ACT Practice Paper2 Solution1

In this page ACT practice paper2 solution1 we are going to see solution of some practice questions.

Question 1:

The traffic lights at three different road crossing change after every 48 seconds, 72 seconds and 108 seconds. If they all change simultaneously at 8:20:00 hours then at what time they will again change simultaneously?

Solution:

To solve this problem we have to find the least number which is divisible by all the three numbers. From this we come to know that we have to find L.C.M. We can find L.C.M using two methods.

(i) Factorization method

(ii) Division method

Method 1:

48 = 2 x 2 x 2 x 2 x 3

72 = 2 x 2 x 2 x 3 x 3

108 = 2 x 2 x 3 x 3 x 3

L.C.M = 2 x 2 x 2 x 2 x 3 x 3 x 3

           = 16 x 27

           = 432

Method 2:

 L.C.M = 2 x 2 x 3 x 3 x 2 x 2 x 1 x 3

             = 432

Therefore the correct answer is 432.So option B is correct.


Question 2:

Simplify the following (796 x 796 - 204 x 204)

Solution:

Instead writing 796 and 204 two times we are going to write it once and we are going to put square for 796 and 204 separately.


                       = (796 x 796 - 204 x 204)

                       = 796² - 204²

Now we are going to compare this with the algebraic identity a² - b².

a² - b² = (a + b) (a - b)

796² - 204² = (796 + 204) (796 - 204)

                 = 1000 (592)

                 = 592000

Therefore the correct answer is 592000.So option B is correct.


Question 3:

If 1.125 x 10^k = 0.001125,then the value of k is

Solution:

1.125 x 10^k = 0.001125

           10^k = 0.001125/1.125

                    = (0.001125/1.125) x (1000/1000)

                    = 1.125/1125

                    = (1.125/1125) x (1000/1000)

                    = (1125/1125 x 10³)

                    = 1/10³

                    = 10⁻³

Therefore the value of k is -3.So option A is correct.


Question 4:

If a + b = 5 and 3 a + 2 b = 20, then find the value of (3 a + b)

Solution:

To find the value of 3 a + b we have to find the values of a and b. For that let us solve the given equations.

 a + b = 5  ----(1)

3 a + 2 b = 20 ---- (2)

(1) x 2 => 2 a + 2 b = 10

               3 a + 2 b = 20

               (-)    (-)    (-)

              ----------------  

              - a = - 10

                a = 10

now we are going to apply the value of "a" in the first or second equation in order to get the value of "b"

             a + b = 5  ----(1)

              10 + b = 5

                     b = 5 - 10

                     b = -5 

Now we can find the value of (3 a + b)

                         = 3 (10) + (-5)

                         = 30 - 5

                         = 25 

ACT practice paper2 solution1 ACT practice paper2 solution1