Quantitative Aptitude

How To Solve Ratio and Proportion Questions for SSC CGL Tier-I 2017

05:53

Ratio And Proportion

What is Ratio?

Ratio is a mathematical term used to compare two similar quantities expressed in the same units. The ratio of two terms ‘x’ and ‘y’ is denoted by x : y. In ratio x : y , we can say that x as the first term or antecedent and y, the second term or consequent.

In general, the ratio of a number x to a number y is defined as the quotient of the numbers x and y i.e. x/y.

Example: The ratio of 25 km to 100 km is 25:100 or 25/100, which is 1:4 or 1/4, where 1 is called the antecedent and 4 the consequent.

Note that fractions and ratios are same; the only difference is that ratio is a unit less quantity while fraction is not.

Compound Ratio

Ratios are compounded by multiplying together the fractions, which denote them; or by multiplying together the antecedents for a new antecedent, and the consequents for a new consequent. The compound of a : b and c : d is

i.e. ac : bd.

Properties of Ratio:

☑ a : b = ma : mb, where m is a constant

☑ a : b : c = A : B : C is equivalent to a / A = b /B = c /C, this is an important property and has to be used in ratio of three things.

☑

i.e. the inverse ratios of two equal ratios are equal. This property is called Invertendo.

☑

i.e. the ratio of antecedents and consequents of two equal ratios are equal. This property is called Alternendo.

☑

This property is called Componendo.

☑

This property is called Dividendo.

☑

This property is called Componendo - Dividendo.

☑

☑The incomes of two persons are in the ratio of a: b and their expenditures are in the ratio of c: d. If the saving of each person be Rs. S, then their incomes are given by-

Example: Annual income of A and B are in the ratio of 5: 4 and their annual expenses bear a ratio of 4: 3. If each of them saves Rs. 500 at the end of the year, then find the annual income.

Dividing a Quantity Into a Ratio

Suppose any given quantity ‘a’ is to be divided in the ratio of m : n.

Then,

Proportion

When two ratios are equal, the four quantities composing them are said to be in proportion.

If a/b=c/d, then a, b, c, d are in proportions.

This is expressed by saying that ‘a’ is to ‘b’ is to ‘c’ is to ‘d’ and the proportion is written as

a : b :: c : d or a : b = c : d

(product of means = product of extremes)

If there is given three quantities like a, b, c of same kind then we can say it proportion of continued.

a : b = b : c the middle number b is called mean proportion. a and c are called extreme numbers.

So, b² = ac. (middle number)² = ( First number x Last number ).

Application: These properties have to be used with quick mental calculations; one has to see a ratio and quickly get to results with mental calculations.

Example:

should quickly tell us that

Q. A certain amount was to be distributed among A, B and C in the ratio 2 : 3 : 4, but was erroneously distributed in the ratio 7 : 2 : 5. As a result of this, B received Rs. 40 less. What is the actual amount?

(a) Rs. 210

(b) Rs. 270

(d) Rs. 280

(e) None of these

Q. Mixture of milk and water has been kept in two separate containers. Ratio of milk to water in one of the containers is 5 : 1 and that in the other container 7 : 2. In what ratio the mixtures of these two containers should be added together so that the quantity of milk in the new mixture may become 80%?

(a) 2 : 3

(b) 3 : 2

(d) 1 : 3

(e) None of these

STATISTICS(basic concept) NOTES for RRB NTPC EXAM

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Statistics

The words ‘Statistics’ appears to have been derived from the latin word ‘status’ meaning a (political) state. In its origin, Statistic was simply the collect of data on different aspects of the life of the people.

Statistics deals with data collected for specific purposes. We can make decisions about the data by analysing and interpreting it.

Central line tendency

Mean: The mean or average of a number of observation is the sum of the value of all the observation divided by the total number of the observations.

It is denoted by the symbol

, read as x bar

Here n is a number of observation.

Example- people were asked about the time in a week they spend indoing social work in their community. They said 10, 7, 13, 20 and 15 hours respectively.Find the mean (or average) time in a week devoted by them in social work.

Sol. The mean =(Sum of all the observations)/(Total number of observations)

=(10 + 7 + 13 + 20 + 15)/5=65/5=13

So, the time spent by these 5 people in doing social work is 13 hours in a week.

The median- the median is that value of the given number of observations, which divide it into exactly two parts. when the data is arranged in ascending or decreasing order.

The median of ungrouped data is calculated as follows:

(i) When the number of observation (n) is odd, the median is the value of the ((n+1)/2)ith observation.

Example: If n = 15, the value of the ((15+1)/2)ith i.e. the 8^th observation will be median

(ii) When the number of observation (n) is even, the median is the mean of the (n/2)ith and (n/2+1)ith observations.

Example: If n = 16 the mean of the value of the (16/2)ith and (16/2+1)ith observations.

Example: The height (in cm) of 9 students of a class follows.

155, 152, 160, 144, 145, 148, 150, 147,149

Find the median of the data

Sol. Arrange the data in ascending data

144, 145, 147, 148, 149, 150, 152, 155, 160

Since the number of students is 9, an odd number.

Median is the height of the = 5^th student which is 149 cm

So, median = 149 cm

Modes: The mode is that value of the observation which occurs most frequently i.e. an observation with the maximum frequency is called the mode.

Example: Find the mode of the following marks (out of 10) obtained by 20 students.

4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9

Sol. We arrange this data in the following form:

2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9,10, 10

Here 9 occurs most frequently i.e. four times. So, the mode is 9.

All About Profit & Loss

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PROFIT & LOSS
Profit and loss are determined by the value of cost price and selling price. Cost price is the price at which an article is purchased and selling price is the price at which article is sold
Profit = selling price - Cost price

Loss = Cost price - Selling price

Percentage profit and loss are always calculated on cost price.

☞If a cost price of m articles is equal to the selling Price of n articles, then Profit percentage

MARKED PRICE
Marked price is also known as the list price. It is the price which is marked on the article.

Where CP = cost price and MP = marked price

DISCOUNT
Shopkeepers devise several ways to attract customers (consumers). Sometimes they sell an article at a price lower than its list price (LP)/marked price (MP). Recall that reduction offered by retailer on the list price is called discount. We may recall that
Discount = MP - SP

Example 1: Marked price of a dining table is Rs 1350. It is sold at Rs. 1188 after allowing certain discount. Find the rate of discount.

Solution:
MP of the dining table = Rs. 1350
SP of the dining table = Rs. 1188
Discount allowed = Rs. (1350 - 1188) = Rs. 162
Discount percent =162/1350×100=12
This the rate of discount is 12%

As we had discussed the Multiplying Factor concept, it is very helpful to calculate the S.P. and C.P.
S.P. = C.P. × M.F.
In case of profit M.F. is greater then 1.

If there is 10% profit, then
S.P. = C.P. × 1.1
M.F. = 1.1
For 15% profit M.F. = 1.5

Let’s take an example
If markup percentage is 30%, and the profit percentage is 17% then find the discount percentage.
Let CP = 100
M.P. = 100 × 1.3 = 130
(M.P. – Marked up price)
S.P. = 100 × 1.17 = 117

Discount %
M.P. × Multiplying factor = S.P
130 × M.F. = 117
M.F = .9
Discount Percentage = 10%

In case of Loss S.P < C.P And M.F. is smaller then 1.

Relation between multiplying factor of, Profit, Mark-up and discount

MF profit = MF mark-up × MF discount.

SUCCESSIVE DISCOUNTS
Sometimes more than one discount are offered by the shopkeeper on a single item or article. When two or more discounts are applicable successively to the list price of an article, they form the discount series.
Suppose a shopkeeper is offering 3 successive discounts of 10%, 20% and 30% then to calculate effective discount we assume that marked price is 100, then final value becomes 0.90 × 0.80 × 0.70 × 100 = 0.54 × 100 = 50.4
Total discount = 49.6%.

☞When there are two successive Profit of x % and y % then the resultant profit per cent is given by

☞If there is a Profit of x% and loss of y % in a transaction, then the resultant profit or loss% is given by

Note- For profit use sign + in previous formula and for loss use – sign.
if resultant come + then there will be overall profit, if it come – then there will be overall loss.

Example 2:
If two articles are sold at same selling price one at 30% profit another at 30% loss then what is his overall percentage profit or loss?

FALSE WEIGHT PROBLEMS
Shown or indicate weight is always equivalent to selling price, and actual/true weight is equivalent to cost price.

☞If a trader professes to sell his goods at cost price, but uses false weights, then

Example 3:
A shopkeeper takes 20%, extra quantity while purchasing the milk, and gives 25% less than the indicated weight while selling the milk. Find the profit percentage of he sells at the cost price only.

Solution:
Suppose the price of milk = 1 Rs per ml shopkeeper takes 120 ml, and pays only Rs. 100
While selling he gives only 75 ml and shows 100 ml.
Total selling price of 120 ml
100/75×120 = 160, hence percentage profit = 60%

Quant Study Notes: Time and work for SSC CGL TIER II

05:46

Time and work

This chapter is based on the concept of direct and inverse variations. We need to understand relation among time, work done and number of employees working.

Assuming that all employees work with the same efficiency, we can conclude that work done is directly proportional to number of employees working and number of days to complete the work is inversely proportional to number of employees working.

☞For example if a person does a work in 10 days then in 1 day he does only one tenth of the work.

☞For example it two men can complete a work working alone in 10 and 20 days respectively, then one day’s work of both the men will be

Thus the total can be completed by both of them in

Important points:

➀ If A can do a piece of work in X days, then A’s one day’s work =1/Xth part of whole work.

➁ If A’s one day’s work =1/Xth part of whole work, then A can finish the work in X days.

➂ If A can do a piece of work in X days and B can do it in Y days then A and B working together will do the same work in

➃ If A, B and C can do a work in X, Y and Z days respectively then all of them working together can finish work in

Example: If X can do a work in 10 day, Y can do the same work in 20 days, Z can do double of the work in 30 days. If all 3 started working together, find the total time required to complete the work.

Solution:

(i) Unitary Method:

(ii) LCM Method:

CONCEPT of MAN DAYS

Assuming that all men work with same efficiency, we can conclude that if the work done is constant then the number of days is inversely proportional to the number of men working. For example, if we say that 20 men can do a work in 30 days, this means that total work is 20x30 man days = 600 man days, which means that the same work can be done by 10 men in 60 days or 60 men can do in 10 days only etc. So for a constant work number

of man days is constant.

where M is the number of men

⇒ MD = constant

Thus we obtain a relationship

If work is not constant then it is directly proportional to both number of men and number of days

Where MD is equivalent to number of man days.

Example: 10 men or 20 women or 30 children, can do a work in 15 days. If 10 men, 12 women and 18 children work on the same work, find the time in which work can be competed.

Solution:

We see that 10 men are equivalent to 20 women that is equivalent to 30 children

Hence 1 man = 2 women = 3 children

Total work is 10 × 15 = 150 man days

10 men, 12 women and 18 children are equivalent to 10 + (12/2) + (18/3) = 22 men.

Hence time taken to complete the work

= 150/22 = 75/11 = 6 (9/11) days.

ALTERNATE WORKING

In the following example, we will discuss how to find the total time taken to complete the work, if the two or more workers are not working together but they are working on alternate days.

Suppose A and B are the two workers working on a project such that A and B can complete a work working alone in 20 and 12 days respectively. Now they are working on alternate day, now to find the total time required to complete the work, there can be two cases:

(a) Starting with ‘A’

(b) Starting with ‘B’

In these types of questions work done on the first day is not same as the work done on the second day but not work done in first two days is same as the work done in the next two days and that is same as the work done in the next two days and so on.

AB AB AB ………

Work done in 2 days

If we assume 2 days to be one cycle and total work is always considered as 1 unit, then approximate number of cycles required to complete the work =15/2=7 (integral number)

Total work done after 7 complete cycles

Hence the remaining work is 1/15

(1) A started the work: In one day A completes 1/20th of the work, hence 1/15th of the work will be completed in more than one day. A will complete 1/20th part and the remaining work will be done by B.

Remaining work =

Now B will complete the remaining work in 1/60×12=1/5 days. Total time taken will be:

(2) B started the work: B will complete the remaining work (1/15 th)of the complete work in

PIPES AND CISTERNS

This is the topic which gives the relation between the time required to fill or empty the tank with the taps opened or closed.

Till the time we have only defined the concept of positive work but in the problems related to pipes and cisterns we have to define the concept of negative work also. Concept of negative work is defined as when the work is done against the requirement.

Example: A tap can fill a tank in 16 minutes and another can empty it in 8 minutes. If the tank is already ½ full and both the taps are opened together, will the tank be filled or emptied? How long will it take before the tank is either filled or emptied completely?

Solution:

If both the pumps are opened together, then the tank will be emptied because the working efficiency of pump empting is more than that of the pump filling it. Thus in 1 min net proportion of the volume of tank filled

(Which means that tank will be emptied 1/6th in one minute.)

Hence in 8 minutes half of the tank will be emptied.

Some Important Tricks

➀ If A and B working together, can finish a piece of work in x days, B and C in y days, C and A in z days, then

☞A, B and C working together will complete the job in

☞ A alone will complete the job in

☞ C alone will complete the job in

➁

☞If A can complete a work in x days and B is k times efficient than A, then the time taken by both A and B, working together to complete the work is

☞If A and B, working together, can complete a work in x days and B is k times efficient than A,

then the time taken by –

A working alone to complete the work in

→ (k + 1) x

B working alone to complete the work is

➂ If A working alone takes ‘a’ days more than A and B, and B working alone takes ‘b’ days more than A and B together, then the number of days taken by A and B, working together, to finish a job is given by