Averages Questions for RRB NTPC Set-3 PDF

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Averages Questions for RRB NTPC Set-3 PDF
Averages Questions for RRB NTPC Set-3 PDF

Averages Questions for RRB NTPC Set-3 PDF

Download RRB NTPC Averages Questions Set-3 PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Download Averages Questions Set-3 PDF

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Question 1: Find the average of first 30 multiples of 7.

a) 94.8

b) 99.6

c) 104.4

d) 108.5

Question 2: The average of 10 numbers is X and one of the number is 38. If 38 is replaced with 19, then what will be the new average?

a) $X – (\frac{1}{2})$

b) $X – (\frac{1}{19})$

c) $X – (\frac{19}{10})$

d) $X – (\frac{19}{9})$

Question 3: The average height of A, B and C is 148 cm. If the average height of A and B is 136 cm and that of B and C is 125 cm, then what is the height (in cm) of B?

a) 56

b) 78

c) 112

d) 130

Question 4: The average of two numbers is 10 and the product of two numbers is 96. Find the two numbers.

a) 3, 32

b) 12, 8

c) 1, 96

d) 24, 4

Question 5: Average age of 4 daughters of a family is 12 years. The average age of daughters and their parents is 26 years. If the mother is 4 years older than the father, then what is the age (in years) of the father?

a) 56

b) 52

c) 48

d) 44

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Question 6: Average of 10 numbers is 14. If each number is multiplied by 6, then what will be the new average?

a) 70

b) 84

c) 56

d) 92

Question 7: Average marks obtained by 19 boys of a college is 56. If the highest marks and lowest marks obtained are removed, then the average reduces by 2. What is the average of the highest and lowest marks obtained?

a) 77

b) 69

c) 73

d) 75

Question 8: What will be the average of all the prime numbers before 19?

a) 8.28

b) 11.5

c) 5.34

d) 9.63

Question 9: What is the average of 90, 250, 240, 204, and 616?

a) 280

b) 210

c) 230

d) 255

Question 10: Average age of 6 boys is 14 years. Average age of 11 girls is 12 years. What is the average age (in years) of all boys and girls?

a) 12.7

b) 14.6

c) 19.3

d) 8.5

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Answers & Solutions:

1) Answer (D)

First thirty multiples of 7 will form an arithmetic progression : 7,14,21,…….,210

First term, $a=7$ and common difference, $d=7$

Sum of $n$ terms = $\frac{n}{2}(a+l)$

=> $S_{30}=\frac{30}{2}(7+210)$

= $15\times217$

Now, average of 30 terms = $\frac{15\times217}{30}$

= $\frac{217}{2}=108.5$

=> Ans – (D)

2) Answer (C)

Average of 10 numbers = $x$

=> Sum of numbers = $10x$

If 38 is replaced with 19, then new average = $\frac{10x-38+19}{10}$

= $\frac{10x-19}{10}$

= $x – (\frac{19}{10})$

=> Ans – (C)

3) Answer (B)

Let height of A, B and C be $a,b,c$ cm respectively.

Average height of A, B and C = $\frac{(a+b+c)}{3}=148$

=> $a+b+c=148\times3=444$ ————(i)

Also, average height of A and B = $\frac{(a+b)}{2}=136$

=> $a+b=272$  ————(ii)

Similarly, $b+c=250$ ———–(iii)

Adding equations (ii) and (iii), => $a+2b+c=522$ ———-(iv)

Subtracting equation (i) from (iv), we get :

=> $b=522-444=78$

$\therefore$ Height of B = 78 cm

=> Ans – (B)

4) Answer (B)

Let the numbers be $x$ and $y$

Average = $\frac{(x+y)}{2}=10$

=> $x+y=10\times2=20$ ————(i)

Product = $xy=96$ ————(ii)

Solving above equations, we get : $x=12$ and $y=8$

=> Ans – (B)

5) Answer (B)

Average age of 4 daughters of a family = 12 years

=> Sum of daughter’s age = $12\times4=48$ years

Average age of daughters and their parents = 26 years

=> Sum of ages of 4 daughters and 2 parents = $26\times6=156$ years

=> Sum of parents age = $156-48=108$ years ————(i)

Let father’s age = $x$ years, => Mother’s age = $(x+4)$ years

Substituting in equation (i),

=> $x+x+4=108$

=> $2x=108-4=104$

=> $x=\frac{104}{2}=52$ years

=> Ans – (B)

6) Answer (B)

Average of 10 numbers is = 14

If each number is multiplied by 6, then the average will also be multiplied by 14.

=> New average = $6\times14=84$

=> Ans – (B)

7) Answer (C)

Average marks obtained by 19 boys of a college = 56

=> Total marks obtained = $56\times19=1064$

Let the highest marks and lowest marks respectively be $x$ and $y$

According to ques,

=> $\frac{1064-x-y}{17}=56-2$

=> $1064-(x+y)=54\times17=918$

=> $(x+y)=1064-918=146$

Dividing both sides by 2, we get :

=> $\frac{(x+y)}{2}=\frac{146}{2}=73$

$\therefore$ Average of the highest and lowest marks obtained = 73

=> Ans – (C)

8) Answer (A)

Sum of all the prime numbers before 19

= $2+3+5+7+11+13+17=58$

=> Average = $\frac{58}{7}=8.28$

=> Ans – (A)

9) Answer (A)

Sum of numbers = 90 + 250 + 240 + 204 + 616 = 1400

=> Average = $\frac{1400}{5}=280$

=> Ans – (A)

10) Answer (A)

Average age of 6 boys = 14 years

=> Total age of 6 boys = $6\times14=84$ years

Similarly, total age of 11 girls = $11\times12=132$ years

$\therefore$ Average age (in years) of all boys and girls = $\frac{(84+132)}{(6+11)}$

= $\frac{216}{17}=12.7$

=> Ans – (A)

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