Quant - Number System Ex. 1

Exercise1 + Answers + Explanations

Question:
The sum of the first 100 numbers, 1 to 100 is divisible by

(1) 2, 4 and 8
(2) 2 and 4
(3) 2 only
(4) None of these

Correct Answer - (3)

Explanation:
The sum of the first 100 natural numbers is given by (n(n + 1))/2 = (100(101))/2 = 50(101).
101 is an odd number and 50 is divisible by 2. Hence, 50 (101) will be divisible by 2.

——————————————————-

Question:
What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?

(1) 176
(2) 187
(3) 54043
(4) 748

Correct Answer - (2)

Explanation:
The marbles used to tile the floor are square marbles. Therefore, the length of the marble = width of the marble.As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74. 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

The number of such square marbles required = = 187 marbles.

——————————————————-

What is the remainder when 9^1 + 9^2 + 9^3 + …. + 9^8 is divided by 6?

(1) 3
(2) 2
(3) 0
(4) 5

Correct Answer - (3)

Explanation:
6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.
On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3. Therefore, when each of the 8 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + …. + 3 (8 remainders) = 24. 24 is divisible by 6. Hence, it will leave no remainder.

Hence the final remainder when the expression 9^1 + 9^2 + 9^3 + ….. + 9^8 is divided by 6 will be equal to ‘0′.

——————————————————-

Question:
What is the reminder when 91 + 92 + 93 + …… + 99 is divided by 6?

(1) 0
(2) 3
(3) 4
(4) None of these

Correct Answer - (2)

Explanation:
Any number that is divisible by ‘6′ will be a number that is divisible by both ‘2′ and ‘3′. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of ‘3′). Any number that is divisible by ‘9′ is also divisible by ‘3′ but unless it is an even number it will not be divisibly by ‘6′. In the above case, ‘9′ is an odd number. Any power of ‘9′ which is an odd number will be an odd number which is divisible by ‘9′.

Therefore, each of the terms 91, 92 etc are all divisibly by ‘9′ and hence by ‘3′ but are odd numbers.

Any multiple of ‘3′ which is odd when divided by ‘6′ will leave a reminder of ‘3′. For example 27 is a multiple of ‘3′ which is odd. 27/6 will leave a reminder of ‘3′. Or take 45 which again is a multiple of ‘3′ which is odd. 45/6 will also leave a reminder of ‘3′.

Each of the individual terms of the given expression 91 + 92 + 93 + …… + 99 when divided by 6 will leave a reminder of ‘3′. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.

27/6 will leave a reminder of ‘3′.

——————————————————-

Question:
Find the value of 1.1! + 2.2! + 3.3! + ……+n.n!

(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
Correct Answer - (3)

Explanation:
1.1! can be written as (2!-1!), likewise 2.2! = (3!-2!) , 3.3! = (4!-3!) and n.n! = (n+1)!-n!.
Thus by summing it up we get the result as given below.
1.1! + 2.2! + 3.3! + ……+n.n! = (2!-1!) + (3!-2!) + (4!-3!) +……+ (n+1)!-n! = (n+1)!-1.

Related Articles :