Sunday 24 November 2013

Problems on Permutauions and Combinations



Q.1. In how many different ways can the letters of the word ‘JUDGE’ be arranged in such a way that the vowels always come together?
         a) 160    b) 120    c) 48     d) 124      e) 96
Ans: c
Explanation:
In the word ‘JUDGE’ the vowels are ‘UE’
When vowels are taken together it will be treated as one letter
Then the number ways of arrangement of ‘JDG(UE) = 4!
The two vowels can be arranged in 2 ways i.e. 2!
Therefore, the required number of ways = 4! × 2! = 24 × 2 = 48 ways
Q.2. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
         a) 1024    b) 540     c) 5040    d) 2520     e) 400
Ans: c
Explanation:
The word ‘LOGARITHMS’ contain  10 letters
Number of 4- letter words, which can be formed from this 10 letters = 10P4 = 10 × 9 × 8 × 7
                                                                                                              = 5040 words
Q.3. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
           a) 5470    b) 5040     c) 86400    d) 11760     e) 2660
Ans: d
Explanation:
Required number of ways = 8C5 × 10C6
                                         = 8 × 7 × 10×  7 × 3 = 11760
Q. 4. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?
             a) 5   b) 10     c) 15    d) 20     e) 25
Ans: d
Explanation:
Because the numbers should be divisible by 5, the unit’s digit of each numbers must be 5.
Such number of ways =1
The digit at the 10th place should be any of the remaining 5 digits = 5ways
Then the 100th digit should be any one from the remaining 4 digits = 4 ways
Therefore, the total number of numbers formed as such = 1 × 5 × 4 = 20
Q. 5. The value of 75P2 is
            a) 2550   b) 2775    c) 1555    d) 5550     e) None of these
Ans: d
Explanation:
nPr = n!/(n – r)!
75P2 = 75!/73! = 75 × 74 × 73!/73! = 75 × 74 = 5550
Q. 6. How many word can be formed by using all the letters of the word, 'ALLAHABAD' ?
             a) 2520   b) 3270    c) 3780    d) 1890     e) 7560
Ans: e
Explanation:
In the word 'ALLAHABAD’ there are 4A, 2L, 1H, 1B and 1D = 9
Required number of word = 9!/ 4! × 2! × 1! × 1! × 1! = 9  × 8 × 7 × 6 × 5 × 4!/4! × 2
                                           = 9  × 8 × 7 × 3 × 5 = 7560
Q. 7. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
              a) 71   b) 69   c) 64    d) 56     e) 48
Ans: c
Explanation:
Such a combination will be 1 Black + 2 Non-black or 2 Black + 1 Non-black  or 3 black.
          = (3C1 × 6C2) + (3C2 × 6C1) + 3C3
          = (3 × 6 × 5/2 × 1) + (3 × 2/2 × 1 × 6) + 1
          = 45 + 18 + 1 = 64
Q. 8. There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?
             a) 3600   b) 2400   c) 3200    d) 2800     e) None of these
Ans: a
Explanation:
5 periods can be organized in 6P5 ways
Remaining period can be arranged in 5P1 ways
Therefore, the total number of ways = 6P5  × 5P1
                                                          = (6 × 5× 4 × 3 × 2) × 5 = 3600
Q. 9. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?
          a) 1480   b) 720   c) 360    d) 1420     e) 1680
  
Ans: e
Explanation:
The first two places can be filled in only 1 way using 3 and 5.
Remaining 4 places can be filled with any of the 8 remaining digits in 8P4 ways = 8 × 7 × 6 × 5
                                                                              = 1680
Q. 10. An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?
             a) 64   b) 110   c) 100    d) 80     e) 70 
Ans: d
Explanation:
A chair can be arranged in 10 ways
A table can be arranged in 8 ways
Therefore, one chair and one table can be arranged in 10 × 8 = 80 ways.

Problems on LCM and HCF




Q.1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. What is sum of the numbers?
a)       28       b) 40     c) 64      d) 42      e) 32
Ans: b
Explanation:
Let the two numbers be 2x and 3x respectively.
LCM of 2x and 3x = 6x
Given, 6x = 48
So, x = 8
The required sum = 2x + 3x = 16 + 24 = 40. Ans.
     Q.2. What is the greatest number of four digits which is divisible by 15, 25, 40 and 75 ?
             a) 9800     b) 9600     c) 9400       d) 9200       e) 9000
              Ans: b
             Explanation:
             The greatest 4- digit number = 9999
             LCM of 15, 25, 40 and 75 = 600
            The remainder obtained when 9999 is divided by 600 = 399
            Therefore, the required greatest number = 9999 – 399 = 9600 Ans.
     Q.3. Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:
           a) 40       b) 30        c) 20    d) 10      e) None of these
           Ans: c
            Explanation :
Let the numbers be 2x, 3x and 4x

LCM of 2x, 3x and 4x = 12x

=> 12x = 240

=> x = 24012 = 20

H.C.F of 2x, 3x and 4x = x = 20
     Q.4.  Find the minimum number of square tiles required to pave the floor of a room of 2m 50cm   long and 1m 50cm broad ?
a)      50      b) 750      c) 45       d) 15     e) None of these
          Ans: d
          Explanation:
  HCF of 250 cm and 150 cm is 50 cm, which is the side of the tile

So, the required number of tiles  =  (250 × 150) / (50 × 50)  =  15
  
Q.5. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?
     a) 1108     b) 1683      c) 2007         d) 3363    e) None of these
    Ans: b
   Explanation:
   Just see which of the given choices satisfy the given conditions

Take 3363. This is not even divisible by 9. Hence this is not the answer

Take 1108. This is not even divisible by 9. Hence this is not the answer

Take 2007. This is divisible by 9.

2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer

Take 1683. This is divisible by 9.

1683 ÷ 5 = 336, remainder = 3

1683 ÷ 6 = 280, remainder = 3

1683 ÷ 7 = 240, remainder = 3

1683 ÷ 8 = 210, remainder = 3

Hence 1683 is the answer

    Q.6. Find the HCF of   22×32, 2×34×7

  a) 128       b) 126       c) 146       d) 434    e) 148
  Ans: b
  Explanation:
  HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
                      2 × 3×3 × 7 = 126
  Q.7.  Find the HCF of 54, 288, 360
   a) 18    b) 36      c) 54     d) 108    e) 72
   Ans: a
   Explanation:
   Let’s solve this question by factorization method.
54 =2×33,  288 = 25×32,  360 = 23×32×5

  So HCF will be minimum term present in all three, i.e.
2×32=18

  Q.8.  Reduce  368/575  to the lowest terms.

      a) 30/25      b)  28/29          c) 16/25     d) 11/12    e) 13/15
   Ans: c
   Explanation:
     We can do it easily by in two steps
     Step1: We get the HCF of 368 and 575 which is 23
     Step2: Divide both by 23, we will get the answer 16/25
Q. 9. Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. The numbers are
           a) 12, 24, 36   b) 5, 10, 15    c) 4, 8, 12      d) 6, 8, 13    e) 10, 20, 30
Ans: a
Explanation:
Numbers are 1 × 12, 2 × 12 and 3× 12
                     i.e. 12, 24 and 36
Q.10. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
           a) 4   b) 5    c) 7     d) 9       e) 13
Ans:
Explanation:
Required number = HCF of ( 91 – 43), (183 – 91)   and (183 – 43)
                             = HCF of 48, 92, 140
                             = 4



Friday 27 September 2013

Average



Average
Q. 1. The average of five positive numbers is 213. The average of the first two numbers is 233.5 and the average of last two numbers is 271. What is the third number?
(a) 64   (b) 56  (c) 106  (d) Cannot be determined  (e) None of these
Ans: (b)56
Explanation:
The sum of the five numbers = 5 × 213 =1065
The sum of the first two numbers = 2 × 233.5 = 467
The sum of  the last two numbers = 542
Then the sum of the four numbers = 467 + 542 =1009
So, the third number will be = 1065 – 1009 = 56. Ans.
Q. 2. The average age of a woman and her daughter is 16 years. The ratio of their ages is 7:1 respectively. What is the woman’s age?
(a) 4 years  (b) 28 years  (c) 32 years  (d) 6 years   (e) None of these
Ans: (b) 28 years
Explanation:
Sum of their ages = 2 × 16 = 32
Let 7x and x be their respective ages,  then, 8x = 32  and x = 4
So, the age of the woman = 7x = 7 × 4 = 28 years.Ans.
Q. 3. Find the average of the following set of scores:
        568, 460, 349, 987, 105, 178, 426
(a)442   (b)441  (c)440  (d)439  (e)None of these
Ans: (d) 439
Explanation:
Average = sum of the scores/No. of scores
              = 3073/7 = 439.Ans.
Q. 4. The average age of 54 girls in a class was calculated as 14 years. It was later realized that the actual age of one of the girls in the class was calculated as 13 years. What is the actual average age of the girls in the class? (Rounded off to two digits after decimal).
(a) 10.50 years  (b) 12.50 years  (c) 12 years  (d) 13.95 years  (e) None of these
Ans: (d) 13.95
Explanation:
The sum of the ages of the 54 girls entered as error = 54 × 14 = 756.
 Now, deduct the error i.e. 13 – 10.5 = 2.5
Then the actual sum of the ages = 756 – 2.5 = 753.5
So, the actual average = 753.5/54 = 13.95 years. Ans.
Q. 5.  The average age of a man and his son is 40.5 years. The ratio of their ages is 2 : 1 respectively. What is the man’s age?
(a) 56 years  (b) 52 years  (c) 54 years   (d) Cannot be determined (e) None of these
Ans: (c) 54 years.
Explanation:
Let the age of the son = x
Then, 2x + 1x        i.e. 3x = 40.5 × 2 = 81
    Therefore, x = 27   and so the man’s age = 2x = 2 × 27 = 54 years.  Ans.
Q. 6. A car covers a distance from town A to town B at the speed of 58 kmph and covers the distance from town B to town A at the speed of 52 kmph. What is the approximate average speed of the car?
(a ) 55 kmph  (b) 52 kmph  (c) 48 kmph   (d) 50 kmph   (e) 60 kmph
Ans: (a) 55 kmph
Explanation:
Average speed = sum of the two speeds/2  = 58 + 52 =110/2 = 55 kmph.Ans.
Note:
Otherwise, If  two equal distances are covered at two different speeds of x kmph and y kmph, then, Average speed = 2xy/x + y kmph.
Q. 7. If 36a + 36b = 576, then what is the average of a and b?
(a) 16  (b) 8  (c) 12  (d) 6 (e) None of these
Ans: (b) 8
Explanation:
36 (a+b) = 576,    given,  then, a + b = 576/36 = 16
So, the average of  a and b = a + b/2   = 16/2 = 8. Ans.
Q. 8. The average marks of 65 students in a class was calculated as 150. It was later realized that the marks of one of the students was calculated as 142, whereas his actual average marks were 152.  What is the actual average marks of the group of 65 students? (Rounded off to two digits after decimal)
(a) 151.25   (b) 150.15   (c) 151.10   (d) 150.19    (e) None of these
Ans: (b) 150.15
Explanation:
Increase in total marks = 152 – 142 = 10
Therefore the New average = 150 + 10/65 = 150.15.Ans.
OR
Sum of the   total average marks of 65 students = 65 × 150 = 9750
Here, add the difference of 10 marks = 9760.
Therefore the New average = 9760/65 = 150.15. Ans.
Q. 9. In a class there are 32 boys and 28 girls. The average age of the boys in the class is 14 years and the average age of the girls in the class is 13 years. What is the average age of the whole class?  (Rounded off to two digits after decimal)
(a) 13.50   (b) 13.53  (c) 12.51   (d) 13.42  (e) None of these
Ans: (b) 13.53
Explanation:
The sum of the ages of  32 boys = 32 × 14 = 448
The sum of the ages of 28 girls = 28 × 13 = 364
Therefore, the sum of the ages of the whole class of  60 students =812
The average age of the whole class of 60 students = 812/60 = 13.53. Ans.
Q. 10. The average of 5 consecutive even numbers A, B, C, D and E respectively is 74. What is the product of C and E?
(a) 5928  (b)5616  (c) 5538   (d) 5772   (e) None of these
Ans: (d) 5772
Explanation:
Let the sum of the 5 consecutive even numbers = x + x +2 + x + 4 + x + 6 + x + 8 = 5 × 74 = 370
5x + 20 = 370;   5x = 370 – 50 = 350
Therefore, x = 350/5 =70,   then the  nos. A,B, C, D and E are 70, 72, 74, 76 and 78 respectively.
Product of  C and E = 74 × 78 = 5772. Ans.
Q. 11. Average of four consecutive odd numbers is 106. What is the third number in ascending order?
(a) 107   (b) 111  (c) 113     (d) Cannot be determined  (e) None of these
Ans: (a) 107
Explanation:
Let the sum of the four consecutive odd nos. = x + x +2 + x + 4 + x + 6
Then, 4x + 12 = 4 × 106 = 424
    Therefore, x = 103,  The nos. in ascending order is 103, 105, 107 and 109 and the third number is = 107. Ans.
Q. 12. Of  the three numbers, the average of the first and the second is greater than the average of the second and the third by 15. What is the difference between the first and the third of the three numbers?
(a) 15    (b) 45  (c) 60   (d) Data inadequate  (e) None of these
Ans: (e) None of these
Explanation:
Let x,  y, z be the three numbers,
Then x + y/2 – y + z/2 = 15
x+y – (y+z)/2 = x + y – y – z/2 = 15
So, x – z = 2 × 15 = 30.Ans.
Q. 13. The ratio of roses and lilies in a garden is 3:2 respectively. The average number of roses and lillies is 180. What is the number of lilies in the garden?
(a) 144   (b) 182   (c)216  (d)360  (e) None of these
Ans: (a) 144
Explanation:
Let the nos. of roses and lilies be 3x and 2x respesctively.
Then, 3x + 2x/2 = 180; 5x = 360  and so, x = 72
Therefore, the number of lilies = 2x = 2 × 72 = 144. Ans.
Q. 14. The average monthly income of a family of four earning members was Rs.15130. One of  the daughters in the family got married and left home, so the average monthly income of the family came down to Rs.14660. What is the monthly income of the married daughter?
(a) Rs.15350   (b)Rs.12000   (c)Rs.16540   (d) Cannot be determined   (e) None of these
Ans: (c) Rs.16540
Explanation:
The sum of the incomes o f the four members = 4 × 15130 = Rs.60520
The sum of the incomes of the family except the married daughter = 3 × 14660 = 43980
Therefore, the income of the married daughter = Rs.60520 – Rs.43980 = Rs.16540. Ans.
Q. 15.  The average temperature of Monday, Tuesday, Wednesday and Thursday was 36.5º C and for Tuesday, Wednesday, Thursday and Friday was 34.5º C. If the temperature on Monday was 38º C, find the temperature on Friday.
(a) 34ºC   (b)  36º C  (c) 37.4º C   (d) 32º C  (e) 30º C
Ans: (e) 30º
Explanation:
The sum of temperatures on Monday, Tuesday, Wednesday and Thursday = 36.5 × 4 = 146º C
The sum of the temperatures on Tuesday, Wednesday and Thursday  = 146 – Temperature on Monday    i.e. = 146- 38 = 108º C.
Sum of the temperatures on Tuesday, Wednesday , Thursday and Friday = 4 × 34.5 = 138º C
Therefore, the temperature on Friday = 138 – 108 = 30º C. Ans.
Q.16. The average age of a man and his two sons born on the same day is 30 years. The ratio of the ages of father and one of his sons is 5 : 2 respectively. What is the father’s age?
(a) 50 years   (b) 30 years  (c) 45 years   (d) 20 years  (e)None of these
Ans: (a) 50 years
Explanation:
Sum of their ages = 3 × 30 =90
i.e. 5x + 2x+ 2x = 90        => 9x  = 90  and x = 10
Then, father’s age = 5x = 5 × 10 = 50 years. Ans.
Q. 17. A cricketer has an average score of 49 runs in 24 innings. How many runs must he score in the 25th innings to make his average 50?
(a) 94   (b) 84  (c) 74  (d) 76    (e)None of these
Ans: (c) 74
Explanation:
His total score in 24 innings = 24 × 49 = 1176 runs.
To get an average of 50 in 25th innings his score should be = 25 × 50 =1250
Therefore, the score required to obtain in his 25th innings = 1250 – 1176 = 74. Ans.
Q. 18. The average age of 14 boys in a class is 13 years. If the class teacher’s age is included, the average age is increased by one year. What is the class teacher’s age?
(a) 31 years   (b) 27.5 years    (c) 24 years   (d) 28 years   (e) None of these
Ans:  (d) 28 years
Explanation:
The teacher’s age = New average + Old No. of  persons × difference in average
i.e.   14 + 14 × 1  =  14 + 14 = 28 years.  Ans.
Q. 19. The average of four positive integers is 73.5. The highest integer is 108 and the lowest integer  is 29. The difference between the remaining two integers is 15. Which of the following is the smaller of the remaining two integers?
(a) 80   (b) 86  (c) 73  (d) Cannot be determined  (e)None of these
Ans: (e) None of these
Explanation:
The sum of the four Integers   =  73.5 × 4 = 294
Sum of the highest and the lowest integer = 108 + 29 = 137
And the sum of the remaining two integers = 294 – 137 = 157
Subtract the difference between them i.e. 157 – 15 = 142
Here, 142 is the double of the smallest integer, therefore, the smallest integer = 142/2 = 71. Ans.
Q. 20. The average age of a woman and her daughter is 46 years. The ratio of their ages is 15 : 8 respectively. What will be the respective ratio of their ages after 8 years?
(a) 8 : 5  (b) 10 : 17   (c) 17 : 10    (d) 5 : 8   (e) None of these
Ans: (c) 17 : 10
Explanation:
The sum of their ages, 15x + 8x = 2 × 46 = 92
23x = 92 :.    x= 92/23 = 4
Therefore  their respective ages are 60 and 32 .
After 8 years their ages will be 68 and 40 respectively
So, the required ratio = 68 : 40  = 17 :  10 Ans.
Q. 21. In a class of 75 students, the average age is 23 years. The average age of male students is 25 years and that of female students is 20 years. Then the ratio of male to female students is
(a) 3:2  (b)7:3  (c) 8:7  (d)6:9  (e) 9:6
Ans: (a) 3 : 2
Explanation:
Total age of 75 students = 75 × 23 = 1725
Let x be the no. of boys and y be the no. of girls.
Then, 25x + 20y = 1725   ----- (i)
 And  x + y = 75   -------(ii) × 20 gives, 20x + 20y   = 1500 ---- (iii)
Subtracting (iii) from (ii) we get, x = 45 and y = 30
Required ratio = 45 : 30
                       = 3 : 2.Ans.
Q. 22. The average age of 3 friends is 32 years. If the age of the fourth friend is added, their average age comes to 31 years. What is the age of the fourth friend?
(a) 32 years  (b) 28 years (c) 24 years   (d) 26 years   (e) None of these
Ans: (b) 28 years
Explanation:
The age of the fourth friend = new average + old no. of persons × difference in average
                                            = 31  + 3 × -1 = 31 -3 = 28 years. Ans.
Q. 23. Average salary of 19 workers in an industry Rs.2500. The salary of supervisor is Rs.5550. Find the average salary of all the 20 employees.
(a)Rs.3355.5  (b) Rs.4500.00  (c) Rs.4642.5    (d) Rs.2652.5   (e) None of these
Ans: (d) Rs.2652.5
Explanation:
The total salary of 19 workers = 19 × 2500 = Rs.47500
Total salary of 20 workers including the supervisor = 47500 + 5550 = Rs.53050
Therefore the required average = 53050/20 = Rs.2652.5. Ans.
Q. 24. The average of three even consecutive numbers is 24. What is the summation of the three numbers?
(a) 24  (b) 72   (c) 26   (d) 80    (e) None of these
Ans: (b) 72
Explanation:
Let the three consecutive even numbers be x, x+2 and x + 4
Then, x + x +2 + x+4 /3  =24
i.e.   3x + 6 = 3 × 24
3x = 72-6      => x  = 66/3 = 22
Then , their summation = 22 + 24 + 26 = 72.Ans.
Q. 25. The sum of seven consecutive even numbers of a set is 532. What is the average of first four consecutive even numbers of the same set?
(a) 76   (b) 75  (c) 74  (d) 73   (e) None of these
Ans:  (d) 73
Explanation:
See that, their summation as, x + x +2 + x +4 + x +6 +x + 8 + x + 10 + x + 12 = 532
7x + 42 = 532       and so, x = 70   (first number)
Therefore, the seven consecutive even numbers are 70, 72, 74, 76, 78, 80, 82
So, the required average = 70 + 72 + 74 + 76/4 = 292/4 = 73.Ans.
Q. 26. There are 50 boys in a class. One boy weighing 40 kg goes away and at the same time another boy joins the class. If the average weight of the class is thus decreased by 100 g, find  the weight of the new boy.
(a) 35kg  (b) 43kg  (c) 36kg  (d) 30 kg  (e) None of these
Ans: (a) 35 kg
Explanation:
The weight of the new boy = weight of the boy who left + Number of boys × difference in average.
i.e.  weight of the new boy = 40kg + 50 × - .1kg = 40kg – 5kg = 35 kg. Ans.
Q. 27. Kamlesh bought 65 books for Rs.1050 from one shop and 50 books for Rs.1020 from another. What is the average price he paid per book?
(a) Rs. 36.40   (b) Rs.18.20   (c) Rs. 24   (d)Rs.18  (e) None of these
Ans: (d) Rs.18
Explanation:
The total price = Rs.1050 + Rs.1020 = Rs.2070
Total number of books = 65 + 50 = 115
So, the average price per book = 2070/115 = Rs.18. Ans.
Q. 28. A car covers the first 39 kms. of it’s journey in  45 minutes and covers the remaining 25 kms. in 35 minutes. What is the average speed of the car?
(a)  40 kms/hr   (b) 64 kms./hr  (c) 49 kms./hr  (d) 48 kms./hr   (e) none of these
Ans: (d) 48 kms/hr
Explanation:
Total speed = 39 + 25 kms. = 64 kms.
Total time taken = 45 + 35 = 80 mins.   = 80/60 hrs.
So, the average speed of the car = 64 ÷ 80/60  = 64 × 60/80 = 48 km/hr. Ans.
Q. 29. The sum of five numbers 260. The average of the first two numbers is 30 and the average of the last two numbers is 70. What is the third number?
(a) 33  (b)60  (c)75  (d) Cannot be determined  (e) None of these
Ans: (b) 60
Explanation:
The sum of  the first two numbers = 30 × 2 = 60
The sum of the last two numbers = 70 × 2 = 140
Therefore, the third number = 260 – (60 + 140) = 260 – 200 = 60. Ans.
Q. 30. The average weight of 12 boys is 35 kgs. If the weight of an adult is added, the average becomes 37 kgm. What is the weight of the adult?
(a) 65kgs   (b) 68 kgs  (c) 62 kgs  (d) 63 kgs  (e) None of these
Ans: (e) None of these
Explanation:
The weight of the adult = New average + No. of boys × difference in average
                                     = 37 + 12 × 2  = 37 + 24 = 61 kgs.Ans.
Q. 31. The average marks in Science subject of a class of 20 students is 68. If the marks of two students were misread as 48 and 65 of the actual marks 72 and 61  respectively, then what would be the correct average?
(a)  68.5   (b)69   (c)69.5   (d)70  (e)66
Ans: (b) 69
Explanation:
The total number of marks = 20 × 68 = 1360
Add difference of marks misread against the actual marks i.e. 20 = 1360 + 20 = 1380
Now the correct average = 1380/20 = 69 Ans.
Q. 32.  The average speed of a car is 75 kms/hr. What will be the average speed of the car if the driver decreases the average speed of the car by 40 percent?
(a) 50 kms/hr  (b) 45 kms/hr  (c) 40 kms/hr   (d) 55 kms/hr   (e)None of these
Ans: (b) 45 kms/hr
Explanation:
40% of 75 kms/hr = 30 kms/hr
Required average = 75 – 30 kms/hr = 45 kms/hr. Ans.
Q. 33. The average marks of a student in seven subjects is 41. After re-evaluation in one subject the marks were changed to 42 from 14 and in remaining subjects the marks remained unchanged. What are the new average marks?
(a) 45  (b) 44   (c)46   (d) 47   (e) None of these
Ans: (a) 45
Explanation:
Present total marks = 41 × 7 = 287
Add the change in marks = 28
Now,  the total marks = 287 + 28 = 315; 
So, the new average marks = 315/7 = 45. Ans.
Q. 34. The body weight of six boys is recorded as 42 kgs.; 72 kgs; 85 kgs; 64 kgs; 54 kgs and 73 kgs.  What is the average body weight of all the six boys?
(a) 64 kgs.   (b) 67 kgs.   (c) 62 kgs.  (d) 65 kgs.   (e) None of these
Ans: (d) 65 kgs.
Explanation:
Required average = 42 + 72 + 85 + 64 + 54 + 73 /6  = 390/6 = 65. Ans.
Q. 35. The rainfall in a city in 5 consecutive years was recorded as 20.54 inches, 33.10 inches, 11.62 inches, 19.20 inches and 21.74 inches. What was the average annual rainfall?
(a) 21.44 inches  (b) 21.24 inches  (c) 20.24 inches  (d) 17.94 inches  (e) None of these
Ans: (b) 21.24 inches
Explanation:
The average annual rainfall = 20.54 + 33.10  + 11.62  + 19.20  + 21.74/5
 = 106.20/5 = 21.24 inches. Ans.
Q. 36. The average score of a cricketer in two matches is 27 and in three other matches is 32. Then find the average score in all the five matches?
(a)  25    (b) 20  (c) 30  (d) 35   (e) None of these
    Ans: (c) 30
Explanation:
Sum of the scores in two matches = 2 × 27 = 54
And sum of the scores in three matches = 3 × 32 = 96
Total score in five matches = 54 + 96 =150
So, the average scores in five matches = 150/5 = 30 Ans.
Q. 37. The average marks of nine students in a group is 63. Three of them scored 78, 69 and 48 marks. What are the average marks of remaining six students?
(a) 63.5   (b) 64  (c) 63   (d) 62.5   (e) None of these
Ans: (e) None of these
Explanation:
The total marks of nine students = 9 × 63 = 567
Sum of the marks of three students = 78 + 69 + 48 = 195
Therefore, the sum of marks of the remaining six students= 567 – 195 = 372
Average marks of remaining six students = 372/6 = 62. Ans.
Q. 38. Out of the three given numbers, the first number is twice the second and thrice the third. If the average of the three numbers is 154. What is the difference between the first and the third number?
(a)126  (b)42  (c) 166 (d)52  (e) None of these
Ans: (e) None of these
Explanation:
Let x be the  first number, then, the second number is x/2 and the third number is x/3
Then , the sum of numbers i.e.  x + x/2 + x/3 = 3 × 154 = 462
ð  11x/6 = 462  and so,   x = 252
ð  Then third number = 252/3 = 84
ð  So, the difference between first and the third number = 252 – 84 = 168 Ans.
Q. 39. The average speed of a bus is three-fifth the average speed of a car which covers 3250 kms. in 65 hours. What is the average speed of the bus?
(a) 30 Kms/hr.  (b) 20 Kms/hr  (c) 35 Kms/hr  (d) 36 Kms/hr  (e) None of these.
Ans: (a) 30 Kms./hr
Explanation:
The average speed of the car = 3250/65 = 50km/hr
Therefore, the average speed of the bus = 3/5th of the average speed of the car = 3 × 50/5
       = 30km/hr.Ans.
Q. 40. The average speed of a car is twice the average speed of a truck. The truck covers 648 kms. in 24 hours. How much distance will the car cover in 15 hours?
(a) 820 kms.  (b) 1014 kms.  (c) 810 kms.  (d) 980 kms  (e) None of these
Ans: (c) 810 kms.
Explanation:
The average speed of the truck = 648/24 = 27 kms /hr
Therefore, the average speed of the car = 2 × 27 = 54 kms/hr
The  distance covered by the car in 15 hours = 54 × 15 = 810 kms.Ans.
Q.41. An aeroplane flies with an average speed of 756 km/hr. A helicopter takes 48 hours to cover twice the distance covered by aeroplane in 9  hours. How much distance will the helicopter cover in 18 hours.?  ( assuming that flights are non-stop and moving with uniform speed)
(a) 5014 km    (b)5140 km   (c) 5130km  (d) 5103 km  (e) none of these
Ans: (d) 5103 km.
Explanation:
The distance covered by the aeroplane in 9 hours = 9 ×756 = 6804 kms.
The distance covered by the helicopter in 48 hours = 2 × 6804 kms.
The distance covered by the helicopter in 1 hour = 2 × 6804/48 kms.
Therefore, the distance covered by the helicopter in 18 hours = 2 × 6804 × 18/48
                                                                                                  = 5103 kms. Ans.
Q. 42. The average age of A,B and C is 26 years. If the average age of A and C is 29 years. What is the age of B?
(a) 26   (b) 20  (c) 29  (d)23  (e) None of these
Ans: (b) 20
Explanation:
The sum of the ages of A, B and C = 3 × 26 = 78
The sum of the ages of  A and C = 2 × 29 = 58
So, age of B =  78 – 58 = 20. Ans.
Q. 43. The average of four positive integers is 124.5. The highest integer is 251 and the lowest integer is 65. The difference between the remaining two integers is 26. Which of the following integers is higher of the remaining two integers?
(a) 78   (b) 102  (c) 100  (d) Cannot be determined  (e) None of these
Ans: (e) None of these
Explanation:
The sum of the four positive integers = 4 × 124.5 = 498
The sum of the highest and lowest integer = 251 + 65 = 316
The sum of the remaining two integers = 498 – 316 = 182
The difference between  these two integers = 26
Subtract 26 from 182 and divide the result by 2, then we get the lowest one of the remaining two integers.
i.e. 182 – 26 = 156;   156/2 = 78   and the higher one of the remaining = 78 + 26 = 104. Ans.
Q. 44. The average height of 21 girls was recorded as 148 cms. If the teacher’s height was added, the average increased by one. What was the teacher’s height?
(a) 156 cms.  (b) 168 cms.  (c) 170 cms.  (d) 162 cms.  (e) none of these
Ans: (c) 170 cms.
Explanation:
The height of the teacher = New average + No. of girls × increase in average
                                         = 149 + 21 × 1 = 149 + 21 =170 cms.Ans.

 Q. 45. The average speed of a tractor is two-fifths the average speed of a car. The car covers 450    km in 6 hours. How much distance will the tractor cover in 8 hours?

a) 210 km b) 240 km   c) 420 km  d) 480 km  e) None of these

Ans: (b) 240 km

Explanation:
The average speed of the car = 450/6 = 75 km/hr
Therefore, the average speed of the tractor = 75 × 2/5 = 30 km/hr
So, the distance covered by the tractor in 8 hours = 30 × 8 = 240 km. Ans.

Q.46. Find the average of the following set of numbers.
          354, 281, 623, 518, 447, 702, 876
a)      538 b) 555 c) 568 d) 513 e) None of these
    Ans: (e) None of these 
  Explanation:
 Required average = 354 + 281 + 623 + 518 + 447 + 702 + 876 = 3801/7 = 543.Ans.
Q. 47.  If  average of 11 consecutive odd numbers is 17, what is the difference between the smallest and the largest number?
(a) 18  (b)  20  (c) 22  (d) 24   (e) None of these
Ans:  (b) 20
 Explanation:
  Let the numbers be x, x + 2, x+ 4, x +6, x+8, x + 10, x+12, x+14, x + 16, x + 18 and x + 20;
 Here, the smallest number = x  and   the largest number = x + 20
Therefore, their difference = x + 20 – x = 20.Ans.
Q. 48. The average of five consecutive odd numbers is 95. What is the fourth number in the descending order?
 (a) 91  (b) 95  (c) 99 (d) 97   (e)None of these
Ans: (e) None of these
Explanation:
Let the sum of the numbers be x + x +2 +  x +4 + x + 6 + x + 8 = 5 × 95 = 475
 5x + 20 = 475        :. x= 91
The fourth number in descending order = x + 2  = 91 + 2 =93. Ans.
Q. 49.  The average of marks obtained in 120 students was 35. If the average of passed candidates was 39 and that of failed candidates is 15, the number of candidates who  passed the examination is:
(a) 100   (b) 110   (c) 120   (d) 80
Ans: (a) 100
Explanation:
The sum of the marks obtained by 120 students = 120 × 35 = 4200
Let ‘x’ be the number of passed candidates
Then, sum of the marks obtained by the passed candidates = x × 39 = 39x
So, sum of the marks obtained by the failed candidates = (120 – x) 15
i.e.  39x + (120 – x ) 15 = 4200
       39x + 1800 – 15x = 4200
      39x -15x = 4200 – 1800 = 2400
      24x = 2400       and  so,   x = 100. Ans.
Q. 50.  The average marks in Hindi subject  of a class of 54 students is 76. If the marks of two students are misread as 60 and 77 of the actual marks 36 and 47 respectively, then what would be the correct average?
 (a) 75.5   (b) 77  (c) 75  (d) 76.5  (e) None of these
 Ans: (c) 75
Explanation:
The sum of the marks of 54 students = 54 × 76 = 4104
Difference of marks misread = 54
Subtracting this difference = 4104 – 54 = 4050
Now, the correct average = 4050/54 = 75. Ans.
Solution to Jessie's problem:
Q. The mean temperature on any day of the week varies from between 35 degree centigrade and 42 degree centigrade. Which of the following can represent maximum average temperature for the week?
(a) 35   (b) 36   (c) 41   (d) Can't be determined    (e) None of these
Ans: E
Explanation:
The required average = (35 + 42)/2 = 38.