Practice Test for NMAT - 9 - CAT MCQ

Based on the given ratios, let the volume of each jug be 45 litres.
Hence, jug 1: juice = 27 litres and water =18 litres jug 2: juice = 25 litres and water = 20 litres jug 3: juice = 9 litres and water = 36 litres When contents of all three jugs are mixed; total juice = 27 + 25 + 9 = 61 litres and total water = 18 + 20 + 36 = 74 litres. Required ratio = 61 : 74 Hence, option 1.

OA and OB bisect chords and pass through the centre of the circle. Hence, OA and OB are perpendicular to chords PQ and QR respectively. Let OA and OB bisect PQ and QR at M and N respectively. OMQ = ONQ = 90° In quadrilateral MONO, MON = AOB = 360° - (90 + 90 + 40)° = 140° ACB = AOB/2 = 140/2 = 70° Hence, option 3.

Total work involved = 20 * 185 = 3700 units. Let one man do 1 unit of work per day. 20 men work for the first 20 days, then 30 men work for the next 20 days, then 40 men work for the next 20 days and so on.
Similarly, 10 men keep getting added every 20 days.
Hence, rather than just adding people, start from the lowest value of days in the options and identify when 3700 units of work is done. 78 = 20 + 20 + 20 + 18
Work done in 78 days = 20(20) + 20(30) + 20(40) + 18(50) = 400 + 600 + 800 + 900 = 2700 units < 3700 units Hence, option 3 can be eliminated.
Next value = 95 = 20 + 20 + 20 + 20 + 15 Work done in 95 days = 20(20) + 20(30) + 20(40) + 20(50) + 15(60) = 400 + 600 + 800 + 1000 + 900 = 3700 units Thus, the goods are transferred in 95 days.Hence, option 4.

The sales volume for each product can be directly calculated from the pie chart.
Consider the S.P. per unit.
Since the S.P. per unit for each product is given in terms of the S.P. for product T, let the S.P. per unit for T be Rs. 100. Hence, the selling price for each product and the sales volume can be tabulated as shown below.

Among the products given in the options, T has the highest revenue. Hence, option 5.
Note: Across products, S has the highest revenue.

Consider the solution to the first question.
Note that the sales volume is constant. However, because the S.R per unit is originally given in terms of ratios, it will keep changing.
In this case, for product R: sales volume = 70000 and revenue = Rs. 1,05,00,000 S.P. per unit = 10500000/70000 = Rs. 150 W.r.t T, the selling price of the products is: R = (3T/5), S = 2T, U = (T/2) and V = (2T/5). S + U + V = 2T + (T/2) + (2T/5) = (29T/10)
Also, T = (5R/3) = (5/3) * 150 = Rs. 250 /. S + U + V = (29/10) x 250 = Rs. 725. Hence, option 3.

Because the percentage difference between the products with highest and second highest revenue is to be found, it is independent of the S.P. of the products. This is because the ratio of the selling price will always remain constant.
Hence, the revenue share will always remain the same for each product.
Consider the solution to the first question.
Based on the table, total revenue = Rs. 3,63,00,000 i.e. Rs. 363 lakhs. The products with the highest and second highest revenue are S (Rs. 80 lakhs) and P (Rs. 75 lakhs) respectively.
Revenue share of S = (80/363) * 100 = 22.04%. Revenue share of P = (75/363) * 100 = 20.66%. Hence, required % = [(22.04 - 20.66)/20.66] * 100 = (1.38/20.66) * 100 = 6.68%. Hence, option 3.

Consider the solution to the first question.
Sales volume for S = 40000 Excluding S, total sales volume = 400000 - 40000 = 360000. Sales volume for T = 60000. Required percentage = (60000/360000) * 100 = 16.67%
 Hence, option 1.

Three-fourths of an hour (in terms of seconds) = (3/4) x 60 * 60 = 2700 seconds.
Number of times the light will flash = 2700/18 = 150. This count starts after the first flash, which was just after the hour began.
Total count = 150 + 1 = 151 Hence, option 3.

There are 18 people in all (10 + 7 + 1).
Total age of these people = 18 x 46 = 828 years. Total age of male teachers = 47 x 10 = 470 years. Total age of female teachers = 48 x 7 = 336 years. Age of boy = 828 - (470 + 336) = 22 years. Hence, option 5.

The probability of a white ball being transferred from bag A is independent on all the events that happen after it.
There are 14 balls in bag A, of which 9 are white. Required probability = 9/14 Hence, option 5.

The sequence of digits repeats after four digits and 900 is a multiple of 4.
Hence, the number is of the form 135713571357.....1357.
Consider 13571357.
It can be written as 13570000 + 1357 = 1357(10000 + 1) = 1357(10⁴ + 10⁰). 

Similarly, 135713571357 can be written as 1357(10⁸ + 10⁴ + 10⁰) Hence, 13571357... 900 digits = 1357(10⁸⁹⁶ + 10⁸⁹² + 10⁸⁸⁸ + ... + 10⁰) . 10⁸⁹⁶ + 10⁸⁹² + 10⁸⁸⁸ + ... + 10⁰ = (10⁴)²²⁴ + (10⁴)²²³ + ... + (10⁴)⁰
= ( 10000)²²⁴ + ( 10000)²²³ + ... + ( 10000)⁰
= (9999 + 1)²²⁴ + (9999 + 1)²²³ + ... + (9999 + 1)⁰

Thus, there are 225 such terms (0 to 224) and the remainder of each of these terms when divided by 9999 is 1.
Also, the remainder when 1357 is divided by 9999 is 1357.
Required remainder = remainder of [1357 * (1 + 1 + 1 + ... 225 times)] divided by 9999 = remainder of (1357 x 225) divided by 9999 = remainder of 305325 divided by 9999 = 5355

Hence, option 2

Consider n = 2. n³ - n = 8 - 2 = 6. The largest number that divides it is 6.
Consider n = 3. n³ - n = 27 - 3 = 24. This is divisible by 6 as well as other larger numbers. Thus, 6 will always divide n³ - n.
Hence, option 3.
Alternatively, n³ - n = n(n² - 1) = (n - 1 )n(n + 1). 

i.e. the product of three consecutive integers.
Hence, at least one of these integers should be even, and at least one should be divisible by 3.
Hence, n³ - n should be divisible by 2 and 3, and therefore by 6.
Hence option 3.

A number is divisble by 8, if its last three digits are divisible by 8.
The available digits are 2, 3 and 5; with repetition.
The 3-digit numbers made from 2, 3 and 5 and divisible by 8 are 352, 232 and 552 i.e. 3 numbers.
Now, each of the remaining 7 seven digits can be chosen in 3 ways from among 2, 3 and 5. Total possible numbers = 3 * 3⁷ = 6561 Hence, option 4.

The FDI outflow figures are given as a percentage of the GDP of the country. Since the actual GDP is not known, the actual FDI figures cannot be found. Hence, the percentage increase in FDI cannot be found.
Hence, option 5.

Assume Singapore’s GDP in 1995 to be 100 billion dollars.
FDI outflow from Singapore in 1995 = 41.8% of 100 = 41.8 billion dollars. FDI from Singapore to India in 1995 = 0.1 billion dollars. Required % = (0.1/41.8) x 100 = 0.24. The GDP of Singapore in 2003 was four times its value in 1995.  Singapore’s GDP in 2003 was 400 billion dollars.
FDI outflow from Singapore in 2003 = 99.5% of 400 = 398 billion dollars. FDI from Singapore to India in 2003 = 0.6 billion dollars. Required % = (0.6/398) x 100 ⇒ 0.15. Hence, the percentage has decreased from 1995 to 2003.
Hence, option 2.

Singapore’s GDP in 2002 = $82 billion. Singapore’s FDI outflow in 2002 = 96.7% of 82 = 79.294 billion. Singapore’s FDI in India and China together = 10.7 + 0.5 = 11.2 billion. Share of India and China = (11.2/79.294) x 100 = 14.12% Hence, option 2.

Solution: United States’s FDI in 2001 is $93.1 billion which is 62.3% of its GDP.
GDP of United State in 2001 = 93.1 x (100/62.3) - $ 150 billion. Total FDI of China and India in 2001 = 46.9 + 3.4 = $ 50.3 billion. Required ratio = 150 : 50.3 ⇒ 3 : 1

Hence, option 3.

Every fourth man was aged above 45 years i.e. 25% of the men were aged above 45 years.
Similarly, 20% of the women were aged above 45 years.
Also, men and women are adults i.e. more than 18 years of age. Hence, all the children are less than or equal to 18 years of age (thus, also being less than 45 years of age). Total population that is less than 45 years old = 420 + (0.75 * 480) + (0.8 * 530) = 420 + 360 + 424 = 1204
Total population of the village in 2011-12 = 420 + 480 + 530 = 1430. Required percentage = (1204/1430) * 100 = 84.2% Hence, option 2.

Here it is necessary to tabulate the actual population and find the population change for each year (which includes increase as well as decrease). Answer elimination is not possible. 

Thus, the highest ever % change in the population is 4.2% (in 2012-13). Hence, option 3.

Since the government budgets an amount exactly equal to the maximum possible outlay, the payable amount needs to be maximised.
This is possible when the number of eligible families is maximised.
Consider 2009-10: There are 430 children. 

To maximise number of class A families, all of them should be part of a 2-child family.
Maximum number of class A families = 430/2 = 215 Hence, one man and one woman each gets allocated to these families.
Number of men left = 470 - 215 = 255 and number of women left = 520 - 215 = 305 Since the number of men is less than the number of women, the count of class B families is dependent on the number of men.
This count is maximised when each of these men is part of a class B family.
Maximum number of class B families = 255 Total budget for 2009-10 = 3000(215) + 2000(255) = Rs. 11,55,000 Consider 2012-13: Men = 540, Women = 500 and Children = 450.

Maximum number of class A families = 450/2 = 225 Number of men left = 540 - 225 = 315 and number of women left = 500 - 225 = 275 Here, since the number of women is less than the number of men, the count of class B families is dependent on the number of women. Maximum number of class B families = 275 Total budget for 2012-13 = 3000(225) + 2000(275) = Rs. 12,25,000. % change in budget = [(12.25 - 11.55)/11.55] * 100 = 6.06% Hence, option 4. 

Since there was no child migration, total migration = adult migration. Adult migration = 2% of 1550 = 31 Total adults in the village in 2013-14 = 580 + 510 = 1090

Also, total adults in the village in 2012-13 = 540 + 500 = 1040 Increase in number of adults = 1090 - 1040 = 50. 31 out of these 50 people have come from outside. Hence, the remaining 19 people have been added to the set of adults from within the village itself.
This can only happen if 19 children have crossed 18 years of age in this period and become adults.
Number of children in 2012-13 = 450 Hence, without any child birth or deaths, number of children in 2013-14 should be 450 - 19 = 431.
However, actual number of children in 2013-14 = 460 Since there is no child migration or child death, this extra count is solely due to child births.

Number of child births in the given perios = 460 - 431 = 29 Hence, option 3.

Let the length of A be / cm and breadth be b cm.
Area of A = / b ... (i) Length and breadth of B is (/ - 4) cm and 12 cm respectively.
Area of B = ( / - 4) x 12 ... (ii) Length of C is (/ - 4) + 6 = (/ + 2) cm and breadth of C is 12 - 4 = 8 cm Area of C = (/ + 2) x 8 ... (iii)

Since each rectangle has the same area, (/ - 4) x 12 = (/ + 2) x 8 3 / - 12 = 2 /+ 4

So  / = 1 6.

Area of each rectangle = (/ - 4) * 12 = 12 * 12 =144 cm². Hence, option 2.

Solution: Let the original rate of interest be r%.
Rs. 1,450 is lent for 12 months at r% while Rs. 725 is lent for 4 months (i.e. 1/3 years) at 2r%. Total S.l. = (1450 x 1 x r)/100 + (725 x 1/3 x 2r)/100 = 14.5r+ 4.83r = 19.33r.  Since the total S.l. earned is Rs. 67; 19.33r= 67. So  r= 3.47% Hence, option 5.

Using Statement I alone: Abhiram scored 152 marks and failed by 98 marks. Passing marks = 152 + 98 = 250 Since the maximum marks are not known, the passing percentage cannot be found.
Hence, the question cannot be answered using statement I alone.
Using Statement II alone: The maximum marks of the test are 448 more than what Abhiram has scored.
Since the marks scored by Abhiram and passing marks are unknown, the passing percentage cannot be found.
Hence, the question cannot be answered using statement II alone.

Using statements I and II together: Maximum marks = 152 + 448 = 600 Since maximum marks and passing marks are known, the passing percentage can be found.
Thus, the question can be answered using statements I and II together but not by using either statement alone.
Hence, option 4.

Using statement I alone: Though the hypotenuse is known, the length and breadth are not known.
Hence, the area cannot be found.
Thus, the question cannot be answered using statement I alone.
Using Statement II alone: The perimeter of the triangle is four times its base.
If the length of base is B cm, height is H cm and Hypotenuse is K cm then, H + B + K = 4B 
H + K = 3B ... (ii).

However, by itself the equation is not enough to get the value of H and B.
Hence, the area cannot be found.
Thus, the question cannot be answered using statement II alone.
Using statements I and II together: K = 5
H + 5 = 3B. 

Since the triangle is a right triangle; H² + B² = 25. (3B - 5)² + B² = 25 
9B² - 30B + 25 + B² = 25. 
10B² - 30B = 0 

B = 0 or 10 Since B cannot be 0, B = 10

Since the base is known, the height can be found, and hence the area can be found.
Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 4.

The train crosses another train in both cases. f=(I₁+I₂)/(x±y)
where, t = time taken, I₁and l₂ = length of the two trains, x and y = speed of the two trains and the ‘±’ sign depends on the direction of the two trains.
Using statement I alone: t,I₁ and l₂ are known but x and y are both unknown.
Hence, the speed of the first train cannot be found.
Thus, the question cannot be answered using statement I alone.

Using statement II alone: t, I₁ and y are known but l₂ and x are unknown.
Hence, the speed of the first train cannot be found.
Thus, the question cannot be answered using statement II alone.
Using both statements I and II together: Since both cases deal with separate data altogether, they cannot be considered together. Hence, the question cannot be answered on the basis of the two statements.
Hence, option 5.

Using Statement I alone: Since the radius of C₂ is not known, the area cannot be found.
Hence, the question cannot be answered using statement I alone.
Using Statement II alone: The ratio of the radii is known, but neither of the individual radii is known.
Hence, the area cannot be found.
Hence, the question cannot be answered using statement II alone.
Using statements I and II together: By combining the statements, the two radii are found as 12 cm and 24 cm respectively. Hence, the area of the circles, and the area of the shaded regio can be found.

Hence, option 4.

Correct Answer :- B

Explanation : The question cannot be answered with any of the statements because Sn is not given in any of them.

So option B is correct.

Using statement I alone: Many possibilities arise from I.
Hence, the question cannot be answered using statement I alone.
Using statement II alone: Many possibilities arise from II too.
Hence, the question cannot be answered using statement II alone.
Combining statements I and II together:

This condition will be satisfied exactly once between 2 a.m. and 3 a.m., between 3 a.m. and 4 a.m. and between 4 a.m. and 5 a.m.
Hence, the unique time cannot be found.
Hence, the statement cannot be answered on the basis of the two statements.
Hence, option 5.