CAT Practice Test - 31 - CAT MCQ

Let the age of son be x and the age of father be y
According to question,
x + y = 60 .....(1)
and
y - 6 = 5(x - 6)
⇒ y - 6 = 5x - 30
⇒ 5x - y = 24 .....(2)
Solving equation (1) and (2)
6x = 84
⇒ x = 14 years
Age of son after 6 yesrs
= (14 + 6) years = 20 years

Let A = 100x + 10y + z
⇒ B = 100z + 10y + x
B - A = 99 (z - x)
For B - A to be divided by 7, z - x has to be divisible by 7. Only possibility is z = 9, x = 2
∴ Biggest number A can be 299

let x be no. of male and y be no. of female operators.
We have 40x + 50y = 1000
x = 25 - (5*y/4)
=> y = 7,12
So y can be 8 or 12
If y = 8 then x = 15
If y = 12 then x = 10
Cost is minimum in 2nd case when no. of males are 10.

I. x² - 11x + 28 = 0
⇒ x² - 7x - 4x + 28 = 0
⇒ x(x - 7) - 4(x - 7) = 0
⇒ x = 7, 4
Also,
II. 2y² + 12y + 18 = 0
⇒ 2y² + 6y + 6y + 18 = 0
⇒ 2y(y + 3) + 6(y + 3) = 0
⇒ y = -3, -3
Hence, x > y

I. x² - 2x - 8 = 0
⇒ x² - 4x + 2x - 8 = 0
⇒ x(x - 4) + 2(x - 4) = 0
⇒ x = 4, -2
Also,
II. y² - 5y - 6 = 0
⇒ y² - 6y + y - 6 = 0
⇒ y(y - 6) + 1(y - 6) = 0
⇒ x = 6, - 1
Hence, x < y

The perimeter of the field =(13m+9m+20m+30m) = 72m

Let the seven consecutive integers be x−3,x−2,x−1,x,x+1,x+2,x+3
Given that average of first five integers is n
Therefore, n = [(x−3)+(x−2)+(x−1)+x+(x+1)]/5
⟹5x−5 = 5n
⟹x−1 = n
⟹x = n+1 ------(1)
Average of seven integers is  [(x−3)+(x−2)+(x−1)+x+(x+1)+(x+2)+(x+3)]/7
⇒ 7x/7 = x = n+1 (from (1))

In a right angled triangle, the length of the median to the hypotenuse is half the length of the hypotenuse.

y=x³ + x² + 5
y = x² + x + 5
We equate each other.
x³ + x² + 5 = x² + x + 5
Subtract x² both the side
x³ + 5 =  x + 5
Subtract 5 both the side
x³  =  x
x³ - x = 0
x(x² - 1) = 0
x(x + 1)(x - 1) = 0
We get
x = 0    &   x - 1 = 0   & x + 1 = 0
x = 0, x = 1 & x = -1
We get the three values of x in interval [-2,2]
So, both the curves cuts each other exactly three times.
 

Let Rs x be the amount that the elder daughter got at the time of the will. Therefore, the younger daughter got (3500000 - x)
The elder daughter’s money earns interest for (21 - 16) = 5 years @ 10% p.a simple interest
The younger daughter’s money earns interest for (21 - 8.5) = 12.5 years @ 10% p.a simple interest
As the sum of money that each of the daughters get when they are 21 is the same,

Clearly, the minimum does not exist.

Suppose A contains 100 litre water,
B contains 100 litre milk,
C contains 100 litre acid.
10% from A (i.e., 10 litre) is poured to B. Then,
Quantity of milk in B 
= 100 litre
Total quantity of the mixture in B 
= 110 litre.
Concentration of milk in B = 100/110
= 10/11
10 litre from B is poured to C. Then,
Quantity of milk in C 
= (10×10/11)= 100/11 litre.
Total quantity of the mixture in C = 110 litre.
Concentration of milk in C 
= (100/11)/110
= 10/121
10 litre from C is poured to A. Then,
Quantity of milk in A 
= 10×(10/121) = 100/121 litre.
Total quantity of the mixture in A = 100 litre.
Concentration of milk in A 
= (100/121)/100
= 1/121

I am assuming that the question is (ab)*(ab) = ccb.
Ok so ab as well as (ab)*(ab) also has its unit's digit as b as (ab)*(ab) = ccb. This can only be possible when b is one of these: 1, 5, 6, 0.
25 is the only 2 digit number whose square 625 is 3 digit and is greater than 300. 15,35,45… don't satisfy the conditions.
Similarly 26*26=676 is the only one which satisfies the above conditions.
Similarly 20*20= 400 and 30*30= 900 are the only squares satisfying the above conditions.
Similarly 21*21= 441 is the only square satisfying the conditions.
Out of 625,676,400,900 and 441, only 441 is of the form ccb as the first digit and the second digit are same.
Hence ccb= 441 and b=1.
 

Since in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5
So the no. may be 31 or 91 
Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1

a₁ = 1.
a_n+1 =  3a_n - 2 + 4n
a₂ = 3 – 2 + 4 = 5
a₃ = 15 – 2 + 8 = 21
a₄ = 63 – 2 + 12 = 73.
Now, our options involve 3n followed by some addition or subtraction.
So a₂ = 5, can be written as 31 + 2 and 32 - 4.
a₃ = 21 can be written as 32 + 12 and 33 – 6.
a₄ = 63 can be written as 33 + 36 and 34 – 8.
We see that the subtraction trend is repeating, as 3n – 2*n. So, a100 is 3¹⁰⁰ – 200.

Let the marks obtained in five subjects be 6x, 7x, 8x, 9x and 10x. 
Total marks obtained= 40x
Max. Marks of the five subjects = 40x/0.6 [40x is 60% of total marks] 
Max. Marks in each subject = 40x/0.6*5 = 13.33x
Hence, % of each subject = 6x*100/13.33 = 45.01%
Or, 7x*100/13.33 = 52.51
In same way other percentage are 60.01%, 67.52%, 75.01%.
Hence, number of subjects in which he gets more than 50% marks = 4.

 In a decagon, by joining one vertex to the remaining vertices you can have 8 triangles. If you are considering all the vertices independently you will have a total of 8*10 = 80 triangles.

Let the number be x,
13 kg sugar costs 195. So 1 kg costs 15.
17 kg rice costs 544, so 1 kg costs 32.
21 kg wheat costs 336, so 1 kg costs 16.
Hence 21 kg sugar + 26 kg rice + 19 kg wheat = (21 × 15) + (26 × 32) + (19 × 16) = 315 + 832 + 304 = Rs. 1451/-.

we know (a+b)³=a³+3a²b+3ab²+b³
which can be written as (a+b)³=a³+b³+3ab(a+b).
=》a³+b³ = (a+b)³- 3ab(a+b).
similarly
α³+β³= (α+β)³ - 3αβ(α+β).
here The first term is,  ax^2  its coefficient is  a .
The middle term is,  bx  its coefficient is  b.
The last term, "the constant", is  c .
we know in a quadratic equation α+β = -b/a.
and αβ = c/a.
so, substitute these values in (α+β)³ - 3αβ(α+β).
=》(-b/a)³ - 3(c/a)(-b/a).
=》-b³/a³+3bc/a^2.
by taking the LCM OF BOTH,WE GET.
(-b³+3abc)/a³.
therefore α³+β³ = (-b³+3abc)/a³.
 

Let both of them meet after T min.
4500 m are covered by Suresh in 60m
In T min = 4500T/60
In T min Suresh's wife will cover = 3750T/60
Given, 4500T/60 + 3750T/60 = 726
=> 8250T = 726
T = 5.28 min

Let the original number of workers in the group be 'x'
Therefore, actual number of workers = x - 7.
We know that the number of manhours required to do the job is the same in both the cases.
Therefore, x (24) = (x - 7).30 
24x = 30x - 210
6x = 210
x = 35.
Therfore, the actual number of workers who worked to complete the job = x - 7 = 35 - 7 = 28.

Let us assume 

By Taking x as common factor.
So, S = [2(1-x) + x)]  

f(x) = 2x + sin x
Differentiating wrt x on both sides, we get
f'(x) = 2 + cos x
−1 ≤ cos x ≤ 1
−1 + 2 ≤ 2 + cos x ≤ 1 + 2
1 ≤ 2 + cos x ≤ 3
So, f'(x) > 0 , for all real values of x 
So, f(x) is one − one
f(x) → ∞ as x→∞ and f(x) → −∞ as x→−∞
So, f(x) is onto.
 

The first part is 1/3rd of the total distance and the second part is 2/3rd of the total distance. He travels at s km/hr speed during the first half and 2s km/hr speed during the second half.
If 3 km is the total distance, then 1 km was travelled at s km/hr and 2 kms was travelled at 2s km/hr speed.
Hence, average speed = Total Distance/Total Time
= [3(1/s + 2/s)]
= 3/(4/2s) = 3s/2
This, however = s + 2s/2
= 3s/2
 which is the arithmetic mean of the speeds of the two parts.