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