这是IBM定量适度的模型文件。本安置文件将介绍IBM安置中要求的能力，并且严格遵循IBM论文中提出的问题的模式。建议解决以下每个问题，以增加清除IBM展示位置的机会。
- 如果5个连续的偶数的平均值为10，请找到中间的数字?
- 4
- 8
- 2个
- 10
回答：
10解释：
Let the five consecutive even numbers be
x-4, x-2, x, x+2, x+4Average of these = ((x-4)+(x-2)+(x)+(x+2)+(x+4))/5 = x
Given average = 10Therefore x = 10
- 两个数字的比例为2：9。如果他们的HCF为19，则数字为：
- 6、27
- 8、36
- 38、171
- 20、90
回答：
38, 171解释：
Let the numbers be 2X and 9X
Then their H.C.F. is X, so X = 19
=> Numbers are (2×19 and 9×19) i.e. 38 and 171 - 三个朋友从早上8:00:00开始一起在环形跑道上奔跑。他们完成一轮赛道所需的时间分别为15分钟，20分钟和30分钟。如果它们连续运转而没有停顿，那么它们将在什么时间第四次在起点再次相遇?
- 上午8:30:00
- 晚上9:00:00
- 下午12:00:00
- 上午12:00:00
回答：
12:00:00 pm解释：
LCM (15, 20, 30) = 60
=> They meet at the starting point after every 60 min, i.e., after every 1 hour.
Therefore, they will meet at the starting point for the fourth time after 4 hours, i.e., at 12:00:00 pm. - 雇用了两个朋友A和B从事工作。最初的截止日期定为24天。两者开始一起工作，但是20天后，A离开了工作，整个工作花了30天才能完成。 B一个人可以在多长时间内完成这项工作?
- 40
- 50
- 60
- 70
回答：
60解释：
Let the total work be 24 units. It is given that A and B together can do the work in 24 days.
=> Combined efficiency of A and B = 24/24 = 1 unit / day
=> Work done in 20 days = 20 units
=> Work left = 24 – 20 = 4 units
Now, this remaining 4 units of work was done by B alone in 10 days.
=> Efficiency of B = 4/10 = 0.4
Therefore, time required by B alone to do the work = 24/0.4 = 60 days - 分别连接到游泳池的两条管道A和B可以分别在20分钟和30分钟内充满游泳池。两者一起打开，但是由于管道A的电动机故障，必须在两分钟后将其关闭，但是B继续工作直到游泳池完全注满为止。查找填充池所花费的总时间。
- 20
- 22
- 25
- 27
回答：
27解释：
Let the capacity of the pool be LCM(20, 30) = 60 units.
=> Efficiency of pipe A = 60 / 20 = 3 units / minute
=> Efficiency of pipe B = 60 / 30 = 2 units / minute
=> Combined efficiency of pipe A and pipe B = 5 units / minute
Now, the pool is filled with the efficiency of 5 units / minute for two minutes.
=> Pool filled in two minutes = 10 units
=> Pool still empty = 60 – 10 = 50 units
This 50 units is filled by B alone.
=> Time required to fill these 50 units = 50 / 2 = 25 minutes
Therefore, total time required to fill the pool = 2 + 25 = 27 minutes - 塞缪尔以25公里/小时的速度从家到办公室，然后以4公里/小时的速度返回。他在5小时48分钟内完成了整个旅程。找出他的家到办公室的距离：
- 20公里
- 18公里
- 15公里
- 25公里
回答：
20 km解释：
Let the speed of travelling to office and back to home be x and y respectively.
So, his average speed is = 2xy / (x+y) = (2 × 25 × 4) / (25 + 4) = 200/29 km/hr
He covers the whole journey in 5 hours 48 minutes = 5=> = 29/5 hrs
Therefore, total distance covered = (200/29 × 29/5) = 40 km
So, the distance from his home to office = 40/2 = 20 km - 船夫向河下游15公里行驶需要3小时45分钟，而河上游5公里则需要2小时30分钟。河流的速度以km / h为单位是多少?
- 2公里/小时
- 1公里/小时
- 6公里/小时
- 4公里/小时
回答：
1 km/h解释：
Downstream:
Time taken = 3 + 45/60 = 3 + 3/4 = 15/4 h.
Distance covered = 15 km.
Downstream Speed = 15 / (15/4) = 4 km/h.
Upstream:
Time taken = 2 + 30/60 = 2 + 1/2 = 5/2 h.
Distance covered = 5 km.
Upstream Speed = 5 / (5/2) = 2 km/h.
We know, speed of stream
= 1/2 (Downstream Speed – Upstream Speed)
= 1/2 (4-2) = 1 km/h. - 约翰的收入比彼得高33.33％。彼得的收入比约翰的收入少百分之几?
- 22％
- 25％
- 26％
- 23％
回答：
25 %解释：
Let John’s income be j and Peter’s income be p. Then,
j = p + p × 33.33% = p + p × 100?3 % = p + p × 1/3 = 4p/3
=> p = 3j/4 = (4 – 1)j/4 = j – j/4 = j – j × 1/4 = j – j × 100?4 % = j – j × 25%.
Therefore, Peter’s earning is less than John’s earning by 25%. - Vinod和Ashok的当前年龄分别为3：4。 5年后，他们的年龄比例分别变为7：9。 Ashok的当前年龄是多少?
- 40年
- 28年
- 32岁
- 36年
回答：
40 years解释：
Let the present age of Vinod and Ashok be 3x years and 4x years respectively.
Then (3x+5) / (4x+5) = 7 / 9=> 9(3x + 5) = 7(4x + 5)
=> 27x + 45 = 28x + 35
=> x = 10
=> Ashok’s present age = 4x = 40 years - 每隔4年出生的4个孩子的总和是36。最小的孩子几岁?
- 2年
- 3年
- 4年
- 5年
回答：
3 years解释：
Let the ages of children be x, (x+4),
(x+8) and (x+12) years.Then x + x + 4 + x + 8 + x +12 = 36
4x + 24 = 36
4x = 12
x = 3
Age of the youngest child = x = 3 years - 如果两个数字的总和为13，且其平方和为85。找到数字了吗? 。
- 6、7
- 5、8
- 4、9
- 3、10
回答：
6, 7解释：
Let the numbers be x and 13-x
Then x2 + (13 – x)2 = 85
=> x2 + 169 + x2 – 26x = 85
=> 2 x2 – 26x + 84 = 0
=> x2 – 13x + 42 = 0
=> (x-6)(x-7)=0Hence numbers are 6 & 7
- 两个数字的HCF为11，而LCM为385。如果两个数字的相差不超过50，那么两个数字的总和是多少?
- 132
- 35
- 12
- 36
回答：
132解释：
Product of numbers = LCM x HCF = 11 x 385 = 4235
Let the numbers be of the form 11m and 11n, such that ‘m’ and ‘n’ are co-primes.
=> 11m x 11n = 4235
=> m x n = 35
=> (m, n) can be either of (1, 35), (35, 1), (5, 7), (7, 5).
=> The numbers can be (11, 385), (385, 11), (55, 77), (77, 55).
But it is given that the numbers cannot differ by more than 50.
Hence, the numbers are 55 and 77.
Therefore, sum of the two numbers = 55 + 77 = 132 - 两个互质数的LCM为117。这些数字的平方和是多少?
- 220
- 1530
- 250
- 22
回答：
250解释：
117 = 3 x 3 x 13
As the numbers are co-prime, HCF = 1.
So, the numbers have to be 9 and 13.
92 = 81
132 = 169
Therefore, required answer = 250 - A和B承担了要在20天内完成的工作。他们开始一起工作，经过12天，C加入了他们，整个工作在15天内完成了。如果只雇用C，C需要多少时间才能完成工作?
- 15
- 12
- 10
- 8
回答：
12解释：
Let the total job be 20 units. It is given that A and B took the job to be completed in 20 days.
=> Combined efficiency of A and B = 20/20 = 1 unit / day
Now, job done in 12 days = 12 units
=> Job Left = 8 units
Now, this remaining 8 units of job have been done by all A, B and C together.
Let the efficiency of C be ‘x’.
=> Combined efficiency of A, B and C = 1+x units/ day
Now, with this efficiency, the job got completed in 3 more days.
=> Job done in 3 days = 3 x (1+x) = 8 units
=> x = 5/3
Therefore, efficiency of C = x = 5/3 units / day
Hence, time required by C alone to do the job = 20/(5/3) = 12 days - 打开三个管道A，B和C，以填充水箱。 A，B和C单独工作分别需要12、15和20分钟。在一起工作4分钟后，A被阻止，又过了1分钟，B也被阻止。 C继续工作直到结束，水箱完全装满。装满水箱总共需要多少时间?
- 6分钟
- 6分15秒
- 6分40秒
- 6分50秒
回答：
6 minutes 40 seconds解释：
Let the capacity of the cistern be LCM(12, 15, 20) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Combined efficiency of pipe A, pipe B and pipe C = 12 units / minute
Now, the cistern is filled with the efficiency of 12 units / minute for 4 minutes.
=> Pool filled in 4 minutes = 48 units
=> Pool still empty = 60 – 48 = 12 units
Now, A stops working.
=> Combined efficiency of pipe B and pipe C = 7 units / minute
Now, the cistern is filled with the efficiency of 7 units / minute for 1 minute.
=> Pool filled in 1 minute = 7 units
=> Pool still empty = 12 – 7 = 5 units
Now, B also stops working.
These remaining 5 units are filled by C alone.
=> Time required to fill these 5 units = 5 / 3 = 1 minute 40 seconds
Therefore, total time required to fill the pool = 4 minutes + 1 minutes + 1 minute 40 seconds = 6 minutes 40 seconds - 如果John以5 km / h的速度行走，他将晚7分钟到达办公室。但是，如果他以6 km / h的速度行走，他会提前5分钟到达办公室。他的办公室离家有多远?
- 9
- 8
- 10
- 6
回答：
6解释：
Let the distance of John’s office from his home be x.
The time difference when covering the distance x at the two different speeds = 5 – (-7) = 12 min = 1/5 hr
=> x/5 – x/6 = 1/5
=> (6x – 5x)/30 = 1/5
=> x = 6.
So, his office is 6 km far from his home. - 快艇在河流上游6公里处行驶，并在33分钟后返回起点。这条河的水流以每小时2公里的速度行驶。快艇在静止水中的速度是多少?
- 25公里/小时
- 21公里/小时
- 26公里/小时
- 22公里/小时
回答：
22 km/h解释：
Let the speed of speedboat in still water be x km/h.
Then, speed downstream = (x + 2) km/h, speed upstream = (x – 2) km/h.
Since it goes 6 km upstream and comes back in 33 minutes, we have
6/(x+2) + 6/(x-2) = 33/60
=> 11x² – 240x – 44 = 0
=> 11x² – 242x + 2x – 44 = 0
=> (x – 22)(11x + 2) = 0
=> x = 22.
Therefore, the required speed = 22 km/h. - 玛丽的工资减少了10％。她的新工资必须增加多少百分比才能获得旧工资?
- 137/9％
- 194/9％
- 100/9％
- 110/9％
回答：
100/9 %解释：
Let her old salary be Rs 100. Then, her new salary = 100 – 10 = Rs 90.
So, to gain her old salary, her new salary must be increased by Rs 10.
Therefore, the required percentage = (10?90) × 100% = 100/9 %.