英文考題講解

Here I’ll explain the test questions one by one:  

Note:  “*”means to multiply;    “/ ”means to divide.   

1. (b) When subtracting 4357-9685, the smaller number must be subtracted from the bigger one, not the other way around. 

4357-9685=(- 1) * (9685-4357)= -5328 

For example, if I borrowed $10  from you yesterday, and I’m paying you back $4 today, how many dollars do I still owe you? 

Note: When I own the money, the number is positive. But if I owe you money, the number is negative.  

The equation is like this:  

 4-10 = (-1)*(10 -4  = -6     

 So, the answer says I owe you $6.   

2. (3 pts x 2 = 6 pts) Calculate with pencil. Show every single step of your calculation. For the division,  round the result to the nearest hundredth.  

95/37 =2.567….  and the hundredth after the decimal point needed to be rounded. Therefore 95/37=2.57

The above two questions (1, 2) use the most basic math skills.  

Next, in every step it’s best that you figure out the reason why you do it.  

3. (a)  2/7+3/4  fractions with different denominators cannot be added, you first find the common factors of the denominators. So when solving the problem, you made sure changing the two fractions into the same denominator and keep the fraction value unchanged, so   

2/7=(2*4)/(7*4)=8/28  

3/4 =(3*7)/(4*7)=21/28.  

So 2/7+3/4= 8/28+21/28=29/28 =1 (1/28).     

 (b) Now let’s analyze  (8/7-55/44). This is the operation of improper fractions, because their numerators are greater than denominators. You can see that both fractions have integer parts. In order to avoid multiplication of large numbers, improper fractions should be converted into their integers; then their real fractions are added together; that is, first change the improper fraction to an integer together with a fraction, then the two integers operate, and the two real fractions operate. 

In short, make the questions concise and avoid complicated calculations.  

8/7=1 +1/7;   

55/44=(11*5)/(11*4) = 5/4=1 +1/4  after eliminating the common factor 11.   

So the steps go like this:    

8/7 - 55/44= (1+1/7) – (1+1/4) = (1-1) +（1/7-1/4）  

=1/7-1/4 = (1*4)/ (7*4)-(1*7)/(4*7)=4/28-7/28  

=(4-7)/28=-3/28   

You see, there is a reason to every step. It’s like you were watching a plant grow inch by inch.   

(c) For the multiplication of (28/27)*(9/35), please pay attention to factorization. You have to eliminate the common factors in the numerator and denominator.  

28=2*2*7,     27=3*9 =3*3*3;  

9=3*3,     35=5*7   

So (28/27)*(9/35) = (28*9) / (27*35) = (4 *7*9) / (9*3*5*7) = 4 / (3*5) = 4/15   

Please note that the numerators and the denominators need to be multiplied separately. Also, reduce first to avoid large number calculations.   

(d) (66/75)/ (11/25): The idea is the same as the above. But the only difference is when doing division, you have to change it to multiplication.   

 (66/75) / (11/25) = (66/75)*(25/11) = (66*25) / (75*11)=(6*11*25 )/(25*3*11 )=6/3=2   

Students could have made better use of your knowledge in math. Again, please remember: avoid multiplication and division of large numbers.  Making things simple always makes life easier, the same is true for math.   

4. In each pair below, circle the greater number.  

It’s a comparison question. There must be a reason for the teacher to have you do it. The principle is:   

(a), only compare the numerators when the denominators are the same. If the numerator is larger, then the fraction is larger;   

 (b), only compare the denominator when the numerator is the same. If the denominator is smaller, the fraction is larger. For example, you have a birthday cake that you would like your friends to share at the party. The more friends you’ve invited, the smaller share each of them will get. If there are only two of them, each can eat half, and if there are four of them present, each can just eat a quarter of the cake. So, the number of your friends is the denominator, and the one cake is the numerator. The more the  denominator, the smaller the fraction.   

(c) When comparing fractions and decimals, convert the decimals into fractions, or vice versa,  to make them either all fractions or all decimals. 

（a), (b), (c) below are the rules, and we’ll use them for other questions too. 

(a) The denominator of 92/107, 93/107are the same, just compare the numerators. the numerator 93 greater than 92, so 93/107 is larger.  

(b) The numerators of 92/107, 92/106 are the same, compare the denominators, use the the-smaller-is-larger rule,106 is less than 107, so 92/106 is larger.   

(c) 92/107, 46/53    

Observe 46*2=92, that is, 46/53=(46*2)/(53*2)=92/106 . So this is the same as question (b)   

(d) 29/56, 0.49     

To make it easier, we can round up 0.49 to 0.5. Because 0.5 is greater than 0.49, and 0.5=1/2=28/56, then it becomes a comparison with the same denominator. Since 29/56 is greater than 28/56= 0.5 which is greater than 0.49, so 29/56 is greater than 0.49.  

 5. Operations with decimal points: The principles are    

(a).The decimal points must be vertically aligned in addition and subtraction. It’s like counting your money. You count dollars, dimes, and cents separately without mixing them up. For example if you made 1 dollar, 2 dimes, and 3 cents today, you can write it as 1.23. On the next day you made 3 dollars, 1 dime and 3 cents, you write it down as 3.13. When actually counting all the money, I am sure you separate them into dollars, dimes and cents. This is why the decimal points must be aligned. 

 (b) In multiplication and division, we must move the space of the points; 

(a) 15.8+2.371, here you forgot to align the points , so the digits got all mixed up. The correct alignment looks like this:   

                                                        15.8  

                                                        +2.371  

                                                       =18.171        

We can reverse the calculation to check if it’s correct: 18.171-2.371=15.8. Yes, it’s correct!           

(b) 630.02 - 34.176    

                                            630.02  

                                           - 34.176  

                                        = 595 .844          

Reverse the calculation to check:  

595.844+34.176=630.02. Correct!  

(c) 1.4X5.0  

Pay attention to where the decimal point is. First do the common multiplication then locate the point location. Doing the common multiplication means to first change the decimal numerals to integers:  

1.4 * 10 = 14    

5.0* 10= 50,  

14*50=700  

Because we multiplied 10*10=100, we need to take out the 100 from the result. That is: the result of 700 divided by 100 —> 700/100= 7.   

The above steps look like this:  

1.4* 5.0 = (1.4*10/10)* (5.0*10//10) = (14/10)*(50/10)  

= (14*50) / (10*10) = 700/100=7.   

The actual operation is very simple.  

First, for each number, we moved the decimal point one space to the right, making it an integer,  and we moved a total of two spaces to the right in the question. So in the product of the multiplication, we need to move the decimal point two spaces back from right to left.                                                           

(e) When doing division of 7.38/0.12,  we first change the denominator to an integer, that means both the numerator and the denominator were multiplied by 100, meaning both points were moved two spaces to the right:    

The denominator:  0.12*100=12,  

The numerator:   7.38*100=738   

We write it like this:   

7.38/0.12=(7.38*100)/(0.12*100)=738 /12 =(123*2*3)/(2*2*3)=123/2=61.5  

Different from multiplication, in division we DO NOT move back the decimal points in the final quotient.

 

6. (a) 12-10 x(99-99/33)+(87x2-174)   

We should remember the most important rule here, that is, first do multiplication and division, then do addition and subtraction. 

That is part of the PEMDAS rule you learned at school.

Here are the specific steps:

99/33=3 , (99-99/33)=99-3=96  

(87x2-174)=174-174=0, so the problem becomes 12-10*96=12-960= -948  

(b) (3x (19+6x (7-9))-2) x (174-(174-29) /29)/ 19-18 = (first part)*(second part)/19-18  

The first part is (3x (19+6x (7-9))-2) and the second part is (174-(174-29) /29)  

For (3x (19+6x (7-9))-2):   

Please also remember the order of operations in PEMDAS here. 

What’s in the innermost parenthesis are operated first, then peeled from the inside to the outside. This is a rule that must be followed. So here (7-9) = -2  

(19+6x (7-9))=19+6*(-2)=19-12=7, the second layer of operation comes out,  

Finish the outside operation:  (3x (19+6x (7-9))-2) =3* 7-2=19.   

In the second part （174-(174-29) /29):   note 174=29*6 

First (174 - 29)/29=(29*6-29)/29=29*(6-1)/29=5

29 is  the common factor of the numerator and the denominator

(174-(174-29)/29)=174-5=169,  

So (first part)*(second part)/19-18=19*169/19-18=169-18=151,

19 is the common factor of the numerator and the denominator.   

7. Calculate with pencil. Sketching on another piece of paper is OK.  

All these questions contain “123”,  so you need to use basic math knowledge and four arithmetic operations to solve them. Please remember to extract the common factor!   

Note: 123=123*1, so 123 is the factor!   

(a) 123 - 123 x 11= 123*1-123*11=123*(1-11)=123*(-10)=-1230     

(b) -123 + (-123) x9=(-123)*1 +(-123)*9=(-123)(1+9)=-123*10=-1230   

(c) 123 - 123 x (-9)=123*1-123*(-9)=123*(1-(-9))=123(1+9)=123*10=1230  

(d) -123 - (-123) x (-9)=(-123)*1-(-123)*(-9)=(-123)(1-(-9))=(-123) *10=-1230   

(e) 123 - 123 / (-123)=123-(123/(-123))=123 – (-1)=123+1=124   

(f) (-123) / 123 - (-123)=((-123)/123)-(-123)=-1-(-123)=-1+123=122   

(g) (-123)/ (-123) + 123 = 1+123=124   

 (h) (-9) x 123 - 123 =123(-9-1)=123*(-10)=-1230   

(I) (-123)/ (12.3 - (-12.3)) = (-123)/(12.3*(1+1))=(-123)/((123/10)*2) =-10/ 2=-5   

 (j) 100 x 123 – 123/(-123/50)=100*123+(123*50/123)=100*123+50=12300+50=12350     

8. (5 pts x 2 = 10 pts) Use the known results to complete the calculations. Show every step of your calculations.  

(a).It is known that  123+234+345+456+567+678+789+890+901=4983 …….. (1)  

Compare (1) and the following formula 123+234+345+456+567+678+789+890? ……….(2)  

What is the sum in (2)?  Here we assume B=123+234+345+456+567+678+789+890,  

Compare (1) and (2)  

On the left side is (1): 123+234+345+456+567+678+789+890+901=B+ 901=4983,  

take out B+901=4983,  

subtract 901 at both ends to get B+901-901=4983-901 ,    

so B=4983-901=4082.  

(b) It is known that (-123) x 234+345+456 + 567+678 + 789+ 890 x 901 = 775943 ………. (1)  

How much is B= (-123) x 234 + 345 + 456 + 567 + 678 + 789 + 890 x 900? ………….(2)  

Compare (1) and (2) the difference is 890*901 in (1) and 890*900 in (2),  

then (1)  890*901=890*(900+1)=890*900+890   here 890*900 the same as the last term of (2)    

so (1) left part = B+890 = 775943,  

minus 890 at both ends is B +890-890=775943-890   

so B=775943-890= 775053.      

9. (8 pts x 3 = 24 pts) Word problems. (Show the details of your calculation.)  

The purpose of the three questions here are the same, i.e. to turn multiple unknowns into a single unknown, and use known conditions to establish equations.  You can fulfill this by patiently analyzing the conditions. 

(a) A shopkeeper had 1080 apples and oranges altogether at the beginning. After selling 60 oranges and 20 apples, the number of apples he had was three times the number of oranges left.   

The first method :  

Let’s say the original number of oranges is A, and the number of apples is 1080-A. Makes sense?  According to the question, after selling oranges A-60, the number of apples is 1080-A-20, which should be three times the number of oranges left.  

That is: the triple of A-60 is 1080-A-20=3*(A-60).

This is an algebraic solution to the equation. It is very simple if you know the method.   

The second method: The four basic arithmetic operations method.

Let’s say the number of oranges after selling is B, and from the known conditions, the number of apples after selling is three times of B , that is 3B. The total number of apples and oranges after selling should be 1080-(20+60), on the other hand the total number of apples and oranges after selling is B+ (3B)=4 B  

So B + 3B=4B= 1080-（20+60）  

So the number of oranges after selling B=(1080-(20+60))/4  

The oranges before selling are A= B +60  

Therefore, the written expression is A= (1080-(20+60))/4 +60. From this we can see the reason for the parentheses in the expression. Computing this formula is similar to the previous one.   

The third method:  To analyze and count step by step. 

The key here is to put it in the simplest way possible, that is, let’s focus on the least quantity with the most known conditions. Which number is the smallest here? The number of apples or oranges before sale is larger than that after the sale, and the condition after the sale is the clearest, that is, the number of apples after the sale is three times the number of oranges,  so the number of oranges after the sale is the smallest.  

Let’s say B is the number of oranges after the sale.  

Calculate while analyzing: First, the total number of oranges and apples sold is 20+60=80,  

The total number of apples and oranges left after selling is 1080-80=1000  

And the remaining apples are 3 times the remaining oranges,  

the remaining apples=3*the remaining oranges=3 B,  

which means 1000=the remaining oranges+the remaining apples  

=B+3B=4B.  

That is 4B = 1000  

So remaining oranges = 1000/4 = 250.  

Original oranges = sold 60 + remaining 250 = 310.  

In order to reduce writing, symbols are used, and B is the number of oranges after the sale. That is what’s written in the formula above.  

   

(b) Katy is 8 years younger than Tess. Nina is 13 years older than Tess. The sum of their ages is 80. How old is each of them?  

According to our prior analysis, who do we know is the smallest in number as far as age is concerned?  

K has the smallest number of age because  

K=T-8,  

N=13+T,  

So we use the age of K to express N and T, N. They are  

T=K+8,  

N=13+ T =13+(K+8)= K+21.  

The most known condition is the sum of ages. That is, K+ T+ N=80, and the left end is represented by K’s age:  

K+T+N=K+(K+8)+(K+21)=3K+29=80. So do the same operation on both sides and subtract 29:  

3K+29-29=80-29=51,  

3K=51 divide both sides by 3: 3K/3=51/3=17, that is, K=17, written in formula K=(80-21-8)/3  

T=K+8=17+8=25  

N=K+21=17+21=38.   

(c) There are 260 sheep and chickens in total on a farm. The animals have 768 legs all together. How many wings do the animals have in total?  

According to the question, the total number of sheep S and that of chickens C is S + C= 260. This is also the number of one leg of all sheep and all chickens, right? And we know that sheep all have four legs, and chickens two legs, so the total number of legs is 768  

That is 4S+2C=768, right? 

All the chicken legs are included in 768. You have to find a way to remove all the chicken legs in order to get only sheep legs. It’s important to remember that chickens have two legs. The total number of one leg of the chickens + the total number of one leg of all sheep is 260, remember? And the number of two chicken legs of all chickens + the number of two sheep legs of all sheep is 260*2, which is 520;  

so 768 (All legs) - 520=248, and this 248 is the number of the remaining two legs of all the sheep.  

( We can use symbols for the above analysis  ——>  

The number of legs: 4S=S+S+S+S, 2C=C+C;  

Remember:  S + C = 260;  

4S+2C= S+S+ S+ S + C+C=(C+S)+ (S+C)+ S+S= 260+260+ 2S=768  

So, 2S= 768-260-260=248     

The number of one leg of all sheep is:  

S=248/2=124,  

which is also the number of heads of all the sheep.   

Writing all steps in one is like this:   

S=(768-(260*2))/2=248/2=124  

The total number of chickens is C=260-S =260-124=136  

The number of chicken wings is equal to 2*C= 136*=272  

Putting the above calculation and analysis process into a formula:   

Number of chicken wings = 2* Number of chickens = 2*(260-number of lamb heads)  

=2*(260-((768-(260*2))/2)).   

Generally speaking, analysis is the key in this kind of questions. It is ok not to create a formula with all the steps. I wrote them down here just to illustrate the actual background of the mixed arithmetic questions. 

Finally, I do have a few more friendly reminders for you:

Analyze each question first, always take a moment to think, try to simplify the seemingly tricky questions;

Make the questions concise and avoid complicated calculations;

In solving word problems, turn multiple unknowns into a single one which usually is the smallest unknown; and, normally we prefer additions and multiplications to subtractions and divisions in the equations;

All math questions by the teacher have an answer to them. Be confident.