DXC Technology Percentages Questions and Answers Quiz-1

Solution : Let the total number of bats on that day be n.
=>0.085 % of n = 34
=> (\frac{0.085}{100}) \times  n = 34
=> n = 34 \times (\frac{100}{0.085})
=> n = 40,000
Therefore, total number of bats manufactured on the day = 40,000

Solution : Let the number be n.
=> n\times \frac{25 }{100}= \frac{n}{4}
=> (\frac{n}{3}) – (\frac{n}{4}) = 8
=> \frac{n}{12} = 8
n = 96
Thus, 96 is the required number.

Solution : We are given that x – y = 100 and 10 % of x = 15 % of y
=> x – y = 100 and (\frac{10}{100}) x = (\frac{15}{100}) y
=> x – y = 100 and 10 x = 15 y
=> x – y = 100 and 2 x = 3 y
=> x – y = 100 and x = 1.5 y
=> 1.5 y – y = 100
=> 0.5 y = 100
=> y = 200
=> x = 1.5 y = 300
Thus, the required numbers are 300 and 200.

Solution : Let the number of registered participants be n.
Number of participants who actually turned up = 75 % of n
Number of valid participations = 98 % of (75 % of n) [because 2% were invalid]
Number of participants defeated by the winner = 75 % of 98 % of (75 % of n) = 9261
=> 0.75 x 0.98 x 0.75 x n = 9261
=> 0.55125 x n = 9261
=> n = 16800
Therefore, number of registered participants = 16800

Solution : We are given that the geek could answer 70 % C++ questions, 40 % C questions and 60 % Java questions correctly and there were total 75 questions : 10 from C++, 30 from C and 35 from Java.
=> C++ questions answered correctly = 70 % of 10 = 7
=> C questions answered correctly = 40 % of 30 = 12
=> Java questions answered correctly = 60 % of 35 = 21
=> Total questions answered correctly = 7 + 12 + 21 = 40
=> Marks secured = 40 x 1 = 40
Now, marks required = 60 % of 75 = 45
=> Shortfall in marks = 45 – 40 = 5
Therefore, the geek missed the interview call by 5 marks.

Solution : Let the retirement money be Rs. 100 n
=> Money given to wife = 40 % of 100 n = 40 n, Balance = 60 n
=> Money given to 3 sons = 3 x (20 % of 60 n) = 3 x 12 n = 36 n, Balance = 24 n
=> Money spent on miscellaneous items = 50 % of 24 n = 12 n, Balance = 12 n
Now, this remaining 12 n is the money deposited in the bank, i.e., Rs. 1,20,000
=> 12 n = 1,20,000
=> n = 10,000
Therefore, geek’s retirement money = 100 n = Rs. 10,00,000

Solution : Let the order amount be Rs. n
=> Commission charged = 5 % of Rs. 10,000 + 4 % of (Rs. n – 10,000) = Rs. 500 + 0.04 n – 400
=> Commission charged = Rs. 100 + 0.04 n
Now, amount remitted = Rs. n – (100 + 0.04 n) = 31,100
=> 0.96 n – 100 = 31,100
=> 0.96 n = 31200
=> n = 32500
Therefore, order amount = Rs. 32,500

Solution : We know that an increase by n % and a successive decrease by n % is equal to an equivalent decrease of (\frac{n}{10})^{2} %.
=> Net decrease or loss of the seller = (\frac{20}{10})^{2}= 4 %

We know that if price of increases by P %, the necessary reduction in consumption to avoid increase in expenditure = [(\frac{P}{(100 + P) }) \times  100] %
Therefore, reduction in consumption required = [(\frac{25}{(100 + 25 )}) \times  100] % =20 %

 Let the fraction be \frac{N}{D}, where N is the numerator and D is the denominator.
=>\frac{0.85 N}{0.90D} = \frac{2}{9}
=>\frac{85 N} {90 D} = \frac{2}{9}
=>\frac{N}{D} = \frac{4}{17}
Therefore, the original fraction is \frac{4}{17}