Teena, Easha and Kavin savings are in the ratio 2: 9: 11.The sum of the savings of all three is Rs.3300. What percent is Teena's saving than that of the other two members?

Let Teena, Easha and Kavin savings be 2x, 9x and 11x.

2x + 9x +11x = 3300 = x --> 150.

Teena’s savings = 2x = 300

Easha’s savings = 9x = 1350

Kavin’ssavings = 11x = 1650

-->\({300 \over {(1350+1650)}}*100=10%\). Teena savings is 10% than that of the two members.

Hence (a).

Find the simplest ratio of 5\({4 \over {9}}\) : 2\({1 \over {3}}\)

The simplest form => 5\({4 \over {9}}\) : 2 \({1 \over {3}}\) = \({49 \over {9}}\) : \({7 \over {3}}\)

= \({49 \over {9}}*{3 \over {7}}\) = 7:3 . Hence (b).

In a chocolate factory having 132 employees, the male and female employees are in the ratio 5:7.If some more female employees were added ratio becomes 11:19.Then the number of female employees exceed the male employees by what number?

Total employees = 132.

Let the male and the female employees be 5x and 7x. => 5x +7x = 132

Then, the number of male and female employees are 55 and 77 respectively.

New ratio of male and female employees à 11:19

11y = 55 à y = 5

Number of female employees = 19y = 95.

Therefore, female employees exceed the male employees by 40. Hence (d).

A starts business with a capital of Rs.2, 80,000. After a few months B also joins with an investment of 4, 20,000. After how many months did B join if they shared profits equally after one and half years?

Let B join the business after x months. Profit is shared in the ratio of their investments

Then, the ratio of profit after one and half years 1:1

(280000 * 18) : (420000 * (18 – x)) = 1:1

420000x = 18(420000 – 280000)

X = 6. Hence (c).

If m:n = 7:2, then find the value of \({6m-5n \over {11m+2n}}\)

Ratio m:n = 7:2

\({6m-5n \over {11m+2n}}\) = \({6(7x)-5(2x) \over {11(7x)+2(2x)}}\)

= \({32 \over {81}}\) Hence (b).

A 50 litres fruit juice mixture contains 10% milk. How much more milk must be added so that milk becomes 25% of the new fruit mixture?

Number of litres of milk in 50 litres of the fruit mixture = 10% of 50 = 5 litres

Let 'x' litres of milk be added to the fruit mixture to make milk 25% of the new fruit mixture. So, the total amount of milk becomes (5 + x) and the total volume of the mixture becomes (50 + x)

Thus, (5 + x) = 25% of (50 + x)

On solving the above equation, we get x = 10 litres. Then 10 litres should be added.Hence (c).

In what ratio must rice at Rs.70 per kg be mixed with rice at Rs.95 per kg so that the cost price of the mixture is Rs.80 per kg?

By alligation method

Therefore, the required ratio = 15:10 = 3:2Hence (b).

How many litres of water must be added to 100 litres of 20% ink solution so as to reduce the ink percentage to 10?

Let the quantity of water to be added be x litres.

Quantity of ink in 100 litres of 20% ink solution = 20 litres

Total quantity of the ink solution after adding water = (100 + x)

Ink percentage in the new solution = \({20 \over {100+x}}\) * 100 = 10

X= 100 litres.Hence (c).

A bag contains 50 paisa, 5 rupee, and 10 rupee coins in the ratio 7:11:3.If the total amount in the bag is Rs.708.Find the number of 50 paisa coins in the bag.

Let the number of 50p, 5 rupees and 10 rupees coins be 7x, 11x and 3x respectively.

7(\({x \over {2}}\) ) + 11 (5x) + 3 (10x) = 708

\({177x \over {2}}\) = 708 à x = 8.

Therefore, number of 50 paises coins = 7x = 56. Hence (d).

Some Amount of money is received by A, B and C each. If B receives 25% greater than A and C receives 25% less than B. Find the proportion of A.

B receives 25% greater than A and C receives 25% less than B

=> B = 1.25A, C = 0.75B = 0.75(1.25A)

= 0.9375A

Ratio will be = A: 1.25A: 0.9375A = 16:20:15

Proportion of A = \({16 \over {51}}\)Hence (a).

Team members P,Q and R are required to prepare equal number of questions for an exam. Q and R prepared 10 and 8 questions respectively. P prepared no questions and is fined Rs.96 in lieu of his nil contribution. How much amount did Q and R get?

Number of questions with P,Q and R = 10 + 8 +0 = 18

They shared questions equally. = \({18 \over {3}}\) = 6 questions.

Number of questions from Q to P = 4 and from R to P = 2

Rs.96 from P should be shared by Q and R in the ratio = 2:1

Q gets =96 * \({2 \over {3}}\) = 64, R gets = 96 * \({1 \over {3}}\) = 32 Hence (d).

In a government exam, Reta gets an aggregate of 75% marks in the six sections in the ratio 7:8:9:11:5:10.The minimum cut off is 69% in each section to qualify. The maximum mark of six sections is same. In how many sections did she qualify?

Reta’s aggregate percentage in six section = 75%

Total marks = 6 * 75 = 450

Let marks in six section be 7x, 8x, 9x, 11x, 5x and 10x respectively

7x + 8x + 9x + 11x +5x + 10x = 450

X = 9

Marks obtained in the six sections = 63, 72, 81, 99, 45 and 90

Pass marks = 69%

She qualified in 4 sections. Hence (c).

A Father, Mother and Son are in the ratio of 9:7:2.After 8 years, Son’s age is 40% of Mother’s age. Find Father’s age before 10 years.

Let f ather, mother and Son’s age be 9x, 7x and 2x respectively

After 8 years Son’s age is 40% of Mother’s age

(2x + 8) =\({2 \over {5}}\) (7x + 8) à x = 6

Father’s age before 10 years = 9x – 10 = 44. Hence (a).

Two fruit juice containers each contain milk and fig juice in the ratio 2:3 and 1:4 respectively. The capacity of first and second fruit juice container is 60 and 65 litres respectively. If the two fruit containers are mixed, find the amount of fig juice in the resultant mixture.

In the first juicer, the ratio = 2:3 Fig juice in first juicer = \({3 \over {5}}\) * 60 = 36 litres. In the second juicer, the ratio = 1:4 Fig juice in second juicer = \({4 \over {5}}\) * 65 = 52 litres. Therefore, Fig juice in the resultant mixture = (36 + 52) = 88 litres.

The cost price of two variety of grains are Rs.180 per kg and Rs.240 per kg respectively. The two varieties are mixed with a third variety in the ratio 1:1:3. If the resultant mixture costs Rs.120 per kg. What is the cost price of third variety per kg?

let the cost of third variety of grain be b.

The quantities of three varieties be a kg, a kg and 3a respectively.

Cost = \({180a+240a+3ab \over {a+a+3a}}\) = 120

\({180a+240a+3ab \over {5a}}\) = 120

b= 60 kg. Hence (d).

12 litres are drawn from a milk can and is then filled with equal amount of water. This operation is repeated two more times. The ratio of the quantity of milk now left in can to that of the water is 27 : 64. Find the capacity of the can?

By iteration method

Final concentration = Initial concentration (1- \({y \over {x}}\) ) ^{n}

\({27 \over {64}}\) = (1 - \({12 \over {x}}\) ) ^{3}

\({3 \over {4}}\) = 1 - \({12 \over {x}}\)

X = 48 litres. Hence (a).

A container has 240 litres of 55% wine. How many litres of 80% wine must be added so that the resultant solution has 70% wine?

\({240 \over {x}}\) = \({2 \over {3}}\)à x = 360 litres. Hence (c).

There are 2 containers of acid. The first contains 66.67% water and the second contains 37.5% water. What quantity of acid must be added from each of the containers so as to obtain 35litres of the solution with water and acid in the ratio of 5:7?

Prop ortion of Acid in first can = 33.33 % = \({1 \over {3}}\)

Proportion of Acid in Second can = 37.5% = \({5 \over {8}}\)

Resultant proportion of acid = \({7 \over {12}}\)

By alligation

Ratio à 6:1

So, the quantity of mixture required from the fir st can = \({6 \over {7}}\) * 35 = 30 litres.

The quantity of mixture required from the seco nd can = \({1 \over {7}}\) * 35 = 5 litres. Hence (b).

A family comprises Mr and Mrs. Ram plus their two daughters. The salaries of Mrs. Ram and the two daughters are in the ratio 4: 3: 2. The salaries of Mr and Mrs. Ram are in the ratio 3: 2. Mr. Ram will pay one third of his salary for home loan. The total expenses are \({4 \over {9}}\)^{th} of total salary of all the family members. Another Rs.12000 they are spending for a monthly jewel scheme. \({1 \over {9}}\)^{th} of total salary is invested in a bank. If the salary of the younger daughter is Rs.18000, Find the percentage of their savings.

Ratio of salaries of Mrs. Ram and her two daughters = 4: 3: 2

Ratio of salaries of Mr and Mrs.Ram = 3: 2

The combined ratio of four members = 6: 4: 3: 2

Let their salary be 6x, 4x, 3x and 2x.

Salary of younger daughter = Rs.18000 => 2x = 18000 à x = 9000

Total salary = 15x = Rs.135000

Mr.Ram ’s salary = 6x = Rs.54000,Mr.Ram will pay one third of his salary for home loan ( i.e. ) = Rs.18000

The total expenses are \({4 \over {9}}\) ^{th} of total salary of family members = \({4 \over {9}}*135000=\) Rs.60000

Another Rs.12000 they are spending for monthly jewel scheme

\({1 \over {9}}\) ^{th} of total salary invest in bank = Rs.15000

Remaining salary = 135000 – (18000 + 60000 + 12000 + 15000) = 30000

Percentage of remaining salary = \({30000 \over {135000}}*100\) = 22.22% . Hence (a).

The ratio of marks obtained by A, B and C in an exam is 10: 7: 13. If the combined average of their percentages is 83.33% and the sum of their marks is 900, find the maximum marks in the exam.

Let A’s marks be 10 x , B’s marks be 7x and C’s marks be 13 x.

Th erefore, sum of their marks = 30x = 900 à x=30

Therefore, A ’s marks = 10x = 300, B’s marks = 7x = 210 and C ’s marks = 13x = 390 .

Combin ed average of their marks = (300+210 + 390) / 3 =300 à 83.33 % (given in question )

Let maximum marks be y.

83.33% of y = 300 à 360

∴ Maximum marks in the exam = 36 0 . Hence (d).

Two containers X and Y having equal capacity contains a mixture of alcohol and water in the ratio 1:3 and 5:1 respectively. One fourth of the mixture in container X is poured out and replaced with the mixture from container Y. Find now the ratio of alcohol and water in container X.

Let the initial capacity in both the containers be a.

Proportion of alcohol and water in X =\({a \over {4}}\), \({3a \over {4}}\)

Proportion of alcohol and water in Y =\({5a \over {6}}\), \({a \over {6}}\)

After removing \({1 \over {4}}\)^{th} of mixture the remaining mixture in container X = \({3a \over {16}}\), \({9a \over {16}}\)

Now, \({1 \over {4}}\)^{th} of mixture in container Y is removed and added to X, then the ratio is

\({3a \over {16}}\) + \({5a \over {24}}\): \({9a \over {16}}\) \(+{a \over {24}}\) = 19: 29

∴ Ratio of alcohol and water in the container x = 19:29. Hence (c).

A milk vendor has 120 litres of pure milk in a container. He removes 10 litres of milk and replaces with water. After that he removes 8 litres of milk solution and replaces with water. Again, he removes another 5 litres of milk solution and replaces with water. Again, he removes another 24 litres of milk solution and replaces with water. Find the amount of water in the final milk solution.

Initial amount of pure milk in the container = 120 litres

After 4 replacements, final amount of milk in the solution

⇒ 120 * (1 – \({10 \over {120}}\) ) (1 – \({8 \over {120}}\) ) (1 – \({5 \over {120}}\) ) (1- \({24 \over {120}})\)

⇒ 120 *(\({11 \over {12}})\) (\({14 \over {15}})({23 \over {24}})\)(\({4 \over {5}})\)

=>78.71 litres

Therefore, final amount of water in the solution = 120 –78.71 = 41.29 litres. Hence (a).

The cost of painting a square shape board varies directly as the square of its area. It will cost Rs.24000 to paint a square shape board of 4m * 4m dimension. What will be the loss percentage to the painting artist if she were to paint 6 boards of dimension 1m * 1m instead of one 4m * 4m board?

Cost x = p(A ^{2} ), where p = constant, A = area

A = 4 * 4 = 16 sq metres, cost = 24000 = p(16)à p = 1500

Then, the relation between the cost and area x = 1500(A) ^{2}

When the dimensions are 1m * 1m, the area is 1 sq metre

Cost of painting one 1m * 1m board = 1500(1) ^{2} = Rs.1500

Cost of painting 6 such boards = 6 * 1500 = 9000

Loss % = \({15000 \over {24000}}\) * 100 = 62.5% . Hence (c).

90 litres of a solution contain Honey and Sugar in the ratio 5: 4. 18 litres of the solution are removed and 3 litres of Sugar solution are added to it. Again, 15 litres of solution are removed. Find the final ratio of Honey and Sugar solution.

Ratio of Honey and Sugar solution in the solution of 90litres = 5:4

Therefore, quantity of Honey⇒ (\({5 \over {9}}\))*90 = 50litres; quantity of Sugar solution⇒ (\({4 \over {9}}\))*90 =40litres

After removing 18 litres of the solution, quantity of Honey in the remaining solution ⇒ (\({5 \over {9}}\))*72=40litres; and quantity of Sugar solution = (\({4 \over {9}}\))*72 = 32litres.

With addition of 3litres of Sugar solution,

Quantity of Sugar solution = 32+3 = 35litres.

∴New ratio of Honey and Sugar solution⇒ 40:35⇒ 8:7.

Even after removing 15 litres of the solution ratio will remains same

∴ Final ratio of Honey and Sugar solution = 8:7.Hence (a).