这是用于能力倾向准备的TCS模型放置论文。这份安置文件将涵盖在TCS招聘活动中提出的才能问题,并且还将严格遵循在TCS面试中提出的问题模式。建议解决以下每个问题,以增加清除TCS面试的机会。
- 确定序列中缺少的数字:2、5,__,19、37、75?
a)16
b)12
c)10
d)9Answer: d) 9
Solution:
2 * 2 + 1 = 5
5 * 2 – 1 = 9
9 * 2 + 1 = 19
19 * 2 – 1 = 37 and so on - 矩形分为面积分别为70、36、20和x的四个矩形。 ‘x’的值是多少?
a)350/7
b)350/11
c)350/9
d)350/13Answer: c) 350/9
Solution:
Since the areas of the rectangles are in proportion we can say,
=> 70/x = 36/20
=> x = 350/9 - 如果VXUPLVH编写为SURMISE,那么SHDVD编写为什么?
a)PEASA
b)PBASA
c)PEBSB
d)以上都不是Answer: a) PEASA
Solution:
It is a question of coding-decoding where,
V is written as S (V – 3 = S)
X is written as U (X – 3 = U)
and so on.
Similarly, SHDVD will be written as PEASA - Aman欠Bipul 50卢比。他同意从星期一开始连续多个天向Bipul支付,每天支付10卢比或20卢比的单笔票据。阿曼可以通过多种方式偿还比普尔。 (如果至少有一天,则给出了两种不同面额的注释,这两种方法被认为是不同的)
a)5
b)6
c)7
d)8Answer: d) 8
Solution:
Aman can pay Bipul in all 10 rupees note in 5 days = 5 * 10 = 50 rupees = 1 way
Aman can pay Bipul in 3 ten rupee note and 1 twenty rupee note = 4!/(3! * 1!) = 4 ways
Aman can pay Bipul in 1 ten rupee note and 2 twenty rupee note = 3!/(1! * 2!) = 3 ways
So in all Aman can pay Bipul in 8 ways. - 萨利姆(Salim)以27卢比的价格购买了一定数量的橙色,价格为M的2倍卢比,其中M为整数。他将这些橙子分成相等的两半,一部分以M卢比以13橙的价格出售,以M卢比以14橙的价格出售。他花了整整卢比,没有卢比,但买了最少数量的橘子。他买了多少?
a)980
b)9828
c)1880年
d)102660Answer: b) 9828
Solution:
Let Salim buy 2x number of oranges.
So he buys 27 oranges at a price of 2M.
He buys 1 orange at a price of 2M/27
or, x oranges cost him Rs. 2Mx/27
Now he sells x oranges at the rate of 13 oranges for Rs. M
So he sells 1 orange at Rs. M/13
and x oranges at Rs Mx/13
Same goes for 14 oranges which are Mx/14,
According to the question, 2Mx/27, Mx/13, Mx/14 are integers
So, x oranges must be divisible by 27, 13 and 14
The lcm of 27, 13 and 14 = 4914 or 2x = 9828 - 在足球比赛中,有16支球队参加,并分为4组。每个小组中的每个团队将互相比赛一次。排名前2的获胜球队将进入下一轮比赛,因此排名前2的前两支球队将参加决赛。那么在那场比赛中将参加多少场最低比赛?
a)40
b)14
c)43
d)50Answer: c) 43
Solution:
Total matches to be played = 4C2 = 6 matches.
So total number of matches played in the first round = 6 * 4 = 24 matches
Now top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.
Total matches played in the second round = 6 × 2 = 12 matches
Now 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matches
From this round, 2 teams progress to the next round. And final will be played between them.
Total matches = 24 + 12 + 6 + 1 = 43 - 有12个字母和正好12个信封。每个信封中随机插入一个字母。在不正确的信封中准确插入1个字母的概率是多少?
a)1
b)0
c)10!
d)这些都不是Answer: b) 0
Solution:
This is a question of very common sense in which,
12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter which is to be inserted into another wrong envelope. So the probability of this is 0. - 地面上的中空空间将被填充。填充的总成本为Rs。 20000。每立方米的填充成本为225卢比。填充空心空间所需的3立方米土壤大小是多少倍?
a)29.62
b)30.32
c)88.88
d)43.64Answer: a) 29.62
Solution:
The total cost of filling = 20, 000
Cost of filling 1 cubic meter = Rs. 225
So cubic meters to be filled = 20, 000/225 = 88.89 meter-cube
Now we need to find the three times of 88.89 to be filled = 88.89/3 = 29.63
So the closest match is 29.62 - 7位数字由所有不同的数字组成。如果将最右边和最左边的数字分别固定为5和6,那么可以找到多少个这样的数字?
a)120
b)30240
c)6720
d)这些都不是Answer: c) 6720
Solution:
If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8
So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways
= 6720 ways - 轿车中有五个轮胎(四个公路轮胎和一个备用轮胎)在行驶40,000公里的过程中将被平均使用。每个轮胎的使用公里数为
a)32000
b)8000
c)4000
d)10000Answer: a) 32000
Solution:
The total km travelled by the sedan = 40, 000 km
Since every tire capacity’s = 40, 000/5 = 8000 km each
So total distance covered by each tire = 8000*4 = 32000 km each will be travelled by each tire after being worn out after every 8000 km.