7. A person covers 800km partly at a speed of 50kmph and partly at a speed of 150kmph, in 10 hours over all. What is the distance covered at the speeds of 150 kmph?
a. 450 km
b. 480 km
c. 510 km
d. 540 km

Answer: A

Explanation:
His average speed for the entire journey is $\displaystyle\frac{{800}}{{10}} = 80$
Now we need to find in what ratio he needs to travel 800 km, partly at 50 kmph, and 150 kmph to get average speed of 80 kmph.
So for every 70 parts of the time travelled at 50kmph, he has to travel 30 parts are 150 kmph.
As the total time is 10 hours he must have travelled $\displaystyle\frac{{30}}{{70 + 30}} \times 10 = 3$ hours
So distance travelled $= 150 × 3 = 450$

8. The cost of oil is Rs.100 per kilogram. After adulteration with another oil, which costs Rs.50 per kilogram, Ram sells the mixture at Rs.96 per kilogram making a profit of 20%. In what rate does he mix the two kinds of oil?
a. 3 : 5
b. 4 : 7
c. 5 : 7
d. 3 : 2

Answer: D

Explanation:
Before applying the alligation rule we need to make sure that all the parameters are in the same units. Here 2 costs prices and 1 selling price was given. So we should convert the selling price into cost price.
Given $C.P \times (100 + 20)\% = 96$
$ \Rightarrow C.P = \displaystyle\frac{{96}}{{120\% }}$ $ \Rightarrow C.P = \displaystyle\frac{{96}}{{\displaystyle\frac{{120}}{{100}}}} \Rightarrow 96 \times \displaystyle\frac{{100}}{{120}} = 80$
Now we apply the alligation rule.

So These two must be mixed in the ratio 3:2.

9. Two solutions of milk and water are combined in the ratio 2:3 by volume. The resultant solution is a 40% milk solution. Find the milk concentration in the first solution if the concentration of milk in the second is 60%?
a. 8
b. 10
c. 12
d. 20

Answer: B

Explanation:
Let the quantities of two solutions are $2k, 3k$ and first solution concentration is $x%$.
Given, Second solution concentration is $60%$ and resultant solution concentration is $40%$.
$ \Rightarrow \dfrac{{2k}}{{3k}} = \dfrac{{20}}{{40 - x}}$
$\require{cancel} \Rightarrow \dfrac{{2\cancel{k}}}{{3\cancel{k}}} = \dfrac{{20}}{{40 - x}}$
$ \Rightarrow 80 - 2x = 60$
$ \Rightarrow x = 10$

10. A shopkeeper sold 45 kgs of goods. If he sells some quantity at a loss of 3% and rest at 17% profit, making 5% profit on the whole, find the quantity sold at profit.
a. 15
b. 18
c. 19
d. 20

Answer: B

Explanation:
Therefore, Ratio of quantities sold at loss and at profit $= 12 : 8 = 3 : 2$
Therefore, Quantity sold at profit = $\displaystyle\frac{{\rm{2}}}{{\rm{5}}} \times 45 = 18$ kgs.

11. A mixture of 70 litres of wine and water contains 10% water. How much water should be added to make 25% water in the resulting mixture?
a. 10
b. 12
c. 14
d. 18

Answer: C

Explanation:

Therefore, The ratio is 75 : 15 = 5 : 1.
Therefore, For every 5 litres of mixture, 1 litre of water is added.
Therefore, For 70 litres, water to be added = $\displaystyle\frac{{\rm{1}}}{{\rm{5}}}$ x 70 = 14 litres.

Alternative Method:
Water = 10% of 70 litres = 7 litres
Wine = 70 - 7 = 63 litres
Now, in the new mixture, water is 25% and wine is 75%.
Hence, in new mixture wine is 3 times of water.
Therefore, Water in new mixture = $\displaystyle\frac{{\rm{1}}}{{\rm{3}}}$ x 63 = 21 litres
Therefore, Water to be added = 21 - 7 = 14 litres

12. A man purchased a horse and a cow for Rs. 5000. He sells the horse at 20% profit and the cow at 10% loss. If he gains 2% on the whole transaction, the cost of the horse is:
a. Rs.2000
b. Rs.1500
c. Rs.1000
d. Rs.800

Answer: A

Explanation:

Therefore, Ratio between cost of a horse and that of a cow = 12 : 18 = 2 : 3.
Therefore, Cost of the horse = $\displaystyle\frac{{\rm{2}}}{{\rm{5}}}$ x 5000 = Rs. 2000

13. In an examination, a student gets 3 marks for every correct answer and loses 1 mark for every wrong answer. If he scores '0' marks in a paper of 100 questions, how many of his answers were correct?
a. 25
b. 30
c. 40
d. 45

Answer: A

Explanation:

Therefore, Ratio of correct and wrong answers = 1 : 3
Therefore, Correct answers = $\displaystyle\frac{{\rm{1}}}{{\rm{4}}}$ x 100 = 25

14. The batting average of a cricket player is 72 runs per inning. In the next 4 inning, he could score only 80 runs and thereby decreases his batting average by 2 runs. What is total number of inning played by him till last match?
a. 104
b. 108
c. 305
d. 406

Answer: A

Explanation:
Average of last 4 inning = $\displaystyle\frac{{{\rm{80}}}}{{\rm{4}}}$ = 20 runs
Average of inning played earlier = 72 runs
New average = 72 - 2 = 70 runs

Ratio of inning played (before and after) = 50 : 2 = 25 : 1
Given, innings played (after) = 4
Therefore, Inning played (before) = 25 x 4 = 100
Therefore, Total inning played = 100 + 4 = 104

Alternative Method:
Average runs in the last 4 inning = $\displaystyle\frac{{80}}{4}$ = 20 runs
Short from previous average = 72 - 20 = 52
Short (total) = 52 x 4
But, short runs (per inning) = 2 runs
Therefore, Total inning played = $\displaystyle\frac{{52 \times 4}}{2}$ = 104 innings

15. The average weight of the students in four sections A, B, C and D is 55 kg. The average weights of the students A, B, C and D are 45 kg, 40 kg, 75 kg and 85kg respectively. If the average weight of the students of section A and D together is 65 kg and that of B and D together 55 kg, then what is the ratio of the number of students in A and C?
a. 5 : 1
b. 6 : 1
c. 2 : 1
d. 8 : 1

Answer: C

Explanation:
If we observe carefully, the total average of A, B, C, D is equal to Average of B and D together. So the average of A and C must be 55.
We apply alligation rule to determine in what ration A and C must be clubbed together to get an average of 55.