Aptitude
1.0.0
在水中，沿流的方向稱為下游。而且，朝向流的方向稱為上游。
如果在靜水中的船速度為u km/hr，而流的速度為v km/hr，則：
速度下游=（u + v）km/hr。
快速上游=（u -v）km/hr。
如果下游的速度為km/hr，並且上游的速度為b km/hr，則：
靜水中的速度= 1/2*（a + b）km/hr。
流率= 1/2*（a -b）km/hr。
利潤，P = SP - CP; sp> cp
損失，L = CP - SP; CP> sp
p％=（p/cp）x 100
l％=（l/cp）x 100
sp = {（100 + p％）/100} x cp
sp = {（100 - l％）/100} x cp
cp = {100/（100 + p％）} x sp
cp = {100/（100 - l％）} x sp
折扣= MP - SP
sp = mp -discou
對於虛假的重量，利潤百分比為p％=（真正的重量 - 虛假體重/假重量）x 100 。
當有兩個成功的利潤說m％和n％時，淨百分比利潤等於（M+N+MN）/100
當利潤為m％並且損失為n％時，淨％的利潤或損失將為： （MN-MN）/100
如果以m％利潤出售產品，然後再次以n％的利潤出售，則產品的實際成本價格為： cp = [100 x 100 x p/（100+m）（100+n）] 。如果損失， cp = [100 x 100 x p/（100-m）（100-n）]
如果p％和l％相等，則P = L， ％損失= P^2/100
要檢查數字乘以5或2，並且比該數字添加（如果mul乘5），則如果mul乘以2（減去）至左數
例如： 532 div by 7
步驟1- mul 2x5 = 10步驟2-忽略單位數字，並添加到左數53 | 2+10 = 63
如果是的，則檢查號碼可除原始數字是否可以排除。
將其乘以2，而添加將其減去。
將單位數字乘以5，如果該數字可除以17，則將該值乘以REST數字乘以REST數字。
例如： 272 iv by 17
步驟1-
mul 2*5 = 10 **步驟2- **
27-10 = 17 17被除以17，因此272被分裂為17。
時鐘的臉部或撥號是一個圓圈，其周長分為60個相等的部分，稱為分鐘空間。
小時的手和微小的手，一個時鐘有兩隻手。較小的手稱為小時或短手，較大的手稱為小手或長手。
在60分鐘內，小小的手在一小部分中獲得了55分鐘的空間。
（在60分鐘內，小時將移動5分鐘的空間，而微小的手將移動60分鐘的空間。實際上，相對於小時手的微小手的空間增益將為60-5 = 55分鐘。）
時鐘的雙手每小時一次重合。
當時鐘的手相吻合或彼此相反時，時鐘的手處於相同的直線上。
當時鐘的兩隻手處於直角時，它們相距15分鐘。
當時鐘的手朝相反的方向時，它們相距30分鐘。
在12小時= 360°的小時手用小時追踪的角度
角度在60分鐘內通過微小的手追踪。 = 360° 。
theta（又名學位）= 30 x h-（11/2）xm
回流角= 360 -theta
1。自然數（n）= 1，2，3，。 。 。 。 2。整數（w）= 0、1、2、3，。 。 。 。 3。 intezers （z）=-∞。 。 。 -2，-1、0、1、2、3，。 。 。 4。有理數（q）= p q≠0的形式的數字。例如：1⁄5、0.46、0.333333它們正在終止，重複。 5。非理性數字（i）= x1⁄n≠intezer形式的數字。 π和e也是非終止的非理性數字，不重複的。
其他類型的數字：甚至數字：完全可以除以2的數字。這些數字的格式為2N。 b。奇數：除以2時給出餘數1的數字。這些數字的格式為2n±1。 c 。素數：可除以1的數字，數字本身是素數。最少的素數為2。 d。複合數字：其數量可除以2個以上的數字。
奇數±奇數=均勻;
偶數±=偶數;
甚至±奇數=奇數
奇數×奇數=奇數;
偶×偶數=偶數;
甚至×奇數=偶數。
奇數（任何數字）=奇數
甚至（任何數字）=
（a -b）2 =（a2 + b2-2ab）
（a + b）2 =（a2 + b2 + 2ab）
（A + B）（A - B）=（A2 - B2）
（A3 + B3）=（A + B）（A2 - AB + B2）
（A3 -B3）=（A -B）（A2 - AB + B2）
（A + B + C）2 = A2 + B2 + C2 + 2（AB + BC + CA）
（A3 + B3 + C3 - 3ABC）=（A + B + C）（AR> 2 + B2 + C2 - AB - AB - BC - AC）
快速提示和技巧
步驟1：簡化給定表達式時，將按順序刪除第一括號：' - –'，'（）'，'{}'，'[]'''[]''
步驟2：操作應嚴格按順序執行：劃分，乘法，加法和減法
Bodmas是用於記住簡化過程的快捷方式。
B：支架
O：訂單（即功率，平方根等）
D：分區（從左到右）
M：乘法（從左到右）
答：加法（從左到右）
S：減法（從左到右）
步驟1：簡化給定表達式時，要按順序刪除第一括號：（），{}，[]
步驟2：操作應嚴格按順序執行：劃分，乘法，加法和減法
**重要公式：**
重要公式
n天然數的平方之和= n（n+1）（2n+1）/2
n自然數的總和= n（n+1）/2
總和ap = nx（a+l）/2
第一個n奇數= n的avg
第一個n偶數= n+1的avg
n naturat編號的avg =（n+1）/2
數字的AVG *總數=總數
基於人群的問題
比率和比例
代碼/解碼
紀錄
Amcat口頭問題
剩餘的負數
資格測試
735/2
735/704
3/704
3/735
解決方案
Option A
Find the LCM of the numerators.
LCM (147, 30) = 1470
Step 2: Find the HCF of denominators.
HCF (64, 44) = 4
So, the LCM of 147/64 and 30/44 is (LCM of Numerators) / (HCF of Denominators) = 1470 / 4 = 735/2
a）5 mph
b）10英里 /小時
c）12英里 /小時
d）20英里 /小時
解決方案
Option C
Let the speed of the river=x mph, then
Time taken row 30 miles upstream and 30 miles downstream = 30/(15-x) + 30/(15+x) = 9/2
= 10/(15-x) + 10/(15+x) = 3/2
= 2[10(15+x) + 10(15-x)] = 3(15-x)²
= 300 + 20x + 300 – 20x = 675 -3x²
x² = 25 or x=5
a）3天
b）4天
c）4.5天
d）5.4天
解決方案
option A
Working 5 hours a day, A can complete the work in 8 days = 5*8 = 40 hours
Working 6 hours a day, B can complete the work in 10 days = 6*10 = 60 hours
(A+B)’s 1 hour’s work = (1/40+1/60)
=(3+2)/120
= 1/24
Hence, A and B can complete the work in 24 hours which is equal to 3 days.
a）6.5升
b）5升
c）4升
D）7.5升
解決方案
Option B
A mixture of 40 liters of milk = 36 liters of Milk and 4 liters of water = 90:10 ratio
Now the new mixture should be in the ratio of 80:20
Hence 80% is equivalent to 36 liters (no addition of milk is done)
100% is (36/80)*100 = 45 liters of milk is present in the new mixture
Thus water shall be added= (45 – 36 – 4) = 5 liters of water
**or**
a）12午夜
b）凌晨3點
c）凌晨6點
d）上午9點
解決方案
Option D
Four different devices beep after every 30 mins, 60 mins, 90 mins and 105 mins.
LCM of 30,60,90 and 105 is 1260.
Which means the devices beep together after every 1260 mins = 1260/60 = 21 hours
Hence 12 noon + 21 hours = 9 a.m
a）21 m
b）26 m
c）28 m
d）29 m
解決方案
Option C
When A travels 100 m, B travels 75 m. Hence A:B = 100:75
When B travels 100 m, C travels 96 m. Hence B:C = 100:96
When B Travels 75 m, C travels (96 x 75)/100 = 72 m
Hence B:C = 75:72.
Therefore, A:B:C = 100:75:72.
So, when A Travels 100 m, C travels 72 m.
Therefore, A beat C by 28 m
a）62
b）68
c）66
d）69
解決方案
Option A
70% students passed in physics = 30% failed in Physics.
65% students passed in Chemistry = 35% failed in Chemistry
Percentage of students failed in both subject = 27%.
Percentage of students failed = 35 + 30 – 27
= 38%.
Percentage of students passed = 100 – 38% = 62%
a）40
b）60
c）100
d）160
解決方案
Option D
15 oxen take 80 days so, 6 oxen take x days
x = 15*80/6 = 200 days
20 oxen also take 80 day. So, 2 cows take y days
y = 20*80/2 = 800 days
Together work will be done in 800*200/(800+200) = 160 days
a）297/10377
b）188/121
c）21/34
D）33/163
解決方案
Solution: Option D
a）6C4
b）6p4
c）4^6
d）6^4
解決方案
C is correct because all 4 can get 4 ticket one by one
a）550
b）450
c）350
D）150
解決方案
Option B
Going through the options,
Taking Cost Price as Rs 450.
Profit = 650 – 450 = 200
Loss = 350 – 450 = 100
Clearly profit is twice the loss incurred.
Hence, Rs 450 is the correct option.
a）7小時30分鐘
b）8小時
c）8小時15分鐘
D）8小時25分鐘
解決方案
Option C
In 1hr, Ronald types = 32/6 pages and Elan types = 40/5 pages
If they work together they will type = 32/6 + 40/5 = 40/3 pages in 1 hr
Time needed to complete the assignment is = (3 x 110)/40 = 33/4 = 8hrs 15mins
Hence, the time required is 8 hrs 15 mins.
a）減少20％
b）增加20％
c）增加了10％
d）減少10％
解決方案
Option D
Let the initial Price = Rs.100 and initial sales = 100
So, the initial revenue = Rs. 10000
Now, the price is reduced to 25% which is equal to Rs.75 and Sales is increased by 20% which is equal to 120.
Now new revenue = 120 x 75 = Rs. 9000
Change in revenue = (10000 – 9000) = Rs.1000 decrease
% decrease = (1000/10000) x 100 = 10%
Hence, the correct option is decrease of 10%.
a）19/21
b）3/7
c）2/21
d）1/3
解決方案
Option A
Probability of getting atleast one nestle chocolate = [(10C1 x 5C1) + 10C2] / 15C2
[(10 x 5) + (10 x 9)/2] / [(15 x 14)/2] = 19/21.
Hence, the required probability is 19/21.
a）7490盧比
b）7350盧比
C）8250盧比
D）8530盧比
e）這些都不
解決方案
Option B
Solution:
Share of Anil : Share of Ruhi : Share of Teena is
2000×8 + 2600×4 : 2800×8 + 3200×4 : 4200×4
33 : 44 : 21
so share of Teena = 21/(33+44+21) × 34300 = Rs 7350
a）3200盧比
b）4500盧比
C）3800盧比
D）3500盧比
e）2800盧比
解決方案：
Option C
Solution:
Rs 3800
Solution:
(7000-x)*8*4/100 = x [ (1 + 10/100)2 – 1] + 226
70*8*4 – 32x/100 = 21x/100 + 226
2240 – 226 = 53x/100
2014 = 53x/100
So, x = Rs 3800
a）16
b）18
c）12
d）10
e）22
解決方案：
Option A
Solution:
20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men
a）3/5
b）2/9
c）1/8
D）3/7
e）這些都不
解決方案：
Option D
Solution
Number of multiples of 3 in 140 = 140/3 = 46
Number of multiples of 7 in 140 = 140/7 = 20
Number of multiples of 3×7= 21 in 140 = 140/21 = 6
So required probability = (46+20 – 6)/140 = 60/140 = 3/7
A.媽媽
B.姐姐
C.侄女
D.孕婦姨媽
解決方案c
a）a
b）b
c）c
d）e
解決方案：
E
Solution:
D is father of A and grandfather of F. So, A is father of F.
Thus. D and A are the two fathers. C is the sister of F So. C is the daughter of A.
Since there is only one mother, it is evident that E is the wife of A and hence the mother of C and F.
So, B is brother of A There are three brothers. So. F is the brother of C.
Clearly, A is E's Husband.
a）媽媽
b）姐姐
c）姨媽
D）祖母
解決方案：
C
Solution:
Only son of Amar's mother's father — Amar's maternal uncle.
So, the girl's maternal uncle is Arnar's maternal uncle. Thus, the girl's mother is Amar's aunt.
a）12
b）13
c）14
d）15
解決方案：
A
Solution:
A+B=B+C+12
so
A=12
解決方案：
P=(3*2*5)/1=30
Q=(4*2*5)/1=40
解決方案：
(150+x)/15=16.
=)150+x=240
=x=90
他們年齡的平均年齡= sum/number =）90/5 = 18
A. 154°
B. 170°
C. 160°
D. 180°
解決方案：
We know that angle traced by hour hand in 12 hrs = 360°
From 8 to 2, there are 6 hours.
Angle traced by the hour hand in 6 hours = 6×360/12= 180°
A. 120°
B. 125°
C. 130°
D. 135°
解決方案：
C
Solution:
Angle traced by hour hand in 12 hrs. = 360°.
Angle traced by it in 11/3 hrs = (360/12 x 11/3)° = 110°.
Angle traced by minute hand in 60 min. = 360°.
Angle traced by it in 40 min. = (360/60x40)°= 240°.
Required angle (240 - 110)° = 130°.
A. 3.6
B. 7.2
C. 8.4
D. 10
解決方案：
B
Solution:
Speed =600/(5 x 60)= 2 m/sec.
Converting m/sec to km/hr (see important formulas section)
= (2 x18/5)km/hr= 7.2 km/hr.
A. 10
B. 8
C. 6
D. 4
解決方案：
C
Opposite direction
speed=60+6=66km/h
time=distance/speed=110/66=5/3 km/h
in m/s 5/3x18/5=6
A. 14.4秒
B. 15.5秒
C. 18.8秒
D. 20.2秒
解決方案 ：
A
Solution :
Let length of each train be x meter.
Then, speed of 1st train = x/18m/sec
Speed of 2nd train = x/12 m/sec
Now,
When both trains cross each other, time taken
=2x/(x/18+x/12)=2x/(2x+3x)/36=(2x X 36)/5x=725=14.4seconds
767 495 359 291 257？
A. 230 B. 240 C. 250 D. 280 E. 260
Soluton：
B
Solution:
797-495=272
495-359=136
So which means it is half of previous diffrences
272/2=136
291-257=32
34/2=17
So subtract 17
257-17=240
50 67 33 84 16？
A. 101 B. 109 C. 107 D. 103 E. 201
解決方案：
A
Solution:
17 is the gap once increase and than decrease follow this order you will get the answer
192
10
38
2
3
解決方案：
3
Solution:
Logic is 2×1 + 1 = 3, 3 × 2 + 4 =10, 10 × 3 + 9 = 39, 39 × 4 + 16 = 172…. So in place of 38, it should be 39.
999980
999990
999984
這些都不是
解決方案：
Greatest six-digit number is 999999. Divide this number by 12 and get remainder as 3. Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12. So the greatest such number will be 999999 – 3 =999996.
5000
4950
4980
4900
這些都不是
解決方案：
Multiples of 3 between 100 and 200 are 102, 105, 108,… ,198.
Here, the first term = 102
last term = 198
Let the number of Multiples of 3 between 100 and 200 = n
W.K.T: Arithmetic Progression Formula:
an = a1 + (n - 1)d
Where, an = last term = 198
a1 = first term = 102
d = common difference = 105 - 102 = 3
---> 198 = 102 + (n - 1) * 3
---> 198 - 102 = (n - 1) * 3
---> 96 = (n - 1) * 3
---> (n - 1) = 96/3 = 32
---> n = 32 + 1
---> n = 33
Formula:
Sum of n terms = Sn = (n/2) * (a + l)
where n = number of elements = 33
a = first term = 102
l = last term = 198
Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)
= (33/2) * 300
= 33 * 150
= 4950
A. 5、15、25
B. 12、15、18
C. 10、15、20
D. -10，-15，-20
解決方案：
Assuming that the numbers are (a – d), a, (a + d) and their sum is 45, we get the middle number as 15. Now, the product (a – d) (a + d) = 200. Solving, we get d = 5. Therefore, the numbers are 10, 15 and 20.
解決方案：
x/(y+1)=1/2
and
(x+1)/y=1
2x-y=1 ....eq (1)
x=y-1 ....eq (2)
by solving eq 1 and 2 we get
x=2 and y=3
A. 80
B. 75
C. 42
D. 53
回答：
Option D
Solution:
As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)
The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}
Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]
Now, On simplifying,
[3 × {4 × (7x + 4) + 1} + 2]
= 84x + 53
We get the final number 53 more than a multiple of 84 Hence, if the number is divided by 84,
The remainder will be 53
a）55/601
b）601/55
c）11/120
D）120/11
解決方案：
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum =1a+1b = a+bab= 55600=11120
解決方案：
5
2、3、5、8、13、20、34
解決方案：
first+second=third
follow this order you get **20** as a answer
196、169、144、121、100、80
解決方案：
80
First - Second=2
A. CMN
B. Uji
C. Vij
D. IJT
解決方案：
C
奧尼
Onilen
Noilen
lnoeni
Onnlie
解決方案：
1
一個。 9年
b。 10年
c。 13年
d。 15年
解決方案：
Correct Option: (c)
We have to find the population of cities A and B after x years.
Step 1: Population of city A = 68000, decreases at the rate of 1200/year
68000 – 1200x
Step 2: Population of city B = 42000, increases at the rate of 800/year
42000 + 800x
Step 3: Find after how many population of cities A and B are equal.
Population of city A = Population of city B
68000 – 1200x = 42000 + 800x
68000 – 42000 = 1200x + 800x
26000 = 2000x
x = 13
一個。 28天
b。 30天
c。 34天
d。 40天
解決方案：
Correct Option: (b)
Step 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x)
Step 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20
Step 3: After subtracting the amount which he forfeited, he receives Rs. 300.
Therefore,
(60 – x) 20 – 10x = 300
1200 – 20x – 10x = 300
900 = 30x
x = 30 days
一個。 11
b。 15
c。 16
d。 18
解決方案：
Correct Option: (b)
Let’s the number of farmers be y.
Step 1: Find number of heads
= (50 hens + 45 goats + 8 horses + y farmers)
= (103 + y)
Step 2: Number of feet
= [(Hens 2 × 50) + (45 × 4) + (8 × 4) + (y × 2)]
= [100 + 180 + 32 + 2y]
= 312 + 2y
Step 3: Find number of farmers
(312 + 2y) – (103 + y) = 224
312 + 2y – 103 – y = 224
y = 15
a）49500
b）49950
c）45000
D）49940
解決方案：
The Correct answer is (B)
Answer with explanation:
The digit 5 has two place values in the numeral, 5 * 105 = 50,000 and 5 * 101 = 50.
∴Required difference = 50000 - 50 = 49950
a）Rs。 180
b）Rs。 204
c）Rs。 210
D）Rs。 220
解決方案：
Option B
CI=P(1+r/100)^t
CI=2500*(1+4/100)^2
CI=2704
So the diffrenece is 204
a）Rs。 5222.2
b）Rs。 5777.7
c）Rs。 6222.2
D）Rs。 6777.7
解決方案：
Option B
80000*12/65000*6=32/13
113/32*20000=5777.7rs
a）星期四
b）星期三
c）星期五
D）星期日
解決方案：
The correct option is (B)
Explanation:
The year 1996 is divisible by 4, so it is a leap year with 2 odd days.
As per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.
a）1.5 km/hr
b）2 km/hr
c）2.5 km/hr
d）1 km/hr
解決方案：
The correct answer is B
Answer with explanation:
Let the speed of stream = X km/hr
Speed of boat = 5 km/hr
Speed upstream = 3km/hr
Apply formula: Speed upstream = speed of boat - speed of stream
∴ 3 = 5 - X
X = 5 - 3 = 2 km/hr
a）24
b）22
c）23
d）21
解決方案：
The Correct answer is (B)
Explanation:
The hands of a clock coincide only once between 11 O' clock and 1 O' clock, so in every 12 hours, the hands of a clock will coincide for 11 times.
∴ In a day or 24 hours, the hands of a clock will coincide for 22 (11+11) times.
a）10個小時
b）12小時
c）14小時
D）16小時
解決方案：
Option B
Water enter in 1 hr=1/6
Water empty in 1 hr=1/12
net=1/6-1/12=1/12
or 12hr
a）8.5 km/s
b）7.5 km/s
c）9.5 km/s
d）6.5 km/s
解決方案：
Option B
sec=4*60=240s
speed=500/240=25/12m/s
in km/s speed is =25/12*18/5=7.5km/s.
解決方案：
Surface area of cube=6a^2
600=6*a^2
a=10
diagonal of cube=sqrt(3)*a
ans=sqrt(3)*10
兩個數字的HCF為23，其LCM的其他兩個因素為13和14。兩個數字中的較大是：
A. 276
B. 299
C. 322
D. 345
解決方案：
Answer: Option C
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
六個鈴鐺開始匯集，分別以2、4、6、8 10和12秒的時間折磨。在30分鐘內，他們共同賺幾次？
A. 4
B. 10
C. 15
D. 16
解決方案：
Answer: Option D
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30 + 1 = 16 times.
2
令n為最大的數字，將劃分1305、4665和6905，在每種情況下都留下相同的剩餘數字。然後n中的數字總和為：
A. 4
B. 5
C. 6
D. 8
解決方案：
Answer: Option A
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
A. 9000
B. 9400
C. 9600
D. 9800
解決方案
Answer: Option C
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.
a）50天
b）60天
c）84天
d）9.333天
解決方案：
Answer: C
Explanation:
Let 5 men can reap a field in x days
So, put the same quantities on the same side.
Men: Days
Now, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.
Inversely proportional means Apti Chain Rule
Apti Chain Rule
i.e., 5: 15 = 28: x
Or, x = (28*15)/ 5
Or, x = 84 days
Hence, 5 men can reap a field in 84 days.
a）16
b）13/5
C）-16/3
d）12
解決方案：
Answer: C
Explanation:
Let log2√2 [1/256] = x
We know that loga y = x is similar to ax = y
So, we can write it as [1/256] = (2√2) x
Or, (2√2) x = [1/28]
Or, [21 * 21/2]x = 1/28
Or, 23x/2 = 2-8
Therefore, 3x/2 = -8
Hence, x = (-8 * 2)/ 3 = -16/3
解決方案：
(6!)/(2!)(2!)=180
解決方案：
You can see number less than 3! are not divisible by 8 so it decide your output
(1!+2!+3!)=9
9%8=1
1 is the answer
A. 2400
B. 2000
C. 1904
D. 1906
E.這些都不
解決方案：
106 x 106 - 94 x 94 = (106)2 - (94)2
= (106 + 94)(106 - 94) [Ref: (a2 - b2) = (a + b)(a - b)]
= (200 x 12)
= 2400.
兩個數字的差異為1365。將較大數字除以較小時，我們得到6個商，而15個作為其餘的。什麼是較小的數字？
A. 240
B. 270
C. 295
D. 360
答案：選項b
解釋：
Let the smaller number be x. Then larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Smaller number = 270
A. 1035
B. 1280
C. 2070
D. 2140
答案：選項A。
解釋：
Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.
Sn = n [2a + (n - 1)d] = 45 x [2 x 1 + (45 - 1) x 1] = 45 x 46 = (45 x 23)
2 2 2
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
= 1035.
Shorcut Method:
Sn = n(n + 1) = 45(45 + 1) = 1035.
2 2
[a] 7小時30分鐘
[b] 8小時
[C] 8小時15分鐘
[D] 8小時25分鐘
解決方案）
C)
Ronald 1 hr work = 32/6=16/3
Elan 1 hr work = 40/5=8
Show both work in an hr=8+16/3=40/3
Show for 110 pages it will take 110/(40/3) or (110 x 3)/40=33/4hr
Since: convert it into hr 4*8=32 1 left in 1 hr 60 min 60/4=15min
Show final answer is 8hr 15 mon
解決方案：
鏈接到解決方案
4^x/4^x + 6^x/4^x = 9^x/4^x
Now,
1 + (3/2)^x=(3/2)^(2x)
Consider (3/2)^x=u
Then,
1 + u = u^2
Simplifying this
0 = u^2 -u -1
By solving we get
u = (1 + sqrt(5))/2
and this equal to
(1 + sqrt(5))/2 = (3/2)^x
Take log both side
and you get 1.187 approx value.
Solution:
circumference of an wheel=πd
=22/7×98
22×14
=308cm =1 revolution
distance covered
1540×100=154000
now,154000÷308
500 rotations
答：2％
B. 2.02％
C. 4％
D. 4.04％
E.這些都不
答案：選項d
100 cm is read as 102 cm.
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2
∴ Percentage error
=(404100×100×100)%=4.04%
Area of the square=484cm
side-22cm
perimeter=22*4=88cm
circumfrence of cirlce is 2*pi*r=88
r=14cm
area=pi*r*r=616cm^2
解決方案
P(A)=1/9
P(B)=1/6
P(C)=26/36=13/18
Apply GP
you get 2/5 ans
