(1) In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A. 6.25
B. 5.5
C. 7.4
D. 5
Answer: Option A – 6.25
Explanation:
- Runs scored in the first 10 overs = 10×3.2=32
- Total runs = 282
- remaining runs to be scored = 282 – 32 = 250
- remaining overs = 40
- Run rate needed = 250/40=6.25
(2) A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs.6500?
A. 4800
B. 4991
C. 5004
D. 5000
Answer: Option B – 4991
Explanation:
- Let the sale in the sixth month = x
- Then (6435+6927+6855+7230+6562+x)/6=6500
- = > 6435 + 6927 + 6855+ 7230 + 6562 + x
- = 6 × 6500 = 39000
- = > 34009 + x = 39000
- = > x = 39000 – 34009 = 4991
(3) The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
A. 1
B. 20
C. 0
D. 19
Answer: Option D – 19
Explanation:
- Average of 20 numbers = 0
- => Sum of 20 numbers/20=0
- => Sum of 20 numbers = 0
- Hence at the most, there can be 19 positive numbers. (Such that if the sum of these 19 positive numbers is x, 20th number will be -x)
(4) The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.
A. 23 years
B. 20 years
C. 24 years
D. 21 years
Answer: Option A – 23 years
Explanation:
- Number of members in the team = 11
- Let the average age of the team = x
- =>(Sum of the ages of all the 11 members of the team)/11=x
- => Sum of the ages of all the 11 members of the team = 11x
- Age of the captain = 26
- Age of the wicket keeper = 26 + 3 = 29
- Sum of the ages of 9 members of the team excluding captain and wicket keeper = 11x – 26 – 29 = 11x – 55
- Average age of 9 members of the team excluding captain and wicket keeper =(11x”55)/9 Given that( 11x”55)/9=(x”1)
- => 11x – 55 = 9(x – 1)
- => 11x – 55 = 9x – 9
- => 2x = 46
- =>x=46/2 = 23 years
(5) The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs.5200. What is the monthly income of A?
A. 2000
B. 3000
C. 4000
D. 5000
Answer: Option C – 4000
Explanation:
- Let the monthly income of A = a
- monthly income of B = b
- monthly income of C = a
- a +b=2×5050 ———— (Equation1)
- b + c=2×6250 ———— (Equation2)
- a +c=2×5200 ———— (Equation3)
- (Equation 1) + (Equation 3) – (Equation 2)
- => ( a +b +a +c) “ (b +c)
- =(2×5050)+(2×5200)”(2×6250)
- => 2a = 2(5050 + 5200 – 6250)
- => a = 4000
- => Monthly income of A = 4000
(6) In Kiran’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?
A. 70 kg
B. 69 kg
C. 61 kg
D. 67 kg
Answer: Option D – 67 kg
Explanation:
- Let Kiran’s weight = x.
- Then According to Kiran, 65 < x < 72 —————— (equation 1)
- According to brother, 60 < x < 70 —————— (equation 2)
- According to mother, xd”68 —————— (equation 3)
- Given that equation 1, equation 2 and equation 3 are correct. By combining these equations, we can write as 65<xd”68
- That is x = 66 or 67 or 68
- average of different probable weights of Kiran = (66+67+68)/3 = 67
7) The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
A. 48.55
B. 42.25
C. 50
D. 51.25
Answer: Option A – 48.55
Explanation:
- The average weight of 16 boys = 50.25
- Total Weight of 16 boys = 50.25×16
- The average weight of the remaining 8 boys = 45.15
- Total Weight of remaining 8 boys = 45.15×8
- Total weight of all boys in the class = (50.25×16) + (45.15×8)
- Total boys = 16+8=24
- Average weight of all the boys = (50.25×16) + (45.15×8)/24
- = (50.25×2)+(45.15×1)/3
- = (16.75×2) + 15.05
- = 33.5 + 15.05 = 48.55
(8) A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday?
A. 290
B. 304
C. 285
D. 270
Answer: Option C – 285
Explanation:
In a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday. Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday
- Average visitors on Sundays = 510
- Total visitors of 5 Sundays = 510×5
- Average visitors on other days = 240
- Total visitors of other 25 days = 240×25
- Total visitors = (510×5)+(240×25)
- Total days= 30
- The average number of visitors per day = { (510×5) (240×25)} / 30
- = { (51×5)+(24×25) }/3
- = (17×5) + (8×25)
- = 85 + 200 = 285
(9) A student’s mark was wrongly entered as 83 instead of 63. Due to that, the average marks for the class got increased by half 1/2. What is the number of students in the class?
A. 45
B. 40
C. 35
D. 30
Answer: Option B – 40
Explanation:
- Let the total number of students = x
- The average marks increased by 1/2 due to an increase of 83-63=20 marks.
- But the total increase in the marks = 1/2×x = x/2
- Hence we can write as x/2=20
- Ex = 20×2 = 40
(10) A family consists of two grandparents, two parents, and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is
A. 32.27 years
B. 31.71 years
C. 28.17 years
D. 30.57 years
Answer: Option B – 31.71 years
Explanation:
- Total age of the grandparents = 67×2
- Total age of the parents = 35×2
- Total age of the grandchildren = 6×3
- Average age of the family = {(67×2)+(35×2)+(6×3)}/7
- = (134+70+18) / 7 = 222/7 = 31.71