averages Model Questions & Answers, Practice Test for ssc chsl tier 1
The average of four consecutive numbers A, B, C and D respectively is 56.5. What is the product of A and C?
Answer: (e)
${x + x + 1 + x + 2 + x + 3}/4$ = 56.5
4x + 6 = 56.5 × 4
4x = 226 – 6 = 220
$x = 220/4$ = 55
A = 55, L = 57
Product = 55 × 57 = 3135
The average age of a lady and her daughter is 28.5. The ratio of their ages is 14 : 5 respectively. What is the daughters age?
Answer: (a)
∴ Average age = 28.5
∴ Total age = 28.5 × 2 = 57
∴ Daughter's age = $5/19 × 57$ = 15 years
If the value of 16a + 16b = 672, what is the average of a and b?
Answer: (c)
16a + 16b = 672
or, 16 (a + b) = 672
∴ a + b = $672/16$ = 42
Required average = ${a + b}/2 = 42/2$ = 21
Eighteen years ago, a father was three times as old as his son. Now the father is only twice as old as his son. Then the sum of the present ages of the son and the father is:
Answer: (d)
Let the present ages of the father and son be 2x and x years respectively.
Then, (2x – 18) = 3 (x – 18) ⇔ x = 36.
∴ Required sum = (2x + x) = 3x = 108 years.
Out of the three given numbers, the first number is twice the second and thrice the third. If the average of the three numbers is 121, what is the difference between the first and the third number?
Answer: (b)
Let the third number be = x
∴ First number = 3x and
second number = ${3x}/2$
According to the question.
or, 3x + ${3x}/2$ + x = 3 × 121
or, ${6x + 3x + 2x}/2$ = 3 × 121
or, ${11x}/2$ = 3 × 121
∴ $x = {3 × 121 × 2}/11$ = 66
∴ Third number = 66
Required difference = 3x – x = 2x = 2 × 66 = 132
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