IQ TEST: Math, Mixture, Solutions, Concentration, Coin Problems

1. It takes 3 parts of cement and 4 parts of sand to make a specific mixture. How many containers of cement is required to make a mixture of 28 containers?

find:

x = number of containers of cement

solution:

TOTAL = sum of parts

7/7 = 3/7 + 4/7  ---> denominator (7) = 3 + 4

mixture = cement + sand

7/7(28) = 3/7(28) + 4/7(28)

x = 3/7(28)

x = 12


2. How many liters of a 20% solution should be added to 40 liters of 80% solution to make a mixture containing 50% of concentration?

find:

v1 = number of liters of 20% solution

solution:

v1*concentration1 + v2*concentration2 = Vtotal*finalconcentration

v1(0.2) + 40(0.8) = (v1 + 40)(0.5)

0.2v1 + 32 = 0.5v1 + 20

0.3v1 = 12

v1 = 40


3. How many quarters, dimes, and nickels are there if their total is $3.60 and there are a total of 21 coins and the number of nickels is twice the number of dimes.

find:

n = number of nickels

d = number of dimes

q = number of quarters


given:

cents = 360

n = 2d ---> equation1

n + d + q = 21 ---> equation2


solution:

TOTAL = sum of parts

360 = 5n + 10d + 25q ---> equation3


substituting n = 2d in equation2

n + d + q = 21

2d + d + q = 21

3d + q = 21

q = 21 - 3d ---> equation4


substituting n = 2d in equation3

360 = 5n + 10d + 25q

360 = 5(2d) + 10d + 25q

360 = 10d + 10d + 25q

360 = 20d + 25q  ---> equation5


substituting equation4 in equation5

360 = 20d + 25q

360 = 20d + 25(21 - 3d)

360 = 20d + 525 - 75d

55d = 525 - 360

55d = 165

d = 3

n = 2d = 2(3) = 6

q = 21 - 3d = 21 - 3(3) = 12


checking:

d cents = 3 * 10 = 30 cents

n cents = 6 * 5 = 30 cents

q cents = 12 * 25 = 300 cents

total = 30 + 30 + 300 = 360