Test: CAT Quantitative Aptitude- 1 (October 7) - CAT MCQ

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.

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.

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.

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

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

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

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.

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

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