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Modal questions for TCS Placement Papers Latest Pattern - Questions on Percentages

          Sample questions for Placement Questions - Percentages 

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       
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


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


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 $
                     
  =>     a2  + b2 = 2ab
                       =>   (a-b)2 =0
                       =>   a=b
              Therefore b is 100% of a
                    

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
Answer : A
Explanation :
Let number be x
 16% of x + 42 =100% of x
ð  84 % of x = 42
ð  $\frac{84}{100} \times $ x = 42  => x = $\frac{100}{84} \times $ 42
On Solving we get   x =50


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 :
1st 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
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

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
Therefore number of full time women workers=60% of 30=18
Number of full time men workers=30-18=12
          Percentage of full time men workers = $\frac{12}{30}\times $ 100=40%


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


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
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


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
Answer : A
Explanation :
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%


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
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)

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%
Answer : A
Explanation :
We can solve this problem using Set theory
In set theory, we have a formula&nbsp $ n(A\cup B ) \, = n(A)\,+\,n(B)\, -n(A\cap B ) $ ;
 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%

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

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
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%

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




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
 
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..

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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