Option(B) is correct

Let the age be x and y years now.

Then, x−y=20 -------- (i)

and (x−5)=5(y−5) ------- (ii)

On solving both equation we get:

x=30 and y=10

Solution:

Let the number be x.
As per what he did,
(x+12)/6 = 112
x+12 = 672
x = 672-12 = 660.
So, the correct answer = (660/6)+ 12
= 110+12
= 122.

Option(C) is correct

The number of times the father would go to the zoological garden = Number of ways of selection of 3 children taken at a time is

=8C3=56

Number of times a child will go to the zoological garden = Number of times he is accompanied by two other

=1×7C2=21

The number of times a child will not go to the zoological garden:

=56−21=35

Solution:
Option(B) is correct

Machines A and B together will produce 30,000 m of cloth in 1 hour.

Machine B alone can produce 20,000 m cloth in 1 hour.

Therefore, Machine A can produce 10,000 m cloth in 1 hour.

So, in 10 hour Machine A can produce 1,00,000 m of cloth.

Solution:
Option(D) is correct

Let A's age be x and B's age be y.

x+y=43 -------- (i)

(x+11)=76(y+11)

⇒6x−7y=11 -------- (ii)

On solving both the equations, we get:

y=19 years

Solution:
Option(A) is correct

B+8=C
A−8=C–3
A+6=2D
B+D=50
solving these we get A=40

Option(C) is correct

Let age of Chetan in 2002 = x
So,

2002−x90=x

⇒x=22
So, Chetan’s age in 2006 is,

=22+4=26 yrs

Solution:
Option(C) is correct

Let the first installment be 'a' and the common difference between any two consecutive installments be 'd'

Using the formula for the sum of an A.P.
S=n2[2a+(n−1)d]

We have,
3600⇒180And 2400⇒160=402[2a+(40−1)d]=2a+39d=302[2a+(30−1)d]=2a+29d(1)(2)

On solving both the equations we get:
d=2 and a=51