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Previous Year Average TCS Questions For SSC CGL, CHSL, CPO 2019

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Previous Year Average Questions with Solutions date wise For SSC CGL, CHSL, CPO

Q1. The average of twelve numbers is 42. The average of the last five numbers is 40 and that of the first four numbers is 44. The 6th number is 6 less than the fifth and 5 less than the 7th number. The average of the 5th and the 7th numbers are: 

बारह संख्याओं का औसत 42 है | अंतिम पांच संख्याओं का औसत 40 तथा पहली चार संख्याओं का औसत 44 है | छठी संख्या पाँचवीं से 6 कम है तथा 7वीं संख्या से 5 कम है | 5वीं और 7वीं संख्या का औसत है : 

SSC CGL - 4 June 2019 (Morning)

(a) 44

(b) 44.5

(c) 43

(d) 43.5

Q2. The average weight of a certain number of students in a class is 68.5 kg. If 4 new students having weights 72.2 kg, 70.8kg, 70.3kg and 66.7 kg join the class, then the average weight of all the students increases by 300 g. The number of students in the class, initially is:

किसी कक्षा में छात्रों की एक निश्चित संख्या का औसत वज़न 68.5 किलो ग्राम है | यदि 4 नए छात्र कक्षा में आ जाते हैं, जिनका वज़न क्रमशः 72.2 किलो ग्राम, 70.8 किलो ग्राम, 70.3 किलो ग्राम तथा 66.7 किलो ग्राम है, तो सभी छात्रों का औसत वज़न 300 ग्राम से बढ़ जाता है | आरंभ में कक्षा में छात्रों की संख्या कितनी थी ?   

SSC CGL - 4 June 2019 (Afternoon)

(a) 21

(b) 16

(c) 11

(d) 26

Q3. Three numbers are such that if the average of any two of them is added to the third number, the sums obtained are 168, 174 and 180 respectively. What is the average of the original three numbers?

तीन संख्याएँ इस प्रकार हैं कि यदि इनमें से किसी भी दो संख्या के औसत को तीसरी संख्या में जोड़ा जाए, तो प्राप्त होने वाले योगफल क्रमशः 168, 174 और 180 होते हैं | इन तीन प्रारंभिक संख्याओं का औसत ज्ञात करें | 

SSC CGL - 4 June 2019 (Evening)

(a) 86

(b) 87

(c) 89

(d) 84

Q4. The total number of students in section A and B of a class is 110. The number of students in section A is 10 more than that of section B. The average score of the students in B, in a test, is 20% more than that of students in A. If the average score of all the students in the class is 72, then what is the average score of the students in A?

किसी कक्षा के खंड A और खंड B के छात्रों की कुल संख्या 110 है | खंड A में छात्रों की संख्या खंड B के छात्रों की संख्या से 10 अधिक है | किसी परीक्षा में B के छात्रों का औसत प्राप्तांक A के छात्रों के औसत प्राप्तांक से 20% अधिक है | यदि सभी छात्रों का औसत प्राप्तांक 72 है, तो A के छात्रों का औसत प्राप्तांक ज्ञात करें |  

SSC CGL - 7 June 2019 (Afternoon)

(a) 66

(b) 68

(c) 63

(d) 70

Q5. Four different positive numbers are written in ascending order. One-third of the average of all the four numbers is 19 less than the greatest of these numbers. If the average of the first three numbers is 12, the greatest number among the given numbers is: 

चार अलग-अलग धनात्मक संख्याओं को आरोही क्रम में लिखा गया है | सभी चार संख्याओं की औसत का एक-तिहाई इनमें से सबसे बड़ी संख्या से 19 कम है | यदि पहली तीन संख्याओं का औसत 12 है, तो इनमें से सबसे बड़ी संख्या ज्ञात करें | 

SSC CGL - 7 June 2019 (Evening)

(a) 25

(b) 22

(c) 24

(d) 21

Q6. The average marks of 40 students were found to be 68. If the marks of two students were incorrectly entered as 48 and 64 instead of 84 and 46 respectively, then what is the correct average?

40 छात्रों के औसत अंक 68 पाए गए | यदि दो छात्रों के अंक भूलवश 84 एवं 46 के बजाय क्रमशः 48 और 64 के रूप में शामिल किये गए हैं, तो सही औसत ज्ञात करें | 

SSC CGL - 10 June 2019 (Afternoon)

(a) 68.25

(b) 68.15

(c) 68.45

(d) 68.35

Q7. In a class of 40 students, 45% are girls and the remaining are boys. If the average of the girls marks is 54 and that of the boys is 46, what is the average of the whole class? 

40 छात्रों की एक कक्षा में, 45% लड़कियाँ हैं तथा शेष लड़के हैं | यदि लड़कियों के अंकों का औसत 54 और लड़कों के अंकों का औसत 46 है, तो पूरी कक्षा का औसत अंक ज्ञात करें | 

SSC CGL - 11 June 2019 (Afternoon)

(a) 49.8

(b) 49.7

(c) 49.6

(d) 49.5

Q8. The average of 27 numbers is zero. Out of them, how many may be greater than zero, at the most? 

27 संख्याओं का औसत 0 है | इनमें से, अधिक से अधिक कितनी संख्याएँ शून्य से अधिक सकती हैं ? 

SSC CGL - 13 June 2019 (Evening)

(a) 0

(b) 15

(c) 26

(d) 20

Q9. 10 years ago, the average age of a family of five members was 38 years. Now, two new members join, whose age difference is 8 years. If the present average age of the family is the same as it was 10 years ago, what is the age (in years) of the new younger member? 

10 वर्ष पहले, पांच सदस्यों के एक परिवार की औसत उम्र 38 वर्ष थी | अब दो नए सदस्य शामिल हो गए हैं, जिनकी उम्र में 8 वर्ष का अंतर है | यदि परिवार की वर्तमान औसत उम्र उतनी ही है जितनी यह 10 साल पहले थी, तो नए छोटे सदस्य की उम्र ज्ञात करें | 

SSC CHSL - 4 July 2019 (Afternoon)

(a) 15

(b) 9

(c) 10

(d) 17

Q10. The average of a number and its reciprocal is 4. The average of its cube and its reciprocal is equal to किसी संख्या तथा उसके पारस्परिक ( reciprocal ) का औसत 4 है | इसके घन और और उसके पारस्परिक का औसत किसके बराबर होगा? 

SSC CHSL - 9 July 2019 (Evening)

(a) 256

(b) 142

(c) 288

(d) 244

Q11. The average of the first 1234 _______ numbers is equal to 1234. 

Fill in the blank. 

पहली 1234 ______ संख्याओं का औसत 1234 के बराबर होता है |

रिक्त स्थान की पूर्ति करें |  

SSC CHSL - 10 July 2019 (Morning)

(a) odd/ विषम

(b) even/ सम 

(c) prime/ अभाज्य 

(d) natural/ प्राकृतिक 

Q12. Fill in the blank.

रिक्त स्थान की पूर्ति करें | 

The average of the first 101 ______ numbers is equal to 102. 

पहली 101 ______ संख्याओं का औसत 102 के बराबर होगा |  

SSC CHSL - 10 July 2019 (Afternoon)

(a) natural/ प्राकृतिक

(b) odd/ विषम 

(c) even/ सम 

(d) perfect square/ पूर्ण वर्ग 

Q13. The difference between the average of first ten prime numbers and the first ten prime numbers of two digits is: 

पहली दस अभाज्य संख्याओं के औसत और पहली दस दो अंकों की अभाज्य संख्याओं के औसत में अंतर होगा :

SSC CHSL - 10 July 2019 (Evening)

(a) 14.5

(b) 16.5

(c) 12.5

(d) 13.5

Q14. The average of 1088 real numbers is zero. At most how many of them can be negative?

1088 वास्तविक संख्याओं का औसत शून्य है | उनमें अधिक से अधिक कितनी संख्याएँ ऋणात्मक हो सकती हैं ? 

SSC CHSL - 11 July 2019 (Morning)

(a) 100

(b) 88

(c) 544

(d) 1087

Q15. The average age of four brothers is 15 years. If their father is included, the average is increased by 5 years. The age of the father (in years) is: 

चार भाइयों की औसत आयु 15 वर्ष है | यदि उनके पिता को शामिल कर लिया जाए, तो औसत 5 वर्ष बढ़ जाता है | पिता की उम्र ( वर्ष में ) है :  

SSC CHSL - 5 July 2019 (Afternoon)

(a) 35

(b) 40

(c) 38

(d) 36

Q16. The average of the squares of numbers 1 to 5 is: 

1 से 5 तक की संख्याओं के वर्गों का औसत ज्ञात करें | 

SSC CPO - 16 March 2019 (Afternoon)

(a)11

(b)5

(c)8

(d)9

Q17. The average of a and b is 36. The average of b and c is 42. What is the difference between c and a? 

a तथा b का औसत 36 है | b तथा c का औसत 42 है | c तथा a के बीच कितना अंतर् है?

SSC MTS - 6 August 2019 (Afternoon)

(a) 18

(b) 12

(c) 16

(d) 14

Q18. What is the average of the first 8 multiples of 6 among the natural numbers?

प्राकृतिक संख्याओं में 6 के पहले 8 गुणजों का औसत क्या है? 

SSC MTS - 7 August 2019 (Evening)

(a) 24

(b) 26

(c) 27

(d) 28

Q19. The average of all the prime and composite numbers upto 100 is:

100 तक की सभी अभाज्य और विभाज्य संख्याओं का औसत है : 

SSC MTS - 9 August 2019 (Morning)

(a) 51

(b) 50

(c) 50.5

(d) 49.5

Q20. The average weight of 12 articles is 18 kg. Addition of another new article reduces the average weight by 500g. What is the weight of the new article?

12 वस्तुओं का औसत वज़न 18 किलो ग्राम है | एक अन्य नयी वस्तु को शामिल करने से औसत वज़न 500 ग्राम कम हो जाता है | नयी वस्तु का वज़न ज्ञात करें | 

SSC MTS - 9 August 2019 (Morning)

(a) 11.5 kg

(b) 15.0 kg

(c) 11.0 kg

(d) 10.1 kg

Q21. Given four different numbers, the average of first three numbers is four times the fourth number and the average of all the four numbers is 52. What is the average of the first three number?

चार अलग-अलग संख्याएँ दी गयी हैं, पहली तीन संख्याओं का औसत चौथी संख्या से चार गुना है तथा सभी चार संख्याओं का औसत 52 है | पहली तीन संख्याओं का औसत क्या है ? 

SSC CHSL - 3 July 2019 (Afternoon)

(a) 39

(b) 65

(c) 70

(d) 64

Q22. The average of all prime numbers 21 to 50 is (round off to one decimal number) 

21 से 50 तक की सभी अभाज्य संख्याओं का औसत ज्ञात करें |  (एक दशमलव संख्या तक पूर्णांक करें ) 

SSC CPO - 16 March 2019 (Morning)

(a) 35.9

(b) 34.8

(c) 33.7

(d) 32.9

Q23. What is the average of 59, 63, 68, 77, 74 and 73, when each number is divided by 23 ?  

59, 63, 68, 77, 74 तथा 73 का औसत कितना है, जब प्रत्येक संख्या को 23 से विभाजित किया जाता है ?

SSC MTS - 6 August 2019 (Afternoon)

(a) 63/23

(b) 67/23

(c) 3

(d) 46

Q24. What is the average of the first 15 odd numbers among the natural numbers?

प्राकृतिक संख्याओं में पहली 15 विषम संख्याओं का औसत क्या है ? 

SSC MTS - 8 August 2019 (Evening)

(a) 18

(b) 15

(c) 16

(d) 17

Q25. What is the average of the first 15 whole numbers?

प्रथम 15 पूर्ण संख्याओं का औसत ज्ञात करें | 

SSC MTS - 8 August 2019 (Morning)

(a) 8

(b) 7

(c) 9

(d) 10

Q26. Out of four numbers, the average of the first three is 16 and that of the last three is 15. If the last number is 21 then the first number is: चार संख्याओं में से पहली तीन संख्याओं का औसत 16 है और अंतिम तीन संख्याओं का औसत 15 है | यदि अंतिम संख्या 21 है, तो पहली संख्या है : 

SSC MTS - 13 August 2019 (Evening)

(a) 28

(b) 22

(c) 21

(d) 24

Here are the solutions to the above questions. Some questions have alternate solutions.

Sol 1. (b)

Let the sixth number is k , 5th number is k+6 and 7th number is k+5.

According to the question 

12 x 42 = (4 x 44)+(k+k+6+k+5)+(5 x 40)

504 = 176+3k+11+200

⇒k = 39

Desired average = =44.50

Alternate :

Average of the last 5 numbers is less than the total average by 2 and the average of the first four numbers is more than total average by 2. This will be managed by other  numbers.

According to the question

(42 x 3) = (4 x 2) + k + (k+6) + (k+5) + {5 x (-2)}

117 = 3k

k = 39

Desired average ==44.50 

Sol 2. (b)

Let the initial number of students = k 

According to the question

68.5k + 72.2 + 70.8 + 70.3 + 66.7 = 68.8 (k+4)

68.5k + 280 = 68.8k +275.2

k = 16

Alternate :

Weights of 4 new students is 3.7, 2.3, 1.8 kg more and 1.8 kg less than the average weight of students while the average weight of the class increases by 0.3 kg. All this will be managed by the new students’ weight.

According to the question

(0.3)(k+4) = 3.7+2.3+1.8-1.8

0.3k + 1.2 = 6

k = 16

Sol 3. (b)

Let the numbers are a, b and c.

According to the question

168+174+180

a+b+c =261

Desired average = 261/3= 87

Sol 4. (a)

Let the number of students in class B = k

⇒ k+k+10 = 110

⇒ k = 50

20% = 1/5

Let the average score of the students of class A = 5k and the average score of the students of class B = 6k

According to the question 

60(5k) + 50(6k) = 110 x 72

⇒ 300k + 300k = 7920

⇒ k = 13.2

the average score of the students of class A = 5 x 13.2 = 66

Sol 5. (c)

Let the average of all the four numbers = A and the last number = k

According to the question

A/3=k - 19

⇒ A = 3k -57              

3k - A = 57                                  ……..(1)

And Sum of first three numbers = 12 x 3 = 36

⇒ =A

4A - k = 36                     …..(2)

Multiply equation (1) by 4 and add in equation (2)

⇒ 12k-4A+4A-k = 228+36

⇒ k = 24

Sol 6.  (c)

Total marks of the students = 68 x 40 = 2720

Actual total marks of the students = 2720-48-64+84+46 = 2738

Desired average =2738/40= 68.45

Alternate :

Increase in total marks = 84+46-48-64 = 18

Increase in average = 18/40= 0.45

Desired average = 68+0.45 = 68.45

Sol 7. (c) 

Ratio Girls to Boys  = 45 % : 55%

                                = 9 : 11

Let the number of girls = 9 unit

And number of boys = 11 unit 

Total marks of the class = (54 x 9) + (11 x 46) = 1984

Desired average = = 49.6

Alternate :

Ratio Girls to Boys  = 45 % : 55%

                               = 9 : 11

Let the ratio of y : z = k : m Then,

Where D is the difference of the number of boys and girls

D = 54-46 = 8

k : m = 9 : 11

We know that 54-z = y+46 = x

x = 54 - 4.4 = 3.6 +46 = 49.6

Here, x is nothing but the average marks of the class.

Sol 8. (c)

Average of 27 numbers is zero, clearly the sum of 27 numbers is equal to zero.

So there can be total 26 number more than zero and the 27th number will be equal to the sum of 26 numbers but with the negative sign.{(i.e 27th number = -(sum of 26 numbers)}

Sol 9. (b)

Let the age of the younger member = k.

According to the question

= 38

2k+8 = 38(7-5)-50

k = 9

Sol 10. (d)

Let the number = k 

Its reciprocal = 1k

According to the question

Sol 11 (a)

Average of n odd numbers is always n.

Sol 12. (c) 

Average of n even numbers is always (n+1). 

Sol 13. (d)

First 10 prime numbers = 2,3,5,7,11,13,17,19,23,29

First 10 2 digit prime numbers = 11,13,17,19,23,29,31,37,41,43

Desired average == 13.5

Sol 14. (d)

Average of 1088 numbers is zero, clearly the sum of 1088 numbers is equal to zero. So there can be total 1087 number less than zero and the 1088th number will be equal to the sum of 1087 numbers but with the positive sign.{(i.e 1088th number = (sum of 1087 numbers)}

Sol 15. (b)

Sum of the ages of four brothers = 15 x 4 = 60 

Let the age of father = k

According to the question

= 20

⇒ k = 40

Alternate :

Average age of four brothers = 15

Increase in total average = 5

Increase in total age = 5 x 5 = 25

This age increase is balanced by father so age of father must be = 15+25 = 40

Sol 16.(a)

Average of the squares of ‘n’ natural numbers 

= 11

Alternate :

Required average =   = 11

Sol 17. (b)

Sum of a and b (a+b) = 2 x 36 = 72

Sum of b and c (b+c) = 2 x 42 = 84

Desired difference = 84-72 = 12

Sol 18. (c)

Sum of first 8 multiples of six = 6 (1+2+3+4+5+6+7+8)= 216

Desired average = 216/8= 27

Sol 19. (a)

Sum of numbers from 1 to 100 = = 5050

Now, excluding the number 1 each and every number beginning from 2 to 100 are prime or either composite.

Sum of all the prime and composite numbers upto 100 = 5050-1 = 5049

Desired average = 5049/99= 51

Sol 20. (a)

Total weight of 12 articles = 12 x 18 = 216

Let the weight of new articles = k

According to the question 

= 18-0.500

216 + k = 17.5 x 13

⇒k = 11.5

Alternate :

Average weight of the articles = 18

Reduction in average weight = 0.5 kg

Total reduction = 13 x 0.5 = 6.5 kg

This decrease in weight is balanced by the third article = 18-6.5 = 11.5

Sol 21.(d)

Let the average of first three numbers = 4A

the 4th number = A

Sum of first three numbers = 4A x 3 = 12A

According to the question

= 52

A = 16

Average of first three numbers = 4 x 16 = 64

Sol 22. (a)

All the prime numbers between 21 to 50 = 23, 29, 31, 37, 41, 43 and 47

⇒ Desired average = = 35.85 ≈ 35.9

Sol 23. (c)

Average of the given numbers = = 69

When each number is divided by 23, average = 69/23= 3

Sol 24. (b)

Average of ‘n’ odd numbers is always ‘n’.

Sol 25. (b)

First 15 whole numbers are 0,1,2,3,4 ….. 14

Sum of the numbers from 1-14 =

105

Desired average = 105/15= 7

Sol 26. (d)

Sum of first three = 3 x 16 = 48

Sum of last three = 3 x 15 = 45

First term - Last term= 48-45 = 3

First term = 3 + 21 = 24

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