Aptitude
1.0.0
在水中，沿流的方向称为下游。而且，朝向流的方向称为上游。
如果在静水中的船速度为u km/hr，而流的速度为v km/hr，则：
速度下游=（u + v）km/hr。
快速上游=（u -v）km/hr。
如果下游的速度为km/hr，并且上游的速度为b km/hr，则：
静水中的速度= 1/2*（a + b）km/hr。
流率= 1/2*（a -b）km/hr。
利润，P = SP - CP; sp> cp
损失，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
对于虚假的重量，利润百分比为p％=（真正的重量 - 虚假体重/假重量）x 100 。
当有两个成功的利润说m％和n％时，净百分比利润等于（M+N+MN）/100
当利润为m％并且损失为n％时，净％的利润或损失将为： （MN-MN）/100
如果以m％利润出售产品，然后再次以n％的利润出售，则产品的实际成本价格为： cp = [100 x 100 x p/（100+m）（100+n）] 。如果损失， cp = [100 x 100 x p/（100-m）（100-n）]
如果p％和l％相等，则P = L， ％损失= P^2/100
要检查数字乘以5或2，并且比该数字添加（如果mul乘5），则如果mul乘以2（减去）至左数
例如： 532 div by 7
步骤1- mul 2x5 = 10步骤2-忽略单位数字，并添加到左数53 | 2+10 = 63
如果是的，则检查号码可除原始数字是否可以排除。
将其乘以2，而添加将其减去。
将单位数字乘以5，如果该数字可除以17，则将该值乘以REST数字乘以REST数字。
例如： 272 iv by 17
步骤1-
mul 2*5 = 10 **步骤2- **
27-10 = 17 17被除以17，因此272被分裂为17。
时钟的脸部或拨号是一个圆圈，其周长分为60个相等的部分，称为分钟空间。
小时的手和微小的手，一个时钟有两只手。较小的手称为小时或短手，较大的手称为小手或长手。
在60分钟内，小小的手在一小部分中获得了55分钟的空间。
（在60分钟内，小时将移动5分钟的空间，而微小的手将移动60分钟的空间。实际上，相对于小时手的微小手的空间增益将为60-5 = 55分钟。）
时钟的双手每小时一次重合。
当时钟的手相吻合或彼此相反时，时钟的手处于相同的直线上。
当时钟的两只手处于直角时，它们相距15分钟。
当时钟的手朝相反的方向时，它们相距30分钟。
在12小时= 360°的小时手用小时追踪的角度
角度在60分钟内通过微小的手追踪。 = 360° 。
theta（又名学位）= 30 x h-（11/2）xm
回流角= 360 -theta
1。自然数（n）= 1，2，3，。 。 。 。 2。整数（w）= 0、1、2、3，。 。 。 。 3。intezers （z）=-∞。 。 。 -2，-1、0、1、2、3，。 。 。 4。有理数（q）= p q≠0的形式的数字。例如：1⁄5、0.46、0.333333它们正在终止，重复。 5。非理性数字（i）= x1⁄n≠intezer形式的数字。 π和e也是非终止的非理性数字，不重复的。
其他类型的数字：甚至数字：完全可以除以2的数字。这些数字的格式为2N。 b。奇数：除以2时给出余数1的数字。这些数字的格式为2n±1。c 。素数：可除以1的数字，数字本身是素数。最少的素数为2。d。复合数字：其数量可除以2个以上的数字。
奇数±奇数=均匀;
偶数±=偶数;
甚至±奇数=奇数
奇数×奇数=奇数;
偶×偶数=偶数;
甚至×奇数=偶数。
奇数（任何数字）=奇数
甚至（任何数字）=
（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
步骤1：简化给定表达式时，将按顺序删除第一括号：' - –'，'（）'，'{}'，'[]'''[]''
步骤2：操作应严格按顺序执行：划分，乘法，加法和减法
Bodmas是用于记住简化过程的快捷方式。
B：支架
O：订单（即功率，平方根等）
D：分区（从左到右）
M：乘法（从左到右）
答：加法（从左到右）
S：减法（从左到右）
步骤1：简化给定表达式时，要按顺序删除第一括号：（），{}，[]
步骤2：操作应严格按顺序执行：划分，乘法，加法和减法
**重要公式：**
重要公式
n天然数的平方之和= n（n+1）（2n+1）/2
n自然数的总和= n（n+1）/2
总和ap = nx（a+l）/2
第一个n奇数= n的avg
第一个n偶数= n+1的avg
n naturat编号的avg =（n+1）/2
数字的AVG *总数=总数
基于人群的问题
比率和比例
代码/解码
日志
Amcat口头问题
剩余的负数
资格测试
735/2
735/704
3/704
3/735
解决方案
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 /小时
c）12英里 /小时
d）20英里 /小时
解决方案
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
解决方案
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
解决方案
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点
c）凌晨6点
d）上午9点
解决方案
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
解决方案
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
解决方案
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
解决方案
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
解决方案
Solution: Option D
a）6C4
b）6p4
c）4^6
d）6^4
解决方案
C is correct because all 4 can get 4 ticket one by one
a）550
b）450
c）350
D）150
解决方案
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小时30分钟
b）8小时
c）8小时15分钟
D）8小时25分钟
解决方案
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）减少20％
b）增加20％
c）增加了10％
d）减少10％
解决方案
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
解决方案
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卢比
b）7350卢比
C）8250卢比
D）8530卢比
e）这些都不
解决方案
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卢比
b）4500卢比
C）3800卢比
D）3500卢比
e）2800卢比
解决方案：
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
解决方案：
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）这些都不
解决方案：
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.妈妈
B.姐姐
C.侄女
D.孕妇姨妈
解决方案c
a）a
b）b
c）c
d）e
解决方案：
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）妈妈
b）姐姐
c）姨妈
D）祖母
解决方案：
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
解决方案：
A
Solution:
A+B=B+C+12
so
A=12
解决方案：
P=(3*2*5)/1=30
Q=(4*2*5)/1=40
解决方案：
(150+x)/15=16.
=)150+x=240
=x=90
他们年龄的平均年龄= sum/number =）90/5 = 18
A. 154°
B. 170°
C. 160°
D. 180°
解决方案：
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°
解决方案：
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
解决方案：
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
解决方案：
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
解决方案 ：
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
解决方案：
A
Solution:
17 is the gap once increase and than decrease follow this order you will get the answer
192
10
38
2
3
解决方案：
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
这些都不是
解决方案：
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
这些都不是
解决方案：
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
解决方案：
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.
解决方案：
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
解决方案：
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum =1a+1b = a+bab= 55600=11120
解决方案：
5
2、3、5、8、13、20、34
解决方案：
first+second=third
follow this order you get **20** as a answer
196、169、144、121、100、80
解决方案：
80
First - Second=2
A. CMN
B. Uji
C. Vij
D. IJT
解决方案：
C
奥尼
Onilen
Noilen
lnoeni
Onnlie
解决方案：
1
一个。 9年
b。 10年
c。 13年
d。 15年
解决方案：
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
一个。 28天
b。 30天
c。 34天
d。 40天
解决方案：
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
一个。 11
b。 15
c。 16
d。 18
解决方案：
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
解决方案：
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
解决方案：
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
解决方案：
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
解决方案：
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
解决方案：
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
解决方案：
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个小时
b）12小时
c）14小时
D）16小时
解决方案：
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
解决方案：
Option B
sec=4*60=240s
speed=500/240=25/12m/s
in km/s speed is =25/12*18/5=7.5km/s.
解决方案：
Surface area of cube=6a^2
600=6*a^2
a=10
diagonal of cube=sqrt(3)*a
ans=sqrt(3)*10
两个数字的HCF为23，其LCM的其他两个因素为13和14。两个数字中的较大是：
A. 276
B. 299
C. 322
D. 345
解决方案：
Answer: Option C
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
六个铃铛开始汇集，分别以2、4、6、8 10和12秒的时间折磨。在30分钟内，他们共同赚几次？
A. 4
B. 10
C. 15
D. 16
解决方案：
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为最大的数字，将划分1305、4665和6905，在每种情况下都留下相同的剩余数字。然后n中的数字总和为：
A. 4
B. 5
C. 6
D. 8
解决方案：
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
解决方案
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
解决方案：
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
解决方案：
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
解决方案：
(6!)/(2!)(2!)=180
解决方案：
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.这些都不
解决方案：
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.
两个数字的差异为1365。将较大数字除以较小时，我们得到6个商，而15个作为其余的。什么是较小的数字？
A. 240
B. 270
C. 295
D. 360
答案：选项b
解释：
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：选项A。
解释：
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小时30分钟
[b] 8小时
[C] 8小时15分钟
[D] 8小时25分钟
解决方案）
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
解决方案：
链接到解决方案
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.这些都不
答案：选项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
解决方案
P(A)=1/9
P(B)=1/6
P(C)=26/36=13/18
Apply GP
you get 2/5 ans
