Quantitative-Aptitude-Data-Interpretation
Quantitative-Aptitude-Data-Interpretation

Quantitative Quiz 37 - Competitive Exams - IBPS/SSC/BANKING EXAMS

Directions (1-5): Study the information carefully to answer the question that follow-
In a sport event there are 5 sports viz. Hockey, Cricket, Tennis, Badminton and Baseball. There is a total number of 800 players in the sports event. The ratio between female and male players is 1 : 3 respectively.

Twenty five p ercent of the total players are in Cricket. There are 110 badminton players. 10 percent of the total players are in Tennis. Hockey players are double the number of badminton players. Remaining players are in Baseball. 30 percent of cricket players are female. Half the female cricketers are equal to female badminton players. 10 percent of total hockey players are equal to the number of female players in Tennis. There are equal numbers of female in Hockey and Baseball.

1.  What is the respective ratio between the female players in Hockey and the male players in Badminton?
(1) 20 : 13                                     (2) 11 : 20
(3) 13 : 20                                     (4) 11 : 23
(5) None of these

2.   What is the total number of males in Hockey, Cricket and Baseball together?
(1) 464                                          (2) 454
(3) 462                                          (4) 432
(5) None of these

3.   Number of females players in Baseball is what percentage of male players in Hockey?
(1) 25                                             (2) 34
(3) 24                                             (4) 15
(5) None of these

4.   What is the difference between the male players in Baseball and the total number of players in Tennis?
(1) 58                                             (2) 76
(3) 56                                             (4) 68
(5) None of these

5.  In which sports female players are maximum and male players are minimum respectively?
(1) Cricket and Badminton   (2) Cricket and Hockey
(3) Baseball and Cricket         (4) Cricket and Tennis
(5) Tennis and Hockey

Answer

1-2,     2-3,         3-1,         4-5,         5-4

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Percentage Problems Shortcut and Tricks for Competitive Exams Free PDF Download

Dear Aspirants,
                         We are sharing PDF document of Percentage Problems Tricks notes for Competitive Exams like SSC, SSC CGL, RRB, SBI PO, SBI Clerk, IBPS PO­, IBPS Clerk­ , IBPS RRB, Railway and other Exams. This pdf will help you to upcoming competitive exams.  You are advised to download PDF Document of Percentage Problems Shortcuts Tricks by go through following Link.


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Simplification Tricks Hand Written Notes PDF Download for SSC, Bank PO & Clerk Exams

Dear Aspirants, Today we are sharing e-pdf of “Simplification Tricks Hand Written Notes PDF” for competitive Examinations. This Notes is very useful for all types of competitive exams like SSC CGL, CHSL, CPO, MTS, Bank PO, Clerk, IBPS, LIC, NICL, RRB, Railway & other public service exams. You are advised to download PDF from the link provided given below.

Simplification Tricks Hand Written Notes PDF Download for SSC, Bank PO & Clerk Exams








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Important Formulas - Decimal Fraction- Tricks & Shortcuts

1. Decimal Fractions:
Fractions in which denominators are powers of 10 are known as decimal fractions.
Thus,     1/10       = 1 tenth = .1;             1/100  = 1 hundredth = .01;                                                      
99/100  = 99 hundredths = .99;  7/1000  = 7 thousandths = .007, etc.;
                                               
2. Conversion of a Decimal into Vulgar Fraction:
Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.
Thus, 0.25 = 25/100 = 1/4 ;       2.008 = 2008/1000 = 251/125.
                                                                                                                               
3. Annexing Zeros and Removing Decimal Signs:
Annexing zeros to the extreme right of a decimal fraction does not change its value. Thus, 0.8 = 0.80 = 0.800, etc.

If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.

Thus,     1.84/2.99             =             184/299                =             8/13.
                                                                                               
4. Operations on Decimal Fractions:

i. Addition and Subtraction of Decimal Fractions: The given numbers are so placed under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way.

ii. Multiplication of a Decimal Fraction By a Power of 10: Shift the decimal point to the right by as many places as is the power of 10.
Thus, 5.9632 x 100 = 596.32;   0.073 x 10000 = 730.

iii. Multiplication of Decimal Fractions: Multiply the given numbers considering them without decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers.
Suppose we have to find the product (.2 x 0.02 x .002).
Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) = 6.
  .2 x .02 x .002 = .000008

iv. Dividing a Decimal Fraction By a Counting Number: Divide the given number without considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend.

Suppose we have to find the quotient (0.0204 Õ 17). Now, 204 Õ 17 = 12.

Dividend contains 4 places of decimal. So, 0.0204 Õ 17 = 0.0012

v. Dividing a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number.

Now, proceed as above.

Thus,     0.00066/0.11      =  0.00066 x 100/0.11 x 100           = 0.066/11 = .006
                                                                                               
5. Comparison of Fractions:
Suppose some fractions are to be arranged in ascending or descending order of magnitude, then convert each one of the given fractions in the decimal form, and arrange them accordingly.
Let us to arrange the fractions 3/5, 6/7 and 7/9 in descending order.
                                                                                               
Now,     3/5         = 0.6,     6/7         = 0.857,                7/9         = 0.777...
                                                                                               
Since, 0.857 > 0.777... > 0.6. So,  6/7         >             7/9         >             3/5         .
                                                                                               
6. Recurring Decimal:
If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal.

n a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set.

Thus,     1/3 = 0.333... = 0.3;          22/7       = 3.142857142857.... = 3.142857.
                                                               
Pure Recurring Decimal: A decimal fraction, in which all the figures after the decimal point are repeated, is called a pure recurring decimal.

Converting a Pure Recurring Decimal into Vulgar Fraction: Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.
Thus, 0.5 = 5/9;   0.53 = 53/99; 0.067 = 67/999, etc.
                                                                                               
Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal.
Eg. 0.1733333.. = 0.173.

Converting a Mixed Recurring Decimal Into Vulgar Fraction: In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated. In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits.
Thus, 0.16 = (16 – 1)/ 90 = 15/90 = 1/6;   0.2273 =                (2273 – 22)/ 9900 = 2251/9900.
                                                                                                                                                               
7. Some Basic Formulae:
i.              (a + b)(a - b) = (a2 - b2)
ii.             (a + b)2 = (a2 + b2 + 2ab)
iii.            (a - b)2 = (a2 + b2 - 2ab)
iv.           (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
v.            (a3 + b3) = (a + b)(a2 - ab + b2)
vi.           (a3 - b3) = (a - b)(a2 + ab + b2)
vii.          (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)
viii.         When a + b + c = 0, then a3 + b3 + c3 = 3abc.





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Important Formulas - Pipes and Cistern - Tricks & Shortcuts

1. Inlet:
A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

Outlet:
A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

2. If a pipe can fill a tank in x hours, then:
part filled in 1 hour =      1/x.

3. If a pipe can empty a tank in y hours, then:
part emptied in 1 hour = 1/y.

4. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then
the net part filled in 1 hour    =   ((1/x) – (1/ y)).
               
5. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour  =  ((1/y) – (1/ x)).

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Important Formulas - Alligation or Mixture- Tricks & Shortcuts

1. Alligation:
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.

2. Mean Price:
The cost of a unit quantity of the mixture is called the mean price.

3. Rule of Alligation:
If two ingredients are mixed, then



4. Suppose a container contains x of liquid from which y units are taken out and replaced by water.

After n operations, the quantity of pure liquid
    


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Important Formulas - Profit and Loss - Tricks & Shortcuts

Cost Price:
The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

Selling Price:
The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

Loss:
If S.P. is less than C.P., the seller is said to have incurred a loss.

IMPORTANT FORMULAE
1) Gain = (S.P.) - (C.P.)
2) Loss = (C.P.) - (S.P.)
3) Loss or gain is always reckoned on C.P.
4)Gain Percentage: (Gain %)
  Gain % = (Gain x 100)/C.P
5) Loss Percentage: (Loss %)
   Loss % = (Loss x 100)/C.P.
6) Selling Price: (S.P.)
   SP = ((100 + Gain %)/100) x C.P
7) Selling Price: (S.P.)
    SP = ((100 - Loss %)/100)x C.P.
8) Cost Price: (C.P.)
   C.P. = (100/(100 + Gain %)) x S.P.
9) Cost Price: (C.P.)
    C.P. = (100/(100 - Loss %)) x S.P.
10) If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
11) If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
12) When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:
    Loss % = ((Common Loss and Gain %)/10)^2 = (x/10)^2.
13) If a trader professes to sell his goods at cost price, but uses false weights, then
    Gain % = (Error/(True Value - Error))x 100%.




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Quantitative Aptitude & Data Interpretation

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Pipes and cisterns tricks & Shortcuts pdf Download

  Dear aspirants,
                             we are sharing today e-pdf of “Pipes and Cisterns Tricks and Shortcuts” Study material for all type of competitive exams. This notes will help you to competitive exams like SSC, SSC CGL, IBPS, SBI, RRB, Clerk, PO, and other competitive exams.  You are all advised to download pipes and Cisterns shortcuts by go through following link.



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Important Formulas - Surds and Indices - Tricks & Shortcuts

1. Laws of Indices:
i. am x an = am + n
ii. am/ an =  am - n
iii. (am)n = amn
iv. (ab)n = anbn
v.  (a/b)n = (an )/(bn)

2. Surds:
Let a be rational number and n be a positive integer such that a(1/n) = nth root of a
Then, is called a surd of order n.

3. Laws of Surds:
i.  nth root of a = a(1/n)

ii. nth root of ab = nth root of a * nth root of b
iii. nth root of (a/b) = (nth root of a)/ (nth root of b)
iv.  (nth root of a)n=a
v. mth root of(nth root of a)) = mnth root of a
vi. (nth root of a)m = nth root of(am)


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Important Formulas - Permutation and Combination - Tricks & Shortcuts

1.  Factorial Notation:
Let n be a positive integer. Then, factorial n, denoted n! is defined as:
n! = n(n - 1)(n - 2) ... 3.2.1.

Examples:
We define 0! = 1.
4! = (4 x 3 x 2 x 1) = 24.
5! = (5 x 4 x 3 x 2 x 1) = 120.

2. Permutations:
The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:
i. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
ii. All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)

3. Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
nPr = n(n - 1)(n - 2) ... (n - r + 1) =    n!/(n - r)!

Examples:
i. 6P2 = (6 x 5) = 30.
ii. 7P3 = (7 x 6 x 5) = 210.
iii. Cor. number of all permutations of n things, taken all at a time = n!.

4. An Important Result:
If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is =    n!/ ((p1!).(p2)!.....(pr!))

5. Combinations:
Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Examples:i. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note: AB and BA represent the same selection.
ii. All the combinations formed by a, b, c taking ab, bc, ca.
iii. The only combination that can be formed of three letters a, b, c taken all at a time is abc.
iv. Various groups of 2 out of four persons A, B, C, D are:1111111
AB, AC, AD, BC, BD, CD.
v. Note that ab ba are two different permutations but they represent the same combination.

6.  Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr =       n!/( (r!)(n - r)!)  = (n(n - 1)(n - 2) ... to r factors)/ r!            .
               
Note:
i. nCn = 1 and nC0 = 1.
ii. nCr = nC(n - r)

Examples:
i.   11C4 = (11 x 10 x 9 x 8)/ (4 x 3 x 2 x 1) = 330.
ii.   16C13 = 16C(16 - 13) = 16C3 = (16 x 15 x 14)/ 3! = (16 x 15 x 14)/ (3 x 2 x 1) =  560.




                
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Important Formulas - Volume and Surface Area - Tricks & Shortcuts

1.       CUBOID
Let length = l, breadth = b and height = h units. Then
                                 i.      Volume = (l x b x h) cubic units.
                               ii.      Surface area = 2(lb + bh + lh) sq. units.
                              iii.      Diagonal = √(l2 + b2 + h2 )units.            (Square root of (l2 + b2 + h2 ))

2.       CUBE
Let each edge of a cube be of length a. Then,
                                 i.      Volume = a3 cubic units.
                               ii.      Surface area = 6a2 sq. units.
                              iii.      Diagonal = 3a units.

3.       CYLINDER
Let radius of base = r and Height (or length) = h. Then,
                                 i.      Volume = (πr2h) cubic units.
                               ii.      Curved surface area = (2πrh) sq. units.
                              iii.      Total surface area = 2πr(h + r) sq. units.

4.       CONE
Let radius of base = r and Height = h. Then,
                                 i.      Slant height, l = √(h2 + r2 )units.     (Square root of (h2 + r2 ))
                               ii.      Volume =(1/3)( πr2h) cubic units.
                              iii.      Curved surface area = (πrl) sq. units.
                             iv.      Total surface area = (πrl + πr2) sq. units.

5.       SPHERE
Let the radius of the sphere be r. Then,
                                 i.      Volume = (r3 )/3 cubic units.
                               ii.      Surface area = (4πr2) sq. units.

6.       HEMISPHERE
Let the radius of a hemisphere be r. Then,
                                 i.      Volume =  (2/3) πr3cubic units.
                               ii.      Curved surface area = (2πr2) sq. units.
                              iii.      Total surface area = (3πr2) sq. units.
                       Note: 1 litre = 1000 cm3.




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