NMAT Quantitative Skills MCQ Quiz - 2 - CAT MCQ

L et the speed o f R eem a, Seem a and H in a be R, S and H respectively .
When Reemac overs xm , See ma and Hin acover (x -15) m and (x - 18) m .
As speed is directly proportional to distance,

When See m a covers x m , the n H in acover s (x -5) m .

From (i) and (ii),

(x - 15) (x - 5) = x (x - 18)

x² - 20x + 75 = x² - 18x
2x = 75

∴  x = 37.5 m

Hence, option 2.

We need to find the highest number n, such that it divides 1406, 974 and 638 and leaves the same remainder.
If two numbers have the same remainder, then the difference of the two numbers will also be divisible by n.
In order to find out the highest number which leaves the same remainder we have to take the HCF of differences between the numbers (taken two at a time) as follows:

1406-974 = 432

974-638 = 336

1406-638 = 768

HCF of 432, 336 and 768 is 48.

Hence, option 1.

Instead of creating an expression for T_n, consider a random value of n, and check the options. Identify the option that gives the same value of T_n as the terms of the given series.

n =1 implies T₁ i.e. the first term.

For n = 1, T₁ = 1 (from the series)

Options 1 and 2 can directly eliminated as they have a term (n - 1) which will make the entire expression 0 for n = 1 For option 3: n(n + l)/2 = (1 x 2)12 = 1 Hence, this may be a valid expression.

Observe that the expression in option 4 is twice the expression in option 3. Since option 3 matches the required value for T₁, option 4 will not match.

Hence, option 4 can be eliminated.

Since there is an option for ‘None of the above’, check option 3 for other values of n.
For n = 2,

For n = 3

and so on.
Hence, option 3.

There are 50 players in all.

∴ People who play no game + People who play 1 game + People who play 2 games +

People who play 3 games = 50 Let the number of people who play exactly 2 games be x.

∴  0.4x + (12 + 11 + 10) + x + 3 = 50

∴ 1.4x + 36 = 50

∴  x = 10

Out of the 10 people who play exactly two games, 4 play cricket and tennis only.
Hence, the remaining 6 play basketball.

∴  Number of people who play basketball = 12 + 6 + 3 = 21
Hence, option 3.

The values currently given in the Venn diagram add up to 40.

Let the number of people who don’t play any game be n.

Since 21 players play basketball,

21 + l l + 4 + 1 0 + n = 50

∴ n = 4

So, there are 6 players who, in total, play basketball and cricket only or tennis and basketball only.

Since the number of cricket players is to be maximized, all 6 players can be assigned to the set who play basketball and cricket only.

∴ Maximum number o f cricket players = 1 1 + 4 + 3 + 6 = 24

Hence, option 1.

Note that the absolute decrease and not the percentage decrease is required.

The sales revenue for CD has decreased only twice, i.e. between 2001 and 2002 and between 2002 and 2003. Also, the sales revenue for CD in 2004 is more than the sales revenue in 2001.

Hence, options 1, 3 and 4 can be eliminated.
From the year 2001 to 2002, absolute decrease registered for CD = 7 5 - 5 0 = 25

From 2002 to 2003, the absolute decrease = 50 - 0 = 50

∴ The absolute decrease in sales value of CD is maximum from 2002 to 2003.
Hence, option 3.

The absolute difference between the sales revenue of GH and IJ for each year is given below.

For 2000= 100-50 = 50

For 2001 = 125-75 = 50

For 2002= 100-100 = 0

For 2003 = 125-75 = 50

For 2004 = 100 - 75 = 25

∴ The absolute difference between the sales revenue of GH and IJ was the least in the year 2002.
Hence, option 3.

The ratio of the sales revenue of GH to that of EF for each year is shown below.

For 2000: 100/25 = 4 

For 2001 : 125/50 = 2.5 

For 2002: 100/75= 1.33 

For 2003 : 75/50= 1.5

For 2004 : 75/75 = 1

∴ The ratio of the sales revenue of GH to that of EF was maximum in the year 2000.
Hence, option 1.

Note: This question can be answered by direct observation as well. It is evident from the bar chart that the ratio is the highest in 2000.

Average sales turnover per year for company AB = (75+75 + 125 + 75 + 125)/5 = 95 lakhs

Hence, option 5.

Theequationofthecirclecanbewrittenas:

Expressitas(x-a)²+(y-b)²=r²where(a,b)aretheco-ordinatesofthecentreandristheradiusofthecircle.

So,co—ordinatesofthecentreare 

andlengthoftheradiusis 5/6

LetthetangentbedrawnfrompointP.

Distance between P and centre of circle (d)=

The tangent, radius and circle can be drawn as shown below:

Using Pythagoras theorem:

Hence, option 1.

From the figure, O is the center of a circle and CE is perpendicular to OB. 
Let EB = x

∴ OE = 4 - x

Applying Pythagoras theorem to ΔOCE, we get,

Applying Pythagoras theorem to ΔCEB, we get,

But CD = AB - 2x = 8 -1 = 7

Hence, option 3.

Let A and B denote the events that Anokhi and Bihari will qualify the examination respectively.

Hence, option 2.

12² x 7 = 1 4 4 x 7 = 1008

Since the required number is divisible by 144 and 7, it has to be a multiple of 1008 i.e. 1008, 2016, 3024, 4032, 5040 etc

Among the four numbers given, only 4032 satisfies this condition.

Also, 15³ = 3375 and 16³ = 4096

3375 < 4032 < 4096 

Hence, option 2.

Alternatively,
Since 122 = 144 = 16 x 9, the required number also has to be divisible by 9.

For a number to be divisible by 9, sum of its digits = multiple of 9

Observe that 3468 and 3864 will have same sum of digits. 3 + 4 + 6 + 8 = 21; 4 + 0 + 4 + 6 = 14 and4 + 0 + 3 + 2 = 9

Hence, 3468, 3864 and 4046 cannot be divisible by 9; and hence cannot be divisible by 144.

Hence, options 1, 3 and 4 can be eliminated. 4032 is divisible by 9 and hence, may be divisible by 144. 4032/9 = 448

Since 144 = 16 x 9, check if 448 is divisible by 16. 448/16 = 28

Thus, 4032 is divisible by 144.

Also, 4032/7 = 576

Thus, 4032 is divisible by 7.

Finally, 15³ = 3375 and 16³ = 4096

3375 < 4032 < 4096 
Hence, option 2.

 to infinity.

S₁ is in G.P.
Here, a = x and r = x²

S₂ is also in G.P.
Here, a = x² and r = x²

Now, x 1.
For an infinite G.P., x < 1. When x < 1,S₂ < S₁ Thus, quantity A is greater than quantity B.
Hence, option 1.

Using Statement I alone: |x| > [y|

Case 1: If x > y, then definitely |x| > |y|

Case 2: I f x < y , it is still possible to have |x| > |y| e.g. If, x = -5 and y = 3 Here,

x < y

However, |—5| > |3| i.e. |x| > |y|

Thus, the exact relationship between x andy can not be found.
Thus, the question cannot be answered using statement I alone.
Using Statement II alone: W=T

In this case, either x andy are equal in sign or opposite in sign, but equal in magnitude.

∴ x≤y Thus, x can never be greater than y.

Thus, the question can be answered using statement II alone.
Thus, the question can be answered using statement II alone but not by using statement I alone.
Hence, option 2.

Hence, (2x + y)² = &xy Thus, (2x - y)² = 0 Or 2x = y
Hence statement A alone is sufficient to answer the question.

Consider statement B alone: (2x - 9)² = (y - 9)² Thus, 2x — 9 = y —9 or 2 x - 9 = 9 -y Hence, 2x = y or 2x = 18 -y Hence, no definite conclusion is possible.
Hence, we cannot answer the question using statement B alone.
Hence, option 1.

Using statement I alone:
We cannot conclude anything about Qarim as we do not know the direction he is facing.

Thus, statement I alone is not sufficient to answer the question.

Using statement II alone:
We know the number of times Qarim turned and his present direction.
But we don't know the angle at which he is turning.
Thus, statement II alone is not sufficient to answer the question.

Using statement I and II both:
We get the angle at which he is turning also the direction in which he is turning from statement I and his direction at 127839th turn.
As he is facing north after 127839.
After 4 turns in any direction, we will face the same direction as that of the initial.

So, 127839/4 , we get remainder as 3.
He is facing north after 3 anticlockwise turns.
∴ He is facing east initially.
Hence, both statements I and II are necessary to answer.
Hence, option 4.

Marked price of the shirt (i.e.M.R) = Rs. 900 Using statement A alone:

M.R is 20% more than the cost price (C.R)

∴ C. P. = 900/1.2 = Rs. 750

Since the profit and amount of discount is the same, let profit = discount = Rs. x.

Selling price (S.R) = C.R + profit = M.P. - discount 

 ∴ 750 + x = 900 - x

∴ x = Rs. 75

Thus, the discount offered is Rs. 75
Thus, the question can be answered using statement A alone.

Using statement B alone:

Cost price (C.P.) = Rs. 750

Since the discount percent and profit percent are equal, let them be x%.

i.e.

Solving the above equation, x = 100/11

Since this discount percent is calculated on the marked price, the actual discount can be found.

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

Thus, the question can be answered using either statement alone.

Hence, option 3.

Using statement A alone:

Let the present age of Amar and Aditya be 3x and lx years respectively.
Their age after 16 years will be 3x + 16 and lx + 16 respectively.
So, the average of their ages will be 5x + 16.
However, this can take multiple values based on the value of x.
Thus, the question cannot be answered using statement A alone.

Using statement B alone:

Let the age of Amar and Aditya be x and 2x, 2 years from now.
So, their ages after 16 years will be x + 14 and 2x + 14.
So, the average of their ages will be 1.5x + 14.

However, this can take multiple values based on the value of x.
Thus, the question cannot be answered using statement B alone.

Using statements A and B together:

The present ages of Amar and Aditya will be 3x and lx respectively. After two years,

∴ x = 2
So, their present ages are 6 years and 14 years.
Once their present age is known, their age after 16 years can be found and so, the average can be found.
Thus, the question can be answered using both statements together but not by using either statement alone.

Hence, option 5.

Hence, option 3.

Let the volume of the tank be 120 units (i.e. the LCM of 10 and 12).

∴ Pipe A fills = 120/10 = 12 litres per hour Similarly, pipes A and E together fill = 120/12= 10 litres per hour

A and B fill less water per hour than A alone as B removes water.

∴ B empties 12-10 = 2 litres per hour

∴  Time taken by B to empty half the tank = = 3 0 hours

Hence, option 2.

Profit = Manufacturing cost x Mark up percentage

∴ Profit on one unit of Model 1 produced by company B

Calculating similarly for all the models we have,

∴ The maximum profit is on Model 5.
Hence, option 2.

The selling price of each unit of any model =

Comparing the models and the price per unit, we have

In the above models, Model 2 is sold at the least price.
Hence, option 2.

Fabrication cost per unit manufactured = Fabrication cost as percentage of manufacturing cost x Cost price per unit

Model 5 of company A has the least fabrication cost.

Hence, option 1.

In company B, to find the model that consumes the most ‘sub-assemblies’ cost, let the total sales be 100.

The cost for Model 1 = 0.30 x 0.15 x 100 =4.5

The cost for Model 2 = 0.10 x 0.15 x 100 = 1.5

The cost for Model 3 = 0.25 x 0.25 x 100 = 6.25

The cost for Model 4 = 0.20 x 0.30 x 100 = 6

The cost for Model5  = 0.15 x 0.40 x 100 =6

∴ Model 3 consumes the maximum cost in ‘Sub-assemblies’ by value.
Hence, option 3.

Consider an amount P placed at a compound interest at r% per annum.

∴ Amount at the end of 2 years = 

∴ Compound interest for second year =

Now, the amount at the end of 2 years becomes the principal for the third year.

∴ Amount at the end of 3 years = 

∴ Compound interest for third year =

∴ Compound interest for third year = Compound interest for second year X

In general,
Compound interest for (n + 1 )^th year = Compound interest for n^th year X 

∴ 500 = 400 + 4r

∴ r = 25
So, the amount is placed at 25% compound interest.
Hence, option 1.

Since the roots are real, b² - 4ac > 0

∴ (4a)² -4(4a ² - 4a + S ) >0

∴ 16a - 32 > 0

∴ a > 2 .........(I)

Now, let the roots be p and q.
Hence, the equation can be written as (x - p) (x - q).
For f(5),x = 5 Since both the roots are less than 5 , x > p and x > q when x = 5 Hence, f(5) is the product of two positive numbers and is, therefore, positive.

Hence, option 1

Since the triangle has integral sides, convert the given ratio to an integral form by multiplying it by 5 (or 10).

Hence, ratio o f sides = ( 1 x 5 ) : (2.4 x 5 ) : (2.6 x 5) = 5 : 12 : 13 Any triangle with ratio of sides as 5 : 12 : 13 is a right triangle.

So, if the sides (in cm) are 5x, 12x and 13x respectively, 5x and 12x become the base and height (in any order).

Since the area of the triangle is 1080 sq.cm,

∴ 3Ox² = 1080

∴ x² = 36

∴ x = 6

Thus, the perimeter = 5x + 12x + 13x = 3Ox = 180 cm

Hence, option 2.

In a dictionary, words are arranged in the alphabetical order.

If A is the 1^st letter, the remaining 3 places can be filled in 3! ways by the letters D, N and G.

Similarly, the number of words with D as the 1^st letter is 3! and the number of words with G as the 1^st letter is 3!

Now, when N is the 1^st letter, the sequence of words is:

NADG
NAGD
.........
.........
..........

Hence, the position of the word NAGD = 3! + 3! + 3!+2
= 6 + 6 + 6 + 2
= 20

Thus, the word NAGD is at the 20^th position in this dictionary.
Hence, option 4.

Q and R cannot get grade A in 2001 as no grade is repeated in 2001. Also, since Q and R got grade A exactly once, Q got it in 2003 while R got it in 2004.

S got grade A twice, one of which was in 2003. Since S also could not have grade A in 2001 (as explained for Q and R), S got its second grade A in 2002. Also, each product saw its grade fall to D atleast once. Hence, S would have got grade D in 2001.

Since no grade was repeated in 2001, Q and R would have got grades B and C (in no specific order).

Finally, P would have got grade B in 2002 (as there are 5 grade Bs in the whole table).

Thus, the filled-in table is as shown below.

Each year, 100 units of each product and the selling price of a product increases/decreases by 10% with every increase/decrease in grade.

Case 1: Q got grade B in 2001

Hence, Q’s grades are B, B, A, B

∴ Total earnings = 100 x [(120 x 0.9) + (120 x 0.9) + (120 x 0.9 x 1.1) + (120 x 0.9 x 1.1 x 0.9)

= 100 x (108 + 108 + 118.8 + 106.92)

= Rs. 44,172

This is not possible as the total earnings from Q were Rs. 42,758.28 over the period.
Hence, Q got grade C in 2001.
Hence, option 3.

Note: You need not calculate the earnings for grade C, but if verify, you will get the given value.