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** ** Sample questions for Placement Questions - Percentages

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**Sample questions for Placement Papers 2015 Latest Pattern Question 1**

If the price of an item is
decreased by 10% and then increased by 10%, the net effect on the price of the
item is

a) a decrease of 99% b) No change in overall

a) a decrease of 99% b) No change in overall

c) A decrease of 1%
d)
An increase of 1%

**Answer: D****Explanation: 1% decrease**

If a certain value is increased by a% then decreased by a% or vice versa, the net change is always decrease.

The overall percentage change = $-\left(\frac{a^{2}}{100}\right)\; =\, -\left(\frac{10^{2}}{100}\right)\; =\, -1 $

Negative sign indicates decrease

Negative sign indicates decrease

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**Sample questions for
. Placement Papers Solved Questions Latest Pattern Question 2**

P is 6 times as large as Q. By
what percentage, Q is less than P?

a)83.33%
b)65%
c)16.66%
d)12.5%

**Answer : A**

**Explanation :**

Given P is 6 times as large as Q.

P = 6Q => P: Q =6:1

Now Q is $\left(\frac{6-1}{6}\right)\; \times100 \, = 83.33% $ less than P

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**Sample questions for Placement Papers with answers 2015 Latest Pattern Question 3**

If a% of a + b% of b = 2% of
ab, what percentage of a is b?

a)25%
b)50%
c)75%
d)100%

**Answer : D**

**Explanation :**

Given that a% of a + b% of
b = 2% of ab

=>$\frac{a}{100}\times a\,+\,\frac{b}{100}\times b\:=\frac{2}{100}\,\times ab $

=> a

=>$\frac{a}{100}\times a\,+\,\frac{b}{100}\times b\:=\frac{2}{100}\,\times ab $

=> a

^{2}^{ }+ b^{2}^{ }= 2ab
=> (a-b)

^{2}=0
=> a=b

Therefore b is 100% of a

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**Sample questions for Placement Papers 2014 Latest
Pattern Question 4**

Find the number which exceeds
16% of it by 42?

a)50
b)70
c)75
d)100

**Explanation :**

Let number be x

16% of x + 42 =100% of x

Ã° 84 %
of x = 42

Ã° $\frac{84}{100} \times $
x = 42 => x = $\frac{100}{84} \times $ 42

On Solving we get x =50

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**Sample questions for Placement Papers Solved
Questions Latest Pattern Question 5**

When 5% of the
total wheat is lost in grinding, a country can export 9 million tonnes of
wheat, but when6% of the total wheat is lost in grinding it needs to import 2
million tonnes of wheat. What is the total production of wheat in the country?
(in million tonnes)

(a) 1000
b) 900
c) 1100
d)1150

**Answer : A****Explanation :**

**1**

^{st}**method :**

When 5 % of the total wheat
is lost => can export 9 million tons

When 6% of the total wheat is
lost => needs to import 2 million tonnes

1% change results in a
change of 11 million tonnes

1% of total wheat = 11 million tonnes

Total production 100% = 1100 million tonnes

**Alternate Method:**

Let total wheat=x million
tons

Wheat left after 5% lost=$ \frac{95}{100}x $;

The required wheat for country after exporting 9 million tonnes = $ \left(\frac{95}{100}x\right)\,- \: 9 $

Wheat left after 6% lost = $ \frac{94}{100}x $

If there is 6% loss in grinding , there is a need of 2 million tonnes

Wheat left after 5% lost=$ \frac{95}{100}x $;

The required wheat for country after exporting 9 million tonnes = $ \left(\frac{95}{100}x\right)\,- \: 9 $

Wheat left after 6% lost = $ \frac{94}{100}x $

If there is 6% loss in grinding , there is a need of 2 million tonnes

Total wheat required for
country is = $\left(\frac{94}{100}x\right)\:+ \: 2 $

Therefore $\left(\frac{94}{100}x\right)\:+ \: 2 \; =\left(\frac{94}{100}x\right)\:- \: 9 $

On solving x=1100

Therefore $\left(\frac{94}{100}x\right)\:+ \: 2 \; =\left(\frac{94}{100}x\right)\:- \: 9 $

On solving x=1100

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**Sample questions for Placement Papers Solved Latest Pattern Question 6**

In a certain
company, 55% of the workers are men. If 30% of the workers are full time
employees and 60% of these are women, what percentage of the full-time workers
in the company are men?

a)25%
b)35%
c)30%
d)40%

**Answer : A**

**Explanation :**

Let total employees be100

Then Number of Men
workers = 55% of 100= 55

Then Number of Women Workers
= 45

full time workers in the company =30% of 100 =30

60% of full time workers are women

full time workers in the company =30% of 100 =30

60% of full time workers are women

Therefore number of full time
women workers=60% of 30=18

Number of full time men workers=30-18=12

Number of full time men workers=30-18=12

Percentage of full time men workers
= $\frac{12}{30}\times $ 100=40%

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**Sample questions for Placement Papers 2015 Latest Pattern Question 7**

60% of male in
a town and 70% of female in a town are eligible to vote. out of which 70% of
male and 60% of female who are eligible to vote voted for candidate A. what is
the value of votes in % did A get?

a)42%
b)58%
c)Cant be determined
d)71%

**Answer : A**

**Explanation :**

Let the ratio of men and
women are 100 : k

Male eligible votes = 60 and female eligible votes = 70% (k)

Number of males who voted for A = 70% (60) = 42

Number of females who voted for A = 60%(70% (K) = 42% (k)

Percentage of votes got by A = 42+42100(K)60+70100(K)×100=4200+42K6000+70K×100

So this value cannot be determined as the value of K is not known

Male eligible votes = 60 and female eligible votes = 70% (k)

Number of males who voted for A = 70% (60) = 42

Number of females who voted for A = 60%(70% (K) = 42% (k)

Percentage of votes got by A = 42+42100(K)60+70100(K)×100=4200+42K6000+70K×100

So this value cannot be determined as the value of K is not known

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**Sample questions for Placement Papers 2015 Latest Pattern Question 8**

A team won 80%
of the games it played. It played 5 more games of which it won 3 and lost 2.
Its loss percentage changed to 25%. How many games did it play overall?

A. 20 B. 14 C. 16 d. 25

A. 20 B. 14 C. 16 d. 25

**Answer : A****Explanation :**

Let total number of games
initially played be n

Team won 80% of the games
=> Lost games = 20% of n = $\frac{n}{5}$

After that they played 5 more
games in which they lost 2

Now their loss

(n+5) $\times\frac{25}{100} $ = $\frac{n}{5} $ + 2

n=15

Therefore total games the team played=15+5 = 20

n=15

Therefore total games the team played=15+5 = 20

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**Sample questions for Placement Papers 2014 Latest
Pattern Question 9**

In a
horticulture assignment, 200 seeds were planted in plot 1 and 300 were planted
in plot 2. If 57% of the seeds are germinated for plot 1 and 42% from plot 2,
what % of total seeds germinated?

a. 45.5 b. 46.5 c. 48 d. 49.5

a. 45.5 b. 46.5 c. 48 d. 49.5

**Answer : A**

Plot 1 : Total seeds planted =200

Total seeds germinated in plot 1 = 57% of 200 = 114

Plot 2 : Total seeds planted = 300

Total seeds germinated in plot 2 = 42% of 300 =126

Total seeds germinated = 114 + 126 =240

Percentage of total seeds germinated =
$\frac{240}{500} $ x 100 =48%

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**Sample questions for Placement Papers fully solved Latest Pattern Question 10**

How many kg of
pure salt must be added to a 30kg of 2% solution of salt and water to increase
it to a 10% solution

a)1.8kg b)2.66kg c)2.4kg b)4kg

a)1.8kg b)2.66kg c)2.4kg b)4kg

**Answer : A**

**Explanation :**

Amount of salt in 30kg solution = 2% of 30
=0.6 kg

Let x kg of salt is added to the solution,
then % of salt in solution will be 10%

$\frac{x+0.6}{30+x} $ = $ \frac{10}{100}$

=> 60 + 100x = 300 + 10x

=> 90x = 240

=> X = $ \frac{240}{90} $
= 2.66 (
Approximately)

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**Sample questions for Placement Papers fully
solved Latest Pattern Question 11**

If 75 % of a
class answered the first question on a certain test correctly, 55 percent
answered the second question on the test correctly, and 20 percent answered
neither of the questions correctly, what percentage answered both correctly?

a)25% b)50% c)60% d)64%

a)25% b)50% c)60% d)64%

**Answer : A**

**Explanation :**

We can solve this problem
using Set theory

Here 75%
of class answered the first question n(A)=75

55% of the class answered the
second question n(B)=55

20% of the class answered
neither of the questions

Therefore, number of students
who answered at least one question $n(A\cup B )$ = 80

Number of students who
answered both questions correctly $ n(A\cup B ) \, = n(A)\,+\,n(B)\, -n(A\cap B ) $

=>

=> 80= 75 + 55 - - $ n(A\cap B ) $ =>50%

=> 80= 75 + 55 - - $ n(A\cap B ) $ =>50%

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**Sample questions for Placement Papers 2015 – Latest Pattern Question 12**

A tank contains 10,000 gallons
of a solution that is 5 percent sodium chloride by volume. If 2500 gallons of
water evaporate from the tank, the remaining solution will be approximately
what percentage of sodium chloride?

a)1.25%
b)3.75%
c)6.67%
d)11.7%

**Answer : A**

**Explanation :**

Ans: Before evaporation
occurs , sodium chloride in the 10000 gallons of solution

= 5% of 10000 gallons = 500 gallons

2500 gallons of water evaporated from the
tank.

After the evaporation , total solution in the
tank = 10000 – 2500 =7500

Of these 7500 gallons , 500 gallons of Sodium
chloride is there in the solution

Calculating the percentages based on these post evaporation accounts ,

Calculating the percentages based on these post evaporation accounts ,

$\frac{a}{b} \times$ 100 % =

Sodium chloride concentration
= $\frac{500}{7500} \times$ 100 % = 6.67 % (
Approximately)

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**Sample questions for Placement Papers 2014–
Latest Pattern Question 13**

In a certain city, 60% of the
registered voters are congress supporters and the rest are BJP supporters. In
an assembly election, if 75% of the registered congress supporters and 20% of
the registered BJP supporters are expected to vote for candidate A, what
percent of the registered voters are expected to vote for candidate A?

a) 42%
b)53%
c)84%
d)92%

**Answer : A**

**Explanation :**

Let the total registered voters in that city
be 100

Total Congress
supporters are 60% of registered voters = 60% of 100 = 60

Rest of the voters are BJP Supporters =40% are registered voters=40% of 100 = 40

75% of registered congress supporters voted for Candidate A = 75% of 60=45

Rest of the voters are BJP Supporters =40% are registered voters=40% of 100 = 40

75% of registered congress supporters voted for Candidate A = 75% of 60=45

20% of registered BJP supporters voted for
candidate A = 20%
of 40 =8

Total votes got by candidate A = 45 +8 =53

Total percent= 53%

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**Sample questions for Placement Papers – New Pattern
Question 14**

Andalusia has
been promoting the importance of health maintenance. From January 1,1991 to
January 1,1993, the number of people enrolled in health maintenance
organizations increased by 15 percent. The enrollment on January 1,1993 was 45
million. How many million people(to the nearest million) was enrolled in health
maintenance organizations on January 1,1991?

a)38
b)39.13
c)42
d)45

**Answer : A**

**Explanation :**

Ans: When a value N is increased by x%, then
its value after increase N x $ \frac{100+x}{100}$

From January 1,1991 to January 1,1993 , the
enrolment is increased by 15%

Let enrolment in January 1 , 1991 be K

Enrolments in January 1 , 1993 was 45 million

=> K x $ \frac{100+15}{100} $
=45

=> 115K = 45×100 = 39.13

=> 115K = 45×100 = 39.13

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**Sample questions for Placement Papers New
Pattern Question 15**

The percentage increase in the
number of pens sold by a trader from Tuesday to Wednesday is same as the
percentage decrease in the number of pens sold from Monday to Tuesday. On
Wednesday the trader sold 16 pens more than those he sold on Tuesday. If the percentage
decrease in sales from Monday to Tuesday is 20%, how many pens did he sell on
Tuesday?

(1) 20 (2) 96 (3) 100 (4) 80

(1) 20 (2) 96 (3) 100 (4) 80

**Answer : A**

**Explanation :**

Given that the percentage increase in
the number of pens sold from Tuesday to Wednesday is same as the percentage
decrease in the number of pens sold from Monday to Tuesday

The percentage decrease in sales from
Monday to Tuesday is 20%

The increase in the number of pens from
Tuesday to Wednesday is 16

Therefore 20% of the sales on Monday =
16

Then Total pens sold on Tuesday => 100% =
$\frac{16}{20} \times$ x 100 = 80

**Alternative Method :**

Lets the number of pen is
sold on Monday is x.

There is 20% decrease in
sales from Monday to Tuesday

Therefore the number of pen
is sold on Tuesday = $ \frac{80x}{100}$

Now the % increase in the
number of pens sold from Tuesday to Wednesday is 20%.

Then sold on Wednesday is $\frac{80x}{100} \times \frac{120}{100} $...

The increase in sales from Tuesday to Wednesday is 16

=> $ \frac{80x}{100} \times \frac{120}{100}- \frac{80x}{100} $ =16

Therefore Pens sold on Tuesday is 80..

Then sold on Wednesday is $\frac{80x}{100} \times \frac{120}{100} $...

The increase in sales from Tuesday to Wednesday is 16

=> $ \frac{80x}{100} \times \frac{120}{100}- \frac{80x}{100} $ =16

Therefore Pens sold on Tuesday is 80..

For more TCS Placement Questions on Percentages , Sample questions for Placement Papers Percentages - 2

For Concept, formulae and practice problems on percentages , Percentages - Concept and Formulae

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