Practice Test for NMAT - 8 - CAT MCQ

log₃x = 10 
Hence, x = 3¹⁰ 
logx y = 100 
y = x¹⁰⁰ = (3¹⁰)¹⁰⁰ = y = 3¹⁰⁰⁰

Solution: Ramesh tills his plot of land in 10 hours. Therefore, in every hour he tills 10% of the plot.
Suresh tills a plot double in size in 40 hours. So he can till a plot that is equal in size to Ramesh’s plot In 20 hours. Therefore, in every hour Suresh tills 5% of the plot.
Working alternately, in two hours Suresh and Ramesh till 5 + 10 = 15% of the plot.
In 12 hours they till 90% of the plot.
In the 13th hour Suresh tills 5% and in the next half an hour Ramesh tills 5%. So the total time required to till the plot is 13.5 hours.
Hence, option 4.

The cyclicity of the unit’s digit of powers of 3 is 4. 3⁴³⁴ = 3 ^{[(4 X 108)+ 2]}
Thus, the unit’s digit of 3⁴³⁴ is the same as the unit’s digit of 3² i.e. 9.
The cyclicity of the unit’s digit of powers of 4 is 2. 4²⁷³ = 4^{[(2 x 136) + 1]}
Thus, the unit’s digit of 4²⁷³ is the same as the unit’s digit of 4¹ i.e. 4. Therefore, the unit’s digit of 3⁴³⁴ - 4²⁷³ is 9 - 4 = 5.
Hence, option 4.

Among the players who got grade A, E scored the minimum number of runs and in grade B group, J took the minimum number of wickets. Hence, option 3.

Those players who scored more than E against at least three countries have performed better than E. C, D and F scored more runs than E against at least three countries.
Hence, option 2.

From the above table, Indian players scored maximum number of runs and took maximum number of wickets against Sri Lanka. Hence, option 1.

A scored 575 runs. Points for run = 575/5 = 115 He took 262 wickets.  Points for wickets = 262 x 5 = 1310

Total points = 115 + 1310 = 1425 From the table it is clear that, A, B, C, F, G and H earn more than 784 points.
Hence, option 4.

Solution: For 6893a98b to be divisible by 11 the difference between odd positioned digits and even positioned digits must be 0 or a multiple of 11. |(23 + a) - (20 + b)|= |3 + a - b| must either be a multiple of 11 or 0.
Now, as 6893a98b is a multiple of 8, 98b must be a multiple of 8. 98b will be a multiple of 8 only if b = 4.
Hence, |3 + a - b| = |a - 11| must be a multiple of 11 or 0.
But a is a single digit number. Therefore, a = 1 is the only solution that satisfies the condition.

Now, 592200c9d is a multiple of 8, c9d must be a multiple of 8.
So, d must be even.
Since 592200c9d is a multiple of 9 as well; sum of its digit must be either a multiple of 9 or 0. 5+ 9+ 2 + 2 + 9 + c + d = 27 + c + d So 27 + c + d must be a multiple of 9. As 27 is a multiple of 9,c + d must be either a multiple of 9 or 0.

Hence, we have three possible alternatives i.e., c + d = 18 or c + d = 9 or c + d = 0 .
c + d = 18 cannot be true because then d will be odd.
If c + d = 0, then both c and d will be 0 as both are positive digits.
The last three digits will be 090. But since 90 is not a factor of 8, c + d = 0 cannot be true. Only c + d = 9 satisfies the condition. a + b + c + d = 1 + 4 + 9 = 14 Hence, option 5.

Solution: Amir buys the sweets at Rs. x He then marks it up by 20% thus, the effective marked price = 1.2x Now, he has to gain 10% on his transaction.
Thus, his selling price should be 1.1x. Thus, discount given should be 1.2x - 1.1x = 0.1x This discount is given on the marked price.
Hence, discount given is (0.1x * 100)/1.2x = 8.33% Hence, option 3.

S_n = n / 2[2a + (n - 1)d]
∴ 15 / 2[2a + (15 - 1)d] = 40 + 7 / 2[2a + (7 - 1)d]
∴ 15(a + 7d) = 40 + 7(a + 3d)
∴ 15a + 105d = 40 + 7a +21d
∴ 8a + 84d = 40
∴ 2a + 21d = 10
∴ S₂₂ = 22 / 2(2a + 21d) = 11 x 10 = 110
Hence, option 3.

Solution: Let us assume that the initial total investment was Rs. 100.
Jacky invested Rs. 60 whereas Anil invested Rs. 40 for the first 8 months.

Let they received Rs. 18,000 profit after 8 + x months.
After 8 months, Jacky reduces his investment by 30%.
Now, 30% of 60 = (30 x 60)/100 = 18.
Thus, Jacky invests Rs. 42 for the next x months.
This means that Anil invests Rs. 58 for the next x months.Ratio of their investment = {(60 x 8) + (42 * x)} : {(40 x 8) + (58 * x)} As, they share the profit equally, ratio o f their investment after 8 + jc months should be 1 : 1. {(60 x 8) + (42 * x)} = {(40 x 8) + (58 * x)} . So 480 + 42 x = 320 + 58x = 10.

Thus they receive Rs. 18,000 as profit after 18 months.
Hence each person receives Rs. 9,000 for the 18 months.
Hence, their average monthly profit = 9000/18 = Rs. 500 Hence, option 1.

Since the profit percent in 2003 and 2004 is 25% and 40% respectively, ratio of profit and expenditure for 2003 and 2004 are 1 : 4 and 2 : 5 respectively.
Let the profit in 2003 and 2004 be x and 2y respectively.
Expenditure in 2003 and 2004 = 4x and 5y respectively and income in 2003 and 2004 = 5x and 7y respectively.
Since the income is the same in 2003 and 2004, 5x = 7y
∴ x / y = 7 / 5
Ratio of profit of year 2003 to2004 = x / 2y = 7 / 2 x 5 = 7 / 10

Since the profit percent in 2005 was 50%, the profit was half the expenditure.
If the profit in 2005 was x, the expenditure and income were 2x and 3x respectively.
Since the expenditure in 2005 was half the income in 2002, the income in 2002 was 2 x 2x i.e. 4x.Income in 2002 : Income in 2005 = 4x : 3x = 4 : 3 Hence, option 4.

Since the percentage profit in 2007 and 2006 was 25% and 30% respectively, the ratio of profit to expenditure for 2007 and 2006 was 1 : 4 and 3:10 respectively.
Since the expenditure in 2007 was Rs. 4 lakhs, the profit in 2007 would have been Rs. 1 lakh (as profit:expenditure = 1 : 4 in 2007). Profit in 2007 : Profit in 2006 = 2 : 3 Profit in 2006 = (3/2) x 1 = Rs. 1.5 lakhs

Expenditure in 2006 = (10/3) x 1.5 = Rs. 5 lakhs Hence, option 5.  

Let the annual expenditure for any one year be Rs.x. 
Total expenditure = Rs. 6x . For 2002, profit = (20/100) x Expenditure = Rs. 0.2x. For 2002, income = x + 0.2x = Rs. 12x Hence, total income = 1.2x + 1.25x + 1.4x + l.5x + 1.3x + 1.25x = Rs. 7.9x. ROI = 1.9x/6x= 1.3166 This is equivalent to an ROI of 31.67% Hence, option 3.

Net score of Suvidh = 60.  Marks obtained in section III = (30 x 1) - (0.5 x 10) = 25 To get the maximum number of incorrect answers, we must maximize the number of questions attempted incorrectly in section II.
Let Suvidh attempt all the questions in section II incorrectly.
So Marks obtained in section II = - (1 x 40) = - 40

Cumulative marks obtained by Suvidh in section II and section III = 25 + (-40) = - 15

Total marks obtained by Suvidh = 60 Marks obtained by Suvidh in section I = 60 - (- 15) = 75

Let Suvidh answer x questions in section I incorrectly.
Hence number of questions answered correctly in section I = 40 - x. ( ( 40 - x) * 3 ) - ( 1.5 x) = 75.  x = 10
Total number of incorrectly answered questions = 10 + 40 + 10 = 60. 

Hence, option 2.

Time taken to go from port Sakini to port Dakini = 20 hours. Let the distance between port Sakini to port Dakini = 4d. Distance between port Sakini and port Sankini = 1/4(4d). Time taken to go from port Sakini and port Sankini = 1/4(Time taken to go from between port Sakini to port Dakini) = 1/4(20) = 5 hours. Time taken to go from port Sakini to port Sankini and come back = 7 hours. Time taken to go from port Sankini to port Sakini = 7 - 5 = 2 hours Again, distance between port Sankini to Dakini = 4d - d = 3d. 

Time taken to go from port Dakini to port Sankini = 2 x 3 = 6 hours. Time taken to go from port Sankini to port Dakini = 5 x 3 = 15 hours. Time taken to go from port Dakini to port Sankini and come back = 6 + 15 = 21 hours Hence, option 5.

Let the number of students in three groups Group A, Group B and Group C be x, y and z respectively. 

Total points scored by students of Group A, Group B and Group C are 70x, 60y and 75z respectively.
According to the first condition: 60y + 75z = (y + z)65 ;y = 2z ... (i) According to the second condition: 70x + 75z = (x + z)72 So x=1.5z(ii) Average points obtained by students of the three groups together: 
70x + 60y + 75z / x + y + z = ​70(1.5z) + 60(2z) + 75z / 1.5z + 2z + z = 300z / 4.5z = 200 / 3 = 66 x 2/3
Hence, option 3.

Let x = 216³²
Taking log on both the sides,
log x = log 216³²
∴ log x = log (6³)³²
∴ log x = log 6⁹⁶
∴ log x = 96 log 6
log x = 96 x 0.7781 =74.69 Thus, there are 74 + 1 = 75 digits in 21632 Hence, option 4.

From the second table, the project lasts 15 months. 3a + b + c = 15 since a = 2, b + c = 9.

Now, depending on the value of c (i.e. maintenance stage) from the options, b can also take different values.
Hence, the month in which debugging starts will vary accordingly.
However, irrespective of the value of c, the specification stage will be in months 1 and 2 while the design stage will be in months 3 and 4. Cost of specification stage = 40000(2 + 3) = Rs. 2,00,000 and, cost of design stage = 20000(4 + 3) = Rs. 1,40,000 Cost of remaining three stages = 685000 - (200000 + 140000) = Rs. 3,45,000. 

Now, depending on the value of c from the options, calculate the total cost in each case.
Note that the programming stage will always start from month 5.
Case 1: c = 2 and b = 7 Thus, programming stage lasts from months 5 to 11. So  Cost of programming stage = 10000(4 + 5 + 5 + 4 + 4+ 1 + 3)= 10000 x 26 = Rs. 2,60,000. This leaves only Rs. 85,000 for the remaining stages.
Now, debugging will start from month 12 and last for two months. Cost of debugging stage = 15000(3 + 1) = Rs. 60,000.

This leaves only Rs. 25,000 for the last stage, which starts from month 14 and lasts for two months.  Cost of maintenance stage = 10000(1 + 1) = Rs. 20,000. Since this does not tally with the amount that was supposed to be spent for Maintenance (Rs.25,000), this case is invalid.

Case 1: c = 5 and b = 4 Thus, programming stage lasts from months 5 to 8. Cost of programming stage = 10000(4 + 5 + 5 + 4) = 10000 x 18 = Rs. 1,80,000
This leaves only Rs. 1,65,000 for the remaining stages.
Now, debugging will start from month 9 and last for two months. Cost of debugging stage = 15000(4 + 1) = Rs. 75,000 This leaves only Rs. 90,000 for the last stage, which starts from month 11 and lasts for five months. Cost of maintenance stage = 10000(3 + 3 + 1 + 1 + 1) = 10000 x 9 = Rs. 90,000. 

Since this tallies with the amount that was supposed to be spent for Maintenance (Rs.90,000), this case is valid.
Hence, the company provides maintenance for 5 months.
Hence, option 2.

Consider the solution to the previous question.
Cost of Programming stage = Rs. 1,80,000 Hence, option 5.

3a + b + c = 15 Since Programming and Maintenance take two-thirds the total time required for the remaining stages, b + c = (2/3) x (3a) = 2a.  3a + 2a = 15. a = 3,  b + c = 2(3) = 6 However, the split of b and c within this sum is not known.
Hence, the cost incurred on the programming stage cannot be found. Hence, option 5.

Consider the solution to the previous question. a = 3 and b + c = 6 Since, b = 2c, b = 4 and c = 2 Thus, the months in which each stage is carried out are: Specification - March to May.

Design - June to August, Programming - September to December, Debugging - January to March, Maintenance - April to May. Hence, the company would be working on Debugging in February.
Hence, option 1.

EF = EH and EG is the bisector of FH.
Hence, FG = GH Using the Apollonius' theorem, EF² + EH² = 2(EG² + FG²)

2EF² = 2(EG² + FG²) ,
EF² = EG² + FG² , 

Using statement A alone: EF is known but EG is not known.
Hence, FG cannot be found.
Thus, the question cannot be answered using statement A alone.
Using statement B alone: EG is known but EF is not known.
Hence, FG cannot be found.
Thus, the question cannot be answered using statement B alone.
Using statements A and B together:

Since both EF and EG are known, FG can be found.
Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 4.

Let the first three terms of the G.P. be air, a and ar where air is the first term.
Using statement A alone: 
a / r + a + ar = 13 ...(1)
This is a linear equation in two variables. So, the value of a and r cannot be found. Thus, the first term cannot be found.
Thus, the question cannot be answered using statement A alone.
Using statement B alone: 
a / r x a x ar = 27
∴ a3 = 27
∴ a = 3 ...(2)
But the value of r is unknown.
So, the first term cannot be found.
Thus, the question cannot be answered using statement B alone.
Using statements A and B together: Substitute the value of a from (2) in (1).

3 / r + 3 + 3r = 13 
3 + 3r² / r = 10 
∴ 3r² - 10r + 3 = 0
∴ r = 3 or 1 / 3
Based on the value of r, the first three terms of the G.P. are 1,3 and 9 or 9, 3 and 1.
Thus, a unique first term is not found.
Thus, the question cannot be answered on the basis of the two statements. Hence, option 5.

Let pipe A take a hours and pipe B take b hours. 
1 / a + 1 / b = 1 / 24
If the filling capacities of both the pipes are increased, then: 
2 / a + 3 / b = 1 / x
The time taken to fill the tank with new filling capacity will be less than the time taken to fill the tank if the capacity of both the pipes is doubled and greater than the time taken to fill the tank if the capacity of both the pipes is tripled. 
2(1 / a + 1 / b) < 2 / a + 3 / b < 3(1 / a + 1 / b)
2 / 24 < 1 / x < 1 / 8
∴ 8 < x < 12
Using statement A alone: Since x is a prime number, only value that x can take is 11. So the time taken by pipes to fill the tank is 11 hours.
Thus, statement A alone is sufficient to answer the question.
Using statement B alone: Since x is an integer greater than 10, only value that x can take is 11. So the time taken by the pipes to fill the tank is 11 hours.
Thus, statement B alone is also sufficient to answer the question. Hence, option 3.

Using statement A alone: 
p² > q²
p² - q² > 0
(p - q) (p + q) > 0
∴ Two possible cases are: p > q or p < q
Thus, statement A alone is not sufficient to answer the question. Using statement B alone:

|P| > |q| If (p, q) = (10, 8), then |p| > |q| and p > q. 

But if (p, q) = (-10, 8), then |p| > |q| and p < q.
Thus, statement B alone is also not sufficient to answer the question.
Using both the statements together: Even after using both the statements we cannot determine whether p > q.
Thus, combining both the statements also, the question cannot be answered. Hence, option 5.

Solution: Using statement A alone: Since Radha’s speed is not known, the time of completion cannot be found.

Thus, the question cannot be answered using statement A alone.
Using statement B alone: Radha reads without breaks at a uniform speed.
Also, Radha finishes a quarter of the book (i.e. 25%) at 11:45 a.m. and three- quarters of the book (i.e. 75%) at 5:15 p.m. on the same day.
Thus, Radha reads half the book in 5 hours and 30 minutes.
At 5:15 p.m., a quarter of the book is to be read. So, she takes another 2 hours and 45 minutes to read this part.
So, she finishes the book at 5:15 + 2:45 i.e. 8:00 p.m. on the same day.
Thus, the question can be answered using statement B alone. Thus, the question can be answered using statement B alone, but not by using statement A alone.
Hence, option 2.

Solution: Consider statement A: Let age of the kid after n years be x. Then age of father after n years is 4x and that of mother is 3x years. Now the age difference will always remain constant. 4x- 3x = 10. So x = 10
After n years the combined ages of parents will be 7x = 70 years Statement A alone is sufficient to answer the question. 

Consider statement B: After n years the kid is going to be twice as old as she is now. This means that right now the kid is n years old and after n years she will be 2n years old. Then age of father after n years is 4 x 2n and that of mother is 3 x 2n years.
F + M = 8n + 6n = 14n. But the value of n is not known, The sum of their ages cannot be determined using statement B alone. Statement B alone is not sufficient to answer the question.
Hence, option 1. 

Solution: The given set has three odd numbers and three even numbers. Since the required number has to have exactly two odd digits, the remaining three digits are even.
Thus, all three even numbers have to be selected.
Now, the odd digit at the unit’s place can be selected from 1, 7 and 9 in ³C₁ = 3 ways.
The other odd digit of the number can now be selected in ²C₁ = 2 ways.
This odd digit and the three even digits can be arranged among themselves in 4! ways i.e. 24 ways. Total number o f ways = 3 x 2 x24 = 144 Hence, option 2.