# Averages Quiz Set 004

### Question 1

The sales(in rupees) of a karyana store for five consecutive days is 2262, 666, 1740, 3216, 4926. What should be the sale on the sixth day so that the overall average sale is 2281?

A

Rs.876.

B

Rs.882.

C

Rs.870.

D

Rs.888.

Soln.
Ans: a

The total sale on first five days is 2262 + 666 + 1740 + 3216 + 4926 = 12810. Let the sale on 6th day be x. The average for 6 days is: 2281 = \${12810 + x}/6\$ which gives x = \$6 × 2281 - 12810\$ = Rs. 876.

### Question 2

The cost per unit of a commodity in three successive years is Rs.14/unit, Rs.8/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?

A

\$11{17/29}\$.

B

\$12{2/3}\$.

C

\$9{1/2}\$.

D

\$7{3/5}\$.

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 = 8, r3 = 16, we get \$11{17/29}\$. You might be wondering why I derived the formula first. The reason is that sometimes it is better to postpone calculations till the end.

### Question 3

Each of the 4 bulbs of a batch has a life 81 years. If a new bulb with a life 44 years more than the average of all 5 bulbs is added, what is the sum total of the values of all 5 bulbs?

A

460 years.

B

461 years.

C

459 years.

D

458 years.

Soln.
Ans: a

Let the average of all 5 bulbs be x, then the required total is T = 5x. By averages, \$x = {4 × 81 + (x + 44)}/5\$ which is same as \$x × 5 = {4 × 81 + (x + 44)}\$, which is same as \$T = {4 × 81 + (T/5 + 44)}\$, solving for T we get 460 years.

### Question 4

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?

A

\$6{10/17}\$.

B

9.

C

\$6{3/4}\$.

D

\$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.

### Question 5

If the average of p and q is 38, the average of q and r is 56, and of r and p is 72, then what is the value of p?

A

54.

B

55.

C

53.

D

56.

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) = (38 + 56 + 72) = 166. So p = 166 - (q + r) = 166 - (2 × average of q and r) = 166 - 2 × 56 = 54.

Updated on 2017-05-17.

Posted by Parveen(Hoven),
Aptitude Trainer