Mu Sigma Equations Quiz-1

Given that F(x)=(x+2)(x+1)(x−1)(x−2)
Putting x=P, we have F(P)=(P+2)(P+1)(P−1)(P−2)
Since P is a prime number, P is in form 6K±1, where K is the positive integer

F(6K+1)=(6K+3)(6K+2)(6K)(6K−1)
=(36)(2K+1)(3K+1)(K)(6K−1)-------- (1)

F(6K−1)=(6K+1)(6K+2)(6K)(6K−3)
=36(6K+1)(3K+1)(K)(3K−1)-------- (2)

Please note that the value of K≥17 and expression F(6K+1) and F(6K–1) always bear the factor 10.

Hence 360 is the correct choice.

Therefore option (D) is the correct choice

If the Abhishek had x Re 1, y Rs 2 coins and z Rs 10 coins, the total value of coins he had:

=x(1)+y(2)+z(10)=x+2y+10z=160
Since, 6y=x
Thus, 8y+10z=160 i.e 8y is a multiple of 10 i.e. y=5 or y=10

i.e. (x,y,z)=(30,5,12) or (60,10,8)
Thus, the maximum value of \'z\' is 12

x2−3x+2=(x−2)(x−1)
⇒ The factors of the cubic equation is x=2 and x=1
On putting x=2, we get, 8−24+2p+q=0
⇒2p+q=16 -------- (i)

On putting x=1, we get, 1−6+p+q=0

⇒p+q=5 -------- (ii)

Thus on solving these two equation we get:

p=11 and q=−6
So p+q > 0 and pq < 0

Let the number of pineapples, mango and black-forest pastries be p,m and b respectively.

Given, p+m+b=23 -------- (i)

p2+m2+b2=211 -------- (ii)

From equations (i) and (ii), we get:

p=11,m=9 and m=3

The number of members in A on Oct 2, 2007:

=n+4m=n+80n=81n
Number of members in B on Oct 2, 2007:

=n×p×p×p×p
As given in the question, now both organisations have the same number of members.

⇒81n=n×p^4
⇒p=3

Man buys x apples at m price and y oranges at n price, then:

x+y=40-------- (i)

mx+ny=17-------- (iI)

On solving both the equations we get:

(m+n)(x+y)=17+15

m+n=32/40

= Rs 0.80
= 80 paise

Let the duration of Mahesh\'s vacation be n days.
Given that on each day, he had engaged in exactly one of swimming and tennis, also given that he was free on 32 mornings and on 18 evenings and on total 28 days he either went for swimming or tennis.
He was busy on (n−32) mornings and (n−18) evenings

And, (n−32)+(n−18)=28
⇒x=39

Let the length of the side of the property be x m and y be the number of posts bought.

When the space between polls is 8 m,

The number of poles

=x/8+1

When space between poles is 6m,
The number of poles

=x/6+1=y+5

On solving both of the equations we get,

x=120 m and y=16

Number of accidents x=a+by
Where y is the number of overtime hour.

8=a+1000b -------- (i)

5=a+400b -------- (ii)

On solving both the equations, we get:

a=3,b=1200

For, y=0, a=x=3