[Wonder Boy]

@Nayyar got your solution …  what about question 7 . Anyone going to solve it ???

[Wonder Boy]

The answer of both 9) and 10) is C … post your solution @patro .. and yeah n^2 is correct 🙂

[Wonder Boy]

I didn’t got your solution @Nayyar. Though your answer is correct. 🙂 . Can you attach the graph ?

[Wonder Boy]

Ok a few questions from my side :). The race must go on 😉

Q7. The pendulum of a clock takes 7 sec to strike 4 O’clock. How much time will it take to strike 11 O’clock?

a. 18 sec  b. 20 sec   c. 19.25 sec  d. 23.33 sec

 

Q8. Suppose the function f is defined so that f(m)= 3m if m (m-1)² if m>1. If n is negative then f(1-n)= __ ?

a. (2-n)² b. n² c. 3-3n d.3n e.(n-1)²

 

Q9. Find the remainder when 1×2+2×3+3×4+….+99×100 is divided by 101 ?
a. 2 b. 100. c. 0 d. 7

Q10. If p and q are roots of quadratic equation ax² + bx + c = 0 such that unit digit of p and q are same. It is also given that D/a² is not divisible by 10. If n is the remainder when (p + q)/2 is divided by 10, then how many values n can take:- (D is discriminant)
a) 0 b) 1 c) 2 d) 3 e) More than 3

[Wonder Boy]

Q3) ->

4,1,1,1: C(4,1)*C(7,4)*C(3,1)*C(2,1)=840
3,2,1,1:C(4,1)*C(7,3)*C(3,1)*C(4,2)*C(2,1)=5040
2,2,2,1:C(4,3)*C(7,2)*C(5,2)*C(3,2)=2520
Thus,Total:840+5040+2520=8400 Ans (c)

 

Q4) ->

Numbers ending with 5 = 24
Numbers ending with 25 = 6
Numbers ending with 125 = 2
Numbers ending in 3125 = 1 (bit here we will get 2 extra)

So, 24 + 6 + 2 + 2 = 34 Ans (d)

[Wonder Boy]

8!=2⁷x3²x5x7

Now for 8! to be divisible by a²b² , factors which are useful is 2⁶ , 3² .

so now we have to make pairs from 2², 2², 2², 3² 

so the pairs will be [2²x2²](since a>b), [2²x2⁴], [2²x3²], [2⁴x3²], [2⁸x3²], [2²x(2×3)²], [2²x2²x(2×3)²], [2²x(2x2x3)²]

So total 7 sets are possible .

[Wonder Boy]

chk the no. of terms of each number

1 is 1 time ,

3 is 3 times ,

5 is 5 times ..

so start calculating the no. of terms somewhere less than 4800

you will get it by adding 1+3+5 +….. +2n-1 <= 4800

1+3+5+…+2n-1 is sum of odd numbers and is equal to n².

so n² <= 4800.

n = 69 since 69² = 4761

Hence next term i.e. 4762th term to 4900th term is  is 70th odd number i.e 70×2-1 = 139 Ans

[Wonder Boy]

since 144 is the total no. of factors so 144= 2x2x2x2x3x3 ..so total of 6 prime factors… Now no. of co-prime factors = 6c0 + 6c1+ 6c2 + 6c3 + 6 c4 + 6c5 + 6c6 = 64 which consist of similar terms so we divide it by 2 so total required term = 64/2 =32

[Wonder Boy]

c+d=10a ; a+b=10c ;
hence , a+b+c+d=10(a+c)

cd=-11d=> c=-11 [since d =/= 0]
ab=-11b=> a=-11 [since b=/= 0]
hence a+b+c+d = 10 x -22 =-220

[Wonder Boy]

1.) For A , S= 0.2 E Total = 1.2 E so S in %  = (0.2E/1.2 E) x100 = 100/6 = 16.66 %

For E = 1.5 S, Total = 2.5 S so S in % = (S/2.5 S )x100 = 20%

so X =/= Y Sufficient

2.) Insufficient

[Wonder Boy]

We can see here that  the denominators are the product of consecutive integers.

so 1/2 + 1/6 + 1/12 + 1/20 +…… = 1/1.2 + 1/2.3 +1/3.4+ 1/4.5 + ….which of th form 1/n(n+1)

1/n(n+1) = [(n+1)-n]/n(n+1) =1/n-1/(n+1)

so  1/1.2 + 1/2.3 +1/3.4+ 1/4.5 + .. = 1- 1/2 +  1/2-1/3 + 1/3-1/4 + 1/4 – 1/5  +……. upto infinity = 1 Ans 

 

[Wonder Boy]

In such problems having roots in denominator just rationalise .. u will find terms automatically arranged in such a manner that u can cancel max of them leaving 2 or 4 terms.. if there is no root then expressing the terms as difference of two may be a bit difficult depending on question

[Wonder Boy]

1/(root2+root1) = (root2- root1)/(2-1) = root2-root1

so 1/(root2+root1) + 1/(root2+root3) +……… 1/(root120+root121)

= (root2 – root1) + (root3 – root2) + (root4 – root3)……… + (root121 – root120)

= (root2 – root1) + (root3 – root2) + (root4 – root3)……… + (root121 – root120)

= root 121 – root 1 = 11-1 = 10 Ans

[Wonder Boy]

(x^2+6x+7)/(x^2+6x+16)
= (x^2+6x+16 – 9)/(x^2+6x+16)
= 1 – {9/(x^2+6x+16)}
= 1 – {9/((x + 3)^2 + 7)}

This exp will be minimum when {9/((x + 3)^2 + 7)} is maximum
=> ((x + 3)^2 + 7) is minimum, which at x = -3 and value is 7

=> So, minimum value is 1 – 9/7 = -2/7

 

ans by Hemant Yadav

[Wonder Boy]

Let the original number 10X+Y
number formed by reversing its digit =  10Y+ X
now the difference (10Y+X) – (10X+Y) = 9Y-9X = 9 (Y-X)
so the number will be divisibe by 9

[Author]

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