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 motherinlaw”. Who is Matthew to the girl?
Solution:
The motherinlaw is the mother of his wife, so the
only daughter of his motherinlaw 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 unshaded part (area of each small
square is 1 cm^{2}).
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 cm^{2}.
The area of the whole rectangle is 5 * 3 = 15 cm^{2}.
The unshaded area is
15 – 7 = 8 cm^{2}.
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.