Solution:
Let the original price be Rs. $$x$$ ∴ (100 - r)% of (100 + r)% of $$x$$ = 1 $$\eqalign{ & \Rightarrow \frac{{\left( {100 - r} \right)}}{{100}} \times \frac{{\left( {100 + r} \right)}}{{100}} \times x \cr & \Rightarrow x = \frac{{100 \times 100}}{{\left( {100 - r} \right)\left( {100 + r} \right)}} \cr & \Rightarrow x = \frac{{10000}}{{\left( {10000 - {r^2}} \right)}} \cr} $$
13.
In company there are 75% skilled workers and reaming are unskilled. 80% of skilled workers and 20% of unskilled workers are permanent. If number of temporary workers is 126, then what is the number of total workers ?
(A) 480
(B) 510
(C) 360
(D) 377
Solution:
$$\eqalign{ & {\text{Let the number of total workers}} = x \cr & {\text{Number of skilled workers}} \cr & = 75\% \,of\,x = \frac{{75x}}{{100}} = \frac{{3x}}{4} \cr & {\text{No}}{\text{. of unskilled workers}} \cr & = 25\% \,of\,x = \frac{{25x}}{{100}} = \frac{x}{4} \cr & {\text{No}}{\text{. of permanent workers}}, \cr & = {\frac{{80}}{{100}}} \times {\frac{{3x}}{4}} + {\frac{{20}}{{100}}} \times {\frac{x}{4}} \cr & = {\frac{{3x}}{5}} + {\frac{x}{{20}}} \cr & = \frac{{13x}}{{20}} \cr & {\text{No}}{\text{.}}\,{\text{of}}\,{\text{temporary}}\,{\text{workers,}} \cr & = x - {\frac{{13x}}{{20}}} = \frac{{7x}}{{20}} \cr & {\text{Now}}, \cr & \frac{{7x}}{{20}} = 126 \cr & x = 360 \cr} $$
14.
The price of a car is Rs. 325000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
(A) Rs. 32500
(B) Rs. 48750
(C) Rs. 76375
(D) Rs. 81250
Solution:
Amount paid to car owner = 90% of 85% of Rs. 325000 = Rs. $$\left( {\frac{90}{{100}} \times \frac{85}{100} \times 325000} \right)$$ = Rs. 248625 ∴ Required difference : = Rs. (325000 - 248625) = Rs. 76375
15.
In an office, 40% of the staff is female. 70% of the female staff and 50% of the male staff are married. The percentage of the unmarried staff in the office is :
(A) 65%
(B) 42%
(C) 60%
(D) 64%
Solution:
Let the total staff = 100 40% of female, i.e 40 female and 60 male 70% of female staff is married i.e. 70% of 40 female = 28 Number of unmarried female = 40 - 28 = 12 50% of male staff is married i.e 50% of 60 male = 30 Number of unmarried male = 60 - 30 = 30 Total number of unmarried staff = 12 + 30 = 42 % of unmarried staff = $$\frac{42}{100}$$ × 100 = 42%
16.
Population of a town increase 2.5% annually but is decreased by 0.5% every year due to migration. What will be the percentage increase in 2 years?
(A) 5%
(B) 4.04%
(C) 4%
(D) 3.96%
Solution:
Net percentage increase in Population = (2.5 - 0.5) = 2% each year. Let the Original Population of the town be 100. Population of Town after 1 year = (100 + 2% of 100) = 102. Population of the town after 2nd year = (102 + 2% of 102 ) = 104.04 Now, % increase in population = $$\frac{{4.04}}{{100}} \times 100 = 4.04\% $$ Mind Calculation Method: 100 == 2% Up(1st year) ==> 102 == 2%Up(2nd year) ==> 104.04 % population increase in 2 years = 4.04%.
17.
In an election, a total of 500000 voters participated. A candidate got 255000 votes which was 60% of total valid votes. What was the percentage of invalid votes :
(A) 10%
(B) 12%
(C) 15%
(D) $$\frac{300}{17}$$%
Solution:
Let the number of valid votes be x Then, 60% of x = 255000 ⇒ x = $$\left( {\frac{{255000 \times 100}}{{60}}} \right)$$ ⇒ x = 425000 Number of invalid votes : = (500000 - 425000) = 75000 ∴ Required percentage : = $$\left( {\frac{{75000 \times 100}}{{500000}}} \right)$$ % = 15%
18.
A shopkeeper first raises the price of Jewellery by x% then he decreases the new price by x%. After such up down cycle, the price of a Jewellery decreased by Rs. 21025. After a second up down cycle the Jewellery was sold for Rs. 484416. What was the original price of the jewellery.
(A) Rs. 5,26,000
(B) Rs. 6,00,625
(C) Rs. 5,25,625
(D) Rs. 5,00,000
Solution:
Let the initial price = Rs. 10000p Price after first increment = 10000p + 100xp Price after first decrement = 10000p + 100xp - (100px + px2) = 10000p - px2 Now, total decrement, px2 = 21025 . . . . . (1) Price after second increment, = 10000p - px2 + 100xp - $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ Price after second increment, = 10000p - p2 + 100xp - $$\frac{{{{\text{p}}^3}}}{{100}}$$ - 100xp + $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ - px2 + $$\frac{{{\text{p}}{{\text{x}}^4}}}{{10000}}$$ = 10000p - 2px2 + $$\frac{{{\text{p}}{{\text{x}}^2}}}{{10000}}$$ = 484416 . . . . . . (2) On solving equation (1) and (2), We get x = 20 Substituting back we get, p = 5,25,625
19.
One-fourth of sixty percent of a number is equal to two-fifth of twenty percent of another number. What is the respective ratio of the first number to the second number ?
(A) 4 : 7
(B) 5 : 9
(C) 8 : 13
(D) Cannot be determined
Solution:
Answer & Solution Answer: Option E Solution: Let the first number be x and the second number be y $$\frac{1}{4}{\text{ of 60% of }}x = \frac{2}{5}{\text{ of 20% of }}y$$ $$\eqalign{ & \Rightarrow \frac{1}{4} \times \frac{{60}}{{100}} \times x = \frac{2}{5} \times \frac{{20}}{{100}} \times y \cr & \Rightarrow \frac{{3x}}{{20}} = \frac{{2y}}{{25}} \cr & \Rightarrow \frac{x}{y} = \frac{2}{{25}} \times \frac{{20}}{3} \cr & \Rightarrow \frac{x}{y} = \frac{8}{{15}} \cr} $$
20.
What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?
(A) 1%
(B) 14%
(C) 20%
(D) 21%
Solution:
Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69. Number of such number = 14 $$ = \left( {\frac{{14}}{{70}} \times 100} \right)\% = 20\% $$