fdaytalk · 15 a bag contains rs. 90 in coins of denominations of 50 paisa, 25 paisa and 10 paisa....

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Author : J. Maha Laxmaiah Mailid : [email protected]

If the price of commodity increases by X%, then the reduction in the consumption

so as not to increase the expenditure

= x

100 + x× 100

And, in case of decrease

= x

100 − x× 100

Let the population of a city be ‘P’ now and suppose it increases at the rate of R%

per annum, then

Population after n- years

= p × (𝟏 + 𝑹

𝟏𝟎𝟎)𝒏

Population n- years ago

= 𝑷

(𝟏+ 𝑹

𝟏𝟎𝟎)𝒏

If A is R% more than B, then B is less than A by

= R

100 + R∗ 100

If A is R% less than B, then B is more than A by

= R

100 − R∗ 100

If A : B = 3 : 5 and B : C = 4 : 7 then A : B : C is ……..

SOLUTION:

Given that, A : B = 3 : 5 …………… (1)

B : C = 4 : 7 ……………. (2)

A : B : C =?

From equations (1) & (2) ‘B’ is common, so equalize the B with suitable number

(1) × 4 and (2) × 5, we get

12 : 20 : 20 : 35

∴ A : B: C = 12 : 20 : 35 (answer)

If A : B = 1 : 2, B : C = 3 : 4 and C : D = 2 : 3 then, A : B : C : D is …………….

SOLUTION:

Given that, A : B = 1 : 2 ……….. (1)

B : C = 3 : 4 ……….. (2)

A : B : C = ?

From equations (1) & (2) ‘B’ is common, so equalize the B with suitable number

(1) × 3 and (2) ×2 , we get

= 3 : 6 : 6 : 8

A : B : C = 3 : 6 : 8 …………….. (3)

C : D = 2 : 3 …………………. (4)

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From equations (3) & (4) ‘C’ is common, so equalize the C with suitable number

(3) ×1 and (2) ×4 , we get

= 3 : 6 : 8 : 8 : 12

∴ A : B : C : D = 3 : 6 : 8 : 12 (answer)

1 In a particular constituency, 60% of voters cast their votes, out of which

4% were rejected. The winning candidate received 75% of the valid votes and

bagged a total of 6750 votes. The total no. of voters in the constituency was ….

SOLUTION:

Let the total no. of voters in the constituency be X.

Then,

SHORT METHOD:

x × 60

100∗

100−4

100∗

75

100= 6750

X = 15625 (answer)

2 If the side of a square is increased by 20%, by how much % its area is

increased?

SOLUTION:

Original = 100×100 = 10000

Revised = 120×120 = 14400

Increase% = 14400−10000

10000 × 100

= 44% (answer)

OR

Use [x + y + 𝒙𝒚

𝟏𝟎𝟎 ]%

= [20 + 20 + 𝟐𝟎×𝟐𝟎

𝟏𝟎𝟎 ]

= 44% (answer)

NOTE:

Use + sign, when there is increase

Use – sign, when there is decrease

3 One side of a square is increased by 10%. To maintain the same area, how

much % the other side is to be decreased?

SOLUTION:

OLD AREA NEW AREA

100 × 100 110 × x

x= 𝟏𝟎𝟎×𝟏𝟎𝟎

𝟏𝟏𝟎 = 90

𝟏𝟎

𝟏𝟏

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% decrease of other side = 100 − 90 𝟏𝟎

𝟏𝟏

= 9 𝟏

𝟏𝟏 % (answer)

OR

Use % reduction of other side = [𝒙

𝟏𝟎𝟎+𝒙 ]𝟏𝟎𝟎

= 𝟏𝟎

𝟏𝟏𝟎 × 100

= 9 𝟏

𝟏𝟏% (answer)

4 Cost of sugar is increased by 20% and consumption is reduced by 10%.

What is the net effect on the expenditure?

SOLUTION:

Formula, [x + y + 𝒙𝒚

𝟏𝟎𝟎 ]%

Here, 20 - 10 - 𝟐𝟎×𝟏𝟎

𝟏𝟎𝟎 = 8% increase (answer)

OR

SHORT METHOD:

100 × 𝟏𝟐𝟎

𝟏𝟎𝟎×

𝟗𝟎

𝟏𝟎𝟎 - 100

= 8% (answer)

Note: here 20% increase means 𝟏𝟐𝟎

𝟏𝟎𝟎 and 10% decrease means

𝟗𝟎

𝟏𝟎𝟎

Same shortcut will be applicable for numbers, salaries, sales, squares, circles etc...

5 Cost of sugar is decreased by 20%. By how much %, consumption can be

increased so as to maintain the same expenditure?

SOLUTION:

Formula

[ 𝒙

𝟏𝟎𝟎−𝒙 ] ×100% (for decreasing)

= 𝟐𝟎

𝟖𝟎 ×100 = 25% (answer)

[ + sign to be used when there is increase]

[Same formula will hold good while dealing with numbers, squares, circles etc...]

6 A reduction in 20% in the price of sugar enables me to purchase 5 kg

more for Rs. 600. What is the price of sugar per kg before reduction of price?

SOLUTION:

Here, expenditure before increase = expenditure after increase

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Cost × Consumption (before) = Cost × Consumption (after)

100 × x = 80 (x+5)

By solving we get, X = 20

Hence, initial consumption = 20 kg

Initial Cost/Kg = 𝟔𝟎𝟎

𝟐𝟎 = Rs. 30 (answer)

7 The population of a town increased by 10% during the first year and 20%

during the second year. If the original population was 5000, what is the

present population?

SOLUTION:

5000 x

x= 5000 × 𝟏𝟎𝟎+𝟏𝟎

𝟏𝟎𝟎 ×

𝟏𝟎𝟎+𝟐𝟎

𝟏𝟎𝟎 = 6600

[ If there is decrease, we should subtract from 100 in the numerator.]

The same analogy is applicable for numbers, salaries etc.

8 The population of a town is increased by 20% during the first year and

reduced by 10% during the second year. If the present population is 4320.

What was the population 2 years ago?

SOLUTION:

x 4320

X = 4320 × 𝟏𝟎𝟎

𝟏𝟎𝟎−𝟏𝟎 ×

𝟏𝟎𝟎

𝟏𝟎𝟎+𝟐𝟎

X = 4000 (answer)

9 In measuring the sides of a rectangle, one side is taken 10% in excess and

the other side is 20% in deficit. Find the error percent in area calculated from

the measurement.

SOLUTION:

Formula

[x + y + 𝒙𝒚

𝟏𝟎𝟎 ] %

Error % = 10−20−𝟏𝟎× 𝟐𝟎

𝟏𝟎𝟎

= − 12%[Deficit] (answer)

+10% +20%

+20% −10%

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10 An increase of 20% in the price of mangoes enables a person to purchase

4 mangoes less for Rs. 40. The price of 15 mangoes before increase was

SOLUTION:

Expenditure before and after will be the same

100x =120 (x− 4)

X= 24

∴ Cost of 24 mangoes before increase = 40

Cost of 15 mangoes before increase = 𝟒𝟎

𝟐𝟒 × 15

= Rs. 25 (answer)

11 In an examination, it is required to get 65% of the aggregate marks to

pass. A student got 540 marks and failed by 5% of marks. What are the

maximum aggregate marks a student can get?

SOLUTION:

Let the maximum marks be x

𝟔𝟓

𝟏𝟎𝟎 × X = (540+

𝟓

𝟏𝟎𝟎× 𝑿)

𝟔𝟎

𝟏𝟎𝟎 × X = 540

∴ x= 900 (answer)

12 A candidate scores 25% and fails by 60 marks, while another candidate

scores 50% marks, gets 40 marks more than that of the minimum required

marks to pass the examination. The maximum marks for the examination is

SOLUTION:

Let the maximum marks be x

According to question

𝟓𝟎 𝒙

𝟏𝟎𝟎 –

𝟐𝟓 𝒙

𝟏𝟎𝟎 = 60 + 40

By solving we get, X = 400 (answer)

13 An employer reduces the number of his employees in the ratio 9 : 8 and

increases their wages in the ratio 14 : 15. If the original wage bill was Rs.

18900, the ratio in which the wage bill is decreased, is

SOLUTION:

Total Wages = Employees × Wages

. Initial After

Employees 9x 8x

× ×

Wages 14y 15y

------------------------------

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Total wages = 126xy : 120xy

= 21 : 20 (answer)

14 A bag contains 10-paisa and 25-paisa coins in the ratio 17 : 6. If the total

money in the bag is Rs. 112, then find the number of 10-paisa coins.

SOLUTION:

Amount = no.of coins × coin denomination

Denominations 10 paisa 25 paisa

Coins 17 : 6

Amounts 17 × 10 : 25 × 6

= 170 : 150

= 17 : 15

Total amount in the bag = 32x (17 + 15)

32X = Rs. 112 (given)

X = Rs. 3.5

10 paisa coins amount = 17x = Rs. 59.5

∴ No. of 10 paisa coins = 595 (answer)

15 A bag contains Rs. 90 in coins of denominations of 50 paisa, 25 paisa and

10 paisa. If coins of 50 paisa, 25 paisa and 10 paisa are in the ratio of 2 : 3 : 5,

then the number of 25 paisa coins in the bag is

SOLUTION:

Amount = no.of coins × coin denomination

Denominations 50 Paisa 25 Paisa 10 Paisa

Coins 2 : 3 : 5

Amount 50 ×2 : 25 × 3 : 10 × 5

100 : 75 : 50

= 4 : 3 : 2

Total amount = 9x (4 +3 + 2)

9x = Rs. 90 (given)

x = Rs. 10

25 paisa coins amount = 3 x = Rs. 30

∴ No. of 25 paisa coins = 30 × 4

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=120 coins (answer)

16 Two vessels A and B of equal capacities contain mixtures of milk and

water in the ratio of 4 : 1 and 3 : 1 respectively. 25% of the mixture from A is

taken out and added to B. After mixing it thoroughly, 20% of the mixture is

taken out from B and added back to A. The ratio of milk to water in vessel A

after the second operation is …

SOLUTION:

Assume there is 20 liters of the mixture in both the vessels

In vessel A, milk = 16 liters and water = 4 liters

25% from A to B; milk in B = 15 + 4 = 19 liters

Water in B = 5 + 1 = 6 liters

Ratio 19 : 6

Now 20% of amount from vessel B is added back to A

Milk in A = 12 + 𝟏𝟗

𝟓=

𝟕𝟗

𝟓 liters

Water in A = 3 + 𝟔

𝟓 =

𝟐𝟗

𝟓 liters

Hence the ratio is 79 : 21 (Answer)

17 A student scored 20% marks & failed by 32 marks, while another

student who scores 32% marks gets 40 marks more than the minimum

required pass marks. What are the maximum marks for the examination ?

SOLUTION:

Let maximum marks = X


20% of X + 32 = 32% of X – 40

20

100× x + 32 =

32

100× x − 40

X = 600 (answer)

18 If 2 liter of water is evaporated on boiling from 8 liters of sugar solution

containing 5% sugar. Find the % sugar in the remaining solution …

SOLUTION:

Sugar = 5% of 8 => 𝟐

𝟓 liters

After 2 liters evaporated, remaining 8 - 2 = 6 liters

Therefore, sugar % = 𝟐

𝟓 ×

𝟏𝟎𝟎

𝟔

= 𝟔𝟑𝟐 % (answer)

19 If Ram’s income is 40% less than that of Shyam’s. How much percent

Shyam’s income is more than that of Ram’s.?

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SOLUTION:

Formula, X

100−X× 100

= 40

100−40× 100

= 66.6% (answer)

20 A mixture of 90 liters of milk & water contains 20% of water. How much

water must be added to mixture so as to make water content equal to 25%.?

SOLUTION:

Short Method:

Water = 20% of 90 = 18

Milk = 80% of 90 = 72 liters

Let X- liters water added to mixture

Water/milk => (18 + X)/72= 25/75

X = 6 liters (answer)

21 The population of a town is 6000. In the next year, if males increase by

8% and females 10%, the population becomes 6530. Find the no. of males in

the town ….

SOLUTION:

Let males = X & females = 6000 – X


108

100∗ X +

110

100∗ (6000 − X) = 6530

X = 3500 (answer)

22 An alloy contains copper, zinc& nickel in the ratio of 5: 3: 2. The

quantity of nickel in kg that must be added to 100 kg of this alloy to have the

new ratio 5: 2: 3.

SOLUTION:

Let copper: zinc: nickel = 5X: 3X: 2X

5X + 3X + 2X = 100

X = 10

Nickel, 2X = 2 × 10 =20 kg

And, 5X + 2X + 3X = 100

X = 10

Nickel, 3X = 3 × 10 = 30 kg

Required answer => 30 -20 = 10 kg (answer)

23 Two equal vessels are filled with the mixtures of water & milk in the

ratio of 3: 4 and 5: 3 respectively. If the mixtures are poured into a third

vessel, the ratio of water & milk in the third vessel will be ….

SOLUTION:

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Total mixture = X

Vessel 1: water = X × 𝟑

𝟕, Milk = X ×

𝟒

𝟕

Vessel 2: water = X × 𝟓

𝟖, milk = X ×

𝟑

𝟖

Water => 𝟑𝑿

𝟕+

𝟓𝑿

𝟖=

𝟓𝟗𝑿

𝟓𝟔

Milk => 𝟒𝑿

𝟕+

𝟑𝑿

𝟖=

𝟓𝟑𝑿

𝟓𝟔

Therefore, water/milk =>

=

𝟓𝟗𝒙

𝟓𝟔𝟓𝟑𝒙

𝟓𝟔

= 𝟓𝟗

𝟓𝟑 (answer)

24 The ratio of the income to expenditure of a family is 10: 7. If the family

expenses are 10500/-, then the savings of the family is …..

SOLUTION:

Ratio Method:

Income =10 & expenditure = 7 Then, savings => 10 – 7 = 3

If 7 (expenditure) ---------- 10500

3 (savings) ------------------- ?

3 × 𝟏𝟎𝟓𝟎𝟎

𝟕= 𝟒𝟓𝟎𝟎/− (answer)

25 Veerendhar got 273 marks in an examination and scored 5% more than

the pass percent. If Balu got 312 marks, then by what % above the pass marks

did he pass the examination.?

SOLUTION:

Let pass% =X

105/100 × X = 273

X= 260

Difference => 312 – 260 = 52

Required answer = 𝟓𝟐

𝟐𝟔𝟎 × 100

= 20% (answer)

26 Due to inflation, there is an increase of 33 1/3 % in the price of bananas.

If 3 less bananas are available for 24/-, the present rate of the bananas per

dozen is …….

SOLUTION:

Formula, 𝑿 × 𝒀

𝟏𝟎𝟎 × 𝑨

Here, X = 100/3 %, Y= 24, A= 3

=

100

3 ×24

100 ×3

= 2.60/- (answer)

27 A sample of milk contains 5% water. What quantity of pure milk should

be added to 10 liters of milk reduced the water content to 2% …….

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SOLUTION:

Ratio Method:

Water = 5% of 10 = 0.5 liters, Milk = 9.5 liters

According to question 𝑾𝒂𝒕𝒆𝒓

𝑴𝒊𝒍𝒌 =>

𝟎.𝟓

𝟗.𝟓+𝒙=

𝟐

𝟗𝟖

X = 15 liters (answer)

28 In a town, the population was 8000. In one year, male population

increased by 10% and female population increased by 8% but the total

population increased by 9%. The no. of males in the town was……

SOLUTION:

By allegation rule

Men women

10% 8%

9%

1% 1%

Men : women = 1 : 1

Men population = 𝟏

𝟐 ×8000

= 4000 (answer)

29 If A exceeds B by 40%, B is less than by C, 20%. Then A: C ….

SOLUTION:

If B = 100, A =140 &

B = 80/100 ×C => C = 125

A: C = 140: 125

= 28: 25 (answer)

30 A mixture contains wine & water in the ratio 3: 2 and another mixture

contains them in the ratio 4: 5. How many liters of the latter must be mixed

with 3- liters of the former so that the resulting mixture may contain equal

quantities of wine & water…

SOLUTION:

By allegation rule

Mixture-1 Mixture- 2

𝟑

𝟓

𝟒

𝟗

𝟏

𝟐 (1: 1 =

𝟏

𝟐)

𝟏

𝟏𝟖

𝟏

𝟏𝟎

Required ratio => 𝟏

𝟏𝟖 :

𝟏

𝟏𝟎 = 5: 9

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Required answer => 3 : 𝟗

𝟓× 𝟑

= 3: 𝟐𝟕

𝟓 (answer)

31 In what ratio should a 20% methyl alcohol solution be mixed with a 50%

methyl alcohol solution so that the resultant solution has 40% methyl alcohol

in it ?

SOLUTION:

By allegation rule:

Mixture-1 Mixture-2

20% 50%

Resulting mixture

40%

10% 20%

Required ration => 10: 20

= 1: 2 (answer)

33 The price of sugar is reduced by 25% but in spite of the decrease,

Ramesh ends up increase his expenditure on sugar by 20%. What is the

percentage change in his monthly consumption of sugar ………

SOLUTION:

Percentage method:

price expenditure

Initially : 100 100

Later : 75 120

Increasing => 120- 75 = 45

= 45

75× 100

= 60% (answer)

34 A man divides his property so that his son’s share to his wife’s & wife’s

share to his daughter both are in the ratio 3:1. If the daughter gets 10000/- less

than son, the value of the whole property was……..

SOLUTION:

Son: wife: daughter = 3: 1

3:1

i.e. 3 × 3 : 3 × 1 : 1 × 1 = 9 : 3 : 1

If (9- 1) ………….10000 ( from question)

13 (9 + 3 + 1) ………….. ?

= 13

8× 10000

= 16250 (answer)

35 To attract more visitors, zoo authority announces 20% discount on every

ticket which costs 25 paise. For this reason, sale of ticket increases by 28%.

Find the percentage of increase in the no. of visitors …..

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SOLUTION:

Let initially 100 tickets

Initially: 100 × 25 paise = 2500

Later: 128 × 20 paise = 2560 (from question)

Increasing => 2560- 2500 = 60

% increase = 𝟔𝟎

𝟏𝟎𝟎 × 100

= 60% (increase)

36 Between two consecutive years my incomes are in the ratio of 2: 3 and

expenses in the ratio 5: 9. If my income in the second year is 45000/- and my

expenses in the first year is 25000/-. My total savings for the two years is……

SOLUTION:

Income in the 2nd year = 45000/-

Therefore, income in the 1st year => 45000 × 𝟐

𝟑 = 30000/-

Total income in 2 years => 45000 + 30000 = 75000/-

Expenses in the 1st year = 25000/-

Expenses in the 2nd year => 𝟗

𝟓∗ 𝟐𝟓𝟎𝟎𝟎 = 45000/-

Total expenses in 2 years => 25000 +45000 = 70000/-

Total savings => 75000- 70000 = 5000/- (answer)

37 Tea costing 136/- a kilogram is mixed with tea costing 141/- a kilogram in

the ratio 2: 3. The cost of one kilogram of the mixture is………

SOLUTION:

Cost of one kilogram of mixture is

= 𝟏𝟑𝟔 × 𝟐 + 𝟏𝟒𝟏 × 𝟑

𝟐+𝟑

=139/- (answer)

38 72% of the students of a certain class took Biology and 44% took

mathematics. If each student took at least one of Biology or Mathematics and

40 students took both of these subjects, the total no. of students in the class

is…………

SOLUTION:

Total no. of students = X

Percentage of students opting both subjects

72 + 44 - 100 = 16%

X

100∗ 16 = 40 (Given)

X= 250 (answer)

39 Two vessels contain milk and water in the ratio 3: 2 and 7: 3. Find the

ratio in which the contents of the two vessels have to be mixed to get a new

mixture in which the ration oif milk & water is 2:1 ?

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SOLUTION:

By allegation rule:

Milk-1 Milk-2

𝟑

𝟓

𝟕

𝟏𝟎

resulting

𝟐

𝟑

𝟕

𝟏𝟎−

𝟐

𝟑

𝟐

𝟑−

𝟑

𝟓

Required ratio => 𝟏

𝟑𝟎∶

𝟏

𝟏𝟓 = 1: 2 (answer)

40 A & B earn in the ratio 2: 1. They spend in the ratio 5: 3 and save in the

ratio 4: 1. If the total monthly savings of both A & B are 5000/-, the monthly

income of B is …….,

SOLUTION:

Let the incomes of A & B be 2X & X respectively and their expenditures 5Y & 3Y

respectively

A’s savings = 𝟒

𝟓 × 5000 = 4000/-

B’s savings = 𝟏

𝟓 × 5000 = 1000/-

Therefore, 2X – 5Y = 4000 …………… (1)

X – 3Y = 1000 ……………… (2)

From equations 1 & 2

X = 7000/- (answer)

41 An employer reduces the no. of his employees in the ratio 9: 8 and

increases their wages in the ratio 14: 15. If the original wage bill was 18900/-,

find the ratio in which the wage bill id decreased …,

SOLUTION:

Required ratio => 9 × 14 : 8 × 15

= 21: 20 (answer)

42 In a laboratory, two bottles contains mixture of acid & water in the ratio

2: 5 in the first bottle and 7: 3 in the second bottle. The ratio in which the

contents of these two bottles be mixed such that the new mixer has acid &

water in the ratio 2: 3 is ..

SOLUTION:

By allegation rule:

Acid-1 Acid-2

𝟐

𝟕

𝟕

𝟏𝟎

resultant

𝟐

𝟓

𝟕

𝟏𝟎−

𝟐

𝟓

𝟐

𝟓−

𝟐

𝟕

Required ratio = 21: 8 (answer)

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43 A jar contains 10 red marbles & 30 green ones. How many red marbles

must be added to the jar so that 60% of the marble will be red …………

SOLUTION:

Ratio Method:

Red = 10 & Green = 30

Let X- red marbles are added to the jar

Red/Green => 10+X/30 = 60/40

X = 35 (answer)

44 In an examination, A got 25% marks more than B, B got 10% less than

C and C got 25% more than D. If D got 320 marks out of 500, the marks

obtained by A were..

SOLUTION:

Percentage method

A = 𝟏𝟐𝟓

𝟏𝟎𝟎 ×

𝟗𝟎

𝟏𝟎𝟎 ×

𝟏𝟐𝟓

𝟏𝟎𝟎 × 320

A = 450 (answer)

45 The daily wages of worker are increased by 10% but the no. of hours

worked by him per week is decreased by 10%. If originally he was getting

2000/- per week, what per week will he get now ?

SOLUTION:

Percentage method:

Wages hour/week

Initially: 100 × 100 => 2000

Later: 110 × 90 =>?

= 𝟏𝟎𝟎 × 𝟗𝟎 × 𝟐𝟎𝟎𝟎

𝟏𝟎𝟎 × 𝟏𝟎𝟎

= 1980/- (answer)

46 Tea worth 126/- per kg and 135/- per kg is mixed with a third variety in

the ratio 1: 1: 2. If the mixture is worth 153/- per k, the price of the third

variety per kg will be ……..

SOLUTION:

Price of the third variety = X/- per kg

126 ×1 + 135×1 + X×2 = 153×4 (given)

X= 175. 5/- (answer)

47 A container contains 60 kg of milk. From this container 6 kg of milk was

taken out & replaced by water. This process was repeated further two times.

The amount of milk left in the mixture is ……

SOLUTION:

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Using formula

Amount of milk left =

Initial amount [ 1 – 𝒎𝒊𝒍𝒌 𝒕𝒂𝒌𝒆𝒏 𝒐𝒖𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏

𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒂𝒎𝒐𝒖𝒏𝒕]3

= 60 [1 - 𝟔

𝟔𝟎 ]3

= 43.74 kg (answer)

48 A water tank contains 5% salt by weight. X- liters of fresh water is

added to 40 liters of tank water, so that the solution contains 2% salt. The

value of X is ……

SOLUTION:

By allegation rule

Percentage of water in current mixture => 100- 5 =

95%

Percentage of water in output mixture => 100- 2 =

98%

95% 100%

98%

100 - 98 98 - 95

Required ratio = 2: 3

Ratio of mixture and fresh water is 2: 3. If there was

40 liters of mixture already available in the tank, we

need

= 3

2× 40 = 60 liters (answer)

49 A vessel is filled with liquid 3 parts of which are water and 5 parts syrup.

How much of the mixture must be drawn off and replaced with water so that

the mixture may be half water and half syrup …….

SOLUTION:

By allegation rule

Total = 8 parts

Water = 3 parts & syrup = 5 parts

𝟑

𝟖 1

𝟏

𝟐

1- 𝟏

𝟐

𝟏

𝟐−

𝟑

𝟖

Required ratio = 4: 1

Required quantity = 𝟏

𝟓 (answer)

50 The ratio of the numbers of boys and girls in a school was 5: 3, some new

boys & girls were admitted to the school, in the ratio 5: 7. At this, the total no.

of students in the school becomes 1200, and the ratio of boys to girls changed

to 7: 5. The no. of students in the school before new admissions was

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SOLUTION:

Let the original no. of boys & girls be 5X & 3X

respectively and that of new boys & girls be 5Y & 7Y

respectively

Therefore, 5X + 3X + 5Y +7Y = 1200

2X + 3Y = 300 ……………… (1)

And

5x + 5y

3x + 7y=

7

5

X= 6Y ………………… (2)

From equations 1 & 2

X = 120

Therefore, original no. of students, 8X = 8 × 120

= 960 (answer)

51 A man saves 20% of his income. Next year, his income increases & so he

increases his expenditure by 40% but his percentage savings half of his initial

savings. Find the percentage in his income …

SOLUTION:

Let the income be = 100/-

So, he spends= 80/- and saving = 20/-

Now after increases his expenditure

140

100× 80 = 112

Let the new income be X, so his percentage savings

are 10% (from question)

So, 𝑿−𝟏𝟏𝟐

𝑿× 𝟏𝟎𝟎 = 10

X = 124. 44/- (answer)

52 The ratio of water & spirit in the mixture is 1: 3. If the volume of the

solution is increased by 25% by adding spirit only, what is the resultant ratio

of water and spirit ?

SOLUTION:

The ratio of water & spirit = 1: 3

Let the volume of mixture = 100L

Therefore, water = 25L & spirit = 75L

Now adding spirit so that volume increases from

100L to 125L (20% increase given)

Now, new ratio would be 25L: 75L + 25L

= 1: 4 (answer)

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53 The ratio of the first & second class fares between Khammam &

vijaywada Railway stations is 4: 1 and that of the no. of passengers travelling

by first & second classes is 1: 40. If one day 1100/- are collected as total fare,

the amount collected from the first class passenger is ………

SOLUTION:

Let the first class fare = 4X

Second class fare = 1X

And, again let the no. of passengers travelling by first

class is 1Y & second class fare is 40Y

Total fare of first class = 4X × 1Y = 4XY/-

Total fare of second class = 1X × 40Y = 40XY/-

4XY + 40XY = 1100 (given)

XY = 25

Amount collected from first class passenger => 4XY =

4 × 25 = 100/- (answer)

54 In an alloy, the ratio of copper & zinc is 5: 2. If 1.250 kg of zinc is mixed

in 17 kg 500 grams of alloy, then the ratio of copper & zinc will be …

SOLUTION:

In 17 kg 500 grams of alloy

Copper: 17500 = 𝟓

𝟕 = 12500 grams

Zinc: 17500 × 𝟐

𝟕 = 5000 grams

Now, when 1.250 kg (1250 grams) of zinc is mixed

with 17 kg 500 grams of alloy

Then in the new mixed, amount of zinc

= 5000 + 1250 = 6250 grams

So, new ratio of copper & zinc = 12500: 6250

= 2: 1 (answer)

56 60 kg of rice-1 at 32/- per kg is mixed with 48 kg of rice-2 and the

mixture is sold at the average price of 28/- per kg. If there be no profit or loss

due to the sale price, then the price of the second variety of rice is …..

SOLUTION:

By allegation rule

Let the price of second variety of rice be X/- per kg

Rice-1 Rice-2

32 X

Mean

(28)

28 - X 32 - 28

According to question 𝟐𝟖−𝒙

𝟒=

𝟔𝟎

𝟒𝟖

x = 23 (answer)

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57 A vessel is filled with milk & water. 70% of milk and 30% of water is

taken out of the vessel. It is found that the vessel is vacated by 55% and has

240L mixture. Find the quantity of milk in mixture .

SOLUTION:

Here, the percentage value of water & milk that is

taken from the vessel should be taken into the

consideration.

Milk% water%

(30) (70)

(45)

(70 - 45) (45 – 30)

Required ratio = milk/water = 5: 3

Quantity of milk in the remaining mixture = 𝟓

𝟖× 𝟐𝟒𝟎

= 150L (answer)

58 The height of a tree increases every year 1/4 times. If the present height

of the tree is 64 cm, then what will be its height after two years …

SOLUTION:

Height after two years

= 64 × 𝟓

𝟒×

𝟓

𝟒 (1+

𝟏

𝟒=

𝟓

𝟒)

= 100 cm (answer)

59 Six liters of a 20% solution of alcohol in water are mixed with 4- liters of

a 60% solution of alcohol in water. Find the percentage of alcohol in the

mixture ………

SOLUTION:

The percentage of alcohol

= {𝟔 ×

𝟐𝟎

𝟏𝟎𝟎+ 𝟒 ×

𝟔𝟎

𝟏𝟎𝟎

𝟔+ 𝟒} × 100

= 36% (answer)

60 There is 1000 liters of milk in a pot. 100 liters of milk is taken out and

same amount of water is poured into it. Again 200 liters of mixture is taken

out and same amount of water is poured into it and finally 400 liters of

mixture is taken out and same amount of water is poured into it, then what is

the amount of milk in the resulting mixture ….

SOLUTION:

After first process, milk & water = 9: 1

After second process milk left = 800 × 𝟗

𝟏𝟎 = 720

So, milk & water => 720: 280 = 18: 7

After third process milk left => 600 × 𝟏𝟖

𝟐𝟓 = 432

(answer)

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So, milk & water => 720: 280 = 18: 7

After third process milk left => 600 × 𝟏𝟖

𝟐𝟓 = 432

(answer)

61 In 165 liters of mixtures of milk & water, water is only 28%. The

milkman sold 40 liters of this mixture and then he added 30 liters of pure milk

and 13 liters of pure water in the remaining mixture. What is the percentage

of water in the final mixture ?

SOLUTION:

Milkman sold 40 liters of mixture

So, remaining mixture => 165- 40 =125 liters

Quantity of water => 125 × 𝟐𝟖

𝟏𝟎𝟎 = 35 liters

Therefore, quantity of milk = 90 liters

Now, milkman made new mixture in which

Water => 35 + 13 = 48 liters

Milk => 90 + 30 = 120 liters

Percentage of water in the new mixture =

48

48 + 120∗ 100

= 28.57% (answer)

62 A solid diamond falls and breaks into three pieces whose weights are in

the ratio of 2 : 3 : 4. The value of each piece is directly proportional to the

square of their weights. Given that the value of solid diamonds is 24, 300/-,

find the loss due to breakage ……

SOLUTION:

Ratio of each piece of diamond = 2x : 3x : 4x

Solid diamond 2x + 3x + 4x = 9x

Cost of solid diamond = (9x)2 = 81x2

Cost of broken diamond = (2x)2 + (3x)2 + (4x)2 = 29x2

Loss = 81x2 − 29x2 = 52x2

Given, 81x2 = 24300

x2 = 300/-

Therefore, loss = 52 × 300 = 15, 600 (answer)

63 There is a vessel holding 40 litres of milk. 4 litres of milk initially is taken

out from the vessel and 5 litres of water is poured in. After this, 6 litres of

mixture from this vessel is replaced with 7 litres of water. And finally 8 litres

mixture from this vessel is replaced with 9 litres of water. How much of the

milk (in litres) is there in the vessel now?

SOLUTION:

Amount of milk remained

= 40 × (40−4

40) × (

41−6

41) × (

42−8

42)

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= 40 × 36

40×

35

41×

34

42

= 24. 87 litres (answer)

64 Price of an article increases by 26%. A family though increases its

expenditure by 10%, can purchase 8 kg less. What amount of article can be

purchased at original price?

SOLUTION:

Let original price of article per kg = 100x

Increased price = 126x

Let original consumption is ‘y’ kg

Then original expenditure = 100xy

Now, according to question

126x(x – 8) = 100xy × 𝟏𝟏𝟎

𝟏𝟎𝟎

By solving, we get y = 63 kg (answer)

65 A Jar full of milk contains 40% water. A part of this milk is replaced by

another mixture containing 19% water and now the percentage of water is

found to be 26%. The quantity of milk replaced is ………….

SOLUTION:


40

100× (1 − 𝑥) +

19

100× 𝑥 =

26

100

x = 2

3 (answer)

66 A and B have some Guavas divided between themselves. A says to B “If I

give you 25% of the Guavas I have, I will still have 2 more Guavas than you

have.” To this, B says “If you give me Guavas equal to 70% of what I have

now, I will have 4 more Guavas than you have.” What is the total no. of

Guavas that they have ………..

SOLUTION:

Let, A have ‘x’ no. of Guavas

And B have ‘y’ no. of Guavas


𝑥 − 𝑥

4= 𝑦 + 2 +

𝑥

4… … … . . (1)

𝑦 + 7𝑦

10= 𝑥 −

7𝑦

10+ 4 ……… (2)

By solving equations (1) and (2), we get

x = 44 and y = 20

Therefore, total Guavas = 44 + 20

= 64 (answer)

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67 Rajni decided to donate 12% of his monthly salary to an orphanage. On

the day of donation he changed his mind and donated 2, 400/- which was

125% of what he had decided earlier. How much is Rajni’s monthly salary ?

SOLUTION:

Let Rajni’s monthly salary = x/-


𝑥 × 12

100×

125

100= 2400

x = 16, 000/- (answer)

68 Harsha has a certain amount of money with him. He can buy either 60

apples or 40 mangoes. He wants to spend only 70% of his money. So he buys

14 mangoes and some apples. Find out the no. of apples purchased by Harsha

…..

SOLUTION:

Amount to required to buy either 60 apples or 40 mangoes = 100% available

amount

Since 14 is 35% of 40 [14

40× 100 = 35%]

∴ Amount spent on buying 14 mangoes = 35% of available amount

Remaining money = (70% - 35%) of available money = 35% of available amount

No. of apples that can be purchased with 35% of available amount = 35% of 60 => 35

100∗ 60

= 21 apples (answer)

69 The wheat sold by a grocer contained 10% low quality wheat. What

quantity of good quantity wheat should be added to 150 kg of wheat so that

the percentage of low quality wheat becomes 5% …………

SOLUTION:

Let x kg of good quality wheat is added


10% of 150 = 5% of (150 + x)

x = 150 kg (answer)

70 A shopkeeper sells notebooks at the rate of 45/- each and earns a

commission of 4%. He also sells pencil box at the rate of 80/- each and earns a

commission of 20%. How much amount of commission will he earn in two

weeks if he sells 10 notebooks and 6 pencil boxes a day ………….

SOLUTION:

Total commission earned by Shopkeeper

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= (10 × 45 × 4

100 + 6 × 80 ×

20

100) × 14

= 1596/- (answer)

71 The ratio of milk and water in mixture of 90 litre is 7 : 2. If some

amount of mixture is replaced by water, then the ratio of milk to water

becomes 5 : 2. Find the quantity of water added to the mixture ………..

SOLUTION:

Let x litre of water is added to the mixture

Initial quantity of milk = 90 × 7

9= 70 𝑙𝑖𝑡𝑟𝑒𝑠

And that of water = 90 – 70 = 20 litres


70 − 𝑥 × 7

9= 90 ×

5

7

By solving we get, x = 717

49 (𝑎𝑛𝑠𝑤𝑒𝑟)

72 There is a mixture of alcohol and water of 120 litre. The ratio of alcohol

to water is 5 : 3. If 30% mixture is taken out and same amount water is added

then find the new ratio of alcohol and water in the mixture ……..

SOLUTION:

Quantity of mixture left after making change before adding water = 120 × 70

100

= 84 litre

In this mixture quantity of alcohol = 84 × 5

8

= 52. 5 litre

And quantity of water = 84 – 52. 5 = 31. 5 litre

Now, after adding water to the mixture, net quantity of water = 31. 5 + 36 = 67. 5

litre

∴ Required ratio = 52.5

67.5

= 7 : 9 (answer)

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