CAT Practice Test - 33 - CAT MCQ

Let Varun's age = V
Kapil's age = K
V + K = 42
⇒ V = 42 - K

⇒ 11(V + 5) = 15(K + 5)
⇒ 11(42 - K + 5) = 15K + 75
⇒ 462 - 11K + 55 = 15K + 75
⇒ - 11K - 15K = 75 - 517
⇒ - 26K = - 442
⇒ K = 17

The numbers must be reciprocals of each other.

Hence the sum is greater than the product of numbers.

 (x - z)² y is even cannot be true.
x is odd and z is even. Therefore, x - z is odd.
And y is odd.
Therefore, (x - z)² will be odd and (x - z)² y will be odd.

I. x² - x - 12 = 0
⇒ x² - 4x + 3x - 12 = 0
⇒ x(x - 4) + 3(x - 4) = 0
⇒ (x - 4) (x + 3) = 0
∴ x = 4 or -3
II. y² + 5y + 6 = 0
⇒ y² + 3y + 2y + 6 = 0
⇒ y(y + 3) + 2(y + 3) = 0
⇒ (y + 3) (y + 2) = 0
∴ y = -3 or -2
x ≥ y

I. x² - 8x + 15 = 0
⇒ x² - 5x - 3x + 15 = 0
⇒ x(x - 5) - 3(x - 5) = 0
⇒ (x - 3) (x - 5) = 0
∴ x = 3 or 5
II. y² - 3y + 2 = 0
⇒ y² - 2y - y + 2 = 0
⇒ y(y - 2) - 1(y - 2) = 0
⇒ (y - 1) (y - 2) = 0
∴ y = 1 or 2
x > y

A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is

Average time per hour = (72-60) / 72 
= 12/72 
= 1/6

Let the sides of the regular hexagon ABCDEF be s.
The area of ABCDEF = area of 6 equilateral triangles of side, s = 6[s*s*sin 60]/2 = 2.598076211 s² sq units.
Applying cosine for the triangle ABC,
AC² = AB² + BC² – 2*AB*BC*cos120o
= s² + s² – 2s² * (-0.5) = s² + s² + s² 
= 3s²
or AC = s√3
The area of ACE, which is an equilateral triangle of side, s√3
= (s√3*s√3*sin 60)/2 = 1.299038106s² sq units which is half the area of the hexagon.
So the ratio of the area of triangle ACE to hexagon ABCDEF is 1 : 2.
 

Suppose x = 11
x+4 = 15, x-3=8
Hypotenuse = (225+64)1/2
= 17
⇒ 17 - 11 = 6
(6)² + (x-3)² = y²
x-3=8
Hence y = 10

Substitute the extreme values in the equation :
v = 1, u = -0.5, z = -2
Then w = vz/u 
= 1*(-2)/-0.5
= 4
Option (B) satisfies.

 P[1+(r/100)⁵] - P = A
P[1+(r/100)¹¹] - P = B
B/A = (r/100)6
2450/1250 = (r/100)⁶
49/25 = (r/100)⁶
7/5 = (r/100)³
now P=100000
Amount after 15 years be M
M = 100000[1+(r/100)¹⁵]
M = 100000[ 1+ {(r/100)³}⁵]
M = 100000[1+ (7/5)⁵]
M = 537824

The curved surface of the cone inscribed in a given sphere is maximum 
if h= maximum, if h is height of cone
because the curved surface area of cone is equal to πrh
when h is maximum , surface area is maximum even when it is inscribed in a given sphere.

Hence both the types should be added in the ratio of 1 : 1 to obtain the required strength.
Hence 20 litres of first type should be added to the 20 litres of the second type to get the desired solution

if p = 1! then p + 2 = 3 when divided by 2! gives a remainder 1
if p = 1! + (2 × 2!) = 5 then p + 2 = 7 when divided by 3! gives a remainder 1
The result is the same for p = 1! + (2 × 2!) + (3 × 3!) + .....
So the answer for the given sum should be1

Given expressions can be reduced as follows
A = 1/4.000004
B = 1/6.000003
C = 1/6.000002
Among all of them B is smallest.

sum of 1st n squares but without the "6" in the denominator.
And about 237  
The above expression will obviously be divisible by 237
When n= 237 as it will be 237*238*475 In fact even for n = 236 and n = 118 it will be divisible by 237.
 

40% of 30 = 0.4*30 = 12
b) 3/5 of 25 = 1.5*25 = 15
c) 6.5% of 200 = 0.065*200
⇒ 6.5*2 = 13
e) (1/2)^-4 = (2)⁴ = 16

Let, f(x)=tan(3x+5)
By property of Periodic function.
period of f(x)= period of f(x+λ),λϵR
∴ period of tan(3x)= period of tan(3x+λ),λϵR
∴ period of tan(3x+5)= π/3
[Period of tanx=π, ∴ period of tanλx = π/∣λ∣ and period of tan(x)=period of tan(x+a)]

Arranging RANDOM in dictionary order gives us
ADMNOR.
Consider the words starting with A. If the starting letter is fixed as A, then the remaining 5 letters can be rearranged in 5! ways.
Similarly with the words starting with letters D,M,N,O.
Hence 5×5!
=5(120)
=600 ...(i)
Now consider the words starting with R,
R−−−−−
In Dictionary order the first word starting with R will be
RADMNO
Keeping the positions of RAD fixed we can rearrange the letters MNO in 3! ways.
Similarly for RAMDNO.
Hence
600+2(3!)
= 612.
The successive word will be
RANDMO and RANDOM.
Hence the rank of the word RANDOM is 612+2 = 614
 

Total outcome:2⁵ = 32
Outcome of at least two of them show heads:
Total-(1 head)-(0 head)= 32 - 5 - 1 = 26
Outcome of 5 heads = 1

Probability = 1/26

let the roots be a and b 
now
a + b = 2A
ab = G2
quadratic equation is given by
x2 - (sum of roots)x + product of roots = 0
x² - 2Ax + G² = 0

Distance covered by B in 6 sec = (1000/300 * 6)
= 20 m
Therefore, A beats B by 20 m.
For a dead-heat race, A must give B a start of 20 m.

The addition of numerator and denominator should gives a prime number.
3 is a factor of 189 and 183 => A and D eliminated
17 is a factor of 187 and 221 => B and C eliminated
181 is a prime.

The first term after 200 divisible by 7 is 203. The last term before 400 divisible by 7 is 399. The number of terms = ?
=> l = a + ( n - 1 ) d
=> 399 = 203 + ( n - 1 ) 7
=> 399 - 203 = ( n - 1 ) 7
=. 196 = ( n - 1 ) 7
=> 196 / 7 = ( n - 1 )
=> 28 = ( n - 1 )
=> n = 28 + 1 = 29
Hence the number of terms is 29.
Applying Sum formula we get,
S_n = n/2[a+l]
S₂₉ = 29/2[203 + 399]
= 29*301
S₂₉ = 8729

ax² + bx + c
Sum of roots : -b/a => 7
Product = c/a => 6
x2 - (sum of roots)x + Product
x2 - 7x + 6 = 0
x2 - 6x - x + 6 = 0
x(x-6) -1(x-6) = 0
(x-1)(x-6) = 0
x = 1,6
 

1cm = 40 Kms
37.5cm = 37.5×40kms
actual distance between Bombay and Calcutta is 1500kms