Solve the following averages questions for competitive exams:
The average of four consecutive even numbers A, B, C and D is 37. What is the product of A & C ?
(a) 1520
(b) 1368
(c) 1292
(d) 1224
(e) None of these
Answer: (c)
Let the even numbers A, B, C and D be x, x + 2, x + 4 and x+ 6 respectively.
Now, according to the question,
x + x + 2 + x + 4 + x + 6 = 4 \(\displaystyle \times \) 37
\(\displaystyle \Rightarrow \)4x + 12 = 148
\(\displaystyle \Rightarrow \)4x = 148 – 12 = 136
\(\displaystyle \Rightarrow \)\(\displaystyle x=\frac{{136}}{4}=34\)
Therefore, A = 34 and C = (34 + 4 =) 38
A \(\displaystyle \times \) C = 34 \(\displaystyle \times \) 38 = 1292
Alternate Method :
Average of consecutive even or odd numbers = a + (n–1), where a is the smallest number.
So, average of four consecutive even numbers = a + (4 – 1)
or, 37 = a + 3
or, a = 37 – 3 = 34
Therefore, A = 34, B = 36, C = 38, D = 40
\(\displaystyle \Rightarrow \) Product of A and C = 34 \(\displaystyle \times \) 38 = 1292
Sushil scared 103 marks in Hindi, 111 marks in Science, 98 marks in Sanskrit, 110 marks in Maths and 88 marks in English. If the maximum marks of each subject are equal and if Sushil scored 85 percent marks in all the subjects together, find the maximum marks of each subject
(a) 110
(b) 120
(c) 115
(d) 100
(e) None of these
Answer: (b)
Let the sum of maximum marks of all the subjects be ‘x’:
or, \(\displaystyle \frac{{x\times 85}}{{100}}=(103+111+98+110+88)\)
or, x = \(\displaystyle \frac{{510\times 100}}{{85}}=600\)
\(\displaystyle \Rightarrow \)Maximum marks of each subject = \(\displaystyle \frac{{600}}{5}=120\)
The average height of 21 girls was recorded as 148 cm. when the teacher’s height was included, the average of their heights increased by 1 cm. What was the height of the teacher?
(a) 156 cm
(b) 168 cm
(c) 170 cm
(d) 162 cm
(e) None of these
Answer: (c)
Sum of the heights of all the girls = 148 × 21 = 3108 cm
Sum of the heights of the teacher and all the girls
= 149 × 22 = 3278 cm
Teacher’s height = 3278 – 3108 = 170 cm
The average of four consecutive odd numbers A, B, C and D respectively is 40. What is the product of B and D?
(a) 1599
(b) 1591
(c) 1763
(d) 1677
(e) None of these
Answer: (d)
Let the four odd consecutive numbers be
x, x + 2, x + 4, and x + 6.
Also, A = x, B = x + 2, C = x + 4 and D = x + 6
Therefore, 4x + 12 = 4 × 40
or, 4x = 160 – 12 = 148
or, x = \(\displaystyle \frac{{148}}{4}=37\)
Therefore, Numbers are A = 37, B = 39, C = 41, D = 43
\(\displaystyle \Rightarrow \) Product of B and D = 39 \(\displaystyle \times \)43 = 1677
Alternative Method
Let the four consecutive odd numbers be
A, B, C, D
x+1, x+3, x+5, x+7
According to questions,
\(\displaystyle \frac{{x+1+x+3+x+5+x+7}}{4}=40\)
\(\displaystyle \frac{{4x+16}}{4}=40\)
\(\displaystyle 4x+16=160\)
\(\displaystyle 4x=144\)
x=36
The average of four consecutive numbers A, B, C and D is 49.5. What is the product of B and D?
(a) 2499
(b) 2352
(c) 2450
(d) 2550
(e) None of these
Answer: (a)
Let A = x, B = x + 1, C = x + 2, D = x + 3
\(\displaystyle \Rightarrow \) x + x + 1 x + 2 + x = 198
or, 4x + 6 = 198
or, x = 48
\(\displaystyle \Rightarrow \) B = 49, D = 51
Product of B and D = 49 × 51 = 2499
