac{10}{3}}x}{11x}\times 100= 30.30 \%\)

• The third proportional of a proportion is the second term of the mean terms.
• For example, if we have a:b = c:d, then the term 'c' is the third proportional to 'a' and 'b'.

• The required formula is

• Let the third proportion be x , 8:12 = 12:x

  \( \Rightarrow \frac{8}{12} = \frac{12}{x} \)

  \( \Rightarrow \)x = 12 \(\times\) \( \frac{12}{8} \)

  \( \Rightarrow \) \( x = 18. \)

• Let originally the salary of Suman,Samar and Shreshth are \(7x ,5x\) and \(8x\) respectively.

• The increased salary will be , \((150\%\) of \(7x), (140\%\) of \(5x) \) and \( (175\%\) of \(8x)\)

  \( ⇒ \left(\frac{150}{100}\times 7x\right) :\left(\frac{140}{100}\times 5x\right) :\left(\frac{175}{100}\times 8x\right)\)

• Therefore, the required ratio \(= \frac{21}{2}x :7x :14x\)\(⇒21x : 14x : 28x\)

  \(⇒3:2:4\)

• Quantity of the milk in mixture = \[60 \times \frac{7}{12}\] = 35 litres
• Quantity of the water in the mixture = \[60 \times \frac{5}{12}\] = 25 litres

• According to the question, \[\frac{35}{25 + x} = \frac{1}{1}\] 35 = 25 + x

  x = 10 litres

• Let weigtht be 10 gm price =10\(^{3}\)=1000
• Weight of their prices = 4 : 5 : 4
• Their values = 4 gm, 5gm, 4gm
• Their prices =4\(^{3}\)+5\(^{3}\)+4\(^{3}\)=64+125+64=253
• Difference =1000-253=747
• % of difference =\(\frac{ 747}{1000}×100=74.7\%\)