The article **Average Practice Quiz** is useful for candidates preparing for different competitive examinations like **RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.**

The**Average** (arithmetic mean) of a group or set of N numbers is defined as the sum of those numbers divided by N. Here N is the number of values or observations in a set

Average = \(\frac{Sum of N numbers/values}{Numbers of values/observations}\)

A = \(\frac{Sum}{N}\) or Sum = N * Average

The

Average = \(\frac{Sum of N numbers/values}{Numbers of values/observations}\)

A = \(\frac{Sum}{N}\) or Sum = N * Average

**Answer –** Option C

**Explanation –**

Here man covers the same distance, but with two different speeds.

When the distance is same and two different speeds are given for covering that distance, then average speed during the whole journey is given by,

Average speed = \(\frac{2uv}{uv}\)

Here, u = 90 km/hr and v = 60 km/hr

Average speed = \(\frac{2 * 90 * 60}{90 + 60}\) = \(\frac{10800}{150}\) = = 72 km/hr

**2. A man goes up hill with an average speed at 20 kmph and comes down with an average speed of 30 kmph. The distance traveled in both the cases being the same, the average speed for the entire journey is**

**Answer –** Option C

**Explanation –**

Here as given, man goes uphill with an average speed of 20 kmph and comes down with an average speed of 30 kmph.

Thus, u = 20kmph and v = 30 kmph

It is given that distance in both the journey is same.

So, average speed for the entire journey

= \(\frac{2uv}{u + v}\)km/hr

= \(\frac{2 * 20* 30}{20 + 30}\)kmph

= \(\frac{1200}{50}\)km/hr = 24 kmph.

**3. A man travels first 80 km at 20 kmph, next 30 km at 15 kmph and then 80 km at 20 kmph. His average speed for the whole journey (in kmph) is**

**Answer –** Option D

**Explanation –**

Man travels total distance of 80 + 30 + 80 = 190 kilometers.

Times taken

for \({1}^{st}\) phase = \(\frac{80 km}{20 km/hr}\) = 4hr

for \({2}^{nd}\) phase= \(\frac{30 km}{15 km/hr}\) = 2hr

for \({3}^{rd}\) phase= \(\frac{80 km}{20 km/hr}\) = 4hr

So, average speed for the whole journey

= \(\frac{Total dist ance}{Total time taken}\)

= \(\frac{80 + 30 + 80}{4 + 2 + 4}hr\) = \(\frac{190 km}{10 hr}\) = 19 km/hr

**4. A car covers four successive 4 km distances at speeds of 10 k mph, 20 k mph and 60 k mph respectively. Its average speed over total distance is**

**Answer –** Option B

**Explanation –**

Here a car covers four successive 4 km distances at 4 different speeds of 10 kmph, 20 kmph, 30 kmph and 60 kmph

So, time for

for \({1}^{st}\) = \(\frac{4 km}{10 kmph}\) = \(\frac{24}{60}\)hr

for \({2}^{nd}\) = \(\frac{4 km}{20 kmph}\) = \(\frac{12}{60}\)hr

for \({3}^{rd}\) = \(\frac{4 km}{30 kmph}\) = \(\frac{8}{60}\)hr

for \({4}^{th}\) = \(\frac{4 km}{60 kmph}\) = \(\frac{4}{60}\)hr

Total dist ance = 4+4+4+4 = 16 km.

Hence, average speed of car over 16 km

= \(\frac{16 km}{(\frac{24}{60} + \frac{12}{60} + \frac{8}{60} + \frac{4}{60} )hr }\)

\(\frac{16 * 60}{48}\) = 20 km/hr

**5. First 150 miles of his trip, John drove at 50 miles per hour, then due to traffic, he drove at only 20 miles per hour for the next 120 miles. His average speed, in miles per hour, for the entire trip
is **

**Answer –** Option C

**Explanation –**

As given in data, John drove first 150 miles at speed of 50 mile per hour.

So, for first 150 miles he took

\(\frac{150 miles}{50 miles/hr}\) =hours

Next 120 miles, he drove at speed of 20 miles per hour.

So for that 120 miles, he took,

\(\frac{120 miles}{20 miles/hr}\) = 6 hours

Thus, in total he drove

150 + 120 = 270 miles in 6 + 3 = 9 hrs.

So, his average speed for the entire trip

= \(\frac{Total distance John drove

}{Total time he took}\)

= \(\frac{(120 + 150)miles}{3 + 6}hrs\)

= \(\frac{(270)miles}{9}hrs\) = 30 miles/ hr

**6. A car travels 70 kilometers in one hour before some fault happens, then it travels for 120 km at 30 kmph. For the entire trip, average speed is**

**Answer –** Option C

**Explanation –**

Here given that a car travels 70 km in one hour before fault happens means it has covered 70 km distnace for first phase.

After the fault happened, it has cover d the remaining 120 km at speed of 30 km/hr

Hence, total time for trip becomes

= (time for 1st phase) + (time for 2nd phase)

= 1 hr + 4 hrs

= 5 hrs.

And, total distance becomes

= (Distance in 1st phase) + (Distance in 2nd phase)

= 70 km + 120 km

= 190 kilometers.

i.e, Average speed

= \(\frac{Total distance}{Total time taken}\) = \(\frac{(120 + 70)km}{(1 + 4)hr}\)

= 38 km/hr

**7. A motor travels 100 miles at the rate of 40 miles per hour. If it returns the same distance at a rate of 50 miles per hour, the average speed for the entire trip, in miles per hour is**

**Answer –** Option D

**Explanation –**

In first phase, a motor travels 100 miles at rate of 40 miles per hour.

So, time taken during \({1}^{st}\) phase = \(\frac{Distance}{Speed}\)

= \(\frac{100 miles}{40 miles/hr}\) = \(\frac{1}{2}\)hrs

Same as that of \({1}^{st}\) phase, but speed the motor is different. So the time for \({2}^{nd}\) phase will be different.

So the distance during \({2}^{nd}\) phase is 100 miles and speed of motor given is 50 miles/hr.

i.e, Time \({1}^{st}\) phase = \(\frac{Distance}{Speed}\)

= \(\frac{100 miles}{50 miles/hr}\) = 2 hrs

Hence total distance for the entire trip

= (100 +100) = 200 miles and time taken for the trip

= 2 \(\frac{1}{2}\) + 2 = \(\frac{9}{2}\)hrs

i.e, Average speed (entire trip) = \(\frac{Total distance of trip}{Total time taken for trip}\)

= \(\frac{(100 + 100)miles}{\frac{9}{2}}\)

= \(\frac{(400)miles}{9hr}\)

= \(\frac{2(Avg.Speed for A to B) (Avg.Speed for B to A}{(Avg.Speed for A to B) (Avg. Speed for B to A)}\)

= \(\frac{2 (45 * 36)}{45 + 36}\)

= 40 km/hr. …(3)

From results (i) and (ii), we get the difference of average speeds of (journey from A to B) and the whole journey.

i.e, Difference

= (Average speed for A to B) – (Average speed for whole Journey)

= (45– 40)km/hr = 5 km/hr

Hence, average speed from A to B exceeds the average speed for whole trip by 5 km/hr.

**8. At a constant velocity of 30 miles per hour, car A leaves point ‘X’ at 1 p.m.. Car B leaves ‘X’ at 3 p.m. at constant velocity and overtakes A at 5 p.m. The average speed of B will be**

**Answer –** Option D

**Explanation –**

As given, car A leaves point X at 1 p.m. and travels with constant velocity of 30 miles per hour.

Car B Leaves point X at 3 p.m. and travels with constant velocity and overtakes car A at 5 p.m.

So, at 5 p.m. the distance covered by both car A and car B will be equal.

Let \({D}_{1}\) and \({D}_{2}\) would be the distances covered by car A and car B respectively,

Hence 5 p.m. \({D}_{1}\) = \({D}_{2}\)

i.e, Time taken by car A * Speed of by A

= Time taken by car B * Speed of car B

i.e, 30 km/hr * 4 hrs

= \({S}_{2}\) (Speed of car B) * 2 hrs

120 km = \({S}_{2}\) * 2 hrs.

\({S}_{2}\) = 60 km/hr

**9. A car covers a distance in 36 minutes. It runs at 50 kmph on an average. The speed at which the car must run to reduce the time of journey by 6 minutes will be**

**Answer –** Option D

**Explanation –**

The distance covered by a car at 50 kmph is in 36 minutes.

To proceed further, we first find the total distance covered by it.

Now, Distance = Speed * Time

Here, Speed = 50 kmph

= 50 * \(\frac{1000}{3600}\) = \(\frac{500}{36}\)

Distance = \(\frac{500}{36} * 36\) = 500km

Further, we are given the time reduced by 6 minutes,

i.e. 30 minutes to complete the same distance.

i.e, Speed = \(\frac{Distance}{Time}\)

= \(\frac{500}{30}\) = \(\frac{50}{3}\)

= \(\frac{50}{3}\) * \(\frac{3600}{1000}\)km/hr

= 60 km/hr

**10. A car travelled from city P to city R in 30 minutes. The first half of the distance was covered at 50 miles perhour, and the second half was covered at 60 miles per hour. What was the overage speed of the car ?
**

**Answer –** Option D

**Explanation –**

Here, we apply the formula

Distance = speed * time.

The total distance covered in 30 mins.

The first half of the distance was covered at 50 miles per hour.

Time Taken for first half distance = \(\frac{x}{2 * 50}\)

where, x = total distance covered.

The second half of the distance was covered at 60 miles per hour.

i.e, Time Taken for second half distance = \(\frac{x}{2 * 60}\)

i.e, \(\frac{x}{2 * 50}\) + \(\frac{x}{2 * 60}\) = 30

\(\frac{x}{2} {(\frac{1}{50} + \frac{1}{60})}\) = 30

x = \(\frac{30 * 300}{11} * 2\) = \(\frac{\frac{1800}{11}}{30}\)

= \(\frac{1800}{11 * 30}\) = \(\frac{600}{11}\)

**11. A man drives 150 km from A to B in 3 hours 20 minutes. He comes back from B to A in 4 hours 10 minutes. Then, average speed from A to B exceeds the average speed for the entire trip by**

**Answer –** Option C

**Explanation –**

Here distance between A and B is 150 km and a man drives this distance in 3 hours 20 min.

So his speed for the trip from A to B.

Average speed from A to B

\(\frac{Distance}{Time}\) = \(\frac{150 km}{3 + (\frac{20}{60} hr)}\)

= 45 km/hr …(i)

Now, we get average speeds for the trips from A to B and B to A.

Average speed from B to A

\(\frac{Distance}{Time}\) = \(\frac{150 km}{4 + (\frac{10}{60} hr)}\)

= \(\frac{150 km}{4 + (\frac{1}{6} hr)}\) = 36 km/hr …(ii)

Average speed for the whole journey

**12. The average of first 50 natural numbers is**

**Answer –** Option D

**Explanation –**

Sum of first n natural numbers = \(\frac{n (n + 1)}{2n}\)

So, average of first n natural numbers = \(\frac{n (n + 1)}{2n}\) = \(\frac{n + 1}{2}\)

\(\frac{50 + 1}{2}\) = \(\frac{51}{2}\) = 25.5

**13. Mean of \({1}^{2}, {2}^{2}, {3}^{2}, {4}^{2}, {5}^{2}, {6}^{2}, …… {50}^{2}\)**

**Answer –** Option B

**Explanation –**

\({1}^{2}, {2}^{2}, {3}^{2}, {4}^{2}, {5}^{2}, {6}^{2}, …… {n}^{2}\) = \(\frac{n(n + 1) (2n + 1)}{6}\)

\({1}^{2}, {2}^{2}, {3}^{2}, {4}^{2}, {5}^{2}, {6}^{2}, …… {50}^{2}\) = \(\frac{50 * 51 * 101}{6}\) = 42925

i.e, Required average = \(\frac{42925}{50}\) = 858.5

**14. Average of all odd numbers upto 100 is**

**Answer –** Option B

**Explanation –**

Sum of odd numbers upto 100

= 1 + 3 + 5 + 7 +… + 95 + 97 + 99

= (1+99)+(3+97)+(5+95) +… upto 25 pairs

= 100 + 100 + 100 + … (25 times)

= 2500

i.e, Average = \(\frac{2500}{50}\) = 50

**15. Average of 7 consecutive numbers is 33, then largest of these numbers is**

**Answer –** Option A

**Explanation –**

Let the numbers be

x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6.

\(\frac{x + x +1+ x + 2+ x + 3+ x + 4 + x + 5+ x+ 6}{7}\) = 103

\(\frac{7x + 21}{7}\) = 103

\(\frac{7(x + 3)}{7}\) = 103

x + 3 = 103

x = 100

i.e, Largest number = x+ 6

= (100 + 6) = 106

**Answer –** Option C

**Explanation –**

Prime numbers between 30 and 50 are: 31, 37, 41, 43, 47

i.e, required average

\(\frac{31 + 37 + 41 + 43 + 47}{5}\) = \(\frac{199}{5}\) = 39

**2. The average age of 5 members of a committee is the same as it was 3 years ago, because an old member has been replaced by a new member. The difference between the ages of old and new member is**

**Answer –** Option D

**Explanation –**

Average age of the five members should have been increased by 3 years.

Hence, required difference = 3 x 5 = 15 years

**3. The average of 5 numbers is 40. If one number is excluded then the average becomes 38. The excluded number is**

**Answer –** Option A

**Explanation –**

Let the number be x.

5 × 40 = x + 4 × 38

x = 48.

**4. The average of 35 observations is 30. Out of these observations the average of first 18 observations is 30 and the last 18 observations is 40. The 18th observation is**

**Answer –** Option D

**Explanation –**

Let the 18th observation be x.

18 × 30 + 18 × 40 = 35 × 30 + x

x = 210.

**5. The average of first five prime numbers is**

**Answer –** Option D

**Explanation –**

Required average = \(\frac{3 + 5 + 7 + 11}{5}\) = 5.6

**6. The average weight of 100 students is 46 kg. The average weight of boys is 50 kg. If the number of boys is 60, the average weight of girls in kilograms is**

**Answer –** Option B

**Explanation –**

Set the average weight of girls be x kg.

50 * 60 * x * 40* = 100 * 46

x = 40 kg.

**7. The average of first six even whole numbers is**

**Answer –** Option A

**Explanation –**

First six whole numbers form on AP → 0, 2, 4,

Sum of Ap = \(\frac{n}{2} (2a + (n – 1)d) = 30\)

= \(\frac{6}{2} (2 * 0 + (6 – 1)2) = 30\)

= \(\frac{30}{6} = 5\)

**8. The average marks of students of section A an B are respectively 60 and 70. The number of students in section A is 47 and in section B is 53. The average marks of both sections taken together are**

**Answer –** Option B

**Explanation –**

Total marks in Section A = average marks × numbers of students = 60 × 47 = 2820

Similarly, total marks in Section B = 70 × 53 = 3710

i.e, Average = \(\frac{2820 + 3710}{47 + 53}\) = 65.3

**9. The average weight of 100 students is 46 kg. The average weight of girls is 40 kg. If the number of girls is 40, the average weight of boys in kilograms is**

**Answer –** Option C

**Explanation –**

Total weight of class = average weight × number of students

= 46 × 100 = 4600

similarly, total weight of girls = 40 × 40 = 1600

i.e, Total weight of boys = 4600 – 1600 = 3000

Average weight of boys = \(\frac{3000}{number of boys}\)

= \(\frac{3000}{100 – 40}\) = 50

**10. If 27, 12. 24, x are in proportion, then x is equal to**

**Answer –** Option C

**Explanation –**

27 : 12 : : 24 : x

\(\frac{7}{12}\) = \(\frac{24}{x}\)

x = \(\frac{288}{27}\) = 10.67

**11. The average of squares of first five whole numbers is**

**Answer –** Option A

**Explanation –**

Squares of first 5 whole numbers = 0, 1, 4, 9, 16

i.e, \(\frac{0 + 1 + 4 + 9 + 16}{5}\) = 6

**12. The average of first 5 odd prime numbers is**

**Answer –** Option D

**Explanation –**

Average of first 5 add prime Numbers

\(\frac{3 + 5 + 7 + 11 + 13}{5}\) = 7.8

**13. The average of 18 observations is 30 and average of 22 observations is 40. The average of all combined observation is
**

**Answer –** Option A

**Explanation –**

Combined Average = \(\frac{18 * 30 * 22 * 40}{18 + 22}\) = 35.5

**14. The average of first seven whole numbers is**

**Answer –** Option D

**Explanation –**

Average = \(\frac{sum of first 7 whole numbers}{7}\)

\(\frac{0 + 1 + 2 + 3 + 4 + 5 + 6}{7}\) = 3

**15. The average 16 observations is 25 and average of 24 observations is 30. The average of all 40 observation is**

**Answer –** Option D

**Explanation –**

Average = \(\frac{16 * 25 * 24 * 30}{40}\) = 28

**Answer –** Option D

**Explanation –**

12a + 6b = 54

As different values of a and b satisfies the given equation.

i.e, Data is insufficient to find average.

**2. The average score of girls in a class is 75 marks. The average scores of boys in the class is 65 marks. If the average of the class is 68.75 marks, what is the ratio of boys to girls in the class?**

**Answer –** Option D

**Explanation –**

**Answer –** Option A

**Explanation –**

Let avg of last two number be x = 3.95 × 6

Let the average of last 2 number be x.

According to question,

3.95 * 6 = 2 * 3.40 + 2 * 3.85 + 2x

i.e, 2x = 23.7 – 14.5

x = 4.6

**4. The average of marks of 28 students in Maths was 50. 8 students left the school and then the average increased by 5. What is the average of marks obtained by the students who left the school?**

**Answer –** Option A

**Explanation –**

Let avg of students who left = x

55 = \(\frac{28 * 50 – 8x}{20}\)

x = 37.5

**5. If average of 7 consecutive number is 203, then the average of the smallest and second number is**

**Answer –** Option A

**Explanation –**

Assume smallest No. = a

i.e, \(\frac{a + (1 + 2+ 3 + 4 + 5 + 6}{7}\) = 203

a = 200

**6. The average of n numbers is x . If the first number is increased by 1, second by 2, third by 3 and so on, then their average is y. The value of y – x is.**

**Answer –** Option D

**Explanation –**

Increase in sum of numbers = \(\frac{n (n + 1)}{2}\)

Distributing the sum to n numbers

\(\frac{n (n + 1)}{2n}\) = \(\frac{(n + 1)}{2}\)

**7. In a mathematics test in a class the average score of boys is 33 and that of girls is 42. I f the average score of all the students in the class is 38, then what is the percentage of girls in the class?**

**Answer –** Option C

**Explanation –**

i.e, percentage of girls = \(\frac{5}{5 + 4} * 100\) = \(55 \frac{5}{9}%\)

**Answer –** Option C

**Explanation –**

3 : 2

i.e, Percentage of boys in class = (\(\frac{3}{3 + 2} * 100\))%

= 60%

**Answer –** Option C

**Explanation –**

x : y = 11 : 10

\(\frac{2x + 5y}{5x – 4y}\) = \(\frac{ 2 \frac{x}{y} + 5}{ 5 \frac{x}{y} – 4}\)

= \(\frac{ 2 \frac{11}{10} + 5}{ 5 \frac{11}{10} – 4}\)

= \(\frac{\frac{11}{5} + 5}{\frac{11}{2} – 4}\)

= \(\frac{36}{5} * \frac{2}{3} = \frac{24}{5}\)

**Answer –** Option B

**Explanation –**

According to question

\(\frac{nx – (1 + 2 +…+ n)}{n}\) = y

\(\frac{nx – \frac {n + 1}{2}}{n}\)

x – y = \(\frac {n + 1}{2}\)

**11. Four positive integers, when added three at a time give the sums 180,197,208 and 219. Average of these four integers is**

**Answer –** Option A

**Explanation –**

Let the integers be a, b, c, d

a + b + c = 180

b + c + d = 197

c + d + a = 208

d + a + b = 219

_____________________

3(a + b + c + d) = 804

a + b + c + d = 268

i.e, Avg = \(\frac{a + b + c + d}{4}\) = 67

**12. The average of 100 numbers is 100. I f the first number is increase by 1, second by 2, third by 3 and so on, then the average of the numbers so obtained exceeds the original average by**

**Answer –** Option C

**Explanation –**

The increase in average = \(\frac{\frac{100(100 + 1)}{2}}{100}\)

= 50.5

**13. The average score of the students in a class is 65. The average score of boys is 60 and that of girls is 68. The percentage of girls in the class is:**

**Answer –** Option B

**Explanation –**

i.e, Percentage of girls = \(\frac{5}{3 + 5} * 100\)% = 62.5%

**Answer –** Option C

**Explanation –**

Increase in the numbers = 2 + 4 + 6 + 8 + … (till 30 numbers)

\(\frac{30}{2}\)[2 * 2 + (30 – 1)* 2] = 930

Since total increase is 930, the increase in average of 30 numbers = \(\frac{930}{30}\) = 31

**15. The average of marks obtained by 150 students was 40. If the average marks of passed students were 50 and that of failed students was 20. Then the number of st udent s who passed the examination was**

**Answer –** Option C

**Explanation –**

Let number of passed students = P

Let number of failed students = 150 – P

i.e, 150 × 40 = 50P + 20(150 – P)

5P + 2(150 – P) = 600

P = 100