这是 Quantitative Aptitude 的 HCL 模型论文。这份实习论文将涵盖 HCL 实习中要求的能力，并严格遵循 HCL 论文中提出的问题模式。建议解决以下每个问题，以增加清除 HCL 安置的机会。
- 如果连续5个偶数的平均值是10，求中间的数？
- 4
- 8
- 2
- 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 - 上午 8:00:00，三个朋友开始在环形跑道上一起跑步。他们完成一轮赛道所需的时间分别为15分钟、20分钟、30分钟。如果他们不停地奔跑，那么他们第四次在什么时候又会在起点相遇？
- 上午 8:30:00
- 晚上 9:00:00
- 中午 12:00:00
- 中午 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. - 两个朋友 A 和 B 受雇工作。最初的截止日期定为 24 天。两人开始一起工作，但20天后，A离开了工作，整个工作用了30天才完成。 B一个人可以在多少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 - 连接在游泳池上的两条管道 A 和 B 可以分别在 20 分钟和 30 分钟内将游泳池填满。两者一起打开，但由于A管的电机故障，两分钟后不得不关闭，但B继续工作，直到游泳池完全填满。找出填充池所需的总时间。
- 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 - Samuel 以 25 公里/小时的速度从他家到他的办公室，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 - 一个船夫需要 3 小时 45 分钟才能向下游行驶 15 公里，需要 2 小时 30 分钟才能向上游行驶 5 公里。河流的流速是多少公里/小时？
- 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少百分之多少？
- 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现在的年龄比例分别为3:4。 5年后，他们的年龄比分别变为7:9。 Ashok 现在的年龄是多少？
- 40年
- 28岁
- 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岁。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 - 如果两个数字的和是 13，它们的平方和是 85。找出这些数字？ .
- 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 是 11，它们的 LCM 是 385。如果数字相差不超过 50，那么这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 是 117。这两个数的平方和是多少？
- 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 需要在 20 天内完成一项工作。他们开始一起工作，12 天后，C 加入了他们，整个工作在 15 天内完成。如果只雇用 C，C 需要多少时间来完成工作？
- 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。单独工作，A、B 和 C 分别需要 12、15 和 20 分钟。合作4分钟，A被挡，再过1分钟，B也被挡。 C继续工作到最后，水箱被完全填满。填充水箱所需的总时间是多少？
- 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