Test: CAT Logical Reasoning & Data Interpretation- 1 (October 9) - CAT MCQ

There are 5 ladies whose weights are 50 or above
There ages are 50,50, 70,60 and 80

Average = 310/5 = 62

For the highest possible value, the individual values must be equal to their maximum possible values.
x = 897 [number of people in the city]
y = 986 [number of people in the city]
z = 1034 [number of people in the city]
p = q = 564 [number of people in the city]
r = s = 1067 [number of people in the city]
x + y + z + p + q + r + s = 6179.
For the lowest possible value, the individual values must be equal to their minimum possible values.
x = 1 [total number of vehicles already exceeds the total number of people]
y = 1 [total number of vehicles already exceeds the total number of people]
z = 1 [total number of vehicles already exceeds the total number of people]
p and q are interdependent. p + q = 87 [If p+q = 87, only then can at least one person own one vehicle, that is, the number of vehicles = number of people]
r and s are interdependent. r + s = 1 + 1 = 2 [total number of vehicles already exceeds the total number of people]
Hence, x + y + z + p + q + r + s = 1 + 1 + 1 + 87 + 2 = 92
Difference = 6179 - 92 = 6087.

Bike owners in Chennai = 2816 – 707 – 971 – 432 – 342 = 364.

To own at least 3 vehicles, one can own either 3 or 4 vehicles.

I + II + III + IV = 1034

I + 2 II + 3 III + 4 IV = 1987

The difference has to be adjusted among people owning 2, 3 and 4 vehicles. To maximise the sum of people owning 3 and 4 vehicles, we will try to allocate the maximum possible to 3 and the remaining to 4.

1987 – 1034 = 953

2III + 3IV = 953.

III = 475

IV = 1

But, if we observe the values for Chennai, the number of people having a bike is 364 and the number of people having a private jet is 24. Hence, even if we consider that people who own a bike also own a 4-wheeler and a scooter(but not a private jet) and people who own a private jet also own a 4-wheeler and a scooter(but not a bike), we won’t be able to reach the above numbers. We would be able to achieve a maximum value of 24 + 364 = 388.
Let us verify if we can represent the above condition in a 4-set Venn Diagram.

Now, we need to arrange the remaining people who own a 4-wheeler and those who own a scooter.

I + II + III + IV = 1034

IV = 0, because we have already assigned all people who own a Private jet and a bike into III.

III = 388.

I + II = 1034 – 388 = 646

I + 2II + 3III + 4IV = 1987

I + 2II + 3 X 388 = 1987

I + 2II = 823

II = 823 – 646 = 177

I = 469

Hence, we get the following Venn Diagram:

Hence, maximum people who own 3 vehicles = 388.

I + II + III + IV = 986
I + 2 II + 3 III + 4 IV = 1265 + y
Subtracting the first from the second,
II + 2 III + 3 IV = 279 + y
163 + 72 + 3 IV = 279 + y
y = 3 IV - 44
Now, IV > III, so , IV > 36
But IV <= 57
For maximum y, we have to put maximum IV,
y = 3 x 57 - 44 = 171 - 44 = 127

In the following table, I represents people who own only one vehicle, II represents people who own two vehicles, III represents people who own three vehicles and IV represents people who own four vehicles.

We have to calculate the additional value for every city.
Then we try to allocate the maximum to IV for each city. If for a city, this value is more than any particular vehicle value, we take that value.
Hence sum =  326

Thus, in Hyderabad, the excess value is 152. We will try to allot maximum to IV, and automatically we will get the maximum I value.
152 = 3 x 50 + 2 x 1
Hence, 50 people own 4 vehicles and 1 person owns 3 vehicles.
Number of people left with 1 vehicle = 897 - 51 = 846
In Bengaluru, the excess value is 429.
But a maximum of 57 people can own all 4 vehicles. We will try to allocate the rest among III.
429 = 3 X 57 + 2 X 129
But 129 people cannot own 3 vehicles, because people owning a scooter is 150, and 57 + 129 exceeds this value.
Hence, the maximum number of people who can own 3 vehicles = 150 - 57 = 93.
Hence we are left with... 429 - 3 x 57 - 2 x 93 = 72
Hence, 57 people own 4 vehicles and 93 people own 3 vehicles, 72 people own 2 vehicles.
Number of people left with 1 vehicle =986 - 57 - 93 - 72 = 764.
Hence, difference = (846 + 764) - (50 + 57)
1503 is the right answer.

From the given information, we can make the following empty table:

From (ii) and (iii), we get that Q got F grade in match II and match IV.
From (v) and (vi), P got same grade in match II and match III. Now, P could obtain grade B in match II and match III (from (vi)).
Now, in every match, at least one century was scored; thus, R obtained grade A in match II and match IV (from (iv)).
Thus, grade C would be obtained by Q in match I, R in match III and P in match IV (from (iii) and (v)).
Now, remaining rows and columns can be filled giving the final table as follows:

From the table, we can see that the % decline in the rupee was maximum in the month of July among the options and also the stock index of UK was lowest in July among the given months so the maximum decline in the portfolio was in July.

From the given information, we can make the following empty table:

From (ii) and (iii), we get that Q got F grade in match II and match IV.
From (v) and (vi), P got same grade in match II and match III. Now, P could obtain grade B in match II and match III (from (vi)).
Now, in every match, at least one century was scored; thus, R obtained grade A in match II and match IV (from (iv)).
Thus, grade C would be obtained by Q in match I, R in match III and P in match IV (from (iii) and (v)).
Now, remaining rows and columns can be filled giving the final table as follows:

Maximum runs scored by Q in match I = 49
Maximum runs scored by Q in match II = 30
Average of maximum runs scored by Q in match I and match II = (49+30)/2 = 39.5

The percentage increase in the USA market index is the maximum among the three markets in a year.

Total C.P of magazines of Week 1 = 54 × 7.5 + 88 × 8 = 1,109
Total C.P of magazines of Week 2 = 34 × 8 + 88 × 6.75 + 2 × 6.5 = 879
Total C.P of magazines of Week 3 = 72 × 6.5 + 36 × 7.75 = 747
Total C.P of magazines of Week 4 = 84 × 7.75 + 30 × 6 = 831
Total C.P of magazines of Week 5 = 66 × 6 + 42 × 6.52 = 669
Total C.P of magazines of Week 6 = 22 × 6.52 + 82 × 8.50 = 840

As we have to see the maximum value the investment reaches, we need to consider two factors
1. INR per USD - the lower the INR per USD, the greater will be the value
2. Indian Stock Market Index - the higher will be the stock market in India, Greater will be the returns.
As we can see INR to USD is same in July and October that is 68 but the stock market is higher in October,
Hence, October is the right ans.

Total C.P of magazines of Week 1 = 54 × 7.5 + 88 × 8 = 1,109
Total C.P of magazines of Week 2 = 34 × 8 + 88 × 6.75 + 2 × 6.5 = 879
Total C.P of magazines of Week 3 = 72 × 6.5 + 36 × 7.75 = 747
Total C.P of magazines of Week 4 = 84 × 7.75 + 30 × 6 = 831
Total C.P of magazines of Week 5 = 66 × 6 + 42 × 6.52 = 669
Total C.P of magazines of Week 6 = 22 × 6.52 + 82 × 8.50 = 840

Thus, the number of magazines with profit > 20% is 410.

Among the given options since the dollar got weak the most on 1st May 2019 and also the US market index fell below the value of the index on 1st January so therefore the minimum value reaches was in 1st May 2019.

Total C.P of magazines of Week 1 = 54 × 7.5 + 88 × 8 = 1,109
Total C.P of magazines of Week 2 = 34 × 8 + 88 × 6.75 + 2 × 6.5 = 879
Total C.P of magazines of Week 3 = 72 × 6.5 + 36 × 7.75 = 747
Total C.P of magazines of Week 4 = 84 × 7.75 + 30 × 6 = 831
Total C.P of magazines of Week 5 = 66 × 6 + 42 × 6.52 = 669
Total C.P of magazines of Week 6 = 22 × 6.52 + 82 × 8.50 = 840

Lowest profit i.e. $138 was made in week 4.

In 2017-18 only the production of commercial vehicles is less than the domestic sales and exports, so only commercial vehicles are imported.
No. of commercial vehicles imported = 954 - 895 = 59

Total C.P of magazines of Week 1 = 54 × 7.5 + 88 × 8 = 1,109
Total C.P of magazines of Week 2 = 34 × 8 + 88 × 6.75 + 2 × 6.5 = 879
Total C.P of magazines of Week 3 = 72 × 6.5 + 36 × 7.75 = 747
Total C.P of magazines of Week 4 = 84 × 7.75 + 30 × 6 = 831
Total C.P of magazines of Week 5 = 66 × 6 + 42 × 6.52 = 669
Total C.P of magazines of Week 6 = 22 × 6.52 + 82 × 8.50 = 840

Total profit from week 3 to week 6 is $864.72.
Average for the 4 weeks = $216.18

The inventory of two-wheelers which was created in 2014-15 is used up in 2015-16. The production of two-wheelers was more than the domestic and sales of two-wheelers therefore,

No. of two-wheelers in inventory by the end of 2019-20 = (19934 - 19930) + (23155 - 23015) + (24500 - 244661) + (21036 - 20938) = 4+140+39+98= 281

Total C.P of magazines of Week 1 = 54 × 7.5 + 88 × 8 = 1,109
Total C.P of magazines of Week 2 = 34 × 8 + 88 × 6.75 + 2 × 6.5 = 879
Total C.P of magazines of Week 3 = 72 × 6.5 + 36 × 7.75 = 747
Total C.P of magazines of Week 4 = 84 × 7.75 + 30 × 6 = 831
Total C.P of magazines of Week 5 = 66 × 6 + 42 × 6.52 = 669
Total C.P of magazines of Week 6 = 22 × 6.52 + 82 × 8.50 = 840

Difference of profit of week 1 and week 2 = $361 - $240 = $121
Now, 121/361 × 100% = 33.52%

• Total vehicles imported in 2015-16 = (789-787) + (942-934) + (18939- 18830) = 2+8 + 109 = 119
• Total vehicles imported in 2016-17 = (3807-3802) + (822-810) = 5+12 = 17
• Total vehicles imported in 2017-18 = (4037-4020) + (954-895) = 17+59 = 76
• Total vehicles imported in 2018-19 = (4053-4028) = 25
• As we can clearly see 2015-16 has the highest number of imports.