Question Detail

A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17th match

Answer: Option D

Explanation:

Let the average after 17th match is x
then the average before 17th match is x-3

so 16(x-3) + 87 = 17x
=> x = 87 - 48 = 39

1. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students

Answer: Option C

Explanation:

Let the average weight of the 59 students be A.
So the total weight of the 59 of them will be 59*A.

The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs

\begin{aligned}
\frac{59A + 45}{60} = A - 0.2
\end{aligned}

=> 59A + 45 = 60A - 12
=> 45 + 12 = 60A - 59A
=> A = 57

2. A motorist travel to a place 150 km away at an avearge speed of 50 km/hr and returns ar 30 km/hr.His average speed for the whole jouney in km/hr is

36.5 km/hr
37.5 km/hr
35.5 km/hr
34.5 km/hr

Answer: Option B

Explanation:

Average speed will be

\begin{aligned}

\frac{2xy}{x+y} km/hr

= \frac{2(50)(30)}{50+30} km/hr
\end{aligned}

= 37.5 km/hr

3. Find the average of all numbers between 6 and 34 which are divisible by 5

Answer: Option B

Explanation:

\begin{aligned}
Average = (\frac{10+15+20+25+30}{5}) = \frac{100}{5} =20
\end{aligned}

4. The average score of a cricketer for ten matches is 38.9 runs. If the average for first six matches is 42, then average for last four matches is

33.25
32.25
34.25
34.50

Answer: Option C

Explanation:

\begin{aligned}
= \frac{(38.9 \times 10)-(42 \times 6)}{4}
\end{aligned}

\begin{aligned}
= \frac{(1216 - 750)}{4} = 34.25
\end{aligned}

5. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is

Answer: Option A

Explanation:

Number is (5*27) - (4*25) = 135-100 = 35