Average
- The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31.The mean score of remaining 55% is :
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Given , The arithmetic mean of the scores of a group of students in a test = 52.
Let mean score of remaining 55% = y
According to question ,52 = 20 × 80 + 25 × 31 + 55 × y 100
⇒ 5200 = 1600 + 775 + 55y
⇒ 55y = 5200 –1600 – 775
Correct Option: C
Given , The arithmetic mean of the scores of a group of students in a test = 52.
Let mean score of remaining 55% = y
According to question ,52 = 20 × 80 + 25 × 31 + 55 × y 100
⇒ 5200 = 1600 + 775 + 55y
⇒ 55y = 5200 –1600 – 775
⇒ 55y = 2825
∴ y = 2825 ÷ 55 = 51.36 ≈ 51.4%
- There are in all, 10 balls; some of them are red and the others white. The average cost of all balls is ₹ 28. If the average cost of red balls is ₹25 and that of white balls is ₹30, the number of white balls is :
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Here , The average cost of all balls = ₹ 28.
The average cost of red balls = ₹25
and the average cost of white balls = ₹30
Let the number of white balls be y.
∴ Number of red balls = (10 – y)
∴ 10 × 28 = y × 30 + 25 (10 – y)
⇒ 280 = 30y + 250 – 25y = 5y + 250Correct Option: C
Here , The average cost of all balls = ₹ 28.
The average cost of red balls = ₹25
and the average cost of white balls = ₹30
Let the number of white balls be y.
∴ Number of red balls = (10 – y)
∴ 10 × 28 = y × 30 + 25 (10 – y)
⇒ 280 = 30y + 250 – 25y = 5y + 250
⇒ 5y = 280 – 250 = 30
⇒ y = 6
- A company produces an average of 4000 items per month for the first 3 months. How much items, it must produce on an average per month over the next 9 months to average 4375 items per month over the whole year?
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Let average production of a company in 9 months be y items
∴ According to question,4375 = 3 × 4000 + 9 × y 12
⇒ 4375 × 12 = 12 × 1000 + 9 × y
∴ 9 × y = 12 (4375 – 1000) = 12 × 3375
Correct Option: A
Let average production of a company in 9 months be y items
∴ According to question,4375 = 3 × 4000 + 9 × y 12
⇒ 4375 × 12 = 12 × 1000 + 9 × y
∴ 9 × y = 12 (4375 – 1000) = 12 × 3375∴ y = 12 × 3375 9
y = 4500
- The average age of a class of 39 students is 15 years. If the age of the teacher is included, then the average increases by 3 months. Find the age of the teacher.
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Given that , The average age of 39 students = 15 years
If the age of the teacher is included, then the average increases by 3 months , then
Age of teacher = 15 years + total increase = 15 years + (40 × 3) monthsAge of teacher = 15 + 
40 × 3 
years 12
Correct Option: B
Given that , The average age of 39 students = 15 years
If the age of the teacher is included, then the average increases by 3 months , then
Age of teacher = 15 years + total increase = 15 years + (40 × 3) monthsAge of teacher = 15 + 40 × 3 years 12
Hence , Age of teacher = (15 + 10) years = 25 years
- Seven years ago, the average age of A, B and C was 51 years. If A is 3 years older than B and B is 3 years older than C then the present ages of A, B and C are (in years)
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Given that , Seven years ago, the average age of A, B and C = 51 years.
Sum of the present ages of A, B and C = (51 × 3 + 3 × 7) years
Sum of the present ages of A, B and C = (153 + 21) years = 174 years
According to question ,
Now , A = B + 3 = C + 6
⇒ B = C + 3
∴ A + B + C = 174
⇒ C + 6 + C + 3 + C = 174
⇒ 3C = 174 – 9 = 165Correct Option: A
Given that , Seven years ago, the average age of A, B and C = 51 years.
Sum of the present ages of A, B and C = (51 × 3 + 3 × 7) years
Sum of the present ages of A, B and C = (153 + 21) years = 174 years
According to question ,
Now , A = B + 3 = C + 6
⇒ B = C + 3
∴ A + B + C = 174
⇒ C + 6 + C + 3 + C = 174
⇒ 3C = 174 – 9 = 165
⇒ C = 165 ÷ 3 = 55 years
∴ A = C + 6 = 55 + 6 = 61 years
B = C + 3 = 55 + 3 = 58 years
