23 + Interview Questions in AVERAGE IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Of the three numbers, the first is twice the second and the second is twice the third. The average of the reciprocal of the numbers is \(\frac{7}{72}\) . What are the three numbers ?
Answer» Let the third number be x. Then Second number = 2x and first number = 4x Sum of the reciprocals of these 3 numbers = \(\frac{1}{4x}+\frac{1}{2x}+\frac{1}{x}=\frac{1+2+4}{4x}=\frac{7}{4x}\) Given, \(\frac{7}{4x} = 3\times\frac{7}{72}\) ⇒ 4x = 24 ⇒ x = 6 ∴ The three numbers are 4 × 6, 2 × 6, 6, i.e., 24, 12, 6.

1.

Of the three numbers, the first is twice the second and the second is twice the third. The average of the reciprocal of the numbers is \(\frac{7}{72}\) . What are the three numbers ?

Answer»

Let the third number be x. Then

Second number = 2x and first number = 4x

Sum of the reciprocals of these 3 numbers = \(\frac{1}{4x}+\frac{1}{2x}+\frac{1}{x}=\frac{1+2+4}{4x}=\frac{7}{4x}\)

Given, \(\frac{7}{4x} = 3\times\frac{7}{72}\) ⇒ 4x = 24 ⇒ x = 6

∴ The three numbers are 4 × 6, 2 × 6, 6, i.e., 24, 12, 6.

Discussion

2.	The average temperature of Monday to Wednesday was 37°C and of Tuesday to Thursday was 34°C. If the temperature on Thursday was \(\frac{4}{5}\) that of Monday, what was the temperature on Thursday ?
Answer» Let the temperatures on Monday, Tuesday, Wednesday and Thursday be M, T, W and Th respectively. Then, M + T + W = 3 × 37°C = 111°C … (i) T + W + Th = 3 × 34°C = 102°C … (ii) ⇒ Eq (i) – Eq (ii) ⇒ M – Th = 111°C – 102°C = 9°C Also given, Th = \(\frac45\) M ⇒ M – \(\frac45\) M = 9 ⇒ \(\frac{M}5\) = 9 ⇒ M = 45°C ∴ Temperature on Thursday = 4 5 × 45°C = 36°C.

2.

The average temperature of Monday to Wednesday was 37°C and of Tuesday to Thursday was 34°C. If the temperature on Thursday was \(\frac{4}{5}\) that of Monday, what was the temperature on Thursday ?

Answer»

Let the temperatures on Monday, Tuesday, Wednesday and Thursday be M, T, W and Th respectively. Then,

M + T + W = 3 × 37°C = 111°C … (i)

T + W + Th = 3 × 34°C = 102°C … (ii)

⇒ Eq (i) – Eq (ii)

⇒ M – Th = 111°C – 102°C = 9°C

Also given, Th = \(\frac45\) M ⇒ M – \(\frac45\) M = 9 ⇒ \(\frac{M}5\) = 9 ⇒ M = 45°C

∴ Temperature on Thursday = 4 5 × 45°C = 36°C.

Discussion

3.	The average monthly salary of the workers in a workshop is 8500. If the average monthly salary of 7 technicians is Rs 10,000 and the average monthly salary of the rest is Rs 7800, the total number of workers in the workshop is (a) 18 (b) 20 (c) 22 (d) 24
Answer» Answer is: (c) 22. Let the total number of workers in the workshop be x. Then, number of other workers (besides technicians) = (x – 7) Given, (x – 7) × 7800 + 7 × 10000 = 8500x ⇒ 7800x – 54600 + 70000 = 8500x ⇒ 700x = 15400 ⇒ x = \(\frac{15400}{700}\) ⇒ x = 22.

3.

The average monthly salary of the workers in a workshop is 8500. If the average monthly salary of 7 technicians is Rs 10,000 and the average monthly salary of the rest is Rs 7800, the total number of workers in the workshop is (a) 18 (b) 20 (c) 22 (d) 24

Answer»

Answer is: (c) 22.

Let the total number of workers in the workshop be x.

Then, number of other workers (besides technicians) = (x – 7)

Given, (x – 7) × 7800 + 7 × 10000 = 8500x

⇒ 7800x – 54600 + 70000 = 8500x

⇒ 700x = 15400 ⇒ x = \(\frac{15400}{700}\) ⇒ x = 22.

Discussion

4.	The average annual income (in Rs) of certain agricultural workers is S and that of other workers is T. The number of agricultural workers is 11 times that of other workers. Then, the average annual income (in Rs) of all the workers is(a) \(\frac{S+11T}{12}\)(b) \(\frac{S+T}{2}\)(c) \(\frac{11S+T}{12}\)(d) \(\frac{1}{11S}+T\)
Answer» Answer is: (c) Let the number of other workers = x Then, the number of agricultural workers = 11x ∴ Total income of other workers = Rs Tx Total income of agricultural workers = Rs 11Sx ∴ Average income of all the workers = \(\frac{11Sx + Tx}{11x +x}=\frac{x(11S+T)}{12x}=\frac{11S+T}{12}\)

4.

The average annual income (in Rs) of certain agricultural workers is S and that of other workers is T. The number of agricultural workers is 11 times that of other workers. Then, the average annual income (in Rs) of all the workers is(a) \(\frac{S+11T}{12}\)(b) \(\frac{S+T}{2}\)(c) \(\frac{11S+T}{12}\)(d) \(\frac{1}{11S}+T\)

Answer»

Answer is: (c)

Let the number of other workers = x

Then, the number of agricultural workers = 11x

∴ Total income of other workers = Rs Tx

Total income of agricultural workers = Rs 11Sx

∴ Average income of all the workers = \(\frac{11Sx + Tx}{11x +x}=\frac{x(11S+T)}{12x}=\frac{11S+T}{12}\)

Discussion

5.	3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is same today. The present age of the baby is (a) 3 years (b) 2 years (c) 1\(\frac12\)years (d) 1 year
Answer» Answer is: (b) 2 years. Total age of the family three years ago = 17 × 5 = 85 years. Let the present age of the child be x years. Present total age of the family = 85 + 5 × 3 + x = (100 + x) years. Given \(\frac{100+x}{6}=17\) ⇒ 100 + x = 102 ⇒ x = 2 years.

5.

3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is same today. The present age of the baby is (a) 3 years (b) 2 years (c) 1\(\frac12\)years (d) 1 year

Answer»

Answer is: (b) 2 years.

Total age of the family three years ago = 17 × 5 = 85 years.

Let the present age of the child be x years.

Present total age of the family = 85 + 5 × 3 + x = (100 + x) years.

Given \(\frac{100+x}{6}=17\)

⇒ 100 + x = 102 ⇒ x = 2 years.

Discussion

6.	The average of marks of 28 students in Mathematics was 50. Eight students left the school and then this average increased by 5. What is the average of marks obtained by the students who left the school ? (a) 37.5 (b) 42.5 (c) 45 (d) 50.5
Answer» Answer is: (a) 37.5. Total marks of 28 students in Mathematics = 28 × 50 = 1400 New number of students = 28 – 8 = 20 New average of marks in Mathematics = 50 + 5 = 55 ∴ Total marks of 20 students in Mathematics = 20 × 55 = 1100 ∴ Total marks of 8 students who left the school = 1400 – 1100 = 300 ∴ Average marks of these 8 students = \(\frac{300}{8}\) = 37.5.

6.

The average of marks of 28 students in Mathematics was 50. Eight students left the school and then this average increased by 5. What is the average of marks obtained by the students who left the school ? (a) 37.5 (b) 42.5 (c) 45 (d) 50.5

Answer»

Answer is: (a) 37.5.

Total marks of 28 students in Mathematics = 28 × 50 = 1400

New number of students = 28 – 8 = 20

New average of marks in Mathematics = 50 + 5 = 55

∴ Total marks of 20 students in Mathematics = 20 × 55 = 1100

∴ Total marks of 8 students who left the school = 1400 – 1100 = 300

∴ Average marks of these 8 students = \(\frac{300}{8}\) = 37.5.

Discussion

7.	In an examination, a pupil’s average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination ?(a) 8 (b) 9 (c) 10 (d) 11
Answer» Answer is: (d) 11. Let there be n papers in all. Total marks originally for n papers = 63n Total marks with increased average = 65n Given, 65n – 63n = 20 + 2 ⇒ 2n = 22 ⇒ n = 11.

7.

In an examination, a pupil’s average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination ?(a) 8 (b) 9 (c) 10 (d) 11

Answer»

Answer is: (d) 11.

Let there be n papers in all.

Total marks originally for n papers = 63n

Total marks with increased average = 65n

Given, 65n – 63n = 20 + 2 ⇒ 2n = 22 ⇒ n = 11.

Discussion

8.	A company produces on an average 4000 items per month for the first three months. How many items it must produce on an average per month over the next 9 months to average 4375 items per month over the whole year ? (a) 4500 (b) 4600 (c) 4680 (d) 4710
Answer» Answer is: (a) 4500 Total number of items produced for first three months = 4000 × 3 = 12000 Total number of items required to be produced over a period of 12 months = 4375 ×12 = 52500 ⇒ Number of items to be produced over a period of 9 months = 52500 – 12000 = 40500 ∴ Average number of articles per month produced over 9 months = \(\frac{40500}{9}\) = 4500

8.

A company produces on an average 4000 items per month for the first three months. How many items it must produce on an average per month over the next 9 months to average 4375 items per month over the whole year ? (a) 4500 (b) 4600 (c) 4680 (d) 4710

Answer»

Answer is: (a) 4500

Total number of items produced for first three months = 4000 × 3 = 12000

Total number of items required to be produced over a period of 12 months = 4375 ×12 = 52500

⇒ Number of items to be produced over a period of 9 months = 52500 – 12000 = 40500

∴ Average number of articles per month produced over 9 months = \(\frac{40500}{9}\) = 4500

Discussion

9.	The average price of 10 books is Rs 12, while the average price of 8 of these books is Rs 11.75. Of the remaining two books, if the price of one is 60% more than the price of the other, what is the price of each of these two books ? (a) Rs 8; Rs 12 (b) Rs 10; Rs 16 (c) Rs 5; Rs 7.50 (d) Rs 12; Rs 14
Answer» Answer is: (b) Rs 16. Total price of the remaining two books = 10 × 12 – 8 × 11.75 = 120 – 94 = Rs 26 Suppose the price of one book = Rs x Then, price of the other book = x + 60% of x = \(\frac{160}{100}\) x = \(\frac{8x}{5}\) Given, x + \(\frac{8x}{5}\) = 26 ⇒ \(\frac{13x}{5}\) = 26 ⇒ x = \(\frac{26\times5}{13}\) = Rs 10 ∴ Price of the other book = Rs \(\frac{8\times10}{5}\) = Rs 16.

9.

The average price of 10 books is Rs 12, while the average price of 8 of these books is Rs 11.75. Of the remaining two books, if the price of one is 60% more than the price of the other, what is the price of each of these two books ? (a) Rs 8; Rs 12 (b) Rs 10; Rs 16 (c) Rs 5; Rs 7.50 (d) Rs 12; Rs 14

Answer»

Answer is: (b) Rs 16.

Total price of the remaining two books = 10 × 12 – 8 × 11.75 = 120 – 94 = Rs 26

Suppose the price of one book = Rs x

Then, price of the other book = x + 60% of x = \(\frac{160}{100}\) x = \(\frac{8x}{5}\)

Given, x + \(\frac{8x}{5}\) = 26 ⇒ \(\frac{13x}{5}\) = 26

⇒ x = \(\frac{26\times5}{13}\) = Rs 10

∴ Price of the other book = Rs \(\frac{8\times10}{5}\) = Rs 16.

Discussion

10.	The average of the squares of the first ten natural numbers is(a) 40 (b) 50 (c) 47.5 (d) 38.5
Answer» Answer is: (d) 38.5 Average = \(\frac{1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2}{10}=\) \(\frac{1+4+9+16+25+36+79+64+81+100}{10}\) \(=\frac{385}{10}=38.5\)

10.

The average of the squares of the first ten natural numbers is(a) 40 (b) 50 (c) 47.5 (d) 38.5

Answer»

Answer is: (d) 38.5

Average = \(\frac{1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2}{10}=\) \(\frac{1+4+9+16+25+36+79+64+81+100}{10}\)

\(=\frac{385}{10}=38.5\)

Discussion

11.	The average weight of 120 students in the second year class of college is 56 kg. If the average weight of boys and that of girls in the class are 60 kg and 50 kg respectively, then the number of boys and girls in the class are respectively : (a) 72, 64 (b) 38, 64 (c) 72, 48 (d) 62, 58
Answer» Answer is: (c) 48 Let the number of boys be x. Then, The number of girls = (120 – x) Total weight of 120 students in the class = 120 × 56 kg = 6720 kg Total weight of x boys = 60x kg Total weight of (120 – x) girls = (120 – x) × 50 kg = 6000 – 50x Given, 60x + 6000 – 50x = 6720 ⇒ 10x = 720 ⇒ x = 72 ∴ Number of boys= 72, Number of girls = 120 – 72 = 48

11.

The average weight of 120 students in the second year class of college is 56 kg. If the average weight of boys and that of girls in the class are 60 kg and 50 kg respectively, then the number of boys and girls in the class are respectively : (a) 72, 64 (b) 38, 64 (c) 72, 48 (d) 62, 58

Answer»

Answer is: (c) 48

Let the number of boys be x. Then,

The number of girls = (120 – x)

Total weight of 120 students in the class = 120 × 56 kg = 6720 kg

Total weight of x boys = 60x kg

Total weight of (120 – x) girls = (120 – x) × 50 kg = 6000 – 50x

Given, 60x + 6000 – 50x = 6720

⇒ 10x = 720 ⇒ x = 72

∴ Number of boys= 72,

Number of girls = 120 – 72 = 48

Discussion

12.	The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then (a) x = y + z (b) 2x = y + z (c) x = 2y + 2z (d) x = y + 2z
Answer» Answer is: (b) 2x = y + z Sum of six numbers = 6x Sum of three of these 6 numbers = 3y Sum of remaining three numbers = 3z ∴ 3y + 3z = 6x ⇒ 2x = y + z

12.

The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then (a) x = y + z (b) 2x = y + z (c) x = 2y + 2z (d) x = y + 2z

Answer»

Answer is: (b) 2x = y + z

Sum of six numbers = 6x

Sum of three of these 6 numbers = 3y

Sum of remaining three numbers = 3z

∴ 3y + 3z = 6x ⇒ 2x = y + z

Discussion

13.	If the average of a, b, c is M and ab + bc + ca = 0, then the average of a2 , b2 , c2 is (a) M2 (b) 3M2 (c) 6M2 (d) 9M2
Answer» Answer is: (b) 3M² Given, \(\frac{a+b+c}{3}=M\) ⇒ a + b + c = 3M Now, (a + b + c)² = a² + b² + c² + 2 (ab + bc + ca) ⇒ (3M)² = a² + b² + c² + 0 ⇒ a² + b² + c² = 9M² Average of a² + b² + c²= \(\frac{a^2 + b^2 + c^2}{3}=\frac{9M^2}{3}=3M^2\)

13.

If the average of a, b, c is M and ab + bc + ca = 0, then the average of a2 , b2 , c2 is (a) M2 (b) 3M2 (c) 6M2 (d) 9M2

Answer»

Answer is: (b) 3M²

Given, \(\frac{a+b+c}{3}=M\) ⇒ a + b + c = 3M

Now, (a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)

⇒ (3M)² = a² + b² + c² + 0 ⇒ a² + b² + c² = 9M²

Average of a² + b² + c²= \(\frac{a^2 + b^2 + c^2}{3}=\frac{9M^2}{3}=3M^2\)

Discussion

14.	If the average of m numbers is n2 and that of n numbers is m2 , then the average of (m + n) numbers is (a) m – n (b) mn (c) (m + n) (d) m/n
Answer» Answer is: (d) mn Average of m numbers = n² ∴ Sum of m numbers = m × n² = mn² Average of n numbers = m² ∴ Sum of n numbers = n × m² = m²n ∴ Required average = \(\frac{mn^2+nm^2}{m+n}=\frac{mn(n+m)}{m+n}=mn\)

14.

If the average of m numbers is n2 and that of n numbers is m2 , then the average of (m + n) numbers is (a) m – n (b) mn (c) (m + n) (d) m/n

Answer»

Answer is: (d) mn

Average of m numbers = n²

∴ Sum of m numbers = m × n² = mn²

Average of n numbers = m²

∴ Sum of n numbers = n × m² = m²n

∴ Required average = \(\frac{mn^2+nm^2}{m+n}=\frac{mn(n+m)}{m+n}=mn\)

Discussion

15.	The average age of 10 students is 6 years. The sum of ages of 9 of them is 52 years. Find the age of 10th student.
Answer» Let the sum of ages of 10 boys = x no of boys = 10 average = \(\cfrac{x}{10}\) 6 = \(\cfrac{x}{10}\) x = 60 (sum of ages of all boys) sum of ages of a boy = 52 Let the age of 10^th student = y sum of ages of a boys = x - y 52 = 60 - y y = 8 \(\therefore\) Age of 10^th student = 8 years

15.

The average age of 10 students is 6 years. The sum of ages of 9 of them is 52 years. Find the age of 10th student.

Answer»

Let the sum of ages of 10 boys = x

no of boys = 10

average = \(\cfrac{x}{10}\)

6 = \(\cfrac{x}{10}\)

x = 60 (sum of ages of all boys)

sum of ages of a boy = 52

Let the age of 10^th student = y

sum of ages of a boys = x - y

52 = 60 - y

y = 8

\(\therefore\) Age of 10^th student = 8 years

Discussion

16.	The average score of a class of boys and girls in an examination is A. The ratio of boys and girls in the class is 3 : 1. If the average score of the boys is (A + 1), the average score of the girls is (a) (A – 1) (b) (A – 3) (c) (A +1) (d) (A + 3)
Answer» Answer is: (b) (A – 3). Let the number of boys be 3x and the number of girls be x. Then, total number of students = 4x Average score of the class = A ∴ Total score of all the students = 4Ax Average score of all the boys = A + 1 ∴ Total score of all the boys = 3x (A + 1) = 3Ax + 3x ⇒ Total score of all the girls = 4Ax – 3Ax – 3x = Ax – 3x = (A – 3)x ⇒Average score of the girls =\(\frac{(A-3)x}{x}\) = (A – 3).

16.

The average score of a class of boys and girls in an examination is A. The ratio of boys and girls in the class is 3 : 1. If the average score of the boys is (A + 1), the average score of the girls is (a) (A – 1) (b) (A – 3) (c) (A +1) (d) (A + 3)

Answer»

Answer is: (b) (A – 3).

Let the number of boys be 3x and the number of girls be x.

Then, total number of students = 4x

Average score of the class = A

∴ Total score of all the students = 4Ax

Average score of all the boys = A + 1

∴ Total score of all the boys = 3x (A + 1) = 3Ax + 3x

⇒ Total score of all the girls = 4Ax – 3Ax – 3x = Ax – 3x = (A – 3)x

⇒Average score of the girls =\(\frac{(A-3)x}{x}\) = (A – 3).

Discussion

17.	The average of four positive integers is 72.5. The highest integer is 117 and the lowest integer is 15. The difference between the remaining two integers is 12. Which is the higher of these two remaining integers ? (a) 70 (b) 73 (c) 85 (d) 80
Answer» Answer is: (c) 85. Let the higher integer be x. Then the other integer = x – 12 ∴ \(\frac{15+(x-12) + x+117}{4}\) = 72.5 ⇒ 120 + 2x = 72.5 × 4 = 290 ⇒ 2x = 170 ⇒ x = 85.

17.

The average of four positive integers is 72.5. The highest integer is 117 and the lowest integer is 15. The difference between the remaining two integers is 12. Which is the higher of these two remaining integers ? (a) 70 (b) 73 (c) 85 (d) 80

Answer»

Answer is: (c) 85.

Let the higher integer be x.

Then the other integer = x – 12

∴ \(\frac{15+(x-12) + x+117}{4}\) = 72.5

⇒ 120 + 2x = 72.5 × 4 = 290

⇒ 2x = 170 ⇒ x = 85.

Discussion

18.	From a class 24 boys, a boy aged 10 years leaves the class and in his place a new boy is admitted. As a result the average age of the class is increased by 2 months. What is the age of the new boys ?
Answer» Let the average age of the class be x years. Then the total age of the class = 24x years New average age of the class = \((x-\frac{2}{12})\) years = \((x-\frac{1}{6})\) years Let the age of the new boy be y years. Given, \(\frac{24x-10+y}{24}=x-\frac16\) ⇒ 24x-10+y = 24x - 4 ⇒ y = 6 ∴ The age of the new boy is 6 years

18.

From a class 24 boys, a boy aged 10 years leaves the class and in his place a new boy is admitted. As a result the average age of the class is increased by 2 months. What is the age of the new boys ?

Answer»

Let the average age of the class be x years.

Then the total age of the class = 24x years

New average age of the class = \((x-\frac{2}{12})\) years = \((x-\frac{1}{6})\) years

Let the age of the new boy be y years.

Given, \(\frac{24x-10+y}{24}=x-\frac16\)

⇒ 24x-10+y = 24x - 4 ⇒ y = 6

∴ The age of the new boy is 6 years

Discussion

19.	What is the average of squares of consecutive odd numbers between 1 and 13 ?
Answer» The consecutive odd numbers from 1 to 13 = 3, 5, 7, 9, 11 ∴ Required average = \(\frac{3^2+5^2+7^2+9^2+11^2}{5}=\frac{9+25+49+81+121}{5}=\frac{285}{5} =57\)

19.

What is the average of squares of consecutive odd numbers between 1 and 13 ?

Answer»

The consecutive odd numbers from 1 to 13 = 3, 5, 7, 9, 11

∴ Required average = \(\frac{3^2+5^2+7^2+9^2+11^2}{5}=\frac{9+25+49+81+121}{5}=\frac{285}{5} =57\)

Discussion

20.	The average of the two digit numbers, which remain the same when the digits interchange their positions is (a) 33 (b) 44 (c) 55 (d) 66
Answer» Answer is: (c) 55 The required two digit numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99. ∴ Required Average = \(\frac{11+22+33+44+55+66+77+88+99}{9}=\frac{495}{9}=55\)

20.

The average of the two digit numbers, which remain the same when the digits interchange their positions is (a) 33 (b) 44 (c) 55 (d) 66

Answer»

Answer is: (c) 55

The required two digit numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99.

∴ Required Average = \(\frac{11+22+33+44+55+66+77+88+99}{9}=\frac{495}{9}=55\)

Discussion

21.	The average weight of three men A, B and C is 84 kg. D joins them and the average weight of the four becomes 80 kg. If E whose weight is 3 kg more than that of D replaces A, the average weight of B, C, D and E becomes 79 kg. The weight of A is (a) 65 kg (b) 70 kg (c) 75 kg (d) 80 kg
Answer» Answer is: (c) 75 kg Total weight of A + B + C = 84 × 3 = 252 kg Total weight of A + B + C + D = 80 × 4 = 320 kg ∴ Weight of D = 320 – 252 = 68 kg Given, E = D + 3 = (68 + 3) kg = 71 kg Total weight of B + C + D + E = (79 × 4) kg = 316 kg ∴ Weight of B + C = 316 kg – 71 kg – 68 kg = 177 kg Weight of Aonly = 320 kg– 177 kg– 68 kg = 75 kg

21.

The average weight of three men A, B and C is 84 kg. D joins them and the average weight of the four becomes 80 kg. If E whose weight is 3 kg more than that of D replaces A, the average weight of B, C, D and E becomes 79 kg. The weight of A is (a) 65 kg (b) 70 kg (c) 75 kg (d) 80 kg

Answer»

Answer is: (c) 75 kg

Total weight of A + B + C = 84 × 3 = 252 kg

Total weight of A + B + C + D = 80 × 4 = 320 kg

∴ Weight of D = 320 – 252 = 68 kg

Given, E = D + 3 = (68 + 3) kg = 71 kg

Total weight of B + C + D + E = (79 × 4) kg = 316 kg

∴ Weight of B + C = 316 kg – 71 kg – 68 kg = 177 kg

Weight of Aonly = 320 kg– 177 kg– 68 kg = 75 kg

Discussion

22.	The batting average of a cricket player for 40 innings is 50 runs. His highest score in an innings exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Determine his highest score, scored in one innings. (a) 175 (b) 180 (c) 174 (d) 185
Answer» Answer is: (c) 174. Let the lowest score in one innings be x. Then, Highest score in an innings = x + 172 Total runs in 40 innings = 40 × 50 = 2000 Total runs in 38 innings = 38 × 48 = 1824 Given, 2000 – {x + (x+ 172)} = 1824 ⇒ 1828 – 2x = 1824 ⇒ 2x = 4 ⇒ x = 2 ∴ Highest score = 172 + 2 = 174.

22.

The batting average of a cricket player for 40 innings is 50 runs. His highest score in an innings exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Determine his highest score, scored in one innings. (a) 175 (b) 180 (c) 174 (d) 185

Answer»

Answer is: (c) 174.

Let the lowest score in one innings be x.

Then, Highest score in an innings = x + 172

Total runs in 40 innings = 40 × 50 = 2000

Total runs in 38 innings = 38 × 48 = 1824

Given, 2000 – {x + (x+ 172)} = 1824

⇒ 1828 – 2x = 1824 ⇒ 2x = 4 ⇒ x = 2

∴ Highest score = 172 + 2 = 174.

Discussion

23.	The average temperature of the town in the first four days of a month was 58 degrees. The average for second, third, fourth and fifth days was 60 degrees. If the temperatures if the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ? (a) 64 degrees (b) 62 degrees (c) 56 degrees (d) 48 degrees
Answer» Answer is: (a) 64 degrees Sum of temperatures on 1st, 2nd, 3rd and 4th days = (58 × 4) = 232 degrees … (1) Sum of temperatures on 2nd, 3rd, 4th and 5th days = (60 × 4) = 240 degrees …(2) Subtracting eqn. (1) from eqn. (2), we get Temp. on 5th day – Temp. on 1st = 8 degrees Given, Temp. on 1st day : Temp. on 5th day = 7 :8 ∴ 8x – 7x = 8 ⇒ x = 8 ∴ Temperature on 5th day = 64 degrees

23.

The average temperature of the town in the first four days of a month was 58 degrees. The average for second, third, fourth and fifth days was 60 degrees. If the temperatures if the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ? (a) 64 degrees (b) 62 degrees (c) 56 degrees (d) 48 degrees

Answer»

Answer is: (a) 64 degrees

Sum of temperatures on 1st, 2nd, 3rd and 4th days

= (58 × 4) = 232 degrees … (1)

Sum of temperatures on 2nd, 3rd, 4th and 5th days

= (60 × 4) = 240 degrees …(2)

Subtracting eqn. (1) from eqn. (2), we get

Temp. on 5th day – Temp. on 1st = 8 degrees

Given, Temp. on 1st day : Temp. on 5th day = 7 :8

∴ 8x – 7x = 8 ⇒ x = 8

∴ Temperature on 5th day = 64 degrees

Discussion

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