CAT — Averages
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The average number of copies of a book sold per day by a shopkeeper is 60 in the initial seven days and 63 in the initial eight days, after the book launch. On the ninth day, she sells 11 copies less than the eighth day, and the average number of copies sold per day from second day to ninth day becomes 66. The number of copies sold on the first day of the book launch is
Step 1: Find the total number of books sold in the first 7 days. Average for the first 7 days = 60. Total books sold in the first 7 days = 7 × 60 = 420.
Step 2: Find the number of books sold on the 8th day. Average for the first 8 days = 63. Total books sold in the first 8 days = 8 × 63 = 504. Books sold on the 8th day = 504 − 420 = 84.
Step 3: Find the number of books sold on the 9th day. The 9th day sales are 11 less than the 8th day sales. Books sold on the 9th day = 84 − 11 = 73.
Step 4: Find the total number of books sold from the 2nd day to the 9th day. Average from the 2nd day to the 9th day = 66. Number of days = 8. Total books sold from the 2nd day to the 9th day = 8 × 66 = 528.
Step 5: Find the number of books sold on the 1st day. The total books sold from the 2nd day to the 9th day can also be written as: (Total books sold in the first 8 days − Books sold on the 1st day) + Books sold on the 9th day. So, (504 − First day sales) + 73 = 528 → 577 − First day sales = 528 → First day sales = 49.
− The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is
Given:
The average of three distinct real numbers is 28.
If the smallest number is increased by 7 and the largest number is reduced by 10,
● the order of the numbers remains unchanged,
● the new arithmetic mean is 2 more than the middle number,
● the difference between the largest and the smallest numbers becomes 64.
Find the largest of the original three numbers.
Step 1: Let the three numbers be
a < b < c
Since their average is 28,
a + b + c = 84
Step 2: Form the new numbers
After the changes, the numbers become
a + 7, b, c − 10
Their sum is
(a + 7) + b + (c − 10)
= 84 − 3
= 81
Hence, the new average is
81/3 = 27
Step 3: Use the given condition
The new average is 2 more than the middle number.
So,
27 = b + 2
b = 25
Step 4: Use the difference condition
The new difference between the largest and the smallest numbers is 64.
(c − 10) − (a + 7) = 64
c − a = 81
Step 5: Find the largest number
Using
a + b + c = 84
and
b = 25,
a + c = 59
Also,
c − a = 81
Adding the two equations,
2c = 140
c = 70
Final Answer
The largest number in the original set is
70
A company has 40 employees whose names are listed in a certain order. In the year 2022, the average bonus of the first 30 employees was Rs. 40000, of the last 30 employees was Rs. 60000, and of the first 10 and last 10 employees together was Rs. 50000. Next year, the average bonus of the first 10 employees increased by 100%, of the last 10 employees increased by 200% and of the remaining employees was unchanged. Then, the average bonus, in rupees, of all the 40 employees together in the year 2023 was
95000
90000
80000
85000
95000
Given:
There are 40 employees.
In 2022,
● Average bonus of the first 30 employees = Rs. 40000
● Average bonus of the last 30 employees = Rs. 60000
● Average bonus of the first 10 and last 10 employees together = Rs. 50000
In 2023,
● Bonus of the first 10 employees doubles.
● Bonus of the last 10 employees triples.
● Bonus of the remaining 20 employees remains unchanged.
Find the average bonus of all 40 employees in 2023.
Step 1: Find the total bonuses in 2022
First 30 employees
= 30 × 40000
= Rs. 12,00,000
Last 30 employees
= 30 × 60000
= Rs. 18,00,000
First 10 and last 10 together
= 20 × 50000
= Rs. 10,00,000
Step 2: Find the total bonus of all 40 employees in 2022
Adding the totals of the first 30 and last 30 counts the middle 20 employees twice.
Hence,
Total bonus of all 40 employees
= (12,00,000 + 18,00,000 + 10,00,000)/2
= Rs. 20,00,000
Step 3: Find the bonuses of different groups
Let
First 10 = F
Middle 20 = M
Last 10 = L
Then,
F + M = 12,00,000
M + L = 18,00,000
F + L = 10,00,000
Adding the first two equations,
F + 2M + L = 30,00,000
Since
F + L = 10,00,000,
2M = 20,00,000
M = Rs. 10,00,000
Therefore,
F = 12,00,000 − 10,00,000
= Rs. 2,00,000
L = 18,00,000 − 10,00,000
= Rs. 8,00,000
Step 4: Find the total bonus in 2023
First 10 bonus doubles
= 2 × 2,00,000
= Rs. 4,00,000
Middle 20 remains
= Rs. 10,00,000
Last 10 bonus triples
= 3 × 8,00,000
= Rs. 24,00,000
Total bonus
= 4,00,000 + 10,00,000 + 24,00,000
= Rs. 38,00,000
Step 5: Find the average bonus
Average
= 38,00,000/40
= Rs. 95,000
Answer:
A. 95,000
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
21
20
22
19
21
Let the marks of each girl be g and the marks of each boy be b.
The average marks of 4 girls and 6 boys is 24.
So, Total marks = 10 × 24 = 240
Therefore, 4g + 6b = 240 → 2g + 3b = 120
Hence, g = (120 − 3b)/2
The conditions are:
● Each girl's marks are at least the marks of a boy.
● Each girl's marks are at most twice the marks of a boy.
So, b ≤ g ≤ 2b
Substitute g = (120 − 3b)/2.
Using g ≥ b: (120 − 3b)/2 ≥ b → 120 − 3b ≥ 2b → 120 ≥ 5b → b ≤ 24
Using g ≤ 2b: (120 − 3b)/2 ≤ 2b → 120 − 3b ≤ 4b → 120 ≤ 7b → b ≥ 120/7
Hence, 120/7 ≤ b ≤ 24
Now we need the total marks of 2 girls and 6 boys.
Required total = 2g + 6b
Using 2g = 120 − 3b, Required total = (120 − 3b) + 6b = 120 + 3b
Since b can take any real value in the interval 120/7 ≤ b ≤ 24, the required total varies from 120 + 3 × (120/7) = 171.43... to 120 + 3 × 24 = 192
Therefore, the possible integer values are 172, 173, ..., 192
Number of integers = 192 − 172 + 1 = 21
Answer:
A. 21
In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employees is
6:5
4:5
5:4
5:6
4:5
Let total employees = 100.
Manufacturing = 20, Non-manufacturing = 80.
Step 1: Assume total salary
Let total salary = 6 units.
Manufacturing salary = 1 unit, Non-manufacturing salary = 5 units.
Step 2: Find average salaries
Average salary of manufacturing = 1/20
Average salary of non-manufacturing = 5/80 = 1/16
Step 3: Find the required ratio
= (1/20) : (1/16) = 16 : 20 = 4 : 5
The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
45000
50000
54000
40000
54000
Step 1: Total salary = 30 × 60,000 = Rs. 18,00,000
Step 2: After 5% increase in average,
New average = 63,000 → Total increase = 30 × 3,000 = Rs. 90,000
Step 3: This increase comes only from managers' 20% raise.
20% of M = 90,000 → M (total managers' salary) = Rs. 4,50,000
Step 4: Total engineers' salary = 18,00,000 − 4,50,000 = Rs. 13,50,000
Average engineers' salary = 13,50,000 ÷ 25 = Rs. 54,000
− There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is
Step 1: Let the four numbers be a, b, c and d.
First condition: (a + b)/2 = a + 1 → b = a + 2
Step 2: Second condition
(a + b + c)/3 = (a + b)/2 + 2 = a + 3
(2a + 2 + c)/3 = a + 3 → c = a + 7
Step 3: Third condition
(a + b + c + d)/4 = (a + b + c)/3 + 3 = a + 6
(3a + 9 + d)/4 = a + 6 → d = a + 15
Step 4: Find the required difference
The four numbers are a, a+2, a+7, a+15
Difference = (a + 15) − a = 15
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