Practice Test for NMAT - 10 - CAT MCQ

LCM of 28, 49 and 98 = 196 Amount with Disha must be a multiple of 196.
Let Disha has Rs. 196x

Money used by Disha to buy things = Rs. (196 - 14)x = Rs. 182x

Amount of money spent by Disha on Pencil = Rs. (7x * 22) = Rs. 154x
Hence, the remaining amount = Rs. (182 - 154)x = Rs. 28x

Let the number of pencils and erasers Disha bought be p and e respectively. (4x)p + (2x)e = 28x i.e., 2p + e = 14 ...(i) p7 is not possible as in that case ‘e’ becomes zero or negative.

p < 7
As the number of pencils is more than the number of erasers, the possible solutions of equation (i) are: (p = 6, e = 2) and (p = 5, e = 4)
The number of pencils Disha must have bought is either 5 or 6.

Hence, option 1.

For annual compounding, A = P(1 + R)^N

Hence, in the first case,

2P = P(1 + R)⁵ 
(1 + R)⁵ = 2 ... (I)
The rate of interest and principal are the same in the second case; and the principal quadruples itself.

4p = p( 1 + R )^N

2² = (1 + R)^N

Using (I), [(1 + R)⁵]² = (1 + R)^N 

(1 + R)¹⁰ = (1 + R)^N 

N= 10 
Hence, option 1.

Let us assume that Ram was born in the yearX.
Shyam was born in the year (X + 12).
In 1976, their ages were respectively (1976 - X) and (1964 - X).

We are given that 1976 - X = 3(1964 - X) + 2, or X = 1959 Ram died at the age of 48.
Ram was aged 48 in the year 1959 + 48 = 2007.
Hence, option 3.

Considering classes I and II,

x = 20
Hence the total number of students = 2x + 3x + 4x = 9x = 9 * 20 = 180
Hence, option 2.

Line A and line with equation 3x + 4y = 4 are parallel lines and are at a distance 2/5 units.
Slope of the line (with equation 3x + 4y = 4) = Slope of line A, m₁ = -3/4
Slope of line with equation 2x + 5y + 9 = 0, m₂ = -2/5

If m₁ and m₂ are slopes of two lines and 9 is the acute angle between them then,

θ =tan^-1 (7/26) 

Hence, option 1.

Let the ratio in which the cook buys the cheapest to costliest rice be a : b : c So, his average cost per kg is the weighted average of these costs.

20a + 25b + 28c = 23a + 23b + 23c

2b + 5c - 3a = 0

Since this is a single linear equation in three variables, it has multiple solutions.
Therefore, substitute the value of a, b and c from the options to check which combination satisfies this equation.
Option 1: 26 + 5 c - 3a = (2 x 4) + (5 x 5) - (3 x 3) = 24  0

Option 2: 2b + 5c - 3a = (2 x 2) + (5 x 3) - (3 x 5) = 4   0

Option 3: 2b + 5c - 3a = (2 x 1) + (5 x 3) - (3 x 4) = 5   o

Option 4: 2b + 5 c - 3a = (2 x 4) + (5 x 5) - (3 x 7) = 12  0

Option 5: 2b + 5c - 3a = (2 x 2) + (5 x 1) - (3 x 3) = o

Thus, this equation is satisfied only for a : b : c = 3 : 2 : 1 i.e. the ratio given in option 5.
Hence, option 5.

Let the distance between Chasseypur and Nosseypur be d km. Speed of the train = s km/hr

After the chain was pulled, it lost 80% of its speed and reached Chasseypur 50 minutes late.
After the chain was pulled, the train’s speed became 0.2s km/hr

Again, if the chain was pulled at Masseypur, the train would have reached 45 minutes late.

Using equation (i) and (ii) we get,

s = 16 and d = 200

d-20 = 180 km

Distance between Masseypur and Wasseypur = 450 - 180 = 270 km Hence, option 5.

Sam completes 15/16^th of the task in 1 day.

Hence, option 2.

Observe that if you count three digit numbers from the right, you get a series of natural numbers in descending order i.e. 846, 845, 844 and so on.

123124125........ 844845846 = 846 + 845(1000) +
844(1000)² + ....... 123(1000)ⁿ
1000 raised to an odd power leaves a remainder of-1 when divided by 11. 1000 raised to an even power leaves a remainder of +1 when divided by 11. Remainder = 846 + 845(-1) + 844(1) + 843(-1) + .......

= 846 - 845 + 844 - 823 
= 1 + 1 + 1 + ....
Since 123 to 846 is 724 numbers and one ‘1’ is obtained from two numbers, there are 724/2 = 362 1s being added in the remainder

. .-. Remainder = 1 + 1 + 1 + ... 362 times = 362

362 = 11(32)+ 10
Hence, the required remainder is 10.
Hence, option 5.

Let the number be N.
N = Dk + 5; where k is the quotient. 4N = 4Dk + 20 ...(i)

As we get 8 as the remainder when 4N is divided by D,

4N = Dy+ 8 ...(ii)

From (i) and (ii);

D(y - 4k) =12

L.H.S. is divisible by D ⇒R.H.S. is divisible by D.

⇒ D = 1 or 2 or 3 or 4 or 6 or 12

D should also be greater than 5 and 8.
From the given option only 12 satisfies the condition.
Hence, option 1.

As all the three lights remain switched on for 1 sec, we can say that the lights get switched off after every 10, 14 and 17 seconds.
So the time after which they will get switched off simultaneously = LCM(10, 14, 17) = 1190 seconds

Hence, option 4.

Thus only statement II is incorrect.
Hence, option 1.

x + y =7 and xy = 6

x²(1 - y) + 2xy + y² (1 - x) = x² + 2xy + y² - x²y - xy² = (x + y)(x + y -xy ) = 7(7-6) = 7

Since the two roots are positive integers, product of the roots can be 7 only when the roots are 1 and 7.
Sum of roots = 1 + 7 = 8

Hence, option 3.

In a parallelogram, opposite angles are equal. ∠A = ∠C

a + 40 = 2a + 10

a = 30

Similarly, ∠B =∠D = b - 20

Also, in a parallelogram, internally adjacent angles are supplementary i.e. ∠A +∠D = 180

a + 40 + b - 2 0 = 180

70 + ( 6 - 2 0 ) = 180
∠B = 6 - 2 0 = 110°

Hence, option 2.

x² - y 2 = 0

x = ±y
This forms a pair of straight lines having equations x + y = 0 and x - y - 0

Now,
( x - k)² + y² = 1 is a circle with centre (Ac, 0) and radius 1.
Since the centre of this circle is on the x-axis and the radius is 1 unit, k = ±1 Consider k = 1

(x - 1 )² + y² = 1

Consider x = y

The above equation becomes y² - 2y + 1 + y² = 1

2y² - 2y = 0 i.e. y = 0 or y = 1

Hence, the possible real solutions here are (0, 0) and (1,1) Now, consider x = -y

The above equation becomes y² + 2y + 1 + y² = 1

2y² + 2y = 0 i.e. y = 0

or y = -1

Hence, the possible real solutions here are (0, 0) and (1,-1)

Thus, k = 1 gives three possible real solutions (0, 0), (1,1) and (1.-1)

Similarly, k = -1 gives three possible real solutions (0, 0), (-1 , 1) and (-1, -1) Thus, there are two possible values of k.
Hence, option 3.
Note: Using co-ordinate geometry (i.e. equation of pair of lines and equation of circle), solution set can be directly found by plotting a figure as shown below.

2sin²y + cosy - 2 = 0

2(1 - cos²y) + cosy - 2 = 0

2 - 2cos2y + cosy - 2 = 0

2cos²y - cosy = 0

cosy (2cosy - 1) = 0

y = 90° or y = 60° Of these, only 60° is given in the options.

Hence, option 5.

The largest possible sphere can be inscribed in the cone when the sphere touches each side of the cone as shown below.

Radius of the base of cone = r = 6 Slant height = / = 2r= 12

If he is the height of the cone, 12² = 6² + h²

Since the centroid and the center of the circle meet at the same point, OA : OM = 2 : 1

Difference between the volume of cone and sphere

Hence, option 2.

(x³ + y³ + z³ - 3xyz)

= (x + y + z)(x² + y² + z² - xy - yz - zx)

= (x + y + z)[(x + y + z)² - 3(xy + yz + zx)]

= 6[(6)² - 3 (11)]
= 6 x 3 = 18.
Hence, option 3.

The word AIRLINES is an eight-letter word that has four consonants (R, L, N, S) and four vowels (A, I, I, E).
Also, there are four odd positions (and consequently, four even positions).
Since the vowels take the odd positions, the consonants take the even positions.
Four consonants can be placed in four positions in 4! ways i.e. 24 ways

Total number of ways = 24 x 12 = 288 Hence, option 3.

Consider w + x + y+ z=15.
This is similar to dividing 15 similar objects between 4 people i.e. n = 15 and r - 4

Non-negative integral solutions imply that each variable can take a value of 0 as well.
Thus, the number of non-negative integral solutions of this equation is equivalent to dividing 15 similar objects among 4

people such that each person can get 0 objects as well.
Number of ways in which this can be done is:

Similarly, positive integral solutions imply that each variable can take a value of at least 1.
Thus, the number of positive integral solutions of this equation is equivalent to dividing 15 similar objects among 4 people such that each person gets at least 1 object.
Number of ways in which this can be done is:

Thus, the difference in the number of solutions = 816 - 364 = 452
Hence, option 2.

So, if the probability of Ajay and Vijay contradicting each other is found, the odds can also be found.
Let A be the event that Ajay is lying and B be the event that Vijay is lying.
Hence, A and B’ are the events where Ajay and Vijay are respectively speaking the truth.
Probability that they contradict each other = P(A) x P(B’) + P(A’) X P(B)

Thus, no. of favourable outcomes = 11 and no. of unfavourable outcomes = 20-11 =9

Hence, option 1.

1947 = 1600 + 300 + 47

Since 400 years have 0 odd days, the first 1600 years also have 0 odd days. 300 years have 1 odd day.
Number of odd days in the first 1900 years = 0 + 1=1

The period from 1901 to 1946 has 11 leap years and 35 nonleap years.
A leap year has 2 odd days and a non-leap has 1 odd day.
Number of odd days from 1901 to 1946 = 11(2) + 35(1) = 57 = 7(8) + 1 i.e. 1 odd day.

1947 is a non-leap year.
Number of odd days in the first 7 months of 1947 = 3 + 0 + 3 + 2 + 3 + 2 + 3 = 16 = 7(2) + 2 i.e. 2 odd days Number of days from 1st August to 15th August = 15 = 7(2) + 1 i.e. 1 odd day

Total odd days =1 + 1+ 2 + 1= 5

5 odd days from Sunday is Friday.
Hence, 15 August 1947 was a Friday.
Hence, option 1.

Let the number of cars on day 1, day 2, day 3 and day 4 be d₁, d₂, and d₄ respectively.
Using Statement I alone:

d₁ + d₃ = 80

There is no information on d₂. So, the number of cars on day 2 cannot be found.
Thus, the question cannot be answered using statement I alone.
Using Statement II alone:

d₄ = d₁ - 3
There is no information on d₂. So, the number of cars on day 2 cannot be found.
Thus, the question cannot be answered using statement II alone.
Using Statements I and II together: When the two statements are combined, we get a relationship between d_1,d₃ and d₄; but no information on d2. So, the number of cars on day 2 cannot be found.
Thus, the question cannot be answered on the basis of the two statements.
Hence, option 5

Since the height of ΔAED and ΔAEB is the same, their bases are in the same ratio as their area.
DE:EB = 1:2
However, there is no other information about ΔAEB.
Thus, the question cannot be answered using statement A alone.

Using Statement B alone:

Since the height of ΔBEC and ΔAEB is the same, their bases are in the same ratio as their area.

CE:AE = 3:1
However, there is no other information about ΔAEB.
Thus, the question cannot be answered using statement B alone.
Using Statements A and B together: No additional information can be obtained even when both statements are combined.
Thus, the question cannot be answered on the basis of the two statements.
Hence, option 5.

f(t) = a + b + c + d + e + ... + x + y + z
Using statement A alone:

f(1) = a + b + c + d + e + ... + x + y + z = 0
1 is a root of f(α).
Thus, statement A alone is sufficient to answer the question.
Using statement B alone:

f (1) = a + b + c + d + e + ... + x + y + z > 0

1 can never be a root of f(α).
Thus, Statement B alone is sufficient to answer the question.
Hence, option 3.

x² - 3x - 4 = 0

(x + 1 )(x - 4) = 0.
Hence the value of x can either be 4 or - 1.
Since, k is a positive integer and a common root, k = 4.
Even if the other equation has two positive integral roots, the only root common will be k = 4

Thus, the question can be answered using statement A alone.
Using statement B alone:

k²- 2k- 8 = 0

(k - 4)(k + 2) = 0

Hence, the value of k can be 4 or -2.
Since there is no other condition, a unique value of k cannot be found.
Thus, the question cannot be answered using statement B alone.

Thus, the question can be answered using statement A alone but not by using statement B alone.
Hence, option 1.

Let there be n integers.
Hence, AM of these n integers = 1001/n As

AM is also an integer, n should be either 7 or 11 or 13.

Using Statement A alone:

As AM  GM, we have,

1001/n​ 143

1001 ​ 143n

7​ n
n = 7 (n cannot be 1 as for n = 1, the GM is 1001, which is not true.)
Hence, the AM of the given numbers = 1001/7 = 143

Hence, statement A alone is sufficient.
Using Statement B alone:

As all numbers are equal, hence their AM = GM

But we don't know the GM or the count of numbers. Hence we cannot determine the AM of these numbers.
Hence, statement B alone is not sufficient to answer the question.
Hence, option 1.

Although it indicates that the pair had scored 150 runs in 30 overs, it does not provide any information about their scoring pattern. They could have scored the runs at a uniform or non- uniform run rate.
Thus it is not possible to find the number of overs required to score a hundred run partnership.
Hence statement A alone is insufficient.
Using statement B alone:

It states that Sachin’s scoring rate is thrice Sourav’s scoring rate. But it gives no information about their scoring pattern.
Hence, it is not possible to find the number of overs required to score a hundred run partnership.
Thus statement B alone is insufficient.

Using both the statements together: If we take both the statements together, still the scoring pattern and the rate of scoring remain unknown.
Hence, we cannot find when they actually completed their 100 run partnership.
Hence the answer cannot be obtained even by using both the statements together.
Hence, option 5.

The lowest value of volume = 48 cc.
The highest value of volume = 130 cc

.

Hence, option 3.

Find the value of k = (PV)/T for each reading The table lists out the corresponding values of /c(Constant) for each reading:

As can be seen from the above table, reading 4 is an exception to the derived relationship.
Hence, option 4.