Quantative Aptitude (Average)

AVERAGE

Ex.1:Find the average of all prime numbers between 30 and 50?

Sol: there are five prime numbers between 30 and 50.

        They are 31,37,41,43 and 47.

            Therefore the required average=(31+37+41+43+47)/5  ó199/5  ó 39.8.

Ex.2. find the average of first 40 natural numbers?

Sol:    sum  of first n natural numbers=n(n+1)/2;

            So,sum of 40 natural numbers=(40*41)/2  ó820.

            Therefore  the required average=(820/40) ó20.5.

Ex.3. find the average of first 20 multiples of 7?

Sol:    Required average =7(1+2+3+…….+20)/20   ó(7*20*21)/(20*2) ó(147/2)=73.5.

Ex.4. the average of four consecutive even numbers is 27. find the largest of these numbers?

Sol:    let the numbers be x,x+2,x+4 andx+6. then,

(x+(x+2)+(x+4)+(x+6))/4) = 27 

ó(4x+12)/4 = 27

óx+3=27      ó x=24.

Therefore the largest number=(x+6)=24+6=30.

Ex.5. there are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?

Sol:   total weight of(36+44) students=(36*40+44*35)kg  =2980kg.

            Therefore weight of the total class=(2980/80)kg  =37.25kg.

Ex:6.nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?

Sol: Let the average expenditure of all nine be Rs.x

Then 12*8+(x+8)=9x or 8x=104 or x=13.

Total money spent = 9x=Rs.(9*13)=Rs.117.

Ex.7: Of the three numbers, second is twice the first and is also thrice the third.If the average of the three numbers is 44.Find the largest number.

Sol: Let the third number be x.

Then second number = 3x.

First number=3x/2.

Therefore x+3x+(3x/2)=(44*3) or x=24

So largest number= 2^nd number=3x=72.

Ex.8:The average of  25 result is 18.The average of 1^st 12 of them is 14 & that of last 12 is 17.Find the 13^th result.

Sol: Clearly 13^th result=(sum of 25 results)-(sum of 24 results)

=(18*25)-(14*12)+(17*12)

=450-(168+204)

=450-372

=78.

Ex.9:The Average of 11 results is 16, if the average of the 1^st 6 results is 58 & that of the last 63. Find the 6^th result.

Sol: 6^th result = (58*6+63*6-60*11)=66

Ex.10:The average waight of A,B,C is 45 Kg. The avg wgt of A & B be 40Kg & that of B,C be 43Kg. Find the wgt of B.

Sol. Let A,B,c represent their individual wgts.

Then,

A+B+C=(45*3)Kg=135Kg

A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg

B=(A+B)+(B+C)-(A+B+C)

=(80+86-135)Kg

=31Kg.

Ex. 11. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by3 months. Find the age of the teacher.

Sol. Total age of 39 persons = (39 x 15) years

                                                = 585 years.

Average age of 40 persons= 15 yrs 3 months

= 61/4 years.

Total age of 40 persons = (_(61/4 )x 40) years= 610 years.

:. Age of the teacher = (610 - 585) years=25 years.

Ex. 12. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new

man.

Sol. Total weight increased =(1.8 x 10) kg =18 kg.

:. Weight of the new man =(53 + 18) kg =71 kg.

Ex. 13. There were 35 students in a hostel. Due to the admission of 7 new students, ;he expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs 1. Wbat was the original expenditure of the mess?

Sol. Let the original average expenditure be Rs. x. Then,

               42 (x - 1) - 35x=42    ó 7x= 84   ó   x =12.

Original expenditure = Rs. (35 x 12) =Rs. 420. .

14. A batsman makes a score of 87 runs in the 17th inning and thus increases his avg by 3. Find his average after 17th inning.

Sol. Let the average after 17th inning = x.

      Then, average after 16th inning = (x - 3).

:. 16 (x - 3) + 87 = 17x or x = (87 - 48) = 39.

Ex.15. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km perhour. Find the average speed of the train during the whole journey.

Sol. Required average speed = ((2xy)/(x+y)) km / hr

=(2 x 84 x 56)/(84+56)km/hr

 = (2*84*56)/140 km/hr

=67.2 km/hr.

                                                   

Quantative Aptitude (Square Roots & Cube Roots)

SQUARE ROOTS AND CUBE ROOTS

IMPORTANT FACTS AND FORMULAE

Square Root: If x² = y, we say that the square root of y is x and we write, √y = x.

Thus, √4 = 2, √9 = 3, √196 = 14.

Cube Root: The cube root of a given number x is the number whose cube is x. We denote the cube root of x by ³√x.

Thus, ³√8  = ³√2 x 2 x 2 = 2, ³√343 = ³√7 x 7 x 7 = 7 etc.

Note:

1.√xy = √x * √y                     2. √(x/y) = √x / √y  = (√x / √y) * (√y / √y) = √xy / y

SOLVED EXAMPLES

Ex. 1. Evaluate √6084 by factorization method .

Sol.     Method: Express the given number as the product of prime factors.          2    6084  

            Now, take the product of these prime factors choosing one out of              2    3042

            every pair of the same primes. This product gives the square root              3    1521  

            of the given number.                                                                                     3    507

Thus, resolving 6084 into prime factors, we get:                                        13   169

6084 = 2² x 3² x 13²                                                                                                         13                \ √6084 = (2 x 3 x 13) = 78.       

                                                                     

Ex. 2. Find the square root of 1471369.                                                   

Sol.     Explanation: In the given number, mark off the digits              1  1471369 (1213

in pairs starting from the unit's digit. Each pair and                        1             

the remaining one digit is called a period.                                22     47

Now, 1² = 1. On subtracting, we get 0 as remainder.                        44

Now, bring down the next period i.e., 47.                               241      313

Now, trial divisor is 1 x 2 = 2 and trial dividend is 47.                       241

So, we take 22 as divisor and put 2 as quotient.                    2423        7269

The remainder is 3.                                                                                7269

Next, we bring down the next period which is 13.                                  x

Now, trial divisor is 12 x 2 = 24 and trial dividend is

313. So, we take 241 as dividend and 1 as quotient.

The remainder is 72. ­

Bring down the next period i.e., 69.

Now, the trial divisor is 121 x 2 = 242 and the trial

dividend is 7269. So, we take 3as quotient and 2423

as divisor. The remainder is then zero.

Hence, √1471369 = 1213.

Ex. 3. Evaluate: √248 + √51 + √ 169 .

Sol.     Given expression = √248 + √51 + 13 = √248 + √64    = √ 248 + 8 = √256 = 16.

Ex. 4. If a * b * c = √(a + 2)(b + 3) / (c + 1), find the value of 6 * 15 * 3.

Sol.      6 * 15 * 3 = √(6 + 2)(15 + 3) / (3 + 1) = √8 * 18 / 4 = √144 / 4 = 12 / 4 = 3.

Ex. 5. Find the value of √25/16.

Sol.    √ 25 / 16   = √ 25 / √ 16 = 5 / 4

Ex. 6. What is the square root of 0.0009?

Sol.      √0.0009= √ 9 / 1000  = 3 / 100 = 0.03.

Ex. 7. Evaluate √175.2976.

Sol.      Method: We make even number of decimal places              1   175.2976 (13.24  

by affixing a zero, if necessary. Now, we mark off                   1

periods and extract the square root as shown.                     23     75            

                                                                                                      69

\√175.2976 = 13.24                                                         262       629

                                                                                                                    524

                                                                                                     2644       10576

                                                                                                                    10576

                                                                                                                        x

Ex. 8. What will come in place of question mark in each of the following questions?

(i) √32.4 / ?  = 2                       (ii) √86.49 + √ 5 + ( ? )² = 12.3.

            Sol.      (i) Let √32.4 / x = 2. Then, 32.4/x = 4 <=> 4x = 32.4 <=> x = 8.1.

                     

(ii) Let √86.49 + √5 + x² = 12.3.

      Then, 9.3 + √5+x² = 12.3 <=> √5+x²  = 12.3 - 9.3 = 3

      <=> 5 + x² = 9   <=> x² = 9 - 5= 4   <=>   x = √4 = 2.

Ex.9. Find the value of √ 0.289 / 0.00121.

 

Sol.      √0.289 / 0.00121 = √0.28900/0.00121 = √28900/121 = 170 / 11.

 

Ex.10. If √1 + (x / 144) = 13 / 12, the find the value of x.

Sol.      √1 + (x / 144) = 13 / 12 Þ ( 1 + (x / 144)) = (13 / 12 )² = 169 / 144

  Þx / 144 = (169 / 144) - 1

  Þx / 144 = 25/144 Þ x = 25.

Ex. 11. Find the value of √3 up to three places of decimal.

Sol.                

  1    3.000000   (1.732

                                1

27    200

189

                      343      1100

                                  1029

                    3462          7100

6924                          \√3 = 1.732.

Ex. 12. If √3 = 1.732, find the value of √192 - 1 √48 - √75 correct to 3 places

                                                                            2                        

of decimal.                                                                                     (S.S.C. 2004)

Sol.     √192 - (1 / 2)√48 - √75 = √64 * 3 - (1/2) √ 16 * 3  - √ 25 * 3

                                                =8√3 - (1/2) * 4√3 - 5√3

                                                =3√3 - 2√3 = √3 = 1.732

 

Ex. 13. Evaluate: √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)

Sol.      Given exp. = √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)

            Now, since the sum of decimal places in the numerator and denominator under the            radical sign is the same, we remove the decimal.

\        Given exp = √(95 * 85 * 18900) / (17 * 19 * 21) = √ 5 * 5 * 900  = 5 * 30 = 150.

Ex. 14. Simplify: √ [( 12.1 )² - (8.1)²] / [(0.25)² + (0.25)(19.95)]

Sol.      Given exp. = √ [(12.1 + 8.1)(12.1 - 8.1)] / [(0.25)(0.25 + 19.95)]

                       

                              =√ (20.2 * 4) /( 0.25 * 20.2)   = √ 4 / 0.25 = √400 / 25 = √16 = 4.

Ex. 15. If x = 1 + √2 and y = 1 - √2, find the value of (x² + y²).

Sol.      x² + y² = (1 + √2)² + (1 - √2)² = 2[(1)² + (√2)²] = 2 * 3 = 6.

Ex. 16. Evaluate: √0.9 up to 3 places of decimal.

Sol.                 

9    0.900000(0.948

         81

184         900

736

                  1888         16400

                                   15104                             \√0.9 = 0.948

                         

Ex.17. If √15 = 3.88, find the value of √ (5/3).

Sol.      √ (5/3) = √(5 * 3) / (3 * 3)  = √15 / 3 = 3.88 / 3 = 1.2933…. = 1.293.

    

Ex. 18. Find the least square number which is exactly divisible by 10,12,15 and 18.

Sol.      L.C.M. of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 2² * 3² * 5.

            To make it a perfect square, it must be multiplied by 5.

\         Required number = (2² * 3² * 5²) = 900.

Ex. 19. Find the greatest number of five digits which is a perfect square.

(R.R.B. 1998)

Sol.      Greatest number of 5 digits is 99999.

                                                    3    99999(316

      9 

                                                  61      99

                                                            61

626      3899

           3756

                                                             143

\                  Required number == (99999 - 143) = 99856.

Ex. 20. Find the smallest number that must be added to 1780 to make it a perfect

square.

Sol.

                                    4    1780 (42

                                          16

82      180         

164

            16

 

      \               Number to be added = (43)² - 1780 = 1849 - 1780 = 69.

Ex. 21. √2 = 1.4142, find the value of √2 / (2 + √2).

Sol.      √2 / (2 + √2) = √2 / (2 + √2) * (2 - √2) / (2 - √2) = (2√2 – 2) / (4 – 2)

                                 = 2(√2 – 1) / 2 = √2 – 1 = 0.4142.

22. If x = (√5 + √3) / (√5 - √3) and y = (√5 - √3) / (√5 + √3), find the value of (x²  + y²).

Sol.     

x = [(√5 + √3) / (√5 - √3)] * [(√5 + √3) / (√5 + √3)] = (√5 + √3)² / (5 - 3)

   =(5 + 3 + 2√15) / 2 = 4 + √15.        

            y = [(√5 - √3) / (√5 + √3)] * [(√5 - √3) / (√5 - √3)] = (√5 - √3)² / (5 - 3)

               =(5 + 3 - 2√15) / 2 = 4 - √15.

\         x² + y² = (4 + √15)² + (4 - √15)² = 2[(4)² + (√15)²] = 2 * 31 = 62.

Ex. 23. Find the cube root of 2744.

Sol.    Method: Resolve the given number as the product                 2    2744

          of prime factors and take the product of prime                        2    1372

           factors, choosing one out of three of the same                        2      686         

           prime factors. Resolving 2744 as the product of                     7      343

           prime factors, we get:                                                              7  ­      49

                                                                                                                       7

           2744 = 2³ x 7³.                                                                            

\      ³√2744= 2 x 7 = 14.

                                                                                                    

Ex. 24. By what least number 4320 be multiplied to obtain a number which is a perfect cube?

Sol.      Clearly, 4320 = 2³ * 3³ * 2² * 5.

            To make it a perfect cube, it must be multiplied by 2 * 5² i.e,50.