There are 4 consecutive odd numbers (x1, x2, x3 and x4) and three consecutive even numbers (y1, y2 and y3). The average of the odd numbers is 6 less than the average of the even numbers. If the sum of the three even numbers is 16 less than the sum of the four odd numbers, what is the average of x1, x2, x3 and x4?
The average of 4 positive integers is 59. The highest integer is 83 and the lowest integer is 29. The difference between the remaining two integers is 28. Which of the following integers is higher of the remaining two integers ?
(A) 39
(B) 48
(C) 76
(D) Cannot be determined
Solution:
Sum of four integers = 59 × 4 = 236 Let the required integers be x and x -28 Then, x + (x - 28) = 236 - (83 + 29) ⇒ 2x - 28 = 124 ⇒ 2x = 152 ⇒ x = 76 Hence, required integer = 76
4.
The average expenditure of a man for the first five months is Rs. 1200 and for the next seven months is Rs. 1300. If he saves Rs. 2900 in that year, his monthly average income is-
(A) Rs. 1500
(B) Rs. 1600
(C) Rs. 1700
(D) Rs. 1400
Solution:
Average expenditure of a man for the first five month = Rs. 1200 Average expenditure of a man for the next seven month = Rs. 1300 Total annual expenditure of man = Rs. (5 × 1200 + 7 × 1300) = Rs. (6000 + 9100) = Rs. 15100 Man saves = Rs. 2900 His total annual income = Rs. (15100 + 2900) = Rs. 18000 ∴ Average monthly income = $$\frac{18000}{12}$$ = Rs. 1500
5.
In 2011, the arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800. The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800, and the arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800. What is the arithmetic mean of the incomes of the three?
(A) Rs. 4000
(B) Rs. 4200
(C) Rs. 4400
(D) Rs. 4800
Solution:
Let a, b, and c be the annual incomes of Ramesh, Suresh, and Pratap, respectively. Now, we are given that The arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800. Hence, $$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 3800 ⇒ a + b = 2 × 3800 = 7600 The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800. Hence, $$\frac{{{\text{b}} + {\text{c}}}}{2}$$ = 4800 ⇒ b + c = 2 × 4800 = 9600 The arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800. Hence, $$\frac{{{\text{c}} + {\text{a}}}}{2}$$ = 5800 ⇒ c + a = 2 × 5800 = 11,600 Adding these three equations yields: (a + b) + (b + c) + (c + a) = 7600 + 9600 + 11,600 2a + 2b + 2c = 28,800 a + b + c = 14,400 The average of the incomes of the three equals the sum of the incomes divided by 3, $$\eqalign{ & \frac{{{\text{a}} + {\text{b}} + {\text{c}}}}{3} \cr & = \frac{{14,400}}{3} \cr & = {\text{Rs}}{\text{.}}\,4800 \cr} $$
6.
Average of first five odd multiples of 3 is
(A) 12
(B) 16
(C) 15
(D) 21
Solution:
According to the question, First five odd multiples of 3 is = 3, 9, 15, 21, 27 $$\eqalign{ & \therefore {\text{Average}} = \frac{{3 + 9 + 15 + 21 + 27}}{5} \cr & = \frac{{75}}{5} \cr & = 15 \cr} $$
7.
A cricketer had a certain average of runs for his 64 innings. In his 65th innings, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is :
(A) 130
(B) 128
(C) 70
(D) 68
Solution:
Let the average of runs for his 64 innings = x ∴ According to the question, $$\frac{64x + 0}{65}$$ = x - 2 64x = 65x - 130 x = 130 ∴ His new average is = 130 - 2 = 128
8.
The average weight of 3 men A, B, and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg more than that of D, replaces A, then the average weight of B, C, D and E becomes 79 kg. The weight if A is-
(A) 70 kg
(B) 72 kg
(C) 75 kg
(D) 80 kg
Solution:
Let A, B, C, D and E represent their respective weights. Then, A + B + C = (84 × 3) = 252 kgA + B + C + D = (80 × 4) = 320 kg ∴ D = (320 - 252) kg = 68 kg E = (68 + 3) kg = 71 kg B + C + D + E = (79 × 4) = 316 kg Now, = (A + B + C + D) - (B + C + D + E) = (320 - 316) kg= 4 kg ∴ A - E = 4⇒ A = (4 + E)⇒ A = 75 kg
9.
In an exam, the average marks obtained by Jhon in English, Math, Hindi and Drawing were 50. His average mark in Maths, Science, Social Studies and Craft were 70. If the average mark in all seven subjects is 58, his score in Maths was -
(A) 50
(B) 52
(C) 60
(D) 74
Solution:
E + M + H + D = 50 × 4 = 200 M + S.S + S + C = 70 × 4 = 280 $$\overline {{\text{M}} + (M + E + H + D + S.S + S + C) = 480} $$ M + (58 × 7) = 480 M = 480 - 406 M = 74 marks
10.
If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is: