Basic Math Class (Number Part-IV)

Numbers-IV

 SOLVED EXAMPLES

This section of numbers chapter mentioned some solved examples, which will helpful for user to understand that how the number system and formulas works:

Ex. 1. Simplify :   (i) 8888 + 888 + 88 + 8   

                              (ii) 11992 - 7823 - 456 

Sol.   i )  8888                       ii) 11992 - 7823 - 456 = 11992 - (7823 + 456)

                888                                                            = 11992 - 8279 = 3713

                  88                                  7823                         11992

         +         8                               +   456                      -   8279

               9872                                  8279                          3713

Ex. 2: What value will replace the question mark in each of the following equations?

        (i)    ? - 1936248 = 1635773            

       (ii)    8597 - ? = 7429 - 4358

Sol.  

(i) Let x  - 1936248=1635773.

Then, x = 1635773 + 1936248

          x = 3572021.                 

(ii) Let 8597 - x = 7429 - 4358.

Then, x = (8597 + 4358) - 7429 

              = 12955 - 7429 

           x = 5526.

Ex. 3. Simplify : (i) 5793405 x 9999  (ii) 839478 x 625

Sol.

i) 5793405 x 9999

= 5793405(10000-1)

= 57934050000-5793405

= 57928256595

ii) 839478 x 625 

=  839478 x 5⁴ 

=  8394780000 

           16

= 524673750.

                                                 

Ex. 4. Evaluate : (i) 986 x 237 + 986 x 863    (ii) 983 x 207 - 983 x 107

Sol.

(i) 986 x 137 + 986 x 863 

=  986 x (137 + 863) 

=  986 x 1000 

=  986000.

(ii) 983 x 207 - 983 x 107 

=  983 x (207 - 107) 

=  983 x 100 

=  98300. 

Ex. 5. Simplify : (i) 1605 x 1605    ii) 1398 x 1398

Sol.

i) 1605 x 1605 

= (1605)² 

= (1600 + 5)² 

= (1600)² + (5)² + 2 x 1600 x 5

= 2560000 + 25 + 16000 

= 2576025.

(ii) 1398 x 1398 - (1398)² 

= (1400 - 2)²

= (1400)² + (2)² - 2 x 1400 x 2

=1960000 + 4 - 5600 

= 1954404.

Ex. 6. Evaluate : (313 x 313 + 287 x 287).

Sol.

(a² + b²) = 1/2 [(a + b)² + (a- b)²]

(313)² + (287)² = 1/2 [(313 + 287)² + (313 - 287)²] 

= ½[(600)² + (26)²]

= 1/2 (360000 + 676) 

= 180338. 

Ex. 7. Which of the following are prime numbers ?

(i) 241           (ii) 337         (iii) 391           (iv) 571

Sol.

                             

(i) Clearly, 16 > Ö241. 

Prime numbers less than 16 are 2, 3, 5, 7, 11, 13. 

241 is not divisible by any one of them.

 241 is a prime number.

                                   

(ii) Clearly, 19>Ö337. 

Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.          

 337 is not divisible by any one of them.

 337 is a prime number.

(iii) Clearly, 20 > Ö39l". 

Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.

We find that 391 is divisible by 17.

 391 is not prime.

(iv) Clearly, 24 > Ö57T. 

Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.

571 is not divisible by any one of them.

571 is a prime number.

 Ex. 8. Find the unit's digit in the product (2467)163 x (341)72.

Sol. 

Clearly, unit's digit in the given product = unit's digit in 7¹⁵³ x 1⁷².                                 

Now, 74 gives unit digit 1.

7¹⁵²  gives unit digit 1,

\ 7¹⁵³  gives unit digit (l x 7) = 7. Also, 1⁷² gives unit digit 1.

Hence, unit's digit in the product = (7 x 1) = 7.

Ex. 09. Find the unit's digit in (264)¹⁰² + (264)¹⁰³

Sol. 

Required unit's digit = unit's digit in (4)¹⁰² + (4)^103.

Now, 4²  gives unit digit 6.

\(4)¹⁰² gives unjt digit 6.

\(4)103 gives unit digit of the product (6 x 4) i.e., 4.

Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.

 Ex. 10. Find the total number of prime factors in the expression (4)¹¹ x (7)⁵ x (11)².

Sol. 

(4)¹¹ x (7)⁵ x (11)² 

= (2 x 2)¹¹ x (7)⁵ x (11)² 

= 2¹¹ x 2¹¹ x 7⁵ x 11² 

= 2²² x 7⁵ x11²

Total number of prime factors = (22 + 5 + 2) = 29. 

Ex.11. Simplify :    (i) 896 x 896 - 204 x 204

                                (ii) 387 x 387 + 114 x 114 + 2 x 387 x 114

                                (iii) 81 X 81 + 68 X 68-2 x 81 X 68.

Sol.

(i)  Given exp  

=  (896)² - (204)² 

= (896 + 204) (896 - 204) 

= 1100 x 692 

= 761200.

(ii) Given exp  

= (387)²+ (114)²+ (2 x 387x 114)

= a² + b² + 2ab,  where a = 387,b=114

= (a+b)²

= (387 + 114 )² 

= (501)²

= 251001.

(iii) Given exp 

= (81)² + (68)² – 2x 81 x 68 

= a² + b² – 2ab,Where a =81,b=68

=  (a-b)²

= (81 –68)² 

= (13)² 

⁼¹⁶⁹

Ex.12. Which of the following numbers is divisible by 3 ?

(i) 541326                                 (ii) 5967013

Sol.

(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.

Hence, 541326 is divisible by 3.

(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.

Hence, 5967013 is not divisible by 3.

Ex.13.What least value must be assigned to * so that the number 197*5462 is r 9 ?

Sol.

Let the missing digit be x.

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +2) = (34 + x).

For (34 + x) to be divisible by 9, x must be replaced by 2 .

Hence, the digit in place of * must be 2.

Ex. 14. Which of the following numbers is divisible by 4 ?

(i) 67920594                    (ii) 618703572

Sol.

     

(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.

Hence, 67920594 is not divisible by 4.

(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.

Hence, 618703572 is divisible by 4.

Ex.30.How many terms are there in 2,4,8,16……1024?

Sol.

Clearly 2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.

Let the number of terms be n  . 

Then 2 x 2^n-1 =1024 or 2^{n-1 =512 = 2}9.

\n-1=9 or n=10.

 Thank You

Note: In next post we will start H.C.F and L.C.M of Numbers.