It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am
Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 
11:00 am
If they leave one by one every minute, the 45th motorcyclist would have left by 10:45 am to reach B 
at 11:00 am.
Thus, time taken by one motorcyclist to reach B from A = 15 minutes.
Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am
So, the last motorcyclist should have left A by 10:15 am
Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B

We are told that exactly 20 of the 30 integers do not exceed 5.
That means exactly 10 of the 30 integers do exceed 5.
In order to keep the average of the 20 integers as high as possible, we need to keep the average of 
the 10 integers above 5 as low as possible. Since we are dealing with integers, the least value that 
the 10 integers above 5 can take is 6.
So, the sum of the 10 integers = 10 * 6 = 60
So the sum of the remainng 20 integers = Total sum - 60 = 5 * 50 - 60 = 90
Hence the average of the remaining 20 is 90/20 = 4.5

We need to check only for the ones given in the options 
Alva – Hindi = 75 
Average of (80+75+70)/3 = 75 
Deep – Hindi = 90 
Average of (100+90+80)/3 = 90 
Esha – Hindi = 85 
This is not the average of best 3 or 2 subjects 
Hence Alva and Deep.

Now we know that Carl only missed maths and Esha only English. 
One student who missed hindi also missed science 
Exams which can be missed 
Alva – Hindi or Science 
Bithi – Social Science and Science 
Carl – Mathematics 
Deep – Hindi or science 
Esha – English 
Foni – English and Social Science 
Hence 3 

Exams which can be missed 
Alva – Hindi or Science 
Bithi – Social Science and Science 
Carl – Mathematics 
Deep – Hindi or science 
Esha – English 
Foni – English and Social Science 
Either Alva or Deep can be the one who missed both Hindi and Science. Rest we are definite 
Hence 4 .

In this we check for the 4 options 
Carl – Mathematics score = 90 
This is average of best three scores 100 + 80 + 90 = 270/3 = 90 
Alva – Mathematics score = 70 
This cannot be the average of best three or best 2 scores 
Esha – Mathematics score = 95 
This is her highest score hence this is not the possible answer 
Foni - – Mathematics score = 78 
This cannot be the average of best three or best 2 scores 
Hence Answer is Carl. 

Alva and Bithi have scored the highest in English, hence they haven’t missed it. 
Carl and deep have the least score in English, hence they haven’t missed it. 
Esha scored 80 which is the average of (95+85+60)/3 = 80 
Foni scored 83 which is the average of (88+83+78)/3 = 83 

We need to check only for the ones given in the options 
Alva – Hindi = 75 
Average of (80+75+70)/3 = 75 
Deep – Hindi = 90 
Average of (100+90+80)/3 = 90 
Esha – Hindi = 85 
This is not the average of best 3 or 2 subjects 
Hence Alva and Deep.

3 cases:
• 7 is in first place: 3 can go to 3 places, remaining 2 places will have 8 & 7 possibilities each
Hence, possible numbers are 3*8*7 = 168 

• 7 is in second place: 3 can go to 2 places, first place will have 7 possibilities (excluding 0), and
remaining place will also have 7 possibilities
Hence, possible numbers are 2*7*7 = 98
• 7 is in third place: 3 can only go to one place, remaining 2 places have 7 possibilities each
Hence, possible numbers are 1*7*7 = 49
Therefore, total number of possibilities are 168 + 98 + 49 = 315 

Let cost of pencil be ‘x’ and sharpener be ‘x+2’.
Amount spent by Aron = p*x + s*(x+2)
Amount spent by Aditya = 2p*x + (s- 10)( x+2)
Therefore: px + sx + 2s = 2px + sx - 10x + 2s – 20
px - 10x - 20 = 0
x(p-10) = 20
Since p-10 > 0, p(min) = 11
Total pencils = p + 2p = 33 

In April, John bought ‘r’ amount of rice at ‘x’ price, and ‘w’ amount of wheat at ‘y’ price.
Money spent = r * x + w * y
Also, r*x = 450
In May, amount spent = r( 1.2x) + w(1.12y)
= 1.2r*x + 1.12w*y = (r*x + w*y) + 150
0.2r*x + 0.12w*y = 150
0.12w*y = 150 – 0.2*450 = 60
w*y = 60/0.12 = 500
Amount spent on wheat in May = 1.12*500 = 560 

Let’s assume the amount of work done by each person in a day:
John – x
Jack – 2x
If Jack and Jim together take 1/3 of the time that John takes, it means they together complete 3 times the work that John alone completes in a day.
Hence, Jack + Jim = 3x
Jim – x
Assuming the work to be done is 100 units.
Number of days taken by each person is:
John – 100/x
Jack – 50/x
Jim – 100/x
All 3 together – 25/x
Therefore, 100/x = 25/x + 3
100 = 25 + 3x
3x = 75
x = 25
Jim will finish the job alone in 100/x = 100/25 = 4 days 

In the figure, ‘O’ is the centre of circles ‘C1’ and ‘C2’ with radii ‘ x ’ & ‘ y ’ respectively
If d1 – d2 = 2cm, x - y = 1cm
Hence, MB = 1cm
AB is a chord with 6cm length and tangent to C2.
Therefore, AN = NB = 3cm ; ON = y; OB = y + 1
In right triangle ONB:
(ON) 2 + (NB) 2 = (OB) 2
y 2 + 9 = (y+1) 2
9 = 2y + 1
2y + 8
y = 4
Therefore, x = y + 1 = 5cm
Diameter = 2x = 10cm 

Assuming Ram’s speed = ‘a’ and Rahim’s speed = ‘b’.
If they cross each other at time ‘t’:
Distance travelled by Ram = a*t
Distance travelled by Rahim = b*t
Further time taken by Ram to reach B = (Dist. Remaining/Speed) = (b*t)/a = 1min -------①
Similarly, further time taken by Rahim to reach A = (a*t)/b = 4min -------②
Assume ratio (a/b) = x
Therefore, ① becomes: t/x = 1 &
② becomes: t*x = 4
Multiplying both equations, we get: t 2 = 4
Therefore t = 2
Applying value of t, we get x = 2 = a/b

Circumference of T1 = 2π*100 = 200π
Time taken by Ram to reach A again = 200π/15 = 40π/3
Circumference of T2 = 2π*20 = 40π
Time taken by Rahim to reach A again = 40π/5
Ratio of the time taken by Ram & Rahim = (40π/3) /( 40 π/5) = 5/3
Ratio of time is inverse of ratio of distance travelled
Therefore, ratio of rounds travelled by Ram : Rahim = 3:5

Ram makes 3 rounds before he meets Rahim again 

\((x^{2}-5x+7)^{(x+1)}=1\)
2 cases:
• Power is 0
x + 1 = 0
x = -1
• \(x^{2}-5x+7=1\)
\(x^{2}-5x+6=0\)
\(x^{2}-2x-3x+6=0\)
\(x(x-2)-3(x-2)=0\)
\((x-2)(x-3)=0\) 

\(x=2, x=3\)
• \(x^{2}-5x+7=-1\) ; Given power is even
\(x^{2}-5x+8=0\)
No real roots
Hence, 3 integers (-1, 2, 3) satisfy the given equation 

Let side of triangle be ‘a’; Breadth of rectangle = x; Length of rectangle = 3x
Therefore: 3a + 2(x+3x) = 90
3a + 8x = 90 ----------( i )
Also, R = T 2
x *3x = ( Ö 3a 2 /4) 2 

3x 2 = 3a 4 /16
x = a 2 /4
Hence ( i ) becomes:
3a + 8(a 2 /4) = 90
3a + 2a 2 = 90
2a 2 + 3a -90 = 0
Solving the quadratic equation, we get a = 6
x = 9 ; 3x = 27
Hence, longer side = 27 

BC = 3*AB
AB = x, BC = 3x
Let 2 trains be T1 & T2.
From A to B: Speed of T1 = s; Speed of T2 = 2s
From B to C: Speed of T1 = 2s; Speed of T2 = s
Total time taken by T1 (t1) = (x/ s)+( 3x/2s) = 5x/2s
Total time taken by T2 (t2) = (x/2s) + 3x/s) = 7x/2s
Ratio t1:t2 = 5x/2s:7x/2s = 5:7 

g( 20) = f(21) – f(19)
72 = 21 2 + 21a + b – (19 2 + 19a + b)
72 = 40*2 + 2a
2a = -8
a = -4
f(x) = x 2 - 4x + b
Now, f(x) ≥ 0 for all real x
Therefore, D ≤ 0
16 – 4b ≤ 0
16 ≤ 4b
4 ≤ b
Hence, smallest possible value of b is 4

We can summarize the information given in the question as per the above Venn diagram.
We can conclude that: 

a + c = 5
b + c = 7
Hence, b – a = 2
We have to minimize a + b.
Assuming if a = 0
Then b = 2 & c = 5
Therefore, number of students choosing Chemistry as one subject = 18 + 2 = 20

Acc. to the question:
x + 9 = z & y + 1 = z
Therefore, x + 9 = y + 1
y = x + 8
x + y < z + 5
x + x + 8 < x + 9 + 5
x < 6
Now, 2x + y = 2x + x + 8 = 3x + 8
For this to be maximum, x has to be maximum.
Maximum possible value of x = 5.
Therefore, max possible value of 3x + 8 = 3(5) + 8 = 23 

Let common SP be x
For 8 toys, SP = 0.8x
For remaining 4 toys, SP = 0.75*0.8x = 0.6x
Total SP = 8* 0.8x + 4* 0.6x = 2112
6.4x + 2.4x = 2112
8.8x = 2112 

x = 240
Also let Total Cost Price = y
2112 = 1.1y
y = 1920
With no discounts, Total SP = 12*240 = 2880
Profit = 2880-1920 = 960
Profit % = (960/ 1920)* 100 = 50%