[Wonder Boy]

First we consider positive integers pair

following pair satisfy the condition

1,1 = 2 ; 1,2 = 5 ; 1,3 = 10; 1,4 = 17 ; 1,5 = 26

2,1 = 5; 2,2 = 8 ; 2,3= 13 , 2,4 = 20 , 2,5 = 29

3,1 = 10; 3,2 = 13 ; 3,3 = 18 ; 3,4 = 25 ; 3,5 = 34

4,1 = 17; 4,2 = 20; 4,3 = 25; 4,4 = 32

5,1 = 26; 5,2 = 29; 5,3 = 34

 

SO total 22 pairs.

Now considering theor -ve values we get 22 x4 = 88 pairs

also for 0

pairs will be 0,1; 0,2;0,3;0,4;0,5 and 1,0; 2,0; 3,0; 4,0; 5,0 and their – ve values

total 20 pairs

 

and (0,0)

so total = 88 + 20 +1 = 109 Ans

[Wonder Boy]

we can assume x=cos A and y=Sin A
then x+y=Cos A +sin A=sqrt(2) sin(45+A)
max value of sin is 1 and min is -1
hence max of x+y=sqrt(2)
and min is -sqrt(2)
and hence product is -2 Ans.

[Wonder Boy]

The probability of not getting a 4 in a throw = 5/6

The probability of getting a 4 in a throw  = 1/6

Now since result of each throw is not dependent on the result of previous throw.

so probability of getting a 4 in the 6th trial = (5/6)⁵ x 1/6

[Wonder Boy]

Let the roots of equation be u & v.

u + v =-r
uv = 6r
1/u +1/v = -1/6 –> u= -6v/(6+v) = [-6(6+v) +36]/(6+v)

u = -6 +36/(6+v)

For both the roots to be integers
(6+v) should be the factor of 36
values that v can take as
{0,3,6,12,30, -2,-3,-4,-5,-7,-8,-9,-10,-12,-15,-18,-24,-42}

and corresponding values of a .

so, answer is 18

[Wonder Boy]

For Circular permutation : The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is (n-1)!

 

Now since A,B sit together, so considering them as 1 unit.

The no. of ways in which they can sit (leaving C & D condition) = (5-1)! x 2 = 48. (since A,B can interchange places among them and form new arrangement so we multiply by 2)

Now We will find the no. of ways in which both A,B and C,D sit together.

= (4-1)! x 2 x 2 = 24.

Hence required no. of ways in which A,B sit together and C,D do not sit together = 48-24 = 24 Ans.

[Wonder Boy]

Lets factorise 8888 to some smaller numbers.

So 8 + 8 + 8 + 8 = 32 ~ 5

Now as Digital Sum Rule of Multiplication says : The digital sum of the product of two numbers is equal to the digital sum of the product of the digital sums of the two numbers.

so digital sum of digits of 8888⁸⁸⁸⁸ = digital sum of 5⁸⁸⁸⁸

Now 5² = 25 ~ 7

so digital sum of 5⁸⁸⁸⁸ = digital sum of (5²)⁴⁴⁴⁴ = digital sum of 7⁴⁴⁴⁴

Now 7³ = 343 ~ 10 ~ 1 and 7⁴⁴⁴⁴= (7³)¹⁴⁸¹x7

so digital sum of 7⁴⁴⁴⁴ = digital sum of(7³)¹⁴⁸¹x digital sum of 7 = 1¹⁴⁸¹ x7 = 7 Ans

[Wonder Boy]

Here we can see  (4.2*4.2-1.9*1.9) is clearly  (4.2²-1.9²)

We know that a²-b² = (a+b)(a-b)

Hence,  (4.2²-1.9²) = (4.2+1.9)(4.2-1.9) = 6.1 x 2.3

So, (4.2*4.2-1.9*1.9) / 2.3*61 = 6.1 x 2.3 / 2.3×61 = 1/10 Ans

 

[Wonder Boy]

We can clearly see that last 3 digit of numerator and denominator is divisible by 8. Hence the Nr. & Dr. is divisble by 8.

So on dividing by 8 we get

128352/238368 = 16044/29796

Again we can see last 2 terms of Nr. & Dr. is clearly divisible by 4 .

Son on dividing by 4, we get

 16044/29796 = 4011/7449

Add the term of Nr. and Dr. individually , clearly it is divisible by 3.

4011/7449 = 1337/2483

Now we factorize both Nr. and Dr.

1337 = 7*191

2483 = 13*191

so lowest term = 7/13 Ans.

 

 

 

[Wonder Boy]

Digital Sum Rule of Multiplication says : The digital sum of the product of two numbers is equal to the digital sum of the product of the digital sums of the two numbers.

 

so 7³ = 343

so sum of digits of 7³ = 3+4 +3 = 1+0 = 1

–> sum of  digits of 7¹⁰⁰ = sum of digits (7)⁹⁹.7 = sum of digits of (7³)³³.7 = sum of digits of (1)³³.7 = 7 Ans

 

[Wonder Boy]

Here we see , we have 6 equations and 6 variables .. so it seems to be a solvable problem but it would take a lots of time. so surely it would not be a CAT problem . But wait. check the trend of the values and you will find out you can solve it in 30 seconds.

 

The problem says  f(1) = 7, f(2) = 10, f(3) = 13, f(4) = 16, f(5) = 19, f(6) = 22

all have a difference of 3. 7+3=10+3=13+3=16+3=19+3=22

 

so f(x) = 3x+4 (for x=1,2,3,4,5,6)

Since f(x) is a six degree polynomial so we can write f(x) as

f(x) = a(x-1)(x-2)(x-3)(x-4)(x-5)(x-6) + 3x+4 (for x=1 to inf)

Since coefficient of highest power of x = 1, so a = 1.

Hence f(7) = 6.5.4.3.2.1 + 25 = 745.

[Wonder Boy]

We know that for any value p=/=0 , p+1/p belongs to (-inf, -2] U [2 , inf).

Also for p>0,  p+1/p belongs to  [2 , inf).

Here, since a,b>0 so a+b>0 ,

Now x + y = 10/(a+b) + 10(a+b) = 10[(a+b) + 1/(a+b)]

so x + y >= 10×2 (=20)

Hence Option B) Always greater than or equal to 20 is correct.

 

 

[Wonder Boy]

We know that (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

–> 2(xy + yz + zx) = (x + y + z)² – ( x² + y² + z² )

–> 2(xy + yz + zx) = (x + y + z)² – 4

since (x + y + z)² >= 0

so, 2(xy + yz + zx) >= 0-4 –> (xy + yz + zx) >= -2  -> k>=-2

[Wonder Boy]

Let 1st term and common ratio of GP be a & r respectively.

so m = a.r^p+q-1

and n = a.r^p-q-1

mn= a²r^2p-2 — > ar^p-1 = (mn)^1/2 = pth term

m/n = r^p+q-1-p+q+1 –> m/n = r^2q –> r = (m/n)^1/2q

Now ar^p+q-1=m –> ar^q-1.r^p = m –> ar^q-1 =mr^-p = qth term

so qth term = mr^-p = m(m/n)^-p/2q = m^1-p/2qn^p/2q

[Wonder Boy]

Sol1 . For this type of problem, we will not try to solve it with exact procedure.

Instead we will try to take some number which satisfies the first critrea and then try to fulfill 2nd critrea.

 

So here lets take a number which when divided by 296 gives 75 as remainder.

296 + 75 is the number say.

on dividing 296 + 75 by 37 gives 1 as remainder. so answer is 1 .

 

Sol. 2  Let the number be 296n +75 .

so (296n+75)/37 = 8 n + 2 +1/37

Hence remainder is 1.

[Wonder Boy]

A beats B by 160 m means B is 160 m behind A

Also A beats B by 20 seconds i.e. B reach the finishing line after 20 seconds.

 

Hence speed of B = 160/20 = 8 m/s.

Total time taken by B to cover 800 m = 800/8 = 100s.

 

so total time taken by A = 100 – 20 = 80 sec.

Speed of A = 800/80 = 10 m/s.

So A will cover 360 m in 360/10 = 36 seconds.

 

 

 

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