Percentage

1.
Let the weight of alloy A needed be x kg and the weight of pure zinc needed be y kg.
x + y = 400 … (1)
[(35% x + y) / 400] x 100% = 66 1/6% …(2)

Explanation:
35% x => weight of zinc from alloy A.
y => weight of zinc from pure zinc (100% zinc).

From (1):
x = 400 – y … (3)
Substitute (3) into (2):
{[35%(400 – y) + y] / 400} x 100% = 66 1/6%
{[0.35(400 – y) + y] / 400} x 100 = 66 1/6
[(140 – 0.35y + y) / 400] x 100 = 66 1/6
(140 + 0.65y) / 4 = 66 1/6
140 + 0.65y = (66 1/6) x 4
140 + 0.65y = 264 2/3
0.65y = 124 2/3
y = 191 31/39 (or 7480/39)
Substitute (y = 7480/39) into (1):
x + (7480/39) = 400
x = 208 8/39 (or 8120/39)

Therefore, 8120/39 kg of alloy A is needed and 8120/39 kg of pure zinc is needed.

*** The answers of this question sound quite strange...

2.
Let the length of the large cube be y cm.
The surface area of the large cube
= 6y² (cm²) (∵ each cube has 6 faces.)
After cutting the large cube into 8 small cubes, the length of a smaller cube becomes (y/2) cm.
The surface area of all the 8 small cubes
= 8 x 6 x (y/2)² ((∵ 8 cubes, each has 6 faces.)
= 48 (y²/ 4)
= 12y² (cm²)
Percentage change in increase of surface area
= [(12y² - 6y²) / 6y²] x 100%
= (6y² / 6y²) x 100%
= 100%

∴ The answer is B.

Remarks:
Sometimes, you may need to use some ‘extra’ unknowns to solve the questions. These unknowns are usually eliminated before reaching the answers.

If there is any mistake, please inform me.

1. An alloy A contain 35% of zinc by weight. Some alloy A and pure zinc

are melt to form a new alloy that weights 400kg and contains 66 1/16 %

of zinc. What are the weights of alloy A and zinc needed respectively to

form the new alloy?

Sol:

400 × 66 1/16% ＝ 264.25 ....... zinc contained in new alloy

400 － 264.25 ＝ 135.75 ....... non-zinc contained in new alloy

135.75 / ( 1 － 35% )

＝ 135.75/0.65

＝ 2715/13 ....... alloy A

400 － 2715/13

＝ 2485/13 ....... pure zinc

Ans: alloy A ~ 2715/13 kg; pure zinc ~ 2485/13

P.S. 數據非整數, 確認是否 ( 66 ＋ 1/16 )%.

2. If a large cube is cut into eight smaller cubes, then the percentage increase

of surface area is ( A ) 50% ( B ) 100% ( C ) 150% ( D ) 20%

Sol:

Let 2x be length of large cube

then, we get the length of smaller cube is x

2x * 2x * 6

＝ 24x^2 ....... surface area of large cube

x * x * 6 * 8

＝ 48x^2 ....... surface area of smaller cubes

48x^2 － 24x^2

＝ 24x^2 ....... surface area increased

24x^2 ÷ 24x^2

＝ 1

＝ 100% ....... percentage increased

hence, the answer is ( B )

Ans: ( B ) 100%