| 1. The average of the first five integers in ten consecutive integers is n. What is the average of all the ten integers? | Medium |
| A. n+2 B. n+2.5 C. n D. None of these |
View Answer
Answer: Option B
Explanation:
Let the ten integers be (n-2), (n-1), (n), (n+1), (n+2), (n+3), (n+4), (n+5), (n+6), and (n+7), which gives the average of the first five integers as n.
Sum of first five integers = Average*Number = 5n
Average of all 10 integers =(5n+ n+3+ n+4+n+5+n+6+n+7)/10 = (5n+5n+25)/10 = n+ 2.5
| 2. There is a class of 21 students, and the average weight of all the students is 55 kg. If one student departs from the class, causing the average weight of the remaining students to decrease by 0.5 kg, what is the weight of the student who left the class? | Easy |
| A. 45 kg B. 55 kg C. 65 kg D. None of these |
View Answer
Answer: Option C
Explanation:
Total weight = 21*55 = 1155 kg
Let the weight of the student who left the class be x kg
-> (1155 – x)/20 = 55-0.5 = 54.5
-> x = 65 kg
| 3. There is a family with an average weight of 40 kg. When a new baby, weighing 4 kg joins the family, the average weight of the entire family decreases by 15%. What was the number of family members before the baby was born? | Easy |
| A. 3 B. 4 C. 5 D. None of these |
View Answer
Answer: Option C
Explanation:
Let the initial number of family members be x
-> (40x+4)/(x+1) = 40*0.85 = 34
-> x = 5
| 4. Consider an office where the average salary of all employees is Rs. 80000. Among the employees, there are 50 men with an average salary of Rs. 30000 and women with an average salary of Rs. 100000. By what percentage is the number of women more than the number of men? | Medium |
| A. 125% B. 100% C. 150% D. None of these |
View Answer
Answer: Option C
Explanation:
Let the number of women be x
-> [(50)(30000)+(x)(100000)]/(50+x) = 80000
-> x = 125
Difference = 125-50 = 75
Required % = (75/50) * 100 = 150%
| 5. There is a set of 10 scores with an average of 100. However, when we exclude the highest and lowest scores from the set, the average increases to 105. Given that the highest score is 90, what is the value of the lowest score? | Easy |
| A. 50 B. 70 C. 60 D. None of these |
View Answer
Answer: Option B
Explanation:
Let the highest and lowest scores be h and l, respectively
-> Total = 100*10 = 1000
-> [1000-(h+l)]/8 = 105
-> h+l = 1000 – 840 = 160
-> l=160-90 = 70
| 6. Let’s consider three classes: Class I, Class II, and Class III, which took a test. The average score in Class I is 80, in Class II is 75, and in Class III is 90. The average score of all students in Classes I and II combined is 78, while the average score of all students in Classes II and III combined is 85. Determine the average score for all three classes combined. | Medium |
| A. 85.67 B. 83.33 C. 84.54 D. None of these |
View Answer
Answer: Option B
Explanation:
Let the number of students in classes I,II,III be x,y,z respectively
Now, (80x+75y)/(x+y) = 78 -> x:y = 3:2 —– ❶
Also, (75y+90z)/(y+z) = 85 -> y:z = 1:2 —– ❷
From ❶ & ❷, x:y:z = 3:2:4
Required Average = [(3*80)+(2*75)+(4*90)]/(3+2+4) = 83.33
| 7. X and Y establish a partnership, with X contributing the entire capital of Rs. 1000000. The agreement states that the profits will be divided equally between them. Furthermore, Y will pay X interest on the capital at a rate of 10% per annum, and X will provide Y with a monthly payment of Rs. 5000 for managing the business. If Y’s income is one-fifth of X’s income, what is the annual profit? | Medium |
| A. Rs. 100000 B. Rs. 12000 C. Rs. 75000 D. None of these |
View Answer
Answer: Option B
Explanation:
Interest on Rs. 1000000 @ 10% pa = Rs. 100000
Let the yearly profit be Rs. N
Y’s income will be N/2 – 100000 + 60000
X’s income will be N/2 + 100000 – 60000
It is given that Y’s income is one-fifth of X’s income
-> N/2 – 100000 + 60000 = 1/5 (N/2 + 100000 – 60000)
N = 120000
| 8. A and B started a business investing Rs. 300000 and Rs. 120000, respectively. After one year, from the profit of Rs. 300000, B was paid a certain amount for running the business, and the rest was divided proportionately. If A and B received the same amount, how much was B paid for running the business? | Medium |
| A. Rs. 90000 B. Rs. 75000 C. Rs. 20000 D. None of these |
View Answer
Answer: Option A
Explanation:
Money invested in the ratio of 300000: 120000 = 5:2
Let B be paid Rs. x
-> Remaining profit of (300000-x) was split in the ratio of 5:2
-> A’s total profit = 300000/2 = 150000 = 5(300000-x)/7
-> x = 90000
| 9. Consecutive positive integers beginning with 1 are written on a piece of paper. Ralph came along and erased one of these numbers. The average of the remaining numbers is 1270/49. What was the number erased by Ralph, and how many numbers were initially written on the piece of paper? | Difficult |
| A. 5, 50 B. 6, 51 C. 7, 55 D. None of these |
View Answer
Answer: Option A
Explanation:
Let the number of consecutive positive integers be n.
Sum = n(n+1)/2
Average = [n(n+1)/2]/n = (n+1)/2
-> (n+1)/2 ~ 25 (average is close to 25
-> n should be close to 50 —– ❶
Let Ralph erase the number k.
Now if k is subtracted from the sum of the n positive integers, the new average is given as:
[n(n+1)/2 – k]/(n-1) = 1270/49 —– ❷
Comparing L.H.S and R.H.S of ❷, n-1 should be a multiple of 49, as R.H.S cannot be reduced any further —– ❸
From ❶ & ❸, n-1 = 49 -> n = 50
-> [50(51)/2 – k]/49 = 1270/49
k = 5
| 10. The initial average age of a group of n persons is 75. Two individuals, aged between 56 and 52, exit the group. Later, a third person, aged between 80 and 89, joins the group. What is a possible age of the individual who joined the group if the resulting average age of the group is a prime number? | Difficult |
| A. 81 B. 87 C. 86 D. 88 |
View Answer
Answer: Option A
Explanation:
Let the new average age of the group be y, and let the age of the person who joined the group be x. Then,
y = (75n-(56+52)+x)/(n-2+1) = (75n-108+x)/(n-1) = (75(n-1)+(x-33))/(n-1) = 75+(x-33)/(n-1)
Check options.
If x = 81, then
y = 75+ 48/(n-1) For y to be prime, it has to be a natural number.
Therefore, (n-1) has to be a factor of 48
-> n-1 = 2, 3, 4, 6, 8, 12, 16, 24, 48. Factors which yield a value of y which is prime, will decide the age of the person who joined the group. For example, when n-1 = 6, y = 75+(48/6) = 83 (prime). Thus, x=81 is a permissible value.
It can be noted that all other values of x in the options will not yield any prime value of y. For example, when x= 86, y = 75+(53/n-1), 53 has only two factors – 1 and 53, Therefore, n-1 = 1 or 53, but these values are not possible as they don’t generate prime values of y.
Similarly, when x = 87, y = 75+54/(n-1), Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54 and none of them generates a prime value of y.
When x = 88, y = 75+55/(n-1), Factors of 55 = 1, 5, 11 and none of them generates a prime value of y.
Thus, the only eligible value from the options is 81.