Practice Test: Averages - 1

15 Questions MCQ Test Quantitative Aptitude for GMAT | Practice Test: Averages - 1

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Attempt Practice Test: Averages - 1 | 15 questions in 30 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Aptitude for GMAT for Quant Exam | Download free PDF with solutions
QUESTION: 1

There are 7 members in a family whose average age is 25 years. Ram who is 12 years old is the second youngest in the family. Find the average age of the family in years just before Ram was born?

Solution:

In order to find the average age of the family before Ram was born, we need to know the age of the youngest member of the family. Thus, D is the right choice.

QUESTION: 2

The average weight of a class of 10 students is increased by 2 kg when one student of 30kg left and another student joined. After a few months, this new student left and another student joined whose weight was 10 less than the student who left now. What is the difference between the final and initial averages?

Solution:

Change in total weight of 10 students = difference in weight of the student who joined and the student who left
=> weight of first student who left = 30 + (10×2) = 50
weight of the student who joined last = 50 – 10 = 40
Thus change in average weight = (40 – 30)/10 = 1

QUESTION: 3

The average of 15 numbers is 18. If each number is multiplied by 9, then the average of the new set of numbers is:

Solution:

When we multiply each number by 9, the average would also get multiplied by 9.

Hence, the new average = 18 X 9 = 162.

QUESTION: 4

The average number of runs scored by Virat Kohli in four matches is 48. In the fifth match, Kohli scores some runs such that his average now becomes 60. In the 6th innings he scores 12 runs more than his fifth innings and now the average of his last five innings becomes 78. How many runs did he score in his first innings? (He does not remain not out in any of the innings)

Solution:

Runs scored by Kohli in first 4 innings = 48*4 = 192
Average of 5 innings is 60, so total runs scored after 5 innings = 60*5 = 300
Hence runs scored by Kohli in fifth inning = 300 – 192 = 108
It is given that in 6th innings he scores 12 runs more than this, so he must score 120 in the sixth inning. Hence total runs scored in 6 innings = 300+120 = 420
Now average of last five innings is 78, so runs scored in last innings = 390
Hence runs scored in first inning = 420 – 390 = 30.

QUESTION: 5

If a – b : b – c : c – d = 1 : 2 : 3, then what is the ratio of (a + d) : c?

Solution:

Let a – b = x, b – c = 2x and c – d = 3x
Thus,
c = 3x + d
b = 2x + c = 5x + d
a = x + b = 6x + d
Hence,
(a + d )/ c =  (6x + d + d) / (3x + d) =  2/1

QUESTION: 6

The average of a batsman after 25 innings was 56 runs per innings. If after the 26th inning his average increased by 2 runs, then what was his score in the 26th inning?

Solution:

Normal process:
Runs in 26th inning = Runs total after 26 innings – Runs total after 25 innings
= 26 X 58 – 25 X 56

For Easy calculation use:

= (56 + 2) X 26 – 56 X 25 )
= 2 X 26 + (56 X 26 – 56 X 25)
= 52 + 56 = 108

Since the average increases by 2 runs per innings, it is equivalent to 2 runs being added to each score in the first 25 innings. Now, since these runs can only be added by the runs scored in the 26th inning, the score in the 26th inning must be 25 X 2 = 50 runs higher than the average after 26 innings (i.e. new average = 58).

Hence, runs scored in 26th inning:
= New Average + Old innings X Change in average

= 58 + 25 X 2
= 108

QUESTION: 7

The average marks of a group of 20 students on a test is reduced by 4 when the topper who scored 90 marks is replaced by a new student. How many marks did the new student have?

Solution:

Let initial average be x.
Then the initial total is 20x and the New average will be (x – 4),

The new total will be:
20(x – 4) = 20x – 80.

The reduction of 80 is created by the replacement. Hence, the new student has 80 marks less than the student he replaces. Hence, he must have scored 10 marks.

Short Cut:
The replacement has the effect of reducing the average marks for each of the 20 students by 4. Hence, the replacement must be 20 X 4 = 80 marks below the original.

QUESTION: 8

The average of the first ten composite numbers is

Solution:

Required average:

= (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18) / 10
= 112 / 10
= 11.2.

QUESTION: 9

The average weight of 3 boys Ross, Joey and Chandler is 74 kg. Another boy David joins the group and the average now becomes 70 kg. If another boy Eric, whose weight is 3 kg more than that of David, replaces Ross then the average weight of Joey, Chandler, David and Eric becomes 75 kg. The weight of Ross is:

Solution:

David's Weight = (4 x 70) – (3 x 74) = 280 – 222 = 58
Eric’s weight = 58 + 3 = 61

Now, we know that:
Ross + Joey + Chandler + David = 4 x 70 = 280
Joey + Chandler + David + Eric = 75 x 4 = 300.

Hence, Ross’s weight is 20 kg less than Eric’s weight. Ross = 61 - 20 = 41 kg

QUESTION: 10

The mean temperature of Monday to Wednesday was 35 °C and of Tuesday to Thursday was 30 °C. If the temperature on Thursday was 1/2 that of Monday, the temperature on Thursday was ______ .

Solution:

Mon + Tue + Wed = 35*3 = 105  ---------(1)
Tue + Wed + Thu = 30*3 = 90  -------------(2)
Thu = (1/2) Mon  ------------(3)

Eqn (1)-(2):
Mon-Thu = 15 ------------(4)

⇒ Mon - (1/2) Mon = 15
⇒ (1/2) Mon = 15
⇒ Mon =30
⇒ Thu = 30/2=15

QUESTION: 11

The average age of a family of 5 members is 20 years. If the age of the youngest member is 10 years, what was the average age of the family at the birth of the youngest member?

Solution:

At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50
Therefore, average age of the family at the time of birth of the youngest member,
= 50/4 =12.5

QUESTION: 12

The average weight of 10 men is decreased by 2 kg when one of them weighing 140 kg is replaced by another person. Find the weight of the new person.

Solution:

Shortcut:
The decrease in weight would be 20kgs (10people’s average weight drops by 2 kgs). Hence, the new person’s weight = 140 - 20 = 120.

Detailed Solution:

Let weight of 9 men =x.
Weight of new men =y

According to the question:

((x+140)/10) ​− 2 = (x+y​)/10
y = 120

QUESTION: 13

The average age of a group of men is increased by 6 years when a person aged 26 years is replaced by a new person of aged 56 years. How many men are there in the group?

Solution:

When a person aged 26 years, is replaced by a person aged 56 years, the total age of the group goes up by 30 years.

Since this leads to an increase in the average by 6 years, it means that there are 30 / 6 = 5 persons in the group.

QUESTION: 14

The average weight of a class is 54 kg. A student, whose weight is 145 kg, joined the class and the average weight of the class now becomes a prime number less than 72. Find the total number of students in the class now.

Solution:

Let the original number of students in the class be N.
Total weight of the class = 54N
New total weight of the class = 54N + 145
New average weight of the class = (54N + 145)/(N+1) = (54N + 54)/(N+1) + 91/(N+1) = 54 + 91/(N+1).
Since the new average is an integer, (N+1) should be a factor of 91.
If N+1 = 7, the new average becomes 54 + 91/7 = 54 + 13 = 67
and if N+1 = 13, then the new average becomes 54 + 91/13 = 54 + 7 = 61
Both 67 and 61 are prime numbers less than 72. So, we cannot uniquely determine the number of students in the class.

QUESTION: 15

There are n weights in a bag measuring 1kg, 2kg and so on till n kg. These weights are divided into 3 parts. The first part contains the weights 1kg, 4kg, 7kg, and so on. The second part contains the weights 2kg, 5kg, 8kg and so on. The third part contains the remaining weights. The average weights any two of the three parts is equal to the weight present in those parts but the average weight of the remaining one part is not equal to the weight present in that part. Which of the following can be a possible value of n?

Solution:

We know that if in an AP the number of terms in a series is odd, the average of the terms of the series is equal to the middle term of the series. However if the number of terms in the series is even, the average of all the terms of the series is not equal to one of the terms of the series. Hence the three part contain terms 2x+1, 2x+1, 2x or 2x-1, 2x-1, 2x
Hence the total number of parts = 2x+1+2x+1+2x or 2x-1+2x-1+2x = 6x+2 or 6x-2
Among the options, the only number of the form 6x+2 or 6x-2 is 88. Hence 88 can be the required value of n.

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