Test 2

1.

Aamir wants to print some brochures for his business. If he goes for one style, it will cost him Rs. 125 plus Rs. 0.2 per brochure. If he goes for the other style, it will cost him Rs. 25 plus Rs. 0.45 per brochure. For how many brochures will the price for both styles be the same?

Solution:

You can understand from the text of the problem that Aamir is printing a number “x” of brochures for which the price will be the same for both styles. Next, you need to note the price with “P”. The price “P” paid for printing the brochures in the first style is:
P = 125 + 0.2 * x

The price “P” paid for printing the brochures in the second style is:

P = 25 + 0.45 *

As you equal the equations, you obtain:

125 + 0.2 * x = 25 + 0.45 * x

125 – 25 = 0.45 * x – 0.2 * x

100 = 0.25 * x

x = 100 / 0.25 = 400

2.

Pointing to the girl playing cricket, Matthew says, “Her mother is the only daughter of my mother-in-law”. Who is Matthew to the girl?

Solution:

The mother-in-law is the mother of his wife, so the only daughter of his mother-in-law is his wife. The girl is the daughter of his wife, so Matthew is the father of the girl.

3.

Find the missing number:

6	10	10	26
8	4	14	26
8	4	14	26
6	3	17	26
7	5	14	26

Solution:

We observe that the sum of the numbers from each row is 26, so, on the second row, the sum needs to be 26.

8 + 4 + x = 26 , so x = 26 – 8 – 4 = 14

4.

Sita and Ammir participated in college election. Sita got 61 % votes and won by a margin of 176 votes. Find total number of votes.

Solution:

Sita got 61% of the votes, so Aamir got 39% of the votes. The difference between their procents is 61 – 39 = 22%, which represent 176 votes. The total number of votes, 100%, is:

22% . . . . . 176

100% . . . . . x

x = 100 * 176 / 22 = 800 (votes)

5.

Find the area of un-shaded part (area of each small square is 1 cm²).

Solution:

The shaded surface has 5 small squares and 4 triangles, each triangle representing half of a small square. The whole shaded surface has the area of 5 + 4 * 0.5 = 7 cm².
The area of the whole rectangle is 5 * 3 = 15 cm².

The unshaded area is 15 – 7 = 8 cm².

6.

Balvinder can finish a work in 28 minutes. Saina works thrice as fast as Balvinder. How long will it take to finish the work, if Balvinder and Saina work together?

Solution:

You’ll introduce the notion of work speed, which represents the speed with which each of them is working, and you’ll note it with S(B), which is Balvinder’s work speed, and S(S), which is Saina’s work speed. Also, you’ll obtain S(S) = 3 * S(B).

The work is finished by Balvider in 28 minutes.

work = 28 * S(B)

When Balvinder and Saina are working together, the two speeds will sum, and the work is finished in time “t”.

work = t * [S(B) + S(S)]

work = t * [S(B) + 3 * S(B)]

work = t * 4 * S(B), but also work = 28 * S(B)

Now you’ll equal them, and you’ll obtain:

28 * S(B) = t * 4 * S(B)

T = [28 * S(B)] / [4 * S(B)] = 7 , so the work is finished in 7 minutes.

7.

The number of hours left in the day are half of hours already passed. How many hours have already passed in the day?

Solution:

A day has 24 hours. You note the number of hours left with “x”, so the number of hours passed will become 2 * x. Now you obtain the equation:

2 * x + x = 24

3 * x = 24

X = 24 / 3 = 8 , so there are 8 hours left of the day.

You’ll now obtain 24 – 8 = 16 hours that already passed.