Test: CAT Quantitative Aptitude- 1 - CAT MCQ

We can use the formula
Quantity left = Quantity present initially *
Let the quantity removed in each iteration be x litres
Putting all values we get

Solving this equation we can get x = 8 litres
Hence, 8 is the correct answer.

Let the present ages of Akhil and Akash be x and y respectively.

⇒ 5x - 40 = y - 8
⇒ 5x - y = 32
⇒ y = 5x - 32

⇒ 15x + 180 = 7y + 84
⇒ 15x + 180 = 7(5x - 32) + 84
⇒ 15x + 96 = 35x - 224
⇒ 20x = 320
⇒ x = 16
⇒ y = 48
x:y = 16:48 = 1:3
Sum of antecedent and consequent = 1 + 3 = 4

Let Rs. 'x' be the mark price of the fridge. Then, the mark price of the television = Rs. 4x. Ratnesh was promised 20% discount on the fridge and 25% discount on the television. But the salesman interchanged the discounts on both the items. Therefore, Ramesh would have received 25% discount on the fridge whereas he would have received 20% discount on the television.
The actual amount which Ratnesh paid = 0.75 ∗ x + 0.80 ∗ 4x = 3.95x
The amount that he would have paid if he was not cheated = = 0.80*x + 0.75*4x = 3.80x
Hence, the ratio of the amount actually paid by Ratnesh to that he would have paid if he was not cheated 
Therefore, option B is the correct answer. 

The ratio of the books with them is 
LCM of 3, 5, 7 and 11 is 1155
So, they must be having books in the ratio
which is equivalent to 
385 : 231 : 165 : 105
Thus, the minimum number of books they must be having is
385 + 231 + 165 + 105 = 886
Hence, 886 is the correct answer.

Given in the question :
The square bracket represent integer function.
Let us assume x - y = A.         (1)
The second equation mentioned was :

Let 3x-2y = B                          (2)
In the question we were asked for the minimum possible value of y .
Multiplying (1) by and subtracting this from (2) we get :
3x-2y -(3x-3y) = y.
Hence y can be written as B - 3*A 
In order to minimise y we must try to make the difference as low as possible and to do this we minimise the value of B to as low as possible and maximise A so the difference gets to its lowest value.
The minimum value of B = minimum value of square of 3x-2y
Since the integral part of the square root of B is 5. This must be in the range of 
The minimum value it can take is 5 and hence 3x-2y minimum value is 25.
The maximum value of A = maximum of x-y.
Since the integral part of fourth root of A is 2. This must be in the range of 
The maximum value this cannot take is 81. But this takes values greater than 80 and less than 81 also.
So 3A can have a maximum value of 242.
The minimum value of B - 3A is 25 - 242 = -217

The region enclosed by the lines is a triangle in the third quadrant formed by the points (0,-7), (0,0) and (9,0). The number of coordinates in the region with x coordinate are as follows
x=0 ⇒ 8 points,
x=1, 7points,
x=2, 6points,
x=3, 5points,
x=4, 4points,
x=5, 4points and so on.
Total no. of points =41

Consider radius of each smaller circle be r and that of the larger circle be R = 6 cm.
Construct PD such that PD is perpendicular to AB.
In right triangle ADP, AP = DP/cos(APD) = DP/ cos(60) = 2DP = 2r
In right triangle OFA, OF/cos(AOF) = AO ⇒ R/(1/2) = AP+OP  ⇒ 2R=2r+r+R   ⇒ R = 3r
 In right triangle OEP, EP=OPcos(EPO) = (R + r)((√3)/2 = (4R/3)((√3) = 2R/(√3)
PQ=2EP = 4R/√3)
Perimeter = 12R/(√3) = 24√3 cm (R = 6cm)

Let us draw perpendicular from centre of the circle to AB and CD.

We are given that ∠ OED = 30°, therefore ∠ AOE =90°-30° = 60°.
Let 'R' be the radius of the given circle. Then, in right-angle triangle ODF,
⇒ OF² =OD² −FD²
⇒ OF² = R² - 5²
⇒ OF = 
In right-angle triangle OFE,

Similarly, in right-angle triangle OGA,
⇒ OG² = OA² - AG²
⇒ OG = 
In right-angle triangle OGE,

By equating (1) and (2)

⇒ 3*(R²-25)=R²-9
⇒ 2R² =66
⇒ R = √33 cm
Hence, option C is the correct answer.

The number of triangles that can be formed for a given perimeter 'p' is given by when p is even and  when p is odd, where [] is the nearest integer function.
The number of scalene triangles that can be formed for a given perimeter 'p' is given by when p is even and if p is odd.
Number of scalene triangles that can be formed with a perimeter of 45 cm = [42² /48] = [36.75]= 37
Therefore, 37 is the correct answer.

In the triangle BOA, tan 
Therefore, θ=60 degrees.
The shaded area = Area of two triangles - area of section of circle.
As the radius is perpendicular to a tangent, triangle OAB is a right angled triangle, with right angle at B, with area = 1/2 ∗ 5 ∗ 5√3​.
Hence area of both triangles= 2 ∗ 1/2 ∗ 5 ∗ 5√3​=25√3​.
The angle BOA = 
Hence, angle BOC = 2*BOA = 120°
Hence the area of the section = area of circle/3 
Hence area of shaded portion= 

Let us draw a line from point E that is parallel to AG and intersects DB at H.

The triangle DFG is similar to the triangle DEH. Hence,

⇒ GH = 7.5GH=7.5cm. ...(1)
Similarly, the triangle BEH is similar to the triangle BAG. Hence,

 

⇒ BH = 11.25BH=11.25cm. ...(2)
From equation (1) and (2) we can say that BG = 7.5+11.25 = 18.75.
We can see that both the triangles △BGC and △DGC are of same height.
This is because their bases lie on the same line and have a common 3rd vertex. Hence, the perpendicular dropped from that vertex will be the same for both triangles.
Hence, the ratio of the area of the triangles will be same as the ratio of base lengths.Therefore,

Hence, option B is the correct answer.

Let f(x) = |x+3|+|x-3|+|x-6|+|x-5|+|x+5|
f(x) is a linear equation in x at all times. It will be a union of straight lines with inflection points at -3, 3, 6, 5 and -5. Hence, the minimum value of the expression occurs at one of these inflection points. We will calculate the value of f(x) at each of these points and then find out the least possible value of f(x)
f(-3) = 25
f(3) = 19
f(6) = 24
f(5) = 21
f(-5) = 31
19 is the least value.

Number of non-negative integral points on x + 2y = 20 ⇒ (20,0), (18,1),......,(2,9) and (0,10) ⇒ 11 points
Intersection point of 2x + y = 16 and x + 2y = 20 is (4,8)
So, all the points among the above 11 points that have x coefficient less than or equal to 4 are removed.
⇒ 3 points are removed.
Hence, required number of points = 11 - 3 = 8

3x + 4y + z = a --> (1)
2x + 6y + 4z = b --> (2)
x - y - 2z = c --> (3)
Eliminating z from (1) and (2) , (1)x4 - (2) : 12x + 16y + 4z - (2x + 6y + 4z) = 4a - b
10x + 10 y = 4a - b
⇒ x + y = (4a - b)/10 - (4)
Eliminating z from (1) and (3), (1)x2 + 3 = 6x + 8y + 2z + (x - y - 2z) = 2a + c
⇒ x + y = (2a + c)/7 - (5)
Equating equations (4) and (5), we get,
(4a - b)/10 = (2a + c)/7
28a - 7b = 20a + 10c
Thus, 8a = 7b + 10c

The equation can be written as follows:


For the logarithms in the question to be defined, x - 1 > 0 ⇒ x > 1
So, the only possible value of x = 5

∣log _(6x+4)(3x−2)∣=1
3x−2>0
x>2/3........(1)
log _(6x+4) (3x−2)=±1
Case 1
log _(6x+4) (3x−2)=1
∴ 6x+4 = 3x-2
∴ 6x+4=3x−2
x=-2
which is not possible according to the equation 1
Case 2

(6x+4)(3x−2)=1
18x² =9

1/√2 > 2/3
thus only one value of x i.e. 1/√2 can satisfy the equation.

For the equation to be defined, 0 < 18 - 3^x
So, the maximum value of x can be 2.

Taking 3^x as t, then:

Let us take the two roots to be A and B. A and B would be values of 3^x  that satisfy the equation.
However, we need to find the product of 3^x+2 that satisfy the equation. So, we must find the value of (A+2)*(B+2) = AB + 2*(A+B) + 4.
From the quadratic equation t^2-18t+27=0, we know that A+B = 18 and AB=27
So, the required product will be 27+36+4 = 67
So the answer is Option C

The LCM of 3 and 11 is 33. Thus, the number of numbers we find in the first 33 numbers will be the number of numbers that are there in every consecutive set 33 numbers.
Multiples of 3 less than 33 that have odd remainders when divided by 11 : 3,9,12,18,27
In the next set of 33 numbers, the numbers that satisfy the conditions are : 36,42,45,51,60 ie 33+3,33+9,33+12,33+18,33+27 respectively.
Below 1000, there will be which means there will be 30 such sets.
From 33x30 to 1000, ie from 990 to 1000, there are only 2 more numbers that satisfy the conditions ie 993 and 999.
To find sum of all the numbers, first we find the sum of the first series of numbers ie all numbers below 33 that satisfy the condition.
S₁ =3+9+12+18+27=69
The sum of the second series of numbers S₂ = (3+33)+(9+33)+(12+33)+(18+33)+(27+33) = 69 + (5 * 33) = 69+165
Similarly, the sum of the third series of numbers S₃ =69+(165×2)
Thus, sum of the n^th series of numbers S_n =69+[165×(n−1)]
Thus, total sum of the series till the 30^th set of numbers S=(69×30)+[165×(1+2+3.....29)]

We should remember that this series does not take into consideration the last 2 numbers.
Therefore the actual sum S' = 73845 + 993 + 999 = 75837

a + b + c + d = 30, a, b, c, d are integers.
(a – b)2 + (a – c)2 + (a – d)2 would have its minimum value when each bracket has the least possible value.
Let (a, b, c, d) = (8, 8, 7, 7)
The given expression would be 2.
It cannot have a smaller value

Time taken by a person to complete 1 job = 120
Work done by a person in 1 day = 1/120
Work done on 2nd day by 2 persons of same efficiency = 1/120 + 2/120 = 3/120
Work done on 3rd day by 3 persons . of same efficiency = 1/120 + 2/120 + 3/120 = 6/120
Work done on nth day by n persons of same efficiency = (1+2+3+4....... + n)/120
1 Job is completed on nth day. So the work done will be equal to 1:
(1+2+3+4....... + n)/120 = 1
n(n+1)/2 = 120 ( Sum of 1st n natural numbers = n(n+1)/2 )
Substituting the values from the options, we'll get n=15