Quantative Aptitude (Decimal Fraction)

3. DECIMAL FRACTIONS

IMPORTANT FACTS AND FORMULAE

I.    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 hundreths=.99;7/1000=7 thousandths=.007,etc

II.  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.

  

III.  1.  Annexing zeros to the extreme right of a decimal fraction does not change its value
            Thus, 0.8 = 0.80 = 0.800, etc.

2.  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;    0.365/0.584 = 365/584=5

IV.  Operations on Decimal Fractions :

1.  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.

2.  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 = 0.0730 x 10000 = 730.

3.Multiplication of Decimal Fractions : Multiply the given numbers considering
them without the 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 .02 x .002). Now, 2x2x2 = 8. Sum of decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.

4.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 + 17). Now, 204 ^ 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 + 17 = 0.0012.

5. 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*100)/(0.11*100) = (0.066/11) = 0.006V

V.  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.

Suppose, we have 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

VI. 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.

In 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 = 0.3333….= 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 = 53/59  ;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.

e.g., 0.17333 = 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 =  15/19 =  1/6;

                   ____

                0.2273 =  (2273 – 22)/9900 = 2251/9900

VII.  Some Basic Formulae :

1.   (a + b)(a- b) = (a² - b²).

2.   (a + b)² = (a² + b² + 2ab).

3.    (a - b)² = (a² + b² - 2ab).

4.    (a + b+c)² = a² + b² + c²+2(ab+bc+ca)

5.    (a³ + b³) = (a + b) (a² - ab + b²)

6.    (a³ - b³) = (a - b) (a² + ab + b²).

7.    (a³ + b³ + c³ - 3abc) = (a + b + c) (a² + b² + c²-ab-bc-ca)

8.    When   a + b + c = 0, then a³ + b³+ c³ = 3abc

SOLVED EXAMPLES

Ex. 1. Convert the following into vulgar fraction:

            (i) 0.75             (ii) 3.004          (iii)  0.0056

Sol. (i). 0.75 = 75/100 = 3/4    (ii) 3.004 = 3004/1000 = 751/250    (iii) 0.0056 = 56/10000 = 7/1250

Ex. 2. Arrange the fractions 5/8, 7/12, 13/16, 16/29 and 3/4 in ascending order of magnitude.

Sol. Converting each of the given fractions into decimal form, we get :

        5/8 = 0.624, 7/12 = 0.8125, 16/29 = 0.5517, and 3/4 = 0.75

        Now, 0.5517<0.5833<0.625<0.75<0.8125

        \ 16/29 < 7/12 < 5/8 < 3/4 < 13/16

Ex. 3. arrange the fractions 3/5, 4/7, 8/9, and 9/11 in their descending order.

Sol. Clearly, 3/5 = 0.6, 4/7 = 0.571, 8/9 = 0.88, 9/111 = 0.818.

        Now, 0.88 > 0.818 > 0.6 > 0.571

        \ 8/9 > 9/11 > 3/4 > 13/ 16

Ex. 4. Evaluate : (i) 6202.5 + 620.25 + 62.025 + 6.2025 + 0.62025

                             (ii) 5.064 + 3.98 + 0.7036 + 7.6 + 0.3 + 2

Sol.  (i)  6202.5                                                    (ii)  5.064

                620.25                                                         3.98

                  62.025                                                       0.7036

                    6.2025                                                     7.6

           + __  0.62025                                                   0.3

              6891.59775                                                 _2.0___

                                                                                  19.6476

Ex. 5. Evaluate : (i) 31.004 – 17.2368                       (ii) 13 – 5.1967

Sol.  (i)    31.0040                                                       (ii)   31.0000

             – 17.2386                                                            – _5.1967  

13.7654                                                                                                                                  7.8033  

Ex. 6. What value will replace the question mark in the following equations ?

(i)                 5172.49 + 378.352 + ? = 9318.678

(ii)               ? – 7328.96 + 5169.38

Sol.  (i) Let       5172.49 + 378.352 + x = 9318.678

                        Then , x = 9318.678 – (5172.49 + 378.352) = 9318.678 – 5550.842 = 3767.836

        (ii) Let      x – 7328.96 = 5169.38. Then, x = 5169.38 + 7328.96 = 12498.34.

Ex. 7. Find the products: (i) 6.3204 * 100               (ii) 0.069 * 10000

Sol.  (i) 6.3204 * 1000 = 632.04                     (ii) 0.069  * 10000 = 0.0690 * 10000 = 690

Ex. 8. Find the product:

            (i) 2.61 * 1.3                (ii) 2.1693 * 1.4           (iii) 0.4 * 0.04 * 0.004 * 40

Sol.  (i) 261 8 13 = 3393. Sum of decimal places of given numbers = (2+1) = 3.

             2.61 * 1.3 = 3.393.

        (ii) 21693 * 14 = 303702. Sum of decimal places = (4+1) = 5

              2.1693 * 1.4 = 3.03702.

       (iii) 4 * 4 * 4 * 40 = 2560. Sum of decimal places = (1 + 2+ 3) = 6

              0.4 * 0.04 * 0.004 * 40 = 0.002560.

Ex. 9. Given that 268 * 74 = 19832, find the values of 2.68 * 0.74.

Sol.  Sum of decimal places = (2 + 2) = 4

         2.68 * 0.74 = 1.9832.

Ex. 10. Find the quotient:

            (i) 0.63 / 9                    (ii) 0.0204 / 17                         (iii) 3.1603 / 13           

Sol.  (i) 63 / 9 = 7. Dividend contains 2 places decimal.

             0.63 / 9 = 0.7.

        (ii) 204 / 17 = 12. Dividend contains 4 places of decimal.

              0.2040 / 17  = 0.0012.

        (iii) 31603 / 13 = 2431. Dividend contains 4 places of decimal.

               3.1603 / 13 = 0.2431.

Ex. 11. Evaluate :

            (i) 35 + 0.07                            (ii) 2.5 + 0.0005

(iii)             136.09 + 43.9

Sol. (i) 35/0.07 = ( 35*100) / (0.07*100) = (3500 / 7) = 500

      (ii) 25/0.0005 = (25*10000) / (0.0005*10000) = 25000 / 5 = 5000

     (iii) 136.09/43.9 = (136.09*10) / (43.9*10) = 1360.9 / 439 = 3.1

Ex. 12. What value will come in place of question mark in the following equation?

            (i) 0.006 +? = 0.6                    (ii) ? + 0.025 = 80

Sol.  (i) Let 0.006 / x = 0.6, Then, x = (0.006 / 0.6) = (0.006*10) / (0.6*10) = 0.06/6 = 0.01

        (ii) Let x / 0.025 = 80, Then, x = 80 * 0.025 = 2           

Ex. 13. If (1 / 3.718) = 0.2689, Then find the value of (1 / 0.0003718).

Sol.  (1 / 0.0003718 ) = ( 10000 / 3.718 ) =  10000 * (1 / 3.718) = 10000 * 0.2689 = 2689.

                                                                                       ___                ______                  

Ex. 14. Express as vulgar fractions : (i) 0.37    (ii) 0.053     (iii)  3.142857

                __                                                  ___

Sol. (i)  0.37 = 37 / 99 .                       (ii)  0.053 = 53 / 999

                  ______             ______ 

        (iii) 3.142857 = 3 + 0.142857 = 3 + (142857 / 999999) = 3 (142857/999999)

                                                                        _                       __                      _

Ex. 15. Express as vulgar fractions : (i) 0.17           (ii) 0.1254        (iii)  2.536

                _

Sol. (i) 0.17 = (17 – 1)/90 = 16 / 90 =  8/ 45

                    __

       (ii) 0.1254 = (1254 – 12 )/ 9900 = 1242 / 9900 = 69 / 550

      (iii)  2.536 = 2 + 0.536 = 2 + (536 – 53)/900 = 2 + (483/900) = 2 + (161/300) = 2 (161/300)

Ex. 16. Simplify:  0.05 * 0.05 * 0.05 + 0.04 * 0.04 * 0.04

                               0.05 * 0.05 – 0.05 * 0.04 + 0.04 * 0.04

Sol. Given expression  = (a³ + b³) / (a² – ab + b²), where a = 0.05 , b = 0.04

                                     = (a +b ) = (0.05 +0.04 ) =0.09

Quantative Aptitude (H.C.F & L.C.M)

2. H.C.F. AND L.C.M. OF NUMBERS

IMPORTANT FACTS AND FORMULAE

I.  Factors and Multiples : If a number a divides another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.

II. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.): The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.

   There are two methods of finding the H.C.F. of a given set of numbers :

1.  Factorization Method : Express each one of the given numbers as the product of prime factors.The product of least powers of common prime factors gives H.C.F.

2.  Division Method: Suppose we have to find the H.C.F. of two given numbers. Divide the larger   number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F. 

Finding the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers.

Similarly, the H.C.F. of more than three numbers may be obtained.

III.  Least Common Multiple (L.C.M.) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

1.  Factorization Method of Finding L.C.M.: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors,

2.  Common Division Method {Short-cut Method) of Finding L.C.M.: Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers,

IV. Product of two numbers =Product of their H.C.F. and L.C.M.

V. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.

VI. H.C.F. and L.C.M. of Fractions:

1.H C F= H.C.F. of Numerators         2.L C M =      L.C.M  of Numerators__

                L.C.M. of Denominators                           H.C.F. of Denominators

VII. H.C.F. and L.C.M. of Decimal Fractions: In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

VIII. Comparison of Fractions: Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

SOLVED EXAMPLES

Ex. 1. Find the H.C.F. of 2³  X 3²   X 5 X 7⁴, 2²  X 3⁵ X 5²  X 7³,2³ X 5³  X 7²

Sol.    The prime numbers common to given numbers are 2,5 and 7.

H.C.F. = 2² x 5 x7² = 980.

Ex. 2. Find the H.C.F. of 108, 288 and 360.

Sol.    108 = 2² x 3³, 288 = 2⁵ x 3² and 360 = 2³ x 5 x 3².

H.C.F. = 2² x 3² = 36.

Ex. 3. Find the H.C.F. of 513, 1134 and 1215.

Sol.

1134 ) 1215 ( 1

           1134   

               81 ) 1134 ( 14

                       81

                       324

                       324

                        x         

\H.C.F. of 1134 and 1215 is 81.

So, Required H.C.F. = H.C.F. of 513 and 81.

     ______

81 )   513 ( 6

       __486____

             27) 81 ( 3

                 81  

                  0

H.C.F. of given numbers = 27.

Ex. 4. Reduce 391     to lowest terms .

                        667

to lowest terms.

Sol.    H.C.F. of 391 and 667 is 23.

On dividing the numerator and denominator by 23, we get :

391 = 391 ¸ 23 = 17

667    667¸ 23    29

Ex.5.Find the L.C.M. of 2² x 3³ x 5 x 7^{2 ,} 2³ x 3² x 5² x 7⁴  , 2 x 3 x 5^{3 x 7 x 11.}  

Sol. L.C.M. = Product of highest powers of 2, 3, 5, 7 and 11 = 2³ x 3³ x 5³ x 7⁴ x 11

Ex.6. Find the L.C.M. of 72, 108 and 2100.

Sol. 72 = 2³ x 3², 108 = 3³ x 2², 2100 = 2² x 5² x 3 x 7.

 L.C.M. = 2³ x 3³ x 5² x 7 = 37800.

Ex.7.Find the L.C.M. of 16, 24, 36 and 54.

Sol.

2	16	-   24	-   36	-   54
2	8	-   12	-   18	-   27
2	4	-     6	-     9	-   27
3	2	-     3	-     9	-   27
3	2	-     1	-     3	-     9
	2	-     1	-     1	-     3

\ L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.

Ex. 8. Find the H.C.F. and L.C.M. of 2  , 8  , 16 and 10.

                                                                3    9    81        27   

Sol.    H.C.F. of given fractions = H.C.F. of 2,8,16,10   =     2_   

                                                       L.C.M. of 3,9,81,27        81

          L.C.M of given fractions = L.C.M. of 2,8,16,10   =    80_   

                                                      H.C.F. of 3,9,81,27           3

Ex. 9. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.

Sol.     Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.

Without decimal places, these numbers are 63, 105 and 210.

Now, H.C.F. of 63, 105 and 210 is 21.

H.C.F. of 0.63, 1.05 and 2.1 is 0.21.

L.C.M. of 63, 105 and 210 is 630.

L.C.M. of 0.63, 1.05 and 2.1 is 6.30.

Ex. 10. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the numbers.

Sol.    Let the required numbers be 15.x and llx.

Then, their H.C.F. is x. So, x = 13.

The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.

Ex. 11. TheH.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77,find the other.

Sol.    Other number = 11 X 693   = 99

                                          77

                             

Ex. 12. Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.

Sol.    Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.

          495 = 3² x 5 x 11, 900 = 2² x 3² x 5², 1665 = 3² x 5 x 37.

      \H.C.F. = 32 x 5 = 45.

          Hence, required length = 45 cm.

Ex. 13. Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

Sol.    Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032

         _______

1651 )   2032 ( 1 1651

            1651_______

              381 )  1651 ( 4

                         1524_________

                            127 )   381 ( 3

                                       381

                                         0

Required number = 127.

Ex. 14. Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.

Sol .  Required number =  H.C.F. of (132 - 62), (237 - 132) and (237 - 62)

                                      =  H.C.F. of 70, 105 and 175 = 35.

Ex.15.Find the least number exactly divisible by 12,15,20,27.

Sol.

           

3	12	-   15	-   20	-   27
4	4	-   5	-   20	-   9
5	1	-   5	-     5	-   9
	1	-    1	-     1	-   9

Ex.16.Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case

Sol. Required number = (L.C.M OF 6,7,8,9,12) + 1

3	6	-   7	-    8	-   9    -   12
4	2	-   7	-   8	-   3   -   4
5	1	-   7	-     4	-   3   -   2
	1	-    7	-     2	-   3   -   1

\L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.

Hence required number = (504 +1) = 505.

Ex.17. Find the largest number of four digits exactly divisible by 12,15,18 and 27.

Sol. The Largest number of four digits is 9999.

       Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.

      On dividing 9999 by 540,we get 279 as remainder .

   \Required number = (9999-279) = 9720.

Ex.18.Find the smallest number of five digits exactly divisible by 16,24,36 and 54.

Sol. Smallest number of five digits is 10000.

       Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,

       On dividing 10000 by 432,we get 64 as remainder.

    \Required number = 10000 +( 432 – 64 ) = 10368.

Ex.19.Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively.

Sol. Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6.

    \Required number = (L.C.M. of 20,25,35,40) – 6 =1394.

Ex.20.Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .

Sol. L.C.M. of 5,6,7,8 = 840.

   \ Required number is of the form 840k + 3

     Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 \Required number = (840 X 2 + 3)=1683

Ex.21.The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change simultaneously .

Sol. Interval of change = (L.C.M of 48,72,108)sec.=432sec.

       So, the lights will agin change simultaneously after every 432 seconds i.e,7 min.12sec

       Hence , next simultaneous change will take place at 8:27:12 hrs.

Ex.22.Arrange the fractions 17   , 31, 43, 59   in the ascending order.

18      36   45  60

Sol.L.C.M. of 18,36,45 and 60 = 180.

Now, 17 = 17 X 10 = 170  ;   31 =   31 X 5    = 155 ;

          18     18 X 10    180     36      36 X 5        180

          43 = 43 X 4 = 172  ;   59  = 59 X 3    = 177 ;

          45     45 X 4    180     60     60 X 3        180

Since, 155<170<172<177, so, 155 < 170 < 172 < 177

                                                 180    180    180     180

Hence, 31 < 17 < 43 < 59

            36    18     45    60