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Mathematical Aptitude for Bank Exams

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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?

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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) 25Ans: 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.

PermutauionsandCombinations

Problemson LCMand HCF

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Posted by Alaudeen A at 23:38 1 comment:

= (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?


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: eExplanation:

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 waysA table can be arranged in 8 ways

Therefore, one chair and one table can be arranged in 10 × 8 = 80 ways.

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Problems on LCM and HCF

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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 = 6xGiven, 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 = 240 ∕ 12 = 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 1m50cm broad ?



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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 9leaves 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

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.6. Find the HCF of 22×32, 2×34×7


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Posted by Alaudeen A at 23:30 No comments:

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

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, 30Ans: 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

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


Friday, 27 September 2013

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Average

AverageQ. 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 theseAns: (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 theseAns: (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:

http://alaudeenali.blogspot.in/2013/09/average.html


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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 marksof 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


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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 = 448The 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 = 350Therefore, 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 andthe 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.


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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 inthe 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 onFriday.

(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.


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Q. 17. A cricketer has an average score of 49 runs in 24 innings. How many runs must he score in the 25 th 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


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.


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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?


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

12/11/2015 M th ti l A tit d f B k E


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even numbers of the same set?


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 theclass. 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.18Explanation:

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



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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


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



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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?


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


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


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

11x/6 = 462 and so, x = 252

Then third number = 252/3 = 84



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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.? ( assumingthat 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



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p


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


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 = 26Subtract 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



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



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number?


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 theseAns: (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:



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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.


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