DXC Technology Algebra Questions and Answers Quiz-1

Explanation:
a1=139/3
a2=-19
a3=a2-a1=-19-139/3=-196/3
a4=a3-a2=-196/3-(-19)=-139/3
a5=a4-a3=-139/3-(-196/3)=57/3
a6=a5-a4=57/3-(-139/3)=196/3
a7=a6-a5=196/3-57/3=139/3=a1
a8=a7-a6=139/3-196/3=-57/3=-19=a2
The series repeats after every 6 numers.
Thus 62nd number=6*10+2=2nd number= -19

when a= -b
then LHS of 1/a + 1/b + 1/c=1/(a+b+c)
1/a + 1/b + 1/c = 1/c = RHS = 1/(a+b+c)

Then a+b=0

hence

(a+b)(b+c)(c+a)=0

Let total no. of students be x.
=> [x(10*70 + 25*80 + 15*85 + 25* 90 + 25*90)/100]/ x = 86
Here, the average is always 86 and does not depend
upon the no. of students.
Hence the answer is the minimum value that is 20.

as by putting different values of x at constant value of k,
we get values of y which first increases and then decreases
with value of x.

151001 / 11^4 = 10 with remainder = 4591

4591 / 11^3 = 3 with remainder = 598

598 / 11^2 = 4 with remainder = 114

114 / 11^1 = 10 with remainder = 4

Therefore 151001=10(11^4)+3(11^3)+4(11^2)+10(11^1)+4

i.e., (v,w,x,y,z) is (10,3,4,10,4). These values satisfies given
conditions and is one valid option.

because -(2-x)/(2-x)=-1
-1 is constant which is independent of x, so range is -1.

1/12 = 1/z - 1/y

y = 12 * z / ( 12 - z )

0 < z < 12

Only for z = 3 ,4 ,6 ,8 , 9 ,10 and 11 , y is intiger

Hence seven solutions

let a and b are two numbers
so a+b=15 ....(1)
now GM=AM*80/100
or, sqrt(ab)=[(a+b)/2]*80/100
squaring both sides...and after that putting the value of (a+b)...we get.....
ab=36 now go through from option then we get 12*3=36
so answer is (b) 12,3

Let T be total number of toffees.
T mod 87=86. So T=87(k1)-1----(1)
T mod 15=14. So T=15(k2)-1----(2)
Here k1,k2 are constants.
Lcm of 87k1,15k2=435k
So total should be 435k-1 and should be odd.
For k=1=>T=434( It is not odd)
For k=2=>T=869 (Yes,It is Odd).
So Total=869.
So 869 mod 16= 5

For equation to have rational roots, Discriminant of the given
equation should be positive and should have value equal
to perfect square .
Now D= b^2 -4ac = 25 - 4c
So to have D positive and perfect square there could be
infinite value of c.