# Percentage

## Percentage

 Question 1 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
Out of the total production of iron from hematite, an ore of Iron, 20% of the ore gets wasted, and out of the remaining iron, only 25% is pure iron. If the pure iron obtained in a year from a mine of hematite was 80,000 kg, then the quantity of hematite mined from that mine in the year is
 A 4,50,000 kg B 5,00,000 kg C 4,00,000 kg D None of these
Question 1 Explanation:
Let 100 kg of hematite be obtained then 20% of it get wasted that means 80 kg of ore remains.

Pure iron = 25% of remaining ore = 80*25/100 = 20 kg.

20 kg pure Iron is obtained from 100 of hematite.

1 kg pure Iron is obtained from = 100/20 hematite;

Then, 80000 kg pure Iron is obtained from = (100/20)*80000 = 400000 kg hematite.

 Question 2 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
1.14 expressed as a per cent of 1.9 is:
 A 90% B 60% C 6% D 10%
Question 2 Explanation:
\begin{align} & {\text{Required}}\,{\text{Percentage}} \cr & = \frac{{\left( {1.14 \times 100} \right)}}{{1.9}} \cr & = 60\% \cr\end{align}
 Question 3 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid.
 A 98% B 99% C 75% D 77%
Question 3 Explanation:
Let each side of the cuboid be 10 unit initially.

Initial Volume of the cuboid,

= length * breadth * height = 10 *10*10 = 1000 cubic unit.

After increment dimensions become,

Length = (10 + 10% 0f 10) = 11 unit.

Breadth = (10 + 20% of 10) = 12 unit.

Height = (10 + 50 of 10) = 15 unit.

Now, present volume = 11 *12 *15 = 1980 cubic unit.

Increase in volume = 1980 - 1000 = 980 cubic unit.

% increase in volume = (980/1000)*100 = 98%.

Mind Calculation Method:

Change in volume = 98%.

[We can take net percentage change in any order.]

 Question 4 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
In a factory there are three types of machine M1, M2 and M3 which produces 25%, 35% and 40% of the total products respectively. M1, M2 and M3 produces 2%, 4% and 5% defective products, respectively. what is the percentage of non-defective products ?
 A 89% B 97.10% C 96.1% D 86.10%
Question 4 Explanation:
Non-defective products M1 = 25 X 0.98 = 24.5%

Non-defective products M2 = 35 X 0.96 = 33.6%

Non-defective products M3= 40 X 0.95 = 38%

Percentage of non-defective products = 24.5 + 33.6 + 38 = 96.1%

 Question 5 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
A shepherd had n goats in the year 2000. In 2001 the no. of goats increased by 40%. In 2002 the no. of goats declined to 70%. In 2003 the no. of goats grew up 30%. In 2004, he sold 10% goats and then he had only 34,398 goats. The percentage increase of the no. of goats in this duration was :
 A 16.66% B 14.66% C 11.33% D 20%
Question 5 Explanation:
There is no need of the number of goats given i.e. 34,398.

Initially, let there be 100 goats. Then

100 == 40% ↑==> 140 ==30%↓(declined to 70%)==> 98 ==30%↑ ==> 127.4 ==10%↓(sold)==> 114.66

Hence, % increase = 14.66% [As 100 becomes 114.66].

 Question 6 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
If 20% of a = b, then b% of 20 is the same as:
 A 20% of a B 4% of a C 5% of a D None of these
Question 6 Explanation:
\begin{align} & 20\% \,{\text{of}}\,a = b \Rightarrow \frac{{20}}{{100}}a = b. \cr & \therefore b\% \,{\text{of}}\,20 = \left( {\frac{b}{{100}} \times 20} \right) \cr & = \left( {\frac{{20}}{{100}}a \times \frac{1}{{100}} \times 20} \right) \cr & = \frac{4}{{100}}a = 4\% \,{\text{of}}\,a. \cr\end{align}
 Question 7 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
The schedule working hour of a labour in a week if 48 hours and he gets Rs. 480 for that. Over time rate is 25% more than the the basic salary rate. In a week a labour gets Rs. 605, how many hours altogether he works in that week.
 A 49 B 58 C 55 D 52
Question 7 Explanation:

Schedule working hours in week = 48.

Total pay in a week for schedule working hours = Rs. 480.

Pay per hour for schedule working hours = 480/48 = Rs. 10.

Pay per hour for over time = 10 + 25% of 10 = Rs. 12.5.

Total pay in that particular week = Rs. 605.

Extra pay = 605 - 480 = 125.

So, total over time = 125/12.5 = 10 hours.

Thus, total work hour altogether in that week = 48 +10 = 58 hours.

 Question 8 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
In Sabarmati Express, there as many wagons as there are the no. of seats in each wagon and not more than one passenger can have the same berth (seat). If the middlemost compartment carrying 25 passengers is filled with 71.428% of its capacity, then find the maximum no. of passengers in the train that can be accommodated if it has minimum 20% seats always vacant.
 A 1060 B 500 C 980 D 786
Question 8 Explanation:
\begin{align} & {\text{Total number of passenger in each compartment,}} \cr & = \frac{{\left( {25 \times 7} \right)}}{5} = 35 \cr & {\text{Total}}\,{\text{berth}} = {35^2} = 1225 \cr & {\text{Maximum available capacity}} \cr & = \frac{{\left( {1225 \times 80} \right)}}{{100}} = 980\,{\text{seats}} \cr\end{align}
 Question 9 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
The total emoluments of A and B are equal. However, A gets 65% of his basic salary as allowances and B gets 80% of his basic salary as allowances. What is the ratio of the basic salaries of and B?
 A 16 :13 B 5 :7 C 7 :9 D 12 :11
Question 9 Explanation:
\begin{align} & {\text{Let the basic salaries of A and B be x and y respectively}}. \cr & {\text{Now}}, \cr & x + 65\% \,of\,x = y + 80\% \,{\text{of}}\,y \cr & x + \frac{{\left( {65x} \right)}}{{100}} = y + \frac{{\left( {80y} \right)}}{{100}} \cr & \frac{x}{y} = \frac{{180}}{{165}} = 12:11 \cr\end{align}

 Question 10 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
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.2
 A Rs. 6,00,625 B Rs. 525625 C Rs. 5,00,000 D Rs. 5,26,000
Question 10 Explanation:
\begin{align} & {\text{Let the initial price}} \cr & = Rs.\,10000p \cr & {\text{price after first increment}} \cr & = 10000p + 100xp \cr & {\text{price after first decrement}} \cr & = 10000p + 100xp - \left( {100px + p{x^2}} \right) \cr & = 10000p - p{x^2} \cr & {\text{Now,}}\,{\text{total}}\,{\text{decrement}}, \cr & p{x^2} = 21025\,.....\left( 1 \right) \cr & \cr & {\text{price after second increment}}, \cr & 10000p - p{x^2} + 100xp - \frac{{p{x^3}}}{{100}} \cr & {\text{price after second increment}}, \cr & 10000p - {p^2} + 100xp - \frac{{{p^3}}}{{100}} - 100xp + \frac{{p{x^3}}}{{100}} - p{x^2} + \frac{{p{x^4}}}{{10000}} \cr & = 10000p - 2p{x^2} + \frac{{p{x^2}}}{{10000}} \cr & = 484416\,.....\left( 2 \right) \cr & \cr & {\text{on solving equation equn }}\left( {\text{1}} \right){\text{ and }}\left( {\text{2}} \right){\text{, We get}} \cr & x = 20 \cr & {\text{substituting back we get,}} \cr & p = 525625 \cr\end{align}
There are 10 questions to complete.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0}