Averages Quiz Set 018 | Maths Aptitude Mock Test on Averages Set 18 | mock tests for Bank PO, SSB/SSC Exams, SO Clerical Exam, GATE Exam, LIC AO, GIC AO, etc.

**Your Score**
Correct Answers:
Wrong Answers:
Unattempted:

Question 1

Average marks of class of 10 students is 62. What will be the average if each student is given 8 as grace marks?

70.

71.

69.

68.

Soln.
Ans: a

Let the total score of the class before grace marks be x. Then, by average formula $62 = x/10$, which gives x = $62 × 10 = 620$. When grace marks = 8 are added for each of the 10 students, the new total becomes $620 + 10 × 8 = 700$, the new average becomes $700/10 = 70$. TIP: As a shortcut, the new average = old average + grace marks.

Solution | Discuss! | Chapters»

Question 2

The cost per unit of a commodity in three successive years is Rs.16/unit, Rs.4/unit and Rs.7/unit. If the annual spending of a family remains fixed, what is the average cost per unit for all the three combined years together?

$6{10/17}$.

$6{3/4}$.

$5{2/5}$.

Soln.
Ans: a

Let the annual spending be Rs. M. The catch in this question is that the spending remains fixed, so the consumption varies from year to year. We shall calculate the total consumption first. Let r1, r2 and r3 be the rates for the three successive years. Consumption in first year = M/r1. Similarly, we get M/r2 and M/r3. So total consumption is $M/{r1} + M/{r2} + M/{r3}$. Money spent in three years is 3M. So the required average = ${3M}/{M/{r1} + M/{r2} + M/{r3}}$ which simplifies to ${3r1r2r3}/{r1r2 + r2r3 + r3r1}$. Putting r1 = 16, r2 = 4, r3 = 7, we get $6{10/17}$. You might be wondering why I derived the formula first. The reason is that sometimes it is better to postpone calculations till the end.

Solution | Discuss! | Chapters»

Question 3

The cost per unit of a commodity in three successive years is Rs.14/unit, Rs.6/unit and Rs.16/unit. If the annual spending of a family remains fixed, what is the average cost per unit for all the three combined years together?

$9{99/101}$.

12.

$7{1/5}$.

Soln.
Ans: a

Let the annual spending be Rs. M. The catch in this question is that the spending remains fixed, so the consumption varies from year to year. We shall calculate the total consumption first. Let r1, r2 and r3 be the rates for the three successive years. Consumption in first year = M/r1. Similarly, we get M/r2 and M/r3. So total consumption is $M/{r1} + M/{r2} + M/{r3}$. Money spent in three years is 3M. So the required average = ${3M}/{M/{r1} + M/{r2} + M/{r3}}$ which simplifies to ${3r1r2r3}/{r1r2 + r2r3 + r3r1}$. Putting r1 = 14, r2 = 6, r3 = 16, we get $9{99/101}$. You might be wondering why I derived the formula first. The reason is that sometimes it is better to postpone calculations till the end.

Solution | Discuss! | Chapters»

Question 4

If the average of p and q is 24, the average of q and r is 52, and of r and p is 60, then what is the value of p?

32.

33.

31.

34.

Soln.
Ans: a

We have three equations (p + q)/2 = average of pq, (q + r)/2 = average of qr and (r + p)/2 = average of rp. Adding these three we get p + q + r = (average of pq + average of qr + average of rp) = (24 + 52 + 60) = 136. So p = 136 - (q + r) = 136 - (2 × average of q and r) = 136 - 2 × 52 = 32.

Solution | Discuss! | Chapters»

Question 5

The sales(in rupees) of a karyana store for five consecutive days is 1092, 1548, 4638, 3372, 984. What should be the sale on the sixth day so that the overall average sale is 2543?

Rs.3624.

Rs.3630.

Rs.3618.

Rs.3636.

Soln.
Ans: a

The total sale on first five days is 1092 + 1548 + 4638 + 3372 + 984 = 11634. Let the sale on 6th day be x. The average for 6 days is: 2543 = ${11634 + x}/6$ which gives x = $6 × 2543 - 11634$ = Rs. 3624.

Solution | Discuss! | Chapters»