### History quiz

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History quiz

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### Learnpolity

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### Top Important Dates in world history

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#### B.C.

776      : First Olympiad in Greece.

753      : Rome founded.
490      :Battle of Marathon; the Greeks defeated the Iranians/ Persians
327-26 : Invasion of India by Alexander, Battle of Hydaspes
221       : Chin-Hung Ti ‘Univesral Emperor in China, Great Wall of China completed
55         :Invasion of Britain by Julius Caesar the Great Roman General
44        :Assassination of Julius Caesarby Brutus
4          :Birth of Jesus Christ

#### A.D.

29        Crucifixion of Jesus Christ
43 Roman conquest of Britain.
570 Birth of Prophet Muhammad at Mecca
622 Migration of Muhammad from Mecca to Medina (‘Fijird), Beginning of Hijira Era
800 Charlemagne crowned Roman Emperor at St. Peter’s.
871 Accession of Alfred the Greatto the throne of Britain
901 Death of King Alfred the Great.

## Procedure Do the tricks

1. Find the near base
2. divide into two sections
3. left side is cross sum and right side multiplication of two values
4. for right side we need to take result number should be equal to the number of zeros in base

For better knowing we will take simple Example

1) Solve 7*8? It is easy for this but in the point of large numbers it is useful
Ans)    step by step

•  Finding the nearer base
7       near base is 10
8       near base is 10
• Finding the Deviation from Base
7         -3( it comes 10-7=  -3)
8         -2 ( it comes 10-8= -2)
• Next step DO the the third step
7          -3
8         -2
—————————–
Either (7+(-2) or (8+(-3)   |     multliply -3*-2( take result euqal to num of zeros in base)
5     |     6
So ans is  7*8  =  56

2.  Solve 999*996?

Ans)
•      Near Base is 1000
• Finding the Deviation
999          -1
996          -4
• FInd the ans  by cross sum and multiplication
999           -1
996          -4
——————————–
995     |       4( is the right ans but not because num of zeros in base is 3 so we will take 004)
=>             9997           -3
9999           -1
——————————–
9996    |       3(

So the final answer is  999*996   =   995004

3.Find  9997*9999?

Ans)
•      Near Base is 10000
• Finding the Deviation
9997          -3
9999          -1
• FInd the ans  by cross sum and multiplication
=>         9997           -3
9999           -1
——————————–
9996    |       3( is the right answer but not because num of zeros in the base is 3 so we will take 004)
=>        9997           -3
9999           -1
——————————–

9996    |       003

So the final ans is  999*996   =   9996003

### DRDO-CEPTAM has extended date to apply for 21 educational categeries jobs

Name of the Exam :  DRDO-CEPTAM

Vacancy details As per the categeory
Post Code Subject SC ST OBC UR Total Code of posting location
0101  Agriculture  0 H1,L1
0102 Automobile Eng  A2,B2,C2
0103 Botany  D2,H1
0104 Chemical Eng 10  13  A3,H2,J1,N1,P2, V1
0105 Chemistry 3 2 9 10 24 A3,D2,G1,H1,H2, J1,J2,K1,K2,N1, P2
0106 Civil Enginnering 0 0 1 3 4 A1,D2,H2
0107 Computer Science 9 6 24 40 79 A1,A2,A3,B1,B2, B3,D2,H2,J2,K2, P1,P2
0108 Electrical & Electronics Engineering 2 1 4 9 16 B2,B3,C2,H2
0109 Electrical Engineering 5 2 8 20 35 A2,B1,B2,H2,K2, P1,P2,V1
0110 Electronics & Instrumentation 1 0 2 4 7 A1,C1,H2
0111 Electronics or Electronics & Communication or Electronics & Telecommunication Engineering 10 8 26 56 100 B1,B2,C2,D1,D2, H2,J2,K2,M1,P2
0112 Geology 0 0 1 2 3 D2
0113 Instrumentation 1 0 1 3 5 B2,J1,N1,P2
0114 Library Science 0 1 3 7 11 A1,C2,D2,H2,N1, P2
0115 Mathematics 3 0 2 3 8 D2,M2
0116 Mechanical Engineering 30 9 35 66 140 A1,A2,A3,B1,B2, C2,D1,D2,H2,J1, K1,K2,N1,P1,P2, V1

CODES OF POSTING STATIONS: A1=AGRA; A2=AHMEDNAGAR; A3=AMBERNATH(EAST); B1=BALASORE; B2=BENGALURU; B3=BHUBANESWAR; C1=CHANDIGARH; C2=CHENNAI; D1=DEHRADUN; D2=DELHI; G1=GWALIOR; H1=HALDWANI; H2=HYDERABAD; J1=JAGDALPUR; J2=JODHPUR; K1=KANPUR; K2=KOCHI; L1=LEH; M1=MUSSOORIE; M2=MYSORE; N1=NASIK; P1=PANAGARH; P2=PUNE; V1=VISAKHAPATNAM

Age Limit: A) The candidate must be within prescribed age limit on the crucial date of eligibility. However, upper age limit is relaxable for SC, ST, OBC, PWD, ESM, Central Govt. Civilian Employees, Disabled Defence Services Personnel, Persons domiciled in Jammu & Kashmir state during the period from 01-01-1980 to 31-12-1989 as per the rule prescribed by the Government of India. No age relaxation is allowed to SC/ST/OBC candidates applying against unreserved (UR) vacancies. Candidates belonging to PWD, ESM, Central Govt. Civilian Employees, Disabled Defence Services Personnel, Persons domiciled in Jammu & Kashmir state during the period from 01-01- 1980 to 31-12-1989 categories, who are applying against UR vacancies will get age relaxation benefit only for their respective category as above and no additional relaxation will be given for belonging to SC/ST/OBC category (Refer FAQ for further details). B) For relaxation, proforma for certificates may be downloaded (if requireFee required: For Gen: Rs:100/-  remaining categeries : nil
REMUNERATION AND SERVICE CONDITION: Pay, as per 7th CPC Pay matrix, at level 6 and other benefits include dearness allowance, house rent allowance, transport allowance, children education allowance, leave travel concession, medical facilities, CSD facility and other allowances/advances. Recruited candidates will be covered under national pension system (NPS) of the government unless provided otherwise as per Govt. of India rules. DRDO has beautiful well laid out green campuses with residential quarters, general amenities & sports facilities at most of the laboratories/ establishments. The recruited candidates will be governed by the central government rules. The personnel policies in DRDO are well laid down. The selected candidates will be appointed on probation and are liable to serve anywhere within limits of union of India including field locations / remote areas, as and when required, as per Govt. of India rules. These posts are covered under merit based limited flex
RESERVATION/RELAXATION BENEFITS: A) Reservation/ relaxation benefits regarding age, minimum qualifying criteria etc. are applicable to the SC/ST/OBC/PWD etc. candidates applying against vacancies earmarked for them, in accordance with the instructions/orders / circulars, as per extant Govt. of India orders. All candidates applying against unreserved (UR) vacancies will be treated as general candidates. B) Candidates seeking reservation/relaxation benefits must support their claim with duly selfattested copies of relevant certificates issued by Govt./notified competent authority, at the time of submitting online applications and document verification or whenever required by DRDO, else their claim for any relaxation/concession etc. will not be considered and their application will be treated under unreserved category. C) A candidate seeking reservation/ relaxation benefits of OBC must ensure that he/she possess a genuine caste/ community certificate in central Govt. format and does not fall in creamy layer on the crucial date of eligibility. D) Candidates with physical disability of 40% and more only would be considered as person with disability (PWD) and entitled to reservation for PWD. E) It may be noted that, candidature will remain provisional till the veracity of the concerned documents are verified/reverified by the appointing authority.
1. 3.2 APPLICATION FEE, EXEMPTION FROM PAYMENT OF FEE AND MODE OF PAYMENT: A) APPLICATION FEE: Non-refundable application fee of Rs. 100/- (Rupees one hundred only) is to be paid by the candidate. The fee should be paid separately for each post code applied.
2. B) EXEMPTION FROM PAYMENT OF FEE: All women and SC/ST/PWD/ESM candidates are exempted from payment of application fee, as per Govt. of India rules.
3. C) MODE OF PAYMENT: Fee is to be paid online through credit card/debit card/net banking. Fee once paid will not be refunded under any circumstances.
4.  D) Ex-servicemen, who have already secured employment in civil side under Central Government on regular basis after availing of the benefits of reservation given to ex-servicemen for their reemployment, are NOT eligible for fee concession.
3.3 A) EXAMINATION CITIES FOR TIER-I: Candidates are advised to choose any three different cities from the following list in order of preference for Tier-I examination. The option/preference once given by the candidate will be treated as final and irreversible. No request for change of examination city will be entertained. CEPTAM reserves the right to add/delete any examination city and allot the candidates to any examination city other than chosen by candidate depending upon the operational constraints. 01 AGRA 17 IMPHAL 33 MYSORE 02 AHMEDABAD 18 INDORE 34 NAGPUR 03 BALASORE 19 ITANAGAR 35 NASIK 04 BENGALURU 20 JABALPUR 36 PANAJI 05 BHOPAL 21 JAIPUR 37 PATNA 06 BHUBANESWAR 22 JALANDHAR 38 PORT BLAIR 07 BIKANER 23 JAMSHEDPUR 39 PUNE 08 CHANDIGARH 24 JAMMU 40 RAIPUR 09 CHENNAI 25 JODHPUR 41 RAJKOT 10 DEHRADUN 26 KANPUR 42 RANCHI 11 DELHI NCR 27 KOCHI 43 SILIGURI 12 GORAKHPUR 28 KOLKATA 44 THIRUVANANTHAPURAM 13 GUWAHATI 29 LUCKNOW 45 VARANASI 14 GWALIOR 30 MADURAI 46 VIJAYWADA 15 HAMIRPUR 31 MANGALORE 47 VISAKHAPATNAM 16 HYDERABAD 32 MUMBAI B) EXAMINATION CITIES FOR TIER-II: No choice for city is required to be given by the candidates for Tier-II examination. Examination cities for the same will be decided by CEPTAM based on the operational requirement.
3.4 REJECTION CRITERIA: The rejection of applications will be based on following grounds:
• A) Not meeting EQR.
• B) Incomplete or partially filled Application.
• C) Applications without Fees (wherever applicable).
•  D) Applications not received through proper mode/channel.
• E) Applications having blurred/irrelevant photo, signature, thumb impression or other documents.
• F) Underage or overage as on crucial date of eligibility.
• G) Higher qualification viz. M.Sc. or B.Tech. or B.E. or Ph.D. degree etc. as on crucial date of eligibility. .
• H) If a candidate submits more than one application successfully for same post code, then only the latest application with application fee (if applicable) will be considered and other applications will be rejected.

### Basics of Profit and Loss-Part1

First of all welcome to www.job-updates.com  and we are providing all job updates, jobs information, shortcuts or tricks on reasoning, General Knowledge, as well as current affairs.

## Profit and Loss

### Basic Definitions and Formulas

• Cost price (C.P.): Buying price.
• Selling price (S.P.): sales Price.
• Profit/Gain: If the selling price is more than the cost price, the difference between them is the profit occurred.
Formula: Profit/ Gain = S.P. – C.P.
• Loss: If the selling price is less than the cost price, the difference between them is the loss occurred.

Formula: Loss = Cost price (C.P.) – Selling Price (S.P.)
• Profit or Loss is always calculated on the cost price.
• Marked price: This is the price marked as the selling price on an article, also known as the listed price.
• Discount or Rebate: This is the reduction in price offered on the marked or listed price.
Below is the list of some basic formulas used in solving questions on profit and loss:
• Gain % = (Gain / CP) * 100
• Loss % = (Loss / CP) * 100
• SP = [(100 + Gain%) / 100] * CP
• SP = [(100 – Loss %) / 100]*CP
The above two formulas can be stated as,
If an article is sold at a gain of 10%, then SP = 110% of CP.
If an article is sold at a loss of 10%, then SP = 90% of CP.
• CP = [100 / (100 + Gain%)] * SP
• CP = [100 / (100 – Loss%)] * SP

## Profit and Loss: Solved Examples

Question 1: An article is purchased for Rs. 450 and sold for Rs. 500. Find the gain percent.
Solution:
Gain = SP – CP = 500 – 450 = 50.
Gain% = (50/450)*100 = 100/9 %
Question 2: A man sold a fan for Rs. 465. Find the cost price if he incurred a loss of 7%.
Solution:
CP = [100 / (100 – Loss %)] * SP
Therefore, the cost price of the fan = (100/93)*465 = Rs. 500
Question 3: In a transaction, the profit percentage is 80% of the cost. If the cost further increases by 20% but the selling price remains the same, how much is the decrease in profit percentage?
Solution:
Let us assume CP = Rs. 100.
Then Profit = Rs. 80 and selling price = Rs. 180.
The cost increases by 20% → New CP = Rs. 120, SP = Rs. 180.
Profit % = 60/120 * 100 = 50%.
Therefore, Profit decreases by 30%.
Question 4: A man bought some toys at the rate of 10 for Rs. 40 and sold them at 8 for Rs. 35. Find his gain or loss percent.
Solution:
Cost price of 10 toys = Rs. 40 → CP of 1 toy = Rs. 4.
Selling price of 8 toys = Rs. 35 → SP of 1 toy = Rs. 35/8
Therefore, Gain = 35/8 – 4 = 3/8.
Gain percent = (3/8)/4 * 100 = 9.375%
Question 5: The cost price of 10 pens is the same as the selling price of n pens. If there is a loss of 40%, approximately what is the value of n?
Solution:
Let the price of each pen be Re. 1.
Then the cost price of n pens is Rs. n and
the selling price of n pens is Rs. 10.
Loss = n-10.
Loss of 40% → (loss/CP)*100 = 40

Therefore, [(n-10)/n]*100 = 40 → n = 17 (approx)

### basics of simplification for SSC UPSC IBPS

First of all welcome to www.job-updates.com  and we are providing all job updates, jobs information, shortcuts or tricks on reasoning, General Knowledge, as well as current affairs.

## Simplification

Step 1: Parts of the expression enclosed in ‘Brackets’ must be solved first. Inside the brackets, once again BODMAS rules apply!
Step  2: The mathematical operators ‘of’ and ‘order’ must be solved next. ‘Of’ means part of and is solved by substituting with a multiplication sign. ‘Order’ is the same as exponent. Powers are solved after brackets. Powers also include roots.
Step 3: Next, the parts of the equation that contain ‘Division’ and ‘Multiplication’ are calculated.
Step 4: Last but not least, the parts of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated.
Here is an example, to help you understand the BODMAS rule concept.

## EXAMPLE:

What will come in place of question mark (?) in the following question?
240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69)2} = ?
(1) 4111
(2) 3441
(3) 4441
(4) 3841
Ans: (4)
Solution:
Given expression is,
240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69)2} = ?
According to the BODMAS, first we need to solve the Brackets.
Here in this equation first we will solve curly bracket and inside the bracket, we will again follow BODMAS Rule. In the curly bracket once again another bracket appears. So we solve that bracket first. Within these smaller brackets, there is no ‘of’ or ‘order’, so we proceed with the following steps in BODMAS.
We see multiplication, division and subtraction signs. Since multiplication and division are of the same rank, we go left to right to solve this. First we multiply, then we divide. Then we perform the second multiplication. After that, we proceed to the subtraction.
Here are the steps to illuminate this process.
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (264 ÷ 8 × 3 – 69)2} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (33 × 3 – 69)2} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (99 – 69)2} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ 302} = ?
Now we have reached the end of the small brackets. We encounter our first exponent. So we need to solve this ahead of division.
Now solving curly bracket,
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ 900} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of 2 = ?
So far we have cleared all the brackets. Now we move to the next step. We come across our first ‘of’. Here, we simply treat ‘of’ as product i.e. multiplication. But note that, ‘of’ will be solved before a regular multiplication.
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ × 2 = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + 1 = ?
Now, we have cleared our expression of all brackets, exponents and ‘of’s. We now move to the 3rd step. Here we perform all the divisions and multiplications. Note that these two operations are the same rank. So we perform either in the order we come across them in, from left to right.
⇒ 30 × 512 ÷ 4 + 1 = ?
⇒ 15360 ÷ 4 + 1 = ?
⇒ 3840 + 1 = ?
Now we have cleared the expression of all multiplication and division operations as well. All we are left with is addition and subtraction.
⇒ ? = 3841
Hence, the required answer is 3841.

Now, Try It Yourself:
Que. 1
What will come in place of question mark (?) in the following question?
24 + 13 – 5 × ½ of 10 – {45 ÷ (17 – 2)} =?
1.
11
2.
-7
3.
9
4.
18