Practice MCQ Questions and Answer on Average

441.

If the average of 6 consecutive even number is 25, the difference between the largest and the smallest number is :

(A) 8
(B) 10
(C) 12
(D) 14

Solution: According to the question, Sum of 6 consecutive even number is = 25 × 6 = 150 Sn = $\frac{n}{2}$ [2a +(n - 1) × d] ⇒ 150 = $\frac{6}{2}$ [2a + (6 - 1) × 2] ⇒ 150 = $\frac{6}{2}$ (2a + 10) ⇒ 300 = 12a + 60 ⇒ 12 a = 240 ⇒ a = $\frac{240}{12}$ ⇒ a = 20 ∴ Numbers are 20, 22, 24, 26, 28, 30 Difference between largest and smallest is = 30 - 20 = 10 Alternate : Let the 6 consecutive number is = x, x + 2, x + 4, x + 6, x + 8, x + 10 According to the question, Largest number = Average + (n - 1) = 25 + 5 = 30 Smallest number = Average - (n - 1) = 25 - 5 = 20 Difference between largest and smallest number = 30 - 20 = 10

442.

The speed of the train going from Nagpur to Allahabad is 100 km/h while when coming back from Allahabad to Nagpur, its speed is 150 km/h. find the average speed during whole journey.

(A) 125 km/hr
(B) 75 km/hr
(C) 135 km/hr
(D) 120 km/hr

Solution: $\begin{aligned} Average speed, \\ = \frac{2 \times x \times y}{x + y} \\ = \frac{2 \times 100 \times 150}{100 + 150} \\ = \frac{200 \times 150}{250} \\ = 120 km/hr \end{aligned}$

443.

The average monthly income of a family of four earning members was Rs. 15130. One of the daughters in the family got married and left home, so the average monthly income of the family came down to Rs. 14660. What is the monthly income of the married daughter?

(A) Rs. 12000
(B) Rs. 15350
(C) Rs. 16540
(D) Cannot be determined

444.

The average of five numbers is 7. If three new numbers would be added, then the new average comes out to be 8.5. What is the average of those three new numbers ?

(A) 9
(B) 10.5
(C) 11
(D) 11.5

Solution: Sum of 5 numbers = 7 × 5 = 35 Sum of 8 numbers = 8.5 × 8 = 68 Sum of added all three numbers = 68 - 35 = 33 Average of three new numbers = $\frac{33}{3}$ = 11

445.

The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is-

(A) 1 : 2
(B) 2 : 3
(C) 3 : 4
(D) 3 : 5

Solution: Let the ratio be K : 1 Then, ⇔ K × 16.4 + 1 × 15.4 = (K + 1) × 15.8 ⇔ (16.4 - 15.8) K = (15.8 - 15.4) ⇔ K = $\frac{0.4}{0.6}$ ⇔ K = $\frac{2}{3}$ ∴ Required ratio = $\frac{2}{3}$ : 1 = 2 : 3

446.

The average age of 9 students and their teacher is 16 years. The average age of the first four students is 19 years and that of the last five is 10 years. The teacher's age is -

(A) 36 years
(B) 34 years
(C) 30 years
(D) 28 years

447.

B was born when A was 4 years 7 months old and C was born when B was 3 years 4 months old. When C was 5 years 2 months old, then their average age was :

(A) 8 years 9 months
(B) 7 years 3 months
(C) 8 years 7 months
(D) 8 years 11 months

Solution: According to the question, A - B = 4y 7m . . . . . . . (i) B - C = 3y 4m . . . . . . . (ii) (+) (+) (+) A - C = 7y 11m . . . . . . . (ii) Given : When, C = 5 years 2 months ∴ A = 13 years 1 month B = 8 years 6 months ∴ Average of $\frac{A + B + C}{3}$ $= \frac{26 years 9 months}{3}$ = 8 years 11 months

448.

The average of the marks of 25 students in a class, in an examination was calculated to be 19. Later, the teacher realized that the marks of two students were taken as 18 and 19 respectively, instead of 14 and 15. Find the new actual average marks of the class.

(A) 17.43
(B) 16.56
(C) 18.68
(D) 17.65

Solution: Total marks = 25 × 19 = 475 Error = (18 + 19) - (14 + 15) = 37 - 29 = 8 Actual marks = 475 - 8 = 467 Average marks $= \frac{467}{25} = 18.68$

449.

In an engineering college the average salary of all engineering graduates from Mechanical trade is Rs. 2.45 lacs per annum and that of the engineering graduates from Electronics trade is Rs. 3.56 lacs per annum. The average salary of all Mechanical and Electronics graduates is Rs. 3.12 lacs per annum. Find the least number of Electronics graduates passing out from this institute.-

(A) 43
(B) 59
(C) 67
(D) Cannot be determined

Solution: Let the number of Mechanical Engineering graduates be M and the number of Electronics Engineering graduates be E. Then, $\Rightarrow 2.45 M + 3.56 E =$ $3.12 (M + E)$ $\Rightarrow 2.45 M + 3.56 E =$ $3.12 M + 3.12 E$ $\begin{aligned} \Rightarrow 0.44 E = 0.67 M \\ \Rightarrow \frac{M}{E} = \frac{0.44}{0.67} = \frac{44}{67} \end{aligned}$ Since the ratio 44 : 67 is in its simplest form, So least number of Electronics graduates = 67

450.

When 15 is included in a list of natural numbers, their mean is increased by 2. When 1 is included in this new list, the mean of the numbers in the new list is decreased by 1. How many numbers were there in the original list?

(A) 4
(B) 5
(C) 6
(D) 8

Solution: Let there be n numbers in the original list and let their mean be x.Then, sum of n numbers = nx $\begin{aligned} ∴ \frac{n x + 15}{n + 1} = x + 2 \\ \Rightarrow n x + 15 = (n + 1) (x + 2) \\ \Rightarrow n x + 15 = n x + 2 n + x + 2 \\ \Rightarrow 2 n + x = 13. . . . . (i) \end{aligned}$ And, $\begin{aligned} ∴ \frac{n x + 16}{n + 2} = (x + 2) - 1 \\ \Rightarrow n x + 16 = (n + 2) (x + 1) \\ \Rightarrow n x + 16 = n x + n + 2 x + 2 \\ \Rightarrow n + 2 x = 14. . . . . (i i) \end{aligned}$ Solving (i) and (ii), we get: n = 4, x = 5

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