Fdaytalk.com
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
% 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
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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%
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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
------------------------------
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
=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
According to question
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.?
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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
According to question
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:
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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% …….
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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 …..
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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 ?
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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:
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
= 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:
According to question
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
According to question
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)
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
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/-
According to question
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
According to question
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
Fdaytalk.com
Author : J. Maha Laxmaiah Mailid : [email protected]
= (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
According to question
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)