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CAT Practice Test - 30 - CAT MCQ


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30 Questions MCQ Test CAT Mock Test Series and 500+ Practice Tests 2024 - CAT Practice Test - 30

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CAT Practice Test - 30 - Question 1

Ten yrs ago,the ages of the father and the son were in the ratio 6:1 and 10 yrs hence the ratio will become 2:1. The present age of the father is

Detailed Solution for CAT Practice Test - 30 - Question 1

let the present age of father be F and that of his son be S
10 years ago 
F-10 and S-10 
Similarly 10 years, it will be F+10 and S+10
(F-10):(S-10) = 6:1 
6S - F = 50.............(1)
(F+10):(S+10) = 2:1
F - 2S = 10......(2)
From (1) and (2), we get
F = 40 and S = 15

CAT Practice Test - 30 - Question 2

The total number of integer pairs (x,y) satisfying the equation x + y = xy is :

Detailed Solution for CAT Practice Test - 30 - Question 2

Given:  x+y=xy  (x,y are integers)
x(1−y) = −y
x = y(1-y)
x is an integer, so y/(y-1) should be integer
y/(y-1) is integer if y - 1 = +-1
therefore, y = 0 or 2
(0,0) (2,2) are the solution of the equation.
Two solutions are possible.

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CAT Practice Test - 30 - Question 3

then r can not take any value except  except :

Detailed Solution for CAT Practice Test - 30 - Question 3

a/(b+c)=b/(c+a)=c/(a+b)=k 
a=k(b+c)
b=k(c+a)
c=k(a+b)
by property sum of numerator/sum of denominator is also = k
a+b+c/2(a+b+c)=k
k=1/2
If a+b+c=0
a+b=−corb+c=−a
a/b+c = a/−a=−1=k
So k = −1or1/2

CAT Practice Test - 30 - Question 4

The roots of 2x2-6x+3=0 are

Detailed Solution for CAT Practice Test - 30 - Question 4

(-6)2 - 4.2.3
36 -24
= 12
so,root be real ,unequal and rational

CAT Practice Test - 30 - Question 5

In the following questions two equations numbered I and II are given. You have to solve both the equations and


Detailed Solution for CAT Practice Test - 30 - Question 5

CAT Practice Test - 30 - Question 6

Let a, b, c, d be the lengths of consecutive sides of a rectangle. Which of the following can be the area of the rectangle?

Detailed Solution for CAT Practice Test - 30 - Question 6

Area(△PQR) = ab/2
Area(△PQS) = ad/2
Area(△RSP) = cd/2
Area(△QRS) = bc/2
=> 2*ar(rect PQRS) = ab/2 + cd/2 + ad/2 + bc/2
=> 2*ar(rect PQRS) = [(a+c)(b+d)]/2
=> ar(rect PQRS) = [(a+c)(b+d)]/4

CAT Practice Test - 30 - Question 7

Two identical circles intersect so that their centres and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is

CAT Practice Test - 30 - Question 8

A boy appears for 5 papers, each of the same number of maximum marks. He scores in each of the papers in the ratio of 6 : 7 : 8 : 9 : 10. If his average score in all the five papers together was 60%, then in how many papers did he get more than 50% marks ?

Detailed Solution for CAT Practice Test - 30 - Question 8

Let one subject is of 100 marks so total there are 5 subjects
⇒ 100 × 5 = 500.
Now according to the question he secured 60% of those which is 60% of 500 = 300 marks in total.
The ratio between the marks is given as 6 : 7 : 8 : 9 : 10 , so now we can distribute 300 marks according to the ratio.
Total ratio = 40

Similarly , we will get for others as 52.5 , 60 , 62.5 , 75.
Hence , there are 4 subject where he secured more that 50 %

CAT Practice Test - 30 - Question 9

CAT Practice Test - 30 - Question 10

ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?

Detailed Solution for CAT Practice Test - 30 - Question 10

let AB = CD = L, and AD = BC = B
now given QB = 2, we have AQ = L−2
if PD = K (let) we have 
PA = AP = (B−K)
now by given condition we have 3 Triangles of equal areas 
A(PAQ) = A(QBC) = A(PCD)
all the three are right triangles where area = 1/2 product of perpendicular sides 
1/2(L−2)(B−K)
= 1/2(2B) = 1/2(K)(L)
if we muliply by 2 we get 
2B = (K)(L) gives K=2B/L
substituting K=2B/L
in equality (L−2)(B−K) = 2B
(L−2)(B − 2B/L)
or multiply L on both sides gives 
2L = (L−2)2 or
L2 − 6L + 4 = 0
we solve for L we get 
L = [−(−6) ± √(−6)2 - 4(1)(4)]/2 = 3±√5
so we have two values (3+√5) and (3−√5)for L = AB 
now length AQ = AB - 2 or we have 
(3+√5−2)
= 1+√5 which is the positive value allowed for length
Hence the length of AQ = √5+1
 

CAT Practice Test - 30 - Question 11

In the given figure, ACB is a right angled triangle. CD is the altitude. Circles are inscribed within the triangles ACD, BCD. P and Q are the centres of the circles. The distance PQ is :

Detailed Solution for CAT Practice Test - 30 - Question 11

Inscribed circle in a right angled triangle x,y,z are sides of right angled triangle
Tangential distances are equal
So, z = x-r+y-r
2r = x+y-z
In given triangle ABC, CD is altitude
= 1/2 * 15 * 20
= 1/2 * CD * 25
=> CD = 12
(BC)2 = BA * BD
(20)2 = 25 * BD
BD = 400/25
BD = 16
AD = 25 - 16 = 9
PO = (12+9-15)/2 = 3
OQ = (12+16-20)/2 = 4
PQ = 3+4
=> PQ = 7
 

CAT Practice Test - 30 - Question 12

x and y are real numbers satisfying the conditions 2 < x < 3 and –8 < y < –7. Which of the following expressions will have the least value?

Detailed Solution for CAT Practice Test - 30 - Question 12

CAT Practice Test - 30 - Question 13

Amar, Akbar and Anthony invested sums , a,b and c respectively under simple interest at the same rate of interest and for the same period . The amount with Akbar at the end of the period is equal to the interest earned by Amar during that period and the amount with Anthony at the end of the period is equal to the interest earned by Akbar during that period .Which of the following is the relation among a,b and c ?

Detailed Solution for CAT Practice Test - 30 - Question 13

Principle amount of Amar - a
Akbar - b
Anthony - c
As we know that = (Pb * R * T)/100
=> (b * R *T)/100………………..(1)
Similarly, (c * R * T)/100………………………(2)
Divide 1 by 2
a/b = [(b * R *T)/100]/[(c * R * T)/100]
a/b = b/c
ac = b2
 

CAT Practice Test - 30 - Question 14

Let f(x)= 6 - 12x + 9x2 - 2x3 , 1 ≤ x ≤ 4. Then the absolute maximum value of f(x) in the interval is

Detailed Solution for CAT Practice Test - 30 - Question 14

 

f'(x) = −12+18x−6x= −6(x2−3x+2) = −6(x−1)(x−2).
∴f'(x)>0 if1<x<2 andf'(x)<0 if 2<x≤4.
∴f′x>0 if1<x<2 andf′x<0 if 2<x≤4.
∴absolute maximum = the greatest among{f(1),f(2)}
=the greatest among{1,2}=2.

CAT Practice Test - 30 - Question 15

A mixture contains spirit and water is the ratio 3:2.If it contains 3 litres more spirit than water,the quantity of spirit in the mixture is

Detailed Solution for CAT Practice Test - 30 - Question 15

Let the quantity of spirit in the mixture = 3x
Let the quantity of water in the mixture = 2x
ATQ,
3x - 2x = 3
x = 3 litre
Quantity of spirit = 3*3 = 9 litre
 

CAT Practice Test - 30 - Question 16

Let S be the set of integers x such that 
I. 100 ≤ x ≤ 200 
II. x is odd 
III. x is divisible by 3 but not 7 
How many elements does S contain?

Detailed Solution for CAT Practice Test - 30 - Question 16

No' of odd no. divisible by 3 not by 7 & b/w 
100 ≤ x ≤ 200  are 
∴ no' of elements divisible by [3 − (by6) − (by21) + (by42)]
∙ no' of elements divisible by '3' & b/w 100 to 200 
is [200/3] − [100/3]
=> 66 − 33 = 33
o' of elements divisible by '6' & b/w 100 to 200
is [200/6] − [100/6]
=> 33 − 16 = 17
no' of elements divisible by '21' & b/w 100 to 200 is
is [200/21] − [100/21]
=> 9 − 4 = 5
no' of elements divisible by '42' & b/w 100 to 200
is [200/42] − [100/42]
=> 4 − 2 = 2
⇒ no' of elements =33−17−5+2
= 13

CAT Practice Test - 30 - Question 17

Of 128 boxes of oranges, each box contains at least 120 and at most 144 oranges. The number of boxes containing the same number of oranges is at least

Detailed Solution for CAT Practice Test - 30 - Question 17

Each box contains at least 120 and at most 144 oranges. 
So boxes may contain 25 different numbers of oranges among 120, 121, 122, .... 144. 
Lets start counting. 
1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125. 
Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144. 
Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes. 
Hence the number of boxes containing the same number of oranges is at least 6. 

CAT Practice Test - 30 - Question 18

What is the digit in the unit's place of 251 ?

Detailed Solution for CAT Practice Test - 30 - Question 18

Last digit of any number shows a definite pattern.
21 (24n+1) = 2
22 (24n+2) = 4
23 (24n+3) = 8
24 (24n+4) = 6
25 (24n+1) = 32
Cyclicity of 2 is 4 → 2, 4, 8, 6
251 = 24n+3  
So, the digit in unit's place is 8.

CAT Practice Test - 30 - Question 19

Last year Deepika saved 10 percent of his annual earnings. This year he earned 5 percent more than last year and he saved 12 percent of his annual earnings. The amount saved this year was what percent of the amount saved last year?

Detailed Solution for CAT Practice Test - 30 - Question 19

LAST YEAR:
Earnings = 100 (assume).
Amount saved = 0.1*100 =10.

THIS YEAR:
Earnings = 100*1.05 = 105.
Amount saved = 0.12*105 =12.6.

The amount saved this year as a percent of the amount saved last year is (this year)/(last year)*100 
= (12.6/10)*100
= 126%.

CAT Practice Test - 30 - Question 20

The period of 

Detailed Solution for CAT Practice Test - 30 - Question 20

We wish to determine the period of  f(x)=sin(πx/2) + 2cos(πx/3) − tan(πx/4) . First determine the frequencies of each of the  3  periodic functions:
{sin(πx/2) = sin(2πf1x), cos(πx/3) = cos(2πf2x), tan(πx/4) = tan(πf3x)}.
Hence we find  f1=1/4 ,  f2=1/6 , and  f3=1/4 . 
Next we find the minimum absolute  T  such that  f1T ,  f2T , and  f3T  all produce integer solutions.  T=12  is a solution because  f1T=3 ,  f2T=2 , and  f3T=3  are all integers. 
Since  2  and  3  are co-prime it is the minimum absolute solution. 
Thus the period of  f(x)  is  12 .
 

CAT Practice Test - 30 - Question 21

How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?

Detailed Solution for CAT Practice Test - 30 - Question 21

As A and T should occupy the first and last position, the first and last position can be filled in only one way. The remaining 4 positions can be filled in 4! Ways by the remaining words (S,C,E,N,T). hence by rearranging the letters of the word ASCENT we can form 1x4!

CAT Practice Test - 30 - Question 22

A man bets on number 16 on a roulette wheel 14 times and losses each time. On the 15th span he does a quick calculation and finds out that the number 12 had appeared twice in the 14 spans and is therefore, unable to decide whether to bet on 16 or 12 in the 15th span. Which will give him the best chance and what are the odds of winning on the bet that he takes? (Roulette has numbers 1 to 36)

Detailed Solution for CAT Practice Test - 30 - Question 22

Each of the span is an independent event and the outcome of the 15th span will not depend on the outcome of the earlier spans.

CAT Practice Test - 30 - Question 23

A merchant buys two articles for Rs.600. He sells one of them at a profit of 22% and the other at a loss of 8% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss?

Detailed Solution for CAT Practice Test - 30 - Question 23

Let C1 be the cost price of the first article and C2 be the cost price of the second article
Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%
We know, C1 + C2 = 600
The first article was sold at a profit of 22%
Therefore, the selling price of the first article = C1 + (22/100) C1 = 1.22C1
Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2
The total selling price of the first and second article = 1.22C1 + 0.92C2
As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2
Therefore, 1.22C1 + 0.92C2 = C1 + C2 = 600
As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get
1.22C1 + 0.92(600 - C1) = 600
or 1.22C1 - 0.92C1 = 600 - 0.92 × 600
or 0.3C1 = 0.08 × 600 = 48
or C1 = 48/(0.3) = 160
If C1 = 160, then C2 = 600 - 160 = 440
The item that is sold at loss is article 2
The selling price of article 2 = 0.92 × C2 = 0.92 × 440 = 404.80

CAT Practice Test - 30 - Question 24

In a quadratic equation with leading coefficient 1, a student reads the coefficient 16 of x wrongly as 19 and obtain the roots as -15 and -4. The correct roots are

Detailed Solution for CAT Practice Test - 30 - Question 24

Since coefficient of x = 16,
∴ sum of roots = -16
Since constant term = (-15)(-4) = 60,
∴ correct answer is -6, -10

CAT Practice Test - 30 - Question 25

A race track is 1680 m long. In a race between A and B, A won by 30 seconds. In a race between A and C on the same track, A wins by 630 m. In a race between B and C on the same track, B wins by 96 seconds. The time taken by A to run the complete track is

Detailed Solution for CAT Practice Test - 30 - Question 25

A travelled = 1680m
C travelled = 1680 - 630
= 1050m
A beats C by 630m,
In same time, A travelled = 1680m
In same time, C travelled = 1050m
Speeds of A : C => 8 : 5
Therefore, B travelled = 1680m and beats C by 96 sec
In 96 sec, C travelled less than B 
Dis = 96 * 5 
= 480m 
Speed(B)/Speed(C) = 1680/(1680-480)
= 7/5
Hence, A=8m/s B=7m/s C=5m/s
A takes = 1680/8
= 210sec
i.e. 3 min 30 sec

CAT Practice Test - 30 - Question 26

There are blue vessels with volumes v1, v2, ...., vm, arranged in ascending order of volumn, v1 > 0.5 litre and vm < 1 litre. Each of these is full of water initially. The water from each these is emptied into a minimum number of empty white vessels, each having volumn 1 litre. The water from a blue vessel is not emptied into a white vessel unless the white vessel has enough empty volumn to hold all the water of the blue vessel. The number of white vessels required to empty all the blue vessels according to the above rules was n.

Q. Among the four values given below, which is the least upper bound on e, where e is the total empty volume in the white vessels at the n end of the above process ?

Detailed Solution for CAT Practice Test - 30 - Question 26

Using options we find that if we suppose m = 1, option (a) will give the answer as vm and option (c) will give the answer as v1.
But, both of these cannot be our answer, as vm and v1 are the amounts of volume filled
If m = 2, option (b) will give the answer as 2(1 - v2) and option (d) will give the answer as 2(1 - v1)
But actual empty volume is greater than 2(1 - v2)
Therefore, for this condition m(1 - v1) is the only possible answer

CAT Practice Test - 30 - Question 27

If Sn = 27 - 72 + 29 - 69 + 31 - 66 + 33 - 63 . . . up to n terms , then find p, given Sp = 0.

Detailed Solution for CAT Practice Test - 30 - Question 27

Sn = 27 - 72 + 29 - 69 + 31 - 66 ......
S1 = 27 + 29 + 31 +.....n/2
S2 = -72 - 69 - 66 -......n/2
S1 = n/4[54 + (n/2-1)2]
S2 = n/4[-144 + (n/2-1)3]
Sp = 0
Sp => S1 + S2 = 0
⇒ n/4[54 - 144 + n - 2 + 3n/2 - 3] = 0
⇒ 5n/2 = 95
⇒ 5n = 190
⇒ n = 38

CAT Practice Test - 30 - Question 28

If f (x) = log ((1 + x)/(1 - x)), then f (x) is

Detailed Solution for CAT Practice Test - 30 - Question 28

 f(x) = log[(1+x)/(1-x)]
f(-x) = log[(1-x)/(1+x)]
= log[(x+1)/(1-x)]-1
f(-x) = -log[(1+x)/(1-x)]
= -f(x)
f(x) is an odd function.

CAT Practice Test - 30 - Question 29

The real roots of the equation 28 x³ - 9x² + 1 = 0 are

Detailed Solution for CAT Practice Test - 30 - Question 29

Given equation, 28x3−9x2+1=0
⟹28x3−16x2+7x2+4x−4x+1=0
⟹(4x+1)(7x2−4x+1)=0
Solving the quadratic equation, we have x = (2 ± √3i)/7
∴ roots of the given equation are x = -1/4, (2 ± √3i)/7

CAT Practice Test - 30 - Question 30

The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the adjoining graph : 

Q. Manasa makes the 200 km trip from Mumbai to Pune at a steady speed of 60 km per hour. What is the amount of petrol consumed for the journey ?

Detailed Solution for CAT Practice Test - 30 - Question 30

Fuel consumption is given in litre per hour.
It is, therefore, clear from the graph that in travelling 60 km fuel consumption is 4 litre.
Hence in travelling 200 km fuel consumption will be 4/60 × 200 = 13.33 litres

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