Q1: The average weight of a class of 20 students is 45 kgs. A new student whose weight is 40 kgs replaces an old student of this class. Hence, the average weight of the whole class decreases by1 kg. The weight of the replaced student is :
A. 55 kgs
B. 50 kgs
C. 60 kgs
D. None of these
Correct Answer: C
Solution: Let the total weight of the 20 students be $T$. The average weight is 45 kgs, so $T = 20 \times 45 = 900$ kgs. A new student of 40 kgs replaces an old student. Let the weight of the replaced student be $x$ kgs. The new total weight is $900 - x + 40 = 940 - x$. The new average weight is $45 - 1 = 44$ kgs. The number of students remains 20. So, $\frac{940 - x}{20} = 44$
$940 - x = 880$
$x = 940 - 880$
$x = 60$
The weight of the replaced student is 60 kgs.
Q2: The average length of first 3 fingers is 3 inches and the average length of the other 2 fingers i. e., thumb and the index finger is 2.8 inches. If the length of the index fingers is 3 inches then the length of the thumb is :
A. 2 inches
B. 2.6 inches
C. 3 inches
D. None of these
Correct Answer: B
Solution: Let the lengths of the first three fingers be $f_1$, $f_2$, $f_3$. Their average length is 3 inches, so $\frac{f_1 + f_2 + f_3}{3} = 3$. This means $f_1 + f_2 + f_3 = 9$ inches. The average length of the thumb ($T$) and index finger ($I$) is 2.8 inches. We are given that the index finger length is 3 inches ($I = 3$). Therefore, $\frac{T + 3}{2} = 2.8$. Solving for $T$: $T + 3 = 5.6 \implies T = 5.6 - 3 = 2.6$ inches. Therefore, the length of the thumb is 2.6 inches.
Q3: Three types of rice whose rates are Rs. 38, Rs. 43 and Rs. 49 per kg are blended together to make a 15 kg of new blend of rice in which there are 8 kgs, 4 kgs 3 kgs of the respective types of rice. The average price of the new blend of rice is :
A. 41.53
B. 43
C. 40
D. 43.3
Correct Answer: A
Solution: To find the average price of the new blend, we need to calculate the weighted average. The weights are the quantities of each type of rice. Total cost of rice type 1: $8 \text{ kg} \times \text{Rs. } \frac{38}{\text{kg}} = \text{Rs. } 304$
Total cost of rice type 2: $4 \text{ kg} \times \text{Rs. } \frac{43}{\text{kg}} = \text{Rs. } 172$
Total cost of rice type 3: $3 \text{ kg} \times \text{Rs. } \frac{49}{\text{kg}} = \text{Rs. } 147$
Total cost of the blend: $\text{Rs. } 304 + \text{Rs. } 172 + \text{Rs. } 147 = \text{Rs. } 623$
Total weight of the blend: $8 \text{ kg} + 4 \text{ kg} + 3 \text{ kg} = 15 \text{ kg}$
Average price of the blend: $\text{Rs. } \frac{623}{15 \text{ kg}} \approx \text{Rs. } 41.533...$ Rounding to two decimal places, we get Rs. 41.53
Q4: Pankaj went to the post-office at the speed of 60 km/hr while returning for his home he covered the half of the distance at the speed of 10 km/hr, but suddenly he realized that he was getting late so he increased the speed and reached the home by covering rest half of the distance at the speed of 30 km/hr. The average speed of the Pankaj in the whole length of journey is :
A. 5.67 km/hr
B. 24 km/hr
C. 22.88 km/hr
D. 5.45 km/hr
Correct Answer: B
Solution: Let the distance to the post office be $D$ km. Time taken to go to the post office = $\frac{D}{60}$ hours. Time taken to cover the first half of the return journey ($\frac{D}{2}$) at 10 km/hr = $\frac{D/2}{10} = \frac{D}{20}$ hours. Time taken to cover the second half of the return journey ($\frac{D}{2}$) at 30 km/hr = $\frac{D/2}{30} = \frac{D}{60}$ hours. Total distance = $D + D = 2D$ km. Total time taken = $\frac{D}{60} + \frac{D}{20} + \frac{D}{60} = \frac{D + 3D + D}{60} = \frac{5D}{60} = \frac{D}{12}$ hours. Average speed = Total distance / Total time = $\frac{2D}{\frac{D}{12}} = 2D \cdot \frac{12}{D} = 24$ km/hr.
Q5: 123 typists typed 984 papers in 1/15 hour. The number of papers typed per minute by an average typist is :
A. 1
B. 2
C. 3
D. 5
Correct Answer: B
Solution: First, find the total number of papers typed per hour: $984 \text{ papers} \times \frac{15}{1} \text{ hours} = 14760 \text{ papers/hour}$. Next, find the total number of papers typed per minute: $14760 \text{ papers/hour} \times \frac{1 \text{ hour}}{60 \text{ minutes}} = 246 \text{ papers/minute}$. Then, divide the total papers typed per minute by the number of typists to find the number of papers typed per minute by an average typist: $\frac{246 \text{ papers/minute}}{123 \text{ typists}} = 2 \text{ papers/minute}$. Therefore, the number of papers typed per minute by an average typist is 2.
Q6: The cost of the Red, Green and Blue colours per kg is Rs. 20, Rs. 15 and Rs. 18 respectively. Rang Mahal is a renowned building in which these three colours are being used in the ratio of 3 : 2 : 4.The average cost of all the three colours used per kg is :
A. 18
B. 20
C. 17.66
D. Cannot be determined
Correct Answer: A
Solution: Let the cost of Red, Green and Blue colours be $R$, $G$, and $B$ respectively. $R = \text{Rs. } 20/\text{kg}$
$G = \text{Rs. } 15/\text{kg}$
$B = \text{Rs. } 18/\text{kg}$
The ratio of the colours used is 3:2:4. Let the quantities of Red, Green and Blue colours be $3x$, $2x$, and $4x$ kg respectively. Total cost of Red colour = $20 \times 3x = 60x$
Total cost of Green colour = $15 \times 2x = 30x$
Total cost of Blue colour = $18 \times 4x = 72x$
Total cost of all three colours = $60x + 30x + 72x = 162x$
Total quantity of all three colours = $3x + 2x + 4x = 9x$
Average cost = $\frac{\text{Total cost}}{\text{Total quantity}} = \frac{162x}{9x} = \text{Rs. } 18/\text{kg}$
Q7: The average of 9 numbers is 11. If each of these 9 numbers is multiplied by 5 and then 5 is added to each of these resultant numbers, then the new average is :
A. 20
B. 30
C. 60
D. 50
Correct Answer: C
Solution: Let the 9 numbers be $n_1, n_2, \dots, n_9$. Their average is 11, so $\frac{n_1 + n_2 + \dots + n_9}{9} = 11$. This means the sum of the numbers is $9 \times 11 = 99$. Each number is multiplied by 5, resulting in $5n_1, 5n_2, \dots, 5n_9$. Then 5 is added to each, giving $5n_1 + 5, 5n_2 + 5, \dots, 5n_9 + 5$. The sum of these new numbers is $(5n_1 + 5) + (5n_2 + 5) + \dots + (5n_9 + 5) = 5(n_1 + n_2 + \dots + n_9) + 9 \times 5 = 5(99) + 45 = 495 + 45 = 540$. The new average is the sum of the new numbers divided by 9: $\frac{540}{9} = 60$.
Q8: The average score of Dhoni after 48 innings is 48 and in the 49th innings Dhoni scores 97 runs. In the 50th innings the minimum number of runs required to increase his average score by 2 than it was before the 50th innings :
A. 99
B. 149
C. 151
D. Cannot be determined
Correct Answer: B
Solution: Let the total score of Dhoni after 48 innings be $T$. Average score after 48 innings = 48
$\frac{T}{48} = 48$
$T = 48 \times 48 = 2304$
After 49 innings, the total score is $2304 + 97 = 2401$
Average score after 49 innings = $\frac{2401}{49} = 49$
Let $x$ be the score in the 50th innings. The average score after 50 innings should be $49 + 2 = 51$
$$ \frac{2401 + x}{50} = 51 $$
$$ 2401 + x = 51 \times 50 $$
$$ 2401 + x = 2550 $$
$$ x = 2550 - 2401 $$
$$ x = 149 $$
Therefore, the minimum number of runs required in the 50th innings is 149.
Q9: The average age of Sachin and Ganguli is 35 years. If Kaif replaces Sachin, the average age becomes 32 years and if Kaif replaces Ganguli, then the average age becomes 38 years. If the average age of Dhoni and Irfan be half of the average age of Sachin, Ganguli and Kaif, then the average age of all the five people is :
A. 28
B. 32
C. 25
D. None of these
Correct Answer: A
Solution: Let $S$, $G$, and $K$ represent the ages of Sachin, Ganguli, and Kaif respectively. The average age of Sachin and Ganguli is 35 years. Therefore, $\frac{S + G}{2} = 35$, which implies $S + G = 70$. If Kaif replaces Sachin, the average age becomes 32 years: $\frac{K + G}{2} = 32$, implying $K + G = 64$. If Kaif replaces Ganguli, the average age becomes 38 years: $\frac{S + K}{2} = 38$, implying $S + K = 76$. We have a system of three equations:
1) $S + G = 70$
2) $K + G = 64$
3) $S + K = 76$
Subtracting equation (2) from equation (1): $S - K = 6$
Adding this to equation (3): $2S = 82$, so $S = 41$
Substituting $S = 41$ into equation (1): $41 + G = 70$, so $G = 29$
Substituting $S = 41$ into equation (3): $41 + K = 76$, so $K = 35$
The average age of Sachin, Ganguli, and Kaif is $\frac{41 + 29 + 35}{3} = \frac{105}{3} = 35$ years. The average age of Dhoni and Irfan is half of this, which is $\frac{35}{2} = 17.5$ years. Let $D$ and $I$ be the ages of Dhoni and Irfan. Then $\frac{D + I}{2} = 17.5$, implying $D + I = 35$. The average age of all five people is $\frac{41 + 29 + 35 + D + I}{5} = \frac{105 + 35}{5} = \frac{140}{5} = 28$ years.
Q10: Out of these five people (in question no. 11) whose age is the greatest?
A. Sachin
B. Ganguli
C. Kaif
D. Cannot be determined
Correct Answer: A
Solution: This question requires referencing a previous question (question no. 11) which provided the ages of five people. Without the data from question 11, this question cannot be answered. The solution involves comparing the ages given in question 11 and identifying the highest value. Therefore, the answer depends entirely on the data presented in question 11. This question tests the ability to interpret and analyze data presented earlier in the exam.
Q11: In a village the average age of n people is 42 years. But after the verification it was found that the age of a person had been considered 20 years less than the actual age, so the new average, after the correction, increased by 1. The value of n is:
A. 21
B. 20
C. 22
D. None of these
Correct Answer: B
Solution: Let the sum of ages of $n$ people be $S$. The average age is 42 years, so $\frac{S}{n} = 42$. Therefore, $S = 42n$. The age of one person was considered 20 years less than the actual age. This means the initial sum $S$ was 20 less than the correct sum. The correct sum is $S + 20$. The new average is $42 + 1 = 43$ years. So, $\frac{S + 20}{n} = 43$. Substitute $S = 42n$:
$\frac{42n + 20}{n} = 43$
$42n + 20 = 43n$
$n = 20$
Therefore, the value of $n$ is 20.
Q12: The average rainfall in the months of January and February is 6 cm and in the months of March to June is 5 cm and July to October is 10 cm and in the November and December, it is 6 cm. The average rainfall for the whole year is :
A. 7
B. 5.5
C. 7.5
D. None of these
Correct Answer: A
Solution: Let's denote the total rainfall in each period as follows:
January & February: Average rainfall = $6$ cm, Number of months = $2$, Total rainfall = $6 \times 2 = 12$ cm
March to June: Average rainfall = $5$ cm, Number of months = $4$, Total rainfall = $5 \times 4 = 20$ cm
July to October: Average rainfall = $10$ cm, Number of months = $4$, Total rainfall = $10 \times 4 = 40$ cm
November & December: Average rainfall = $6$ cm, Number of months = $2$, Total rainfall = $6 \times 2 = 12$ cm
$$ \text{Total rainfall for the year} = 12 + 20 + 40 + 12 = 84 \text{ cm} $$
$$ \text{Total number of months in a year} = 12 \text{ months} $$
$$ \text{Average rainfall for the whole year} = \frac{\text{Total rainfall}}{\text{Total number of months}} = \frac{84 \text{ cm}}{12 \text{ months}} = 7 \text{ cm} $$
Q13: On an average 300 people watch the movie in Sahu Cinema hall on Monday, Tuesday and Wednesday and the average number of visitors on Thursday and Friday is 250. If the average number of visitors per day in the week be 400, then the average number of people who watch the movie in weekends (i. e., on Saturday and Sunday) is :
A. 500
B. 600
C. 700
D. None of these
Correct Answer: C
Solution: Let's denote the number of people watching the movie on Monday, Tuesday, and Wednesday as $M$, $T$, and $W$ respectively. The average for these three days is 300, so $\frac{M + T + W}{3} = 300$. This means $M + T + W = 900$. The average number of visitors on Thursday and Friday is 250, so let's denote the number of visitors on Thursday and Friday as $Th$ and $F$ respectively. Then $\frac{Th + F}{2} = 250$, which means $Th + F = 500$. The average number of visitors per day for the entire week is 400. The total number of visitors for the week is $400 \times 7 = 2800$. Let's denote the number of visitors on Saturday and Sunday as $S$ and $Su$ respectively. The total number of visitors for the week can be expressed as: $M + T + W + Th + F + S + Su = 2800$. We know $M + T + W = 900$ and $Th + F = 500$. Substituting these values, we get: $900 + 500 + S + Su = 2800$. Therefore, $S + Su = 2800 - 1400 = 1400$. The average number of people who watch the movie on the weekend is $\frac{S + Su}{2} = \frac{1400}{2} = 700$.
Q14: The average weight of 11 players of Indian cricket team is increased by 1 kg, when one player of the team weighing 55 kg replaced by a new player. The weight of the new player is :
A. 55
B. 64
C. 66
D. None of these
Correct Answer: C
Solution: Let the sum of weights of the original 11 players be $S$ kg. The average weight of the 11 players is $\frac{S}{11}$ kg. When one player weighing 55 kg is replaced by a new player, the sum of weights becomes $S - 55 + x$, where $x$ is the weight of the new player. The new average weight is $\frac{S - 55 + x}{11}$. The new average is 1 kg more than the original average. Therefore:
$$ \frac{S - 55 + x}{11} = \frac{S}{11} + 1 $$
$$ S - 55 + x = S + 11 $$
$$ x = 11 + 55 $$
$$ x = 66 $$
The weight of the new player is 66 kg.
Q15: The average age of a family of 6 members 4 year ago was 25 years. Mean while a child was bom in this family and still the average age of the whole family is same today. The present age of the child is :
A. 2
B. 1.5
C. 1
D. Data Inadequate
Correct Answer: C
Solution: Let the sum of ages of the 6 family members 4 years ago be $S$. The average age 4 years ago was 25, so $\frac{S}{6} = 25$ which means $S = 150$. Four years later, the sum of their ages will be $S + 6*4 = 150 + 24 = 174$. A child was born, so now there are 7 members in the family. Let the present age of the child be $x$. The sum of the ages of the 7 family members today is $174 + x$. The average age today is still 25, so $\frac{174 + x}{7} = 25$. Solving for $x$: $174 + x = 25 * 7 = 175$
$x = 175 - 174 = 1$
Therefore, the present age of the child is 1 year.
Q16: Amitabh’s average expenditure for the January to June is Rs. 4200 and he spends Rs. 1200 in January and Rs 1500 in July. The average expenditure for the months of February to July is :
A. 4250
B. 4520
C. 4060
D. None of these
Correct Answer: A
Solution: Let's denote the total expenditure from January to June as $T_{Jan-Jun}$. The average expenditure for January to June is Rs. 4200. Since there are 6 months, the total expenditure is:
$$T_{Jan-Jun} = 4200 \times 6 = 25200$$
Amitabh's expenditure in January is Rs. 1200. Therefore, the total expenditure from February to June is:
$$T_{Feb-Jun} = T_{Jan-Jun} - 1200 = 25200 - 1200 = 24000$$
In July, he spends Rs. 1500. The total expenditure from February to July is:
$$T_{Feb-Jul} = T_{Feb-Jun} + 1500 = 24000 + 1500 = 25500$$
There are 6 months from February to July. Therefore, the average expenditure from February to July is:
$$Average_{Feb-Jul} = \frac{T_{Feb-Jul}}{6} = \frac{25500}{6} = 4250$$
Therefore, the average expenditure for the months of February to July is Rs. 4250.
Q17: The average salary is being paid to all its employees by the Biotech corporation is Rs. 15,500. The average salary of the senior employees is Rs. 18000 per month and the average salary of the junior employees is Rs. 12,000 per month. If there are only two levels of employees viz junior and senior level, then what fraction of the total employees is the junior level employees are :
A. 7 / 10
B. 5 / 12
C. 5 / 10
D. None of these
Correct Answer: B
Solution: Let's use the concept of alligation to solve this problem. Let $x$ be the fraction of senior employees and $1-x$ be the fraction of junior employees. The weighted average salary is given by:
$$15500 = 18000x + 12000(1-x)$$
Solving for $x$:
$$15500 = 18000x + 12000 - 12000x$$
$$3500 = 6000x$$
$$x = \frac{3500}{6000} = \frac{7}{12}$$
Therefore, the fraction of senior employees is $\frac{7}{12}$. The fraction of junior employees is $1 - \frac{7}{12} = \frac{5}{12}$.
Q18: The average of any 5 consecutive odd numbers a, b, c, d, e is:
A. abcde/5
B. bd/3
C. (a+c+e)/5
D. None of these
Correct Answer: D
Solution: Let the five consecutive odd numbers be $n$, $n+2$, $n+4$, $n+6$, $n+8$. Their sum is $5n + 20$. The average is the sum divided by the number of terms, which is $\frac{5n + 20}{5} = n + 4$. Alternatively, the average of consecutive numbers is the middle number. In a sequence of 5 consecutive odd numbers, the middle number is the third one, which is $c$. Therefore, the average is $c$.
Q19: The average age of 30 students of a class is 30 years. When the average age of class teacher is also included, the average age of the whole class increases by 1 year. The age of the class teacher is :
A. 31
B. 60
C. 61
D. None of these
Correct Answer: C
Solution: The total age of 30 students is $30 \text{ students} \times 30 \frac{\text{years}}{\text{student}} = 900 \text{ years}$. When the teacher's age is included, the total number of people becomes 31. The new average age is $30 \text{ years} + 1 \text{ year} = 31 \text{ years}$. The total age of 31 people (30 students + 1 teacher) is $31 \text{ people} \times 31 \frac{\text{years}}{\text{person}} = 961 \text{ years}$. The teacher's age is the difference between the new total age and the old total age: $961 \text{ years} - 900 \text{ years} = 61 \text{ years}$.
Q20: There were five sections in MAT paper. The average score of Pooja in first 3 sections was 83 and the average in the last 3 sections was 97 and the average of all the sections (i. e., whole paper) was 92, then her score in the third section was:
A. 85
B. 92
C. 88
D. None of these
Correct Answer: D
Solution: Let $S_1, S_2, S_3, S_4, S_5$ be the scores in the five sections. Average of first 3 sections = $\frac{S_1 + S_2 + S_3}{3} = 83$
$S_1 + S_2 + S_3 = 3 \times 83 = 249$ (Equation 1)
Average of last 3 sections = $\frac{S_3 + S_4 + S_5}{3} = 97$
$S_3 + S_4 + S_5 = 3 \times 97 = 291$ (Equation 2)
Average of all 5 sections = $\frac{S_1 + S_2 + S_3 + S_4 + S_5}{5} = 92$
$S_1 + S_2 + S_3 + S_4 + S_5 = 5 \times 92 = 460$ (Equation 3)
Substitute Equation 1 into Equation 3:
$249 + S_4 + S_5 = 460$
$S_4 + S_5 = 460 - 249 = 211$ (Equation 4)
Substitute Equation 4 into Equation 2:
$S_3 + 211 = 291$
$S_3 = 291 - 211 = 80$
Therefore, Pooja's score in the third section was 80.
Q21: The average age of 18 pupils of Dronacharya was 25 years. If the age of Dronacharya was also included, the average age of 19 people becomes 26 years. The average age of the Dronacharya at that time was :
A. 33
B. 44
C. 50
D. 51
Correct Answer: B
Solution: Let the sum of the ages of 18 pupils be $S$. The average age of 18 pupils is 25 years. Therefore, $S = 18 \times 25 = 450$ years. When Dronacharya's age is included, the total number of people becomes 19, and the average age is 26 years. The sum of the ages of 19 people is $19 \times 26 = 494$ years. Dronacharya's age = (Sum of ages of 19 people) - (Sum of ages of 18 pupils)
Dronacharya's age = $494 - 450 = 44$ years.
Q22: The average of 7 consecutive odd numbers if the smallest of those numbers is denoted by k :
A. k + 4
B. k + 7
C. k + 6
D. 7k
Correct Answer: C
Solution: Let the 7 consecutive odd numbers be $k$, $k+2$, $k+4$, $k+6$, $k+8$, $k+10$, $k+12$. The sum of these numbers is $7k + 42$. The average of these numbers is $\frac{7k + 42}{7} = k + 6$. Therefore, the average of 7 consecutive odd numbers is $k+6$, where $k$ is the smallest number.
Q23: If the average marks of 1/4th class is 85% and that of 1/3rd class is 70% and the average marks of the rest class is 56%, then the average of the whole class is (for the given subjects) :
A. 67.92%
B. 72.33%
C. 69.16%
D. Cannot be determined
Correct Answer: A
Solution: Let the total number of students in the class be $x$. $\frac{1}{4}$th of the class = $\frac{x}{4}$ students, whose average marks are 85%. $\frac{1}{3}$rd of the class = $\frac{x}{3}$ students, whose average marks are 70%. The rest of the class = $x - \frac{x}{4} - \frac{x}{3} = \frac{(12x - 3x - 4x)}{12} = \frac{5x}{12}$ students, whose average marks are 56%. Total marks of $\frac{1}{4}$th class = $(\frac{x}{4}) * 85 = \frac{85x}{4}$
Total marks of $\frac{1}{3}$rd class = $(\frac{x}{3}) * 70 = \frac{70x}{3}$
Total marks of the rest of the class = $(\frac{5x}{12}) * 56 = \frac{280x}{12} = \frac{70x}{3}$
Total marks of the whole class = $\frac{85x}{4} + \frac{70x}{3} + \frac{70x}{3} = \frac{(255x + 280x + 280x)}{12} = \frac{815x}{12}$
Average marks of the whole class = $\frac{\text{Total marks}}{\text{Total students}} = \frac{\frac{815x}{12}}{x} = \frac{815}{12} \approx 67.92\%$
Q24: The average length of any four fingers of my left hand is 600 mm. Then the average length of all the five fingers of my left hand is :
A. 800 mm
B. 750 mm
C. 480 mm
D. Cannot be determined
Correct Answer: B
Solution: Let the sum of the lengths of any four fingers be $S$. The average length of these four fingers is 600 mm, so $\frac{S}{4} = 600$ mm. Therefore, $S = 2400$ mm. Let the length of the fifth finger be $x$ mm. The sum of the lengths of all five fingers is $S + x = 2400 + x$ mm. The average length of all five fingers is $\frac{2400 + x}{5}$ mm. We don't have the length of the fifth finger, so we cannot find the exact average. However, the question is flawed as it implies a single definitive answer exists without providing the length of the fifth finger. The question needs additional information to be solved. If we assume that the length of all five fingers are equal then we have: Average of 5 fingers = (Average of 4 fingers) $\times \frac{4}{5} = 600$ mm $\times \frac{4}{5} = 480$ mm. However, this is an incorrect approach because the average length of 4 fingers is given, and it doesn't necessitate the fifth finger to also have the same length.
Q25: The average of 7 consecutive numbers which are positive integers is 10. The average of lowest and highest such numbers is :
A. 7
B. 10
C. 15
D. Data Inadequate
Correct Answer: B
Solution: Let the 7 consecutive positive integers be $n$, $n+1$, $n+2$, $n+3$, $n+4$, $n+5$, $n+6$. Their average is given as 10. Therefore, the sum of these numbers is $7 \times 10 = 70$. The sum can also be expressed as $7n + 21$. Equating the two expressions for the sum: $7n + 21 = 70$. Solving for $n$: $7n = 49$, $n = 7$. The lowest number is 7 and the highest number is $7 + 6 = 13$. The average of the lowest and highest numbers is $\frac{7 + 13}{2} = 10$.
Q26: The average of first 100 natural number is :
A. 100
B. 50
C. 50.5
D. 55
Correct Answer: C
Solution: The average of consecutive numbers from 1 to $n$ is given by the formula $\frac{n+1}{2}$. In this case, $n = 100$. Therefore, the average of the first 100 natural numbers is $\frac{100+1}{2} = \frac{101}{2} = 50.5$.
Q27: The average of first 50 odd natural numbers is :
A. 50
B. 55
C. 51
D. 101
Correct Answer: A
Solution: The first 50 odd natural numbers are 1, 3, 5, ..., 99. This is an arithmetic progression with the first term $a = 1$, the common difference $d = 2$, and the number of terms $n = 50$. The sum of an arithmetic progression is given by $S = \frac{n}{2} \times [2a + (n-1)d]$. Therefore, the sum of the first 50 odd numbers is $S = \frac{50}{2} \times [2(1) + (50-1)(2)] = 25 \times [2 + 98] = 25 \times 100 = 2500$. The average is the sum divided by the number of terms: Average $= \frac{S}{n} = \frac{2500}{50} = 50$. Alternatively, the average of consecutive odd numbers is always the middle number or the average of the middle two numbers if there are an even number of terms. In this case, the average of the 25th and 26th odd number is $\frac{49+51}{2} = 50$.
Q28: The average of first 99 even numbers is :
A. 9999
B. 100
C. 9801
D. 9009
Correct Answer: B
Solution: The first 99 even numbers are 2, 4, 6, ..., 198. This is an arithmetic sequence with first term $a = 2$, last term $l = 198$, and number of terms $n = 99$. The sum of an arithmetic sequence is given by $S = \frac{n}{2} (a + l)$. Therefore, the sum of the first 99 even numbers is $S = \frac{99}{2} (2 + 198) = \frac{99}{2} \times 200 = 99 \times 100 = 9900$. The average is the sum divided by the number of terms: Average $= \frac{S}{n} = \frac{9900}{99} = 100$. Alternatively, recognizing that the average of an arithmetic sequence is the average of the first and last terms, we can calculate $\frac{2 + 198}{2} = \frac{200}{2} = 100$.
Q29: The average of a, b and c is 79 and the average of a and c is also 79. Then the value of b is :
A. 0
B. 79
C. -79
D. None of these
Correct Answer: B
Solution: Let $a$, $b$, and $c$ be three numbers. The average of $a$, $b$, and $c$ is given as 79. Therefore, $\frac{a + b + c}{3} = 79$. This implies $a + b + c = 79 \times 3 = 237$. The average of $a$ and $c$ is also given as 79. Therefore, $\frac{a + c}{2} = 79$. This implies $a + c = 79 \times 2 = 158$. Substituting $a + c = 158$ into the equation $a + b + c = 237$, we get:
$158 + b = 237$
$b = 237 - 158$
$b = 79$
Therefore, the value of $b$ is 79.
Q30: The average value of property of Mittal, Ambani and Singhania is Rs. 11111 crore. The property of Singhania is as less as the property of Mittal is greater than the average property of both the Singhania and Mittal. The value of property of Ambani is :
A. 111 crore
B. 11111 crore
C. 3703.7 crore
D. Cannot be determined
Correct Answer: B
Solution: Let $M$, $A$, and $S$ represent the property values of Mittal, Ambani, and Singhania, respectively. The average of their property values is 11111 crore, so $\frac{M + A + S}{3} = 11111$. This means $M + A + S = 33333$. Let $x$ be the difference between Mittal's property and the average of Mittal and Singhania's properties. Then $M = \frac{M+S}{2} + x$, and $S = \frac{M+S}{2} - x$. The problem states that Singhania's property is as less as Mittal's property is greater than the average property of both Singhania and Mittal. This means the difference between Mittal's property and the average of Mittal and Singhania's properties is the same as the difference between Singhania's property and the average of Mittal and Singhania's properties, but in the opposite direction. This is represented by the '$x$' above. We have $M = \frac{M+S}{2} + x$ and $S = \frac{M+S}{2} - x$. Adding these equations gives $M + S = M + S$, which doesn't help us directly. However, we know that $M + A + S = 33333$. Since Singhania's property is less than the average of Mittal and Singhania's properties by $x$, and Mittal's property is greater than the average of Mittal and Singhania's properties by $x$, this means that Mittal's and Singhania's properties are symmetrically distributed around their average. Their average is therefore the same as their average. Let the average of Mittal and Singhania be $Avg(M,S) = \frac{M+S}{2}$. $M = Avg(M,S) + x$
$S = Avg(M,S) - x$
$M + S = 2 \times Avg(M,S)$
Substituting this into the total property equation:
$2 \times Avg(M,S) + A = 33333$
Since the average property is 11111, $Avg(M,S)$ could vary as long as the sum is correct. However, the crucial information is that the properties of Mittal and Singhania are equidistant from their average. Therefore, their average is also 11111. $2 \times 11111 + A = 33333$
$A = 33333 - 22222$
$A = 11111$
Q31: I went to Delhi @ speed of 200 km/hr but suddenly I returned to the samejdace @ speed of 600 km/hr. What is my average speed :
A. 300 km/hr
B. 400 km/hr
C. 366.66 km/hr
D. None of these
Correct Answer: A
Solution: Let the distance to Delhi be $d$ km. Time taken to go to Delhi = $\frac{d}{200}$ hours
Time taken to return from Delhi = $\frac{d}{600}$ hours
Total distance = $d + d = 2d$ km
Total time = $\frac{d}{200} + \frac{d}{600} = \frac{3d + d}{600} = \frac{4d}{600} = \frac{d}{150}$ hours
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{\frac{d}{150}} = 2d \times \frac{150}{d} = 300$ km/hr
Q32: The average of A and B is 400 and the average of C and D is 600 the average of A, B, C and D is :
A. 500
B. 450
C. 525
D. 625
Correct Answer: A
Solution: Let $A$ and $B$ be two numbers. Their average is given as 400. Therefore, $\frac{A + B}{2} = 400$, which implies $A + B = 800$. Let $C$ and $D$ be two numbers. Their average is given as 600. Therefore, $\frac{C + D}{2} = 600$, which implies $C + D = 1200$. To find the average of $A$, $B$, $C$, and $D$, we add all the numbers and divide by 4:
$$ \frac{A + B + C + D}{4} = \frac{800 + 1200}{4} = \frac{2000}{4} = 500 $$
Therefore, the average of $A$, $B$, $C$, and $D$ is 500.
Q33: The average score of Sehwag in 10 innings was 77 runs. In the 11th innings he had scored zero runs. The overall average score of Sehwag in all the 11 innings was :
A. 77
B. 7.7
C. 11
D. None of these
Correct Answer: D
Solution: The total score in the first 10 innings is $77 \text{ runs/inning} \times 10 \text{ innings} = 770 \text{ runs}$. In the 11th innings, he scored $0$ runs. The total score in all 11 innings is $770 \text{ runs} + 0 \text{ runs} = 770 \text{ runs}$. The overall average score in 11 innings is $\frac{770 \text{ runs}}{11 \text{ innings}} = 70 \text{ runs/inning}$.
Q34: The average age of 3 children of Arihant Singh is 12 years and their ratio of ages is 3:4: 5. The average age of the youngest and eldest child is if he had only 3 children :
A. 12
B. 21
C. 8
D. 9
Correct Answer: A
Solution: Let the ages of the three children be $3x$, $4x$, and $5x$. Their average age is given as 12 years. Therefore, $\frac{3x + 4x + 5x}{3} = 12$. This simplifies to $\frac{12x}{3} = 12$, which means $4x = 12$, and $x = 3$. The ages of the children are 9, 12, and 15 years. The average age of the youngest (9) and eldest (15) child is $\frac{9 + 15}{2} = 12$ years.
Q35: The average income of all the Infosys employees is Rs. 20,000 per month. Recently the company announced the increment of Rs. 2,000 per month for all the employees. The new average of all the employees is :
A. 22000
B. 40000
C. 2200
D. Data Inadequate
Correct Answer: A
Solution: The initial average income of Infosys employees is Rs. $20,000$ per month. Each employee receives an increment of Rs. $2,000$. Since the increment is applied to every employee, the average income will also increase by the same amount. Therefore, the new average income will be the initial average plus the increment: $$20,000 + 2,000 = 22,000$$
Q36: The average age of 10 students in a class is 20 years, if a new student is also included, then the new average age of all the students increases by 1 year. The age of the new student is :
A. 21
B. 30
C. 31
D. None of these
Correct Answer: C
Solution: Let the sum of ages of 10 students be $S$. The average age is 20 years, so $\frac{S}{10} = 20$, which means $S = 200$ years. When a new student (let's say with age $x$) joins, the total number of students becomes 11. The new average age is $20 + 1 = 21$ years. Therefore, $\frac{S + x}{11} = 21$. Substituting $S = 200$, we get $\frac{200 + x}{11} = 21$. Multiplying both sides by 11, we have $200 + x = 231$. Solving for $x$, we get $x = 231 - 200 = 31$ years. The age of the new student is 31 years.
Q37: The average weight of 20, four wheelers is 180 kg. If an old car is removed from this group of four wheelers, the new average weight decreases by 2 kg. The weight of the removed car is :
A. 220
B. 218
C. 182
D. None of these
Correct Answer: B
Solution: Let the total weight of 20 four-wheelers be $T$ kg. The average weight of 20 four-wheelers is 180 kg. Therefore, $T = 20 \times 180 = 3600$ kg. When an old car is removed, the number of four-wheelers becomes 19. The new average weight is $180 - 2 = 178$ kg. The total weight of the remaining 19 four-wheelers is $19 \times 178 = 3382$ kg. The weight of the removed car is the difference between the total weight of 20 four-wheelers and the total weight of the remaining 19 four-wheelers. Weight of removed car = $3600 - 3382 = 218$ kg.
Q38: The average price of 3 diamonds of same weights is Rs. 5 crore, where the average price of the two cosdiest diamonds is double the price of the cheapest diamond. The price of the cheapest diamond is :
A. 3 crore
B. 5 crore
C. 1.66 crore
D. Cannot be determined
Correct Answer: A
Solution: Let the prices of the three diamonds be $a$, $b$, and $c$, where $a \le b \le c$. The average price is Rs. 5 crore, so $\frac{a + b + c}{3} = 5 \text{ crore}$. This means $a + b + c = 15 \text{ crore}$. We are given that the average price of the two costliest diamonds is double the price of the cheapest diamond. Therefore, $\frac{b + c}{2} = 2a$. This simplifies to $b + c = 4a$. Substituting this into the first equation, we get $a + 4a = 15 \text{ crore}$, which simplifies to $5a = 15 \text{ crore}$. Solving for $a$, we find $a = 3 \text{ crore}$. Therefore, the price of the cheapest diamond is Rs. 3 crore.
Q39: In the previous question, the price of the costliest diamond is:
A. 5 crore
B. 6 crore
C. 8 crore
D. Cannot be determined
Correct Answer: D
Solution: This question cannot be answered without knowing the context of the "previous question." Data sufficiency questions require evaluating whether given statements provide enough information to answer a target question. The question itself does not provide any statements. Therefore, it's impossible to determine if any information is sufficient to find the price of the costliest diamond. The question is inherently flawed as a data sufficiency problem because it relies on external, unprovided data.
Q40: The average of all the prime and composite numbers upto 100 is :
A. 51
B. 49.5
C. 50.5
D. 55
Correct Answer: A
Solution: First, identify all prime numbers up to 100. These are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. There are 25 prime numbers. Next, identify all composite numbers up to 100. Composite numbers are whole numbers greater than 1 that are not prime. To find them, subtract the prime numbers and 1 from the total numbers (100). Therefore, there are $100 - 25 - 1 = 74$ composite numbers. The sum of integers from 1 to 100 is given by the formula $\frac{n(n+1)}{2}$ where $n=100$. This sum is $\frac{100(101)}{2} = 5050$. However, we need to exclude 1 which is neither prime nor composite. Thus the sum is still 5050. Total numbers = $25 + 74 = 99$
The average is the total sum divided by the total numbers (excluding 1) = $\frac{5050}{99} = 51$
Q41: The average of all the perfect squares upto 100 is :
A. 38.5
B. 1000
C. 100
D. 385
Correct Answer: A
Solution: The perfect squares up to 100 are $1, 4, 9, 16, 25, 36, 49, 64, 81$, and $100$. There are $10$ such numbers. To find the average, we sum these numbers and divide by the count ($10$). The sum is $1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385$. The average is $\frac{385}{10} = 38.5$.
Q42: The average of all the non-negative integers upto 99 is :
A. 50.49
B. 49.5
C. 50.5
D. 99
Correct Answer: B
Solution: The non-negative integers up to 99 are $0, 1, 2, \dots, 99$. This is an arithmetic sequence. The average of an arithmetic sequence is the average of the first and last terms. Therefore, the average is $\frac{0 + 99}{2} = \frac{99}{2} = 49.5$.
Q43: The average of 7, 14, 21, 28 ... 77 is :
A. 7
B. 11
C. 42
D. 66
Correct Answer: C
Solution: The given numbers are multiples of 7: $7 \times 1$, $7 \times 2$, $7 \times 3$, ..., $7 \times 11$. This is an arithmetic progression with the first term $a = 7$, common difference $d = 7$, and last term $l = 77$. To find the number of terms ($n$), we use the formula $l = a + (n-1)d$. $77 = 7 + (n-1)7$. Solving for $n$, we get $n = 11$. The sum of an arithmetic progression is given by $S = \frac{n}{2} \times (a + l)$ or $S = \frac{n}{2} \times (2a + (n-1)d)$. Using the first formula, $S = \frac{11}{2} \times (7 + 77) = \frac{11}{2} \times 84 = 11 \times 42 = 462$. The average is $\frac{S}{n} = \frac{462}{11} = 42$.
Q44: The average weight of A, B,C and D is 40 kg. A new person E is also included in the group, then the average weight of the group is increased by1 kg. Again a new person F replaces A, then the new average of 5 persons becomes 42. The average weight of B, C, D, F is :
A. 42
B. 41.25
C. 42.5
D. None of these
Correct Answer: B
Solution: Let the weights of A, B, C, and D be $a$, $b$, $c$, and $d$ respectively. Their average weight is 40 kg, so $\frac{a+b+c+d}{4} = 40$. Therefore, $a+b+c+d = 160$. When E is included, the average weight increases by 1 kg, becoming 41 kg. So, $\frac{a+b+c+d+e}{5} = 41$. $a+b+c+d+e = 205$. Since $a+b+c+d = 160$, we have $160 + e = 205$, which means $e = 45$ kg. When A is replaced by F, the average weight becomes 42 kg. So, $\frac{b+c+d+e+f}{5} = 42$. $b+c+d+e+f = 210$. We know $e = 45$, so $b+c+d+45+f = 210$. $b+c+d+f = 165$. We want to find the average weight of B, C, D, and F, which is $\frac{b+c+d+f}{4}$. Since $b+c+d+f = 165$, the average weight is $\frac{165}{4} = 41.25$ kg.
Q45: The average of 3 consecutive natural numbers (which are in increasing order) is k. If two more consecutive number, just next the first set of numbers, is added, then the new average becomes :
A. k + 2
B. k + 1
C. (2k + 1) / 2
D. 2k - 1
Correct Answer: B
Solution: Let the three consecutive natural numbers be $n$, $n+1$, and $n+2$. Their average is $k$, so:
$$ \frac{n + (n+1) + (n+2)}{3} = k $$
$$ \frac{3n + 3}{3} = k $$
$$ n + 1 = k $$
$$ n = k - 1 $$
The three numbers are $k-1$, $k$, $k+1$. The next two consecutive numbers are $k+2$ and $k+3$. Now we have five numbers: $k-1$, $k$, $k+1$, $k+2$, $k+3$. Their sum is:
$$ (k-1) + k + (k+1) + (k+2) + (k+3) = 5k + 5 $$
The new average is $\frac{5k + 5}{5} = k + 1$
Therefore, the new average is $k + 1$. Note that the question is incomplete, as it asks for the new average in terms of "$k$". A multiple choice question would need options showing this result.
Q46: The average of any5 consecutive odd natural numbers is k. If two more such numbers, just next to the previous 5 numbers are added, the new average becomes :
A. 2(k + 1)/7
B. 2k - 3
C. 2k + 1
D. k + 2
Correct Answer: D
Solution: Let the 5 consecutive odd natural numbers be $n$, $n+2$, $n+4$, $n+6$, $n+8$. Their average is $k$. Therefore, $\frac{n + (n+2) + (n+4) + (n+6) + (n+8)}{5} = k$
$$5n + 20 = 5k$$
$$n + 4 = k$$
$$n = k - 4$$
Now, two more consecutive odd numbers are added: $n+10$ and $n+12$. The new set of numbers is $n$, $n+2$, $n+4$, $n+6$, $n+8$, $n+10$, $n+12$. The new average is $\frac{7n + 42}{7} = n + 6$. Since $n = k - 4$, the new average is $(k - 4) + 6 = k + 2$.
Q47: The average weight of the 5 officers of a regiment is 42 kg. If a senior officer was replaced by a new officer and thus the average increased by 500 gm, the weight of the new officer is :
A. 44.5
B. 45
C. 42.5
D. Cannot be determined
Correct Answer: D
Solution: Let the total weight of the 5 officers be $5 \times 42 = 210$ kg. The average weight increased by 500 gm ($0.5$ kg) after replacing one officer. The new average weight is $42 + 0.5 = 42.5$ kg. The total weight of the 5 officers after the replacement is $5 \times 42.5 = 212.5$ kg. The weight of the new officer is the difference between the new total weight and the old total weight: $212.5 - 210 = 2.5$ kg. Therefore, the weight of the new officer is $2.5$ kg more than the officer he replaced. We need to find his weight. Let $x$ be the weight of the officer who was replaced. Then the sum of the weights of the other 4 officers is $210 - x$. The new total weight is $(210 - x) + \text{weight of new officer} = 212.5$
Let the weight of the new officer be $y$. Then,
$$(210 - x) + y = 212.5$$
The average weight of the 5 officers is $\frac{210 - x + y}{5} = 42.5$
$$210 - x + y = 212.5$$
$$y = x + 2.5$$
However, we are given that the average weight increased by $0.5$kg which means the total weight increased by $5 \times 0.5 = 2.5$ kg. Therefore, the weight of the new officer is $42 + 2.5 = 44.5$ kg
Q48: The average age of 6 servants in my farm house is 28 years. A new and young servant replaces an old servant, then the new average reduces by 1 year, the age of the new servant is :
A. 26
B. 22
C. 35
D. Cannot be determined
Correct Answer: D
Solution: Let the sum of ages of the 6 servants be $S$. The average age is 28, so $\frac{S}{6} = 28$. Therefore, $S = 28 \times 6 = 168$ years. When an old servant is replaced by a new servant, the new average becomes $28 - 1 = 27$ years. The sum of ages of the 5 remaining servants and the new servant is $27 \times 6 = 162$ years. Let the age of the new servant be $x$ and the age of the old servant be $y$. We have the equation: $S - y + x = 162$. Substituting $S = 168$, we get: $168 - y + x = 162$. This simplifies to $x - y = -6$. We don't know the age of the old servant ($y$), but we can say that the new servant is 6 years younger than the old servant. However, the question asks for the age of the new servant. We need more information to solve this completely. This is a trick question, the correct answer would be based on the assumption that the old servant's age and the new servant's age are part of the original average. The question implies that the reduction of one year in average is solely due to the new servant's age. Therefore $x = 27 \times 1 - (28 \times 6 - 27 \times 6) = 17$ years
Q49: The average age of 6 servants in my farm house is 28 years. A new and young servant replaces an old servant, then the new average reduces by 1 year. If the age of the replaced servant was 31 years, then the age of the new servant is :
A. 25
B. 35
C. 24
D. None of these
Correct Answer: A
Solution: Let the sum of ages of the 6 servants be $S$. The average age of the 6 servants is 28 years. Therefore, $\frac{S}{6} = 28 \implies S = 28 \times 6 = 168$ years. An old servant (age 31 years) is replaced by a new servant. Let the age of the new servant be $x$. The sum of ages of the remaining 5 servants is $S - 31 = 168 - 31 = 137$ years. The new sum of ages of the 6 servants is $137 + x$. The new average age is $28 - 1 = 27$ years. Therefore, $\frac{137 + x}{6} = 27$
$137 + x = 27 \times 6$
$137 + x = 162$
$x = 162 - 137$
$x = 25$ years
The age of the new servant is 25 years.
Q50: The average income of A, B and C is Rs. 12,000 per month and the average income of B,C and D is Rs. 15,000 per month. If the average salary of D be twice that of A, then the average salary of B and C is (in Rs.) :
A. 8000
B. 18000
C. 13500
D. 9000
Correct Answer: C
Solution: Let A's income be $A$, B's income be $B$, C's income be $C$, and D's income be $D$. Given: $\frac{A + B + C}{3} = 12000 \Rightarrow A + B + C = 36000$ (Equation 1)
$\frac{B + C + D}{3} = 15000 \Rightarrow B + C + D = 45000$ (Equation 2)
Also given that $D = 2A$. Subtracting Equation 1 from Equation 2:
$(B + C + D) - (A + B + C) = 45000 - 36000$
$D - A = 9000$
Since $D = 2A$, substituting this into the above equation:
$2A - A = 9000$
$A = 9000$
Now, $D = 2A = 2 \times 9000 = 18000$
Substituting $A = 9000$ in Equation 1:
$9000 + B + C = 36000$
$B + C = 36000 - 9000$
$B + C = 27000$
The average salary of B and C is $\frac{B + C}{2} = \frac{27000}{2} = 13500$
Q51: A has 50 coins of 10 paise denominations. While B has 10 coins of 50 paise denominations. C has 20 coins of 25 paise denominations while D has 25 coins of 20 paise denominations. The average number of paise per person is :
A. 450 paise
B. 500 paise
C. 600 paise
D. Cannot be determined
Correct Answer: B
Solution: First, calculate the total paise each person has:
A: $50 \text{ coins} \times \frac{10 \text{ paise}}{\text{coin}} = 500 \text{ paise}$
B: $10 \text{ coins} \times \frac{50 \text{ paise}}{\text{coin}} = 500 \text{ paise}$
C: $20 \text{ coins} \times \frac{25 \text{ paise}}{\text{coin}} = 500 \text{ paise}$
D: $25 \text{ coins} \times \frac{20 \text{ paise}}{\text{coin}} = 500 \text{ paise}$
Next, calculate the total paise for all four people:
$$ \text{Total paise} = 500 + 500 + 500 + 500 = 2000 \text{ paise} $$
Then, find the average number of paise per person:
$$ \text{Average paise per person} = \frac{\text{Total paise}}{\text{Number of people}} = \frac{2000 \text{ paise}}{4 \text{ people}} = 500 \text{ paise} $$
Therefore, the average number of paise per person is 500.
Q52: A travel agency has three types of vehicles viz. four seater, autorickshaw, 10 seater maxi cab and 20 seater minibus. The rate of each passanger (irrespective of its age or weight or seniority) for the auto rickshaw is Rs. 12 and for the maxicab is Rs. 15 and for the minibus is Rs. 8 for the one round. The average occupancy of the seats is 100%, 80% and 75% respectively. If he has only one vehicle of each kind, then the average earning for one round of each vehicle is :
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