averages Model Questions & Answers, Practice Test for ssc cgl tier 1
ssc cgl tier 1 SYLLABUS WISE SUBJECTS MCQs
Computation of whole numbers, decimals, fractions and relationships between numbers
Percentage
Averages
If the average of 9 consecutive positive integers is 55, then what is the largest integer?
Answer: (b)
Let integers be x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6, x + 7 and x + 8 respectively
According to the question
${x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 + x + 7 + x + 8}/9 = 55$
9x + 36 = 495
9x = 495 – 36 = 459
$x = 459/9 = 51$
Hence, largest integer = 51 + 8 = 59
If the average of A and B is 30, the average of C and D is 20, then which of the following is/are correct? I. The average of B and C must be greater than 25. II. The average of A and D must be less than 25. Select the correct answer using the codes given below.
Answer: (d)
Average of A and B = 30
⇒ ${A + B}/2$ = 30 ⇒ A + B = 60
Average of C and D = 20
⇒ ${C + D}/2$ = 20 ⇒ C + D = 40
Here, we can't find the avg of B and C, A and D
so that Neither I or II are follows.
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 the same today. The present age of the baby is
Answer: (a)
Let total age of family be S years
3 years ago, total age = S – 3 × 5 = S – 15
${S - 15}/5$ = 17
S = 17 × 5 + 15 = 100
Let present age of baby be x years
${S + x}/6$ = 17
100 + x = 17 × 6
x = 102 – 100 = 2 years
A man's average expenditure for the first 4 months of the year was Rs. 251.25. For the next 5 months the average monthly expenditure was Rs. 26.27 more than what it was during the first 4 months. If the person spent Rs. 760 in all during the remaining 3 months of the year, find what percentage of his annual income of Rs. 3000 he saved in the year.
Answer: (b)
251.25*4 + 277.52 * 5 + 760 = 3152.6
The age of a man is three times the sum of the ages of his two sons. Five years hence, his age will be double of the sum of the ages of his sons. The father's present age is
Answer: (a)
Let the father's present age be x and son's age be x 1 and x 2 .
Now, $x = 3(x_1 + x_ 2)$ .....(i)
Also, $x + 5 = 2(x_1 + 5 + x_2 + 5)$
$x + 5 = 2(x_1 + x_2 + 10)$ .....(ii)
Putting value of $(x_1 + x_2) = x/3$ from (i) in equation (ii)
$x + 5 = 2(x/3 + 10) = 45$
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