CAT Practice Test - 30 - CAT MCQ

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

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.

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

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

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

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 %

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
 

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
 

Principle amount of Amar - a
Akbar - b
Anthony - c
As we know that = (P_b * 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
 

f'(x) = −12+18x−6x² = −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.

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
 

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

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. 

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.

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%.

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 .
 

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!

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.

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

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

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

Using options we find that if we suppose m = 1, option (a) will give the answer as v_m and option (c) will give the answer as v₁.
But, both of these cannot be our answer, as v_m 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

S_n = 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]
S_p = 0
S_p => S1 + S2 = 0
⇒ n/4[54 - 144 + n - 2 + 3n/2 - 3] = 0
⇒ 5n/2 = 95
⇒ 5n = 190
⇒ n = 38

 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.

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

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