Free DI and LR - 02 Practice Test - CAT 

The answer is fastest obtained by the process of elimination. Lets take for example, that 360 is the total, then saurav will score 54 runs, this implies that the number of 4’s he scores will not be an integral number $(\frac{140}{3}\times \frac{54}{(4\times 100)})$ which is not integral. Similarly, the other options can be eliminated. 400 will be the total score

Once the total score is obtained as 400 the score table will look as follows.

$\begin{array}{cc}\text{PLAYER}& \text{SCORE}\\ \text{Saurav}& 60\\ \text{Sehwag}& 40\\ \text{Sachin}& 160\\ \text{Rahul}& 80\\ \text{Others}& 60\end{array}$

Number of 6’s scored by Sehwag= $\frac{45}{100}$$\times$40= $\frac{18}{6}$ = 3

The number of 4’s scored by Rahul in the innings is, $\frac{40}{100}$ $\times$ 80 = $\frac{32}{4}$ = 8

The table is obtained as follows

$\begin{array}{cc}\text{PLAYER}& \text{No of 6's}\\ \text{Saurav}& 1\\ \text{Sehwag}& 3\\ \text{Sachin}& 4\\ \text{Rahul}& 2\\ \text{Others}& 4\end{array}$

Total= 14

Out of 11 players, 4 are top-scorers means that 7 players together can score a maximum of 39 each. Score left to distribute among 7 players= 60, can be distributed among 2 players, hence 7-2=5 will the max no. of players who will not score any runs

All vehicles other than others pay a parking fee.
2 wheelers = 14% of 200000 = 28000
3 wheelers = 17% of 50000 = 8500
Cars = 18% of 150000 = 27000
Vans = 16% of 75000 = 12000
Trucks 19% of 55000 = 10450
Total = 85950

Calculate the parking fee for each of the years using the same logic as above 2005 = (16% of 200000) $\times$ 10 + (14% of 50000)$\times$ + (13% of 150000) $\times$ 30 + (15% of 75000) $\times$ 40 + (20% of 55000) $\times$ 50 = 2045000 2002 = (15% of 200000) $\times$ 10 + (16% of 50000)$\times$ 20 + (15% of 150000) $\times$ 30 + (17% of 75000) $\times$ 40 + (18% of 55000) $\times$ 50 = 2140000 2001 = (15% of 200000) $\times$ 10 + (16% of 50000) $\times$ 20 + (19% of 150000) $\times$ 30 + (17% of 75000) $\times$ 40 + (15% of 55000) $\times$ 50 = 2237500 2003 = (14% of 200000) $\times$ 10 + (17% of 50000) $\times$ 20 + (18% of 150000) $\times$ 30 + (16% of 75000) $\times$ 40 + (19% of 55000) $\times$ 50 = 2262500 Hence answer = 2003

Parking fee collected by trucks can be calculated as = 55000 x 50=2750000
Parking fee collected by 2 wheelers can be calculated as= 200000 x 10 = 2000000
Percentage difference in fee collected from trucks different from the parking fee collected by 2 wheelers = (750000/2000000)*100 =37.5%

2 wheelers = (16 % of 200000)$\times$10 = 320000
3 wheelers = (14 % of 50000)$\times$20 = 140000
Total =460000
total parking fee in 2005 (refer the 1${}^{st}$ question of this caselet) = 2045000
Percentage contribution of 2 wheelers and 3 wheelers to the parking fee in 2005 = $\frac{460000}{2045000}$$\times$100=22.5%

Out of 15 candidates, except Alok, Ravi and Pankaj, all the others scored 10 points or A+ grade in test 5. Hence , [2].

Vijay, Sagar, Shekhar, Akshay, Salman and Vinod i.e. 6 candidates got A+ in at least two of the five tests. Paresh may or may not have scored 10 points in at least two out of the five tests. 

Shakhar scored A+ in tests 3, 4 and 5 and Pankaj scored D grade in tests 1, 2 and 3. Ravi may or may not have scored the same grades in three consecutive tests. Thus option [1] may or may not be true.

Ravi, Mohit, Sagar, Shakhar and Vinod did not score D grade in any of the tests.

Thus, option [2] is also true.

The average score of the 15 candidates in tests 2 is $\frac{27}{15}$ = 1.8 or $\frac{30}{15}$ = 2 or $\frac{33}{15}$ = 2.2.

Thus, option [3] can be true.

The total score of all the 15 candidates in Test 3, except Ravi, Akshay and Paresh, is 56. Considering all the possibilities of the points scored by these three in Test 5, none of them add up to 19. Thus the average of 5 cannot be achieved.