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6. The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
a. 50
b. 100
c. 150
d. 216
Read Solution (Total 4)
-
- total no. of multiples of 10 = 1000/10 = 100
now,
10=1+2+3+4
But 20 can not be written as the sum of 4 consecutive no.s
since some of the multiples of 10 can not be written as the sum of 4 consecutive no.s, the answer should be less than total no. of multiples of 10 (i.e. less than 100).
In the options only A. is less than 100.
therefore ans. is option A. - 9 years agoHelpfull: Yes(24) No(1)
- We can write 10 as (1 + 2 + 3 + 4) four consecutive numbers.
Now adding 5 to each of these digits we get:
6 + 7 + 8 + 9 = 30 (again consecutive four digits.)
Again adding 5 to each of the above digits we get:
11 + 12 + 13 + 14 = 50
From the above we observe that 10, 30, 50, 70 ……that are multiple of 10 have consecutive four digits.
Thus we have 50 numbers that can be written in four consecutive digits between 1 and 1000. - 9 years agoHelpfull: Yes(16) No(0)
- We can write 10 = 1 + 2 + 3 + 4. So we have to find how many multiples of 10 can be written in this manner.
Let the first of the four numbers be n. So
n + (n+1) + (n+2) + (n+3) = 10k
4n + 6 = 10k
2n + 3 = 5k
n = 5k−3/2 = 2k – 1 + k–1/2
So n is intezer for k = an odd number. So for k = 1, 3, 5, .... 99 we can write a number as a sum of four consecutive intezers.
So there are 50 numbers. - 9 years agoHelpfull: Yes(4) No(5)
- b)100 is answer
- 9 years agoHelpfull: Yes(2) No(13)
TCS Other Question
5. Find the probability that a 3 digit number formed by using the digits 1, 3, 6, 9 without repetition, is divisible by 4.
a. 1/2
b. 1/3
c. 1/4
d. 1/5
7. A, B, C, D are vertices of a quadrilateral and P is a point inside it, if PA=2, PB=5, PC=6, PD=3 then the maximum possible are of quadrilateral formed by A, B, C, D is?
a. 128
b. 64
c. 32
d. 16