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1.Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is:

Explanation:

Solution:
Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings.
In the 10th inning, he scored 100 runs then average became (x+8). And he scored (x+8)*10 runs in 10 innings.
Now,
$\begin{array}{rl}& ⇒9x+100=10×\left(x+8\right)\\ & \text{or},\phantom{\rule{thinmathspace}{0ex}}9x+100=10x+80\\ & \text{or},\phantom{\rule{thinmathspace}{0ex}}x=100-80\\ & \text{or},\phantom{\rule{thinmathspace}{0ex}}x=20\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{average}=\left(x+8\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=28\phantom{\rule{thinmathspace}{0ex}}\text{runs}\end{array}$

2.The average temperature for Wednesday, Thursday and Friday was 40oC. The average for Thursday, Friday and Saturday was 41oC. If temperature on Saturday was 42oC, what was the temperature on Wednesday?

Explanation:

Solution:
Average temperature for Wednesday, Thursday and Friday = 40oC
Total temperature = 3*40 = 120oC
Average temperature for Thursday, Friday and Saturday = 41oC
Total temperature = 41*3 = 123oC
Temperature on Saturday = 42oC
Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123-120;
Saturday - Wednesday = 3
Wednesday = 42-3 = 39oC.

3.The average of the first five multiples of 9 is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{average}\\ & =\left(\frac{\text{total}\phantom{\rule{thinmathspace}{0ex}}\text{sum}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{multiple}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}9}{5}\right)\\ & =\left(\frac{9+18+27+36+45}{5}\right)\\ & =27\end{array}$

Note that, average of 9 and 45 is also 27.
And average of 18 and 36 is also 27.

4.Find the average of first 97 natural numbers.

Explanation:

Solution:
1st Method:
Average of 1st n natural number is given by
$=\frac{\left(\frac{\left[\text{n}×\left(\text{n}+1\right)\right]}{2}\right)}{\text{n}}$

Average of 1st 97 natural number is given by
$\begin{array}{rl}& \left\{\frac{\left(\frac{\left[97×\left(97+1\right)\right]}{2}\right)}{97}\right\}\\ & =49\end{array}$

2st Method:
These numbers are in AP series so average.

5.The average age of three boys is 15 years. If their ages are in ratio 3:5:7, the age of the youngest boy is

Explanation:

Solution:
$\begin{array}{rl}& \text{Sum}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{ages}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{three}\phantom{\rule{thinmathspace}{0ex}}\text{boys}=\phantom{\rule{thinmathspace}{0ex}}45\phantom{\rule{thinmathspace}{0ex}}\text{years}\\ & \text{Now},\phantom{\rule{thinmathspace}{0ex}}\left(3x+5x+7x\right)=45\\ & Or,\phantom{\rule{thinmathspace}{0ex}}15x=45\\ & Or,\phantom{\rule{thinmathspace}{0ex}}x=3\\ & \text{So,}\phantom{\rule{thinmathspace}{0ex}}\text{age}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{youngest}\phantom{\rule{thinmathspace}{0ex}}\text{boy}\\ & =3x=3×3=9\phantom{\rule{thinmathspace}{0ex}}\text{years}\end{array}$

6.The average of a group of men is increased by 5 years when a person aged of 18 years is replaced by a new person of aged 38 years. How many men are there in the group?

Explanation:

Solution:
Let N be the no. of persons in the group.
Required number of person is given by;
Member in group* aged increased = difference of replacement
N*5 = 38-18
Or, 5N = 20
Or, N = 4.

7.In a boat there are 8 men whose average weight is increased by 1 kg when 1 man of 60 kg is replaced by a new man. What is weight of new comer?

Explanation:

Solution:
Member in group * age increased = difference of replacement
Or, 8*1 = new comer - man going out
Or, new comer = 8+60;
Or, new comer = 68 years.

8.A train covers the first 16 km at a speed of 20 km per hour another 20 km at 40 km per hour and the last 10 km at 15 km per hour. Find the average speed for the entire journey.

Explanation:

Solution:
$\begin{array}{rl}& \text{Average}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\\ & =\frac{\text{total}\phantom{\rule{thinmathspace}{0ex}}\text{distance}\phantom{\rule{thinmathspace}{0ex}}\text{covered}}{\text{total}\phantom{\rule{thinmathspace}{0ex}}\text{time}}\\ & \text{Total}\phantom{\rule{thinmathspace}{0ex}}\text{Distance}\\ & =\left(16+20+10\right)=46\phantom{\rule{thinmathspace}{0ex}}km\\ & \text{Time}\phantom{\rule{thinmathspace}{0ex}}\text{taken}\\ & =\left(\frac{16}{20}\right)+\left(\frac{20}{40}\right)+\left(\frac{10}{15}\right)\\ & =\frac{59}{30}\\ & \text{Average}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\\ & =\frac{46×30}{59}\\ & =23\left(\frac{23}{59}\right)\phantom{\rule{thinmathspace}{0ex}}km/hr\end{array}$

9.A man has 'n' magical eggs whose average weight is 'k' gm. Each of the 'n' eggs produces 'n' eggs next day such that the average weight of 'n' eggs produced is same as that of the parental egg for each 'n' groups individually i.e. each egg produces 'n' eggs of next generation and average weight of all the 'n' eggs of next generation is same as the weight of the mother egg. This process is continued without any change in pattern. What is the total weight of all the eggs of rth generation, where the initial number of eggs with man are considered as the eggs of first generation.

Explanation:

Solution:
The weight is increasing in form of GP so the total weight of eggs in the end of rth will be nrk.

10.A man started his journey from Lucknow to Kolkata, which is 200 km, at the speed of 40 kmph then he went to Banglore which is 300 km, at the speed of 20 kmph. Further he went to Ahmedabad which is 500 km, at the speed of 10 kmph. The average speed of the man is :

$\begin{array}{rl}& \text{Average}\phantom{\rule{thinmathspace}{0ex}}\text{speed},\\ & =\left(\frac{\text{Total}\phantom{\rule{thinmathspace}{0ex}}\text{Distance}}{\text{Total}\phantom{\rule{thinmathspace}{0ex}}\text{time}}\right)\\ & =\left[\frac{\left(200+300+500\right)}{\left\{\left(\frac{200}{40}\right)+\left(\frac{300}{20}\right)+\left(\frac{500}{10}\right)\right\}}\right]\\ & =\frac{1000}{70}\\ & =14\left(\frac{2}{7}\right)\phantom{\rule{thinmathspace}{0ex}}kmph\end{array}$