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Given,

⇒ 

⇒ 

since Y is a negative number, 

Let the initial number of chocolates be 64x.
First child gets 32x + 1 and 32x - 1 are left.
2^nd child gets 16x + 1/2 and 16x - 3/2 are left
3^rd child gets 8x + 1/4 and 8x - 7/4 are left
4^th child gets 4x + 1/8 and 4x - 15/8 are left
5^th child gets 2x + 1/16 and 2x - 31/16 are left.
Given, 2x - 31/16 = 0 ⇒ 2x = 31/16 ⇒ x = 31/32.
∴ Initially the Gentleman has 64x i.e. 64 x 31/32 = 62 chocolates.

Let the number be 'abc'. Then, 2 < a × b × c < 7. The product can be 3,4,5,6.
We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.
We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers. 
We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers. 
Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

The graphs of 2^x + 2^{− x} and 2 − (x − 2)²  never intersect. So, number of solutions=0.

Alternate method:
We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x = 0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x = 2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

if (x² − 7x + 11) = 1 or (x2 − 7x + 11) = -1 and (x² − 13x + 42) is even number
For x = 6,7 the value (x² − 13x + 42) = 0
(x² −7x + 11) = 1 for x = 5,2.
(x² − 7x + 11) = -1 for x = 3,4 and for X = 3 or 4, (x² − 13x + 42) is even number.
∴ {2,3,4,5,6,7} is the solution set of x.
∴ x can take six values. 

Let the base radius be 3r.
Height of upper cone is 9 so, by symmetry radius of upper cone will be r.
Volume of frustum

= 

Volume of upper cone = π .r². 9
Difference

= 

Volume of larger cone 

= π/3 .9r². 27 = 243

The area of the region contained by the lines  | x | − y ≤ 1, y ≥ 0 and y ≤ 1  is the white region.
Total area = Area of rectangle + 2 * Area of triangle = 2 + (1/2 x 2 x 1) = 3
Hence, 3 is the correct answer.

Let ABCD be the rectangle with length 2l and breadth 2b respectively.
Area of the circle i.e. area of painted region = π b²
Given,
4lb - π b² = (2/3) π b²
⇒ 4lb = (5/3)π b²
⇒ 
Given, 

⇒ 

Perimeter of rectangle = 

Hence option A is correct. 

Let the length of radius be 'r'.
From the above diagram,
x² + r² = 6²   ....(i)
(10−x)² + r² = 82 ----(ii)
Subtracting (i) from (ii), we get: 
x = 3.6 ⇒ r² = 36 - (3.6)² ⇒ r² = 23.04
Area of circle = π r²  = 23.04π 
Area of rhombus= 1/2 x d1 x d2 = 1/2 x 12 x 16 = 96.
∴ Ratio of areas = 23.04π / 96 = 6π/25

Let the speed of Car 2 be 'x' kmph and the time taken by the two cars to meet be 't' hours.
In 't' hours, Car 1 travels (60 × t) km while Car 2 travels (x × t) km
It is given that the time taken by Car 1 to travel (x × t) km is 45 minutes or (3/4) hours.

∴ 

Similarly, the time taken by Car 2 to travel (60 × t) km is 20 minutes or (1/3) hours.

∴ 

Equating the values in (i) and (ii), and solving for x:

∴ 

x = 90 kmph
Hence, Option B is the correct answer.

Let the price of desktop and laptop be x,y respectively.
Given,
x + y = 50000...(i)
12.x + 0.9y = 50000(1.02) = 51000 ...(ii)
(ii) - 0.9 (i) gives
0.3x = 6000 ⇒ x = 20000.

Initially the amount of Dye and Water are 16,24 respectively.
To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye ⇒ 16l of water is added.
Now, the Dye and Water arr 16,40 respectively.
After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.
To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye ⇒ 8l of Dye should be added.
Hence , 8 is correct answer. 

Hence C is correct answer.

Let the volume of Metals A,B,C we 3x, 4x, 7x
Ratio weights of given volume be 5y, 2y, 6y
∴ 15xy + 8xy + 42xy = 130 ⇒ 65xy = 130 ⇒ xy = 2.
∴ The weight, in kg. of the metal C is 42xy = 84.

Let a = x + 1/x
So, the given equation is a² − 3a + 2 = 0
So, aa can be either 2 or 1.
If a = 1 , x + 1/x =1 and it has no real roots. 
If a = 2, x + 1/x = 2 and it has exactly one real root which is x = 1
So, the total number of distinct real roots of the given equation is 1

log 36 = log y;
∴ y =36

The difference in the time take to traverse the same distance 'd' at two different speeds is 35 minutes. Equating this: d/8 - d/15 = 35/60
On solving, we obtain d = 10 kms. Let x kmph be the speed at which Amal needs to travel to reach the office in 50 minutes; then
10/x = 50/60 or x = 12 kmph.
Hence, Option B is the correct answer.

Let the total distance be 'D' km and the speed of the train be 's' kmph. The time taken to cover D at speed d is 't' hours. Based on the information: on equating the distance, we get 

On solving we acquire the value of t = 1/4 or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed 's' {distance traveled in terms of s = s/12. Remaining distance in terms of 's'  = s/4 - s/12 = s/6
The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in 15 - 9 = 6mins or {1/10 hours}. Thus, let 'x' kmph be the new value of speed. Based on the above, we get
s/6/x = 1/10 or x = 10s/6
Since the increase in speed needs to be calculated:

 increase.

Hence, Option C is the correct answer.

The four digit even numbers will be of form:
1100, 1122, 1144 ... 1188, 2200, 2222, 2244 ... 9900, 9922, 9944, 9966, 9988
Their sum 'S' will be (1100 + 1100 + 22 + 1100 + 44 + 1100 + 66 +1100 + 88) + (2200+ 2200 + 22 + 2200 + 44 +...)....+ (9900 + 9900 + 22 + 9900 + 44 + 9900 + 66 + 9900 + 88)
⇒ S = 1100*5 + (22 + 44 + 66 + 88) + 2200*5 + (22 + 44 + 66 + 88)....+ 9900*5 +(22 + 44 + 66 + 88)
⇒ S = 5*1100(1 + 2 + 3+...9) + 9(22 + 44 + 66 + 88)
⇒ S = 5*1100*9*10/2 + 9*11*20
Total number of numbers are 9*5 = 45
∴ Mean will be S/45 = 5*1100 + 44 = 5544.
Option D

Let the length of the train be l kmslkms and speed be s kmph. Base on the two scenarios presented, we obtain:

On dividing (ii) by (i) and simplifying we acquire the value of ss as 22kmph. Substituting this value in (i), we have l = 90/3600 x 20 kms {keeping it in km and hours for convenience}
Since we need to find l/s, let this be equal to x. Then, x = 90 x 20/22 = 81.81 ≈ 82 seconds.
Hence, Option B is the correct choice.

Since c < 9, we can have the following viable combinations for b × c = 96 (given our objective is to minimize the sum):
48 × 2 ; 32 × 3 ; 24 × 4 ; 16 × 6 ; 12 × 8
Similarly, we can factorize a × b = 432 into its factors. On close observation, we notice that 18 × 24 and 24 × 4  corresponding to a × b and b × c  respectively together render us with the least value of the sum of a + b + c  = 18 + 24 + 4 = 46
Hence, Option D is the correct answer.