number systems Model Questions & Answers, Practice Test for ibps po prelims

ibps po prelims SYLLABUS WISE SUBJECTS MCQs

Number Systems

Profit & Loss

Question :1

$(x^n – a^n)$ is completely divisible by (x + a), when

Answer: (c)

$(x^n – a^n)$ is always divisible by (x + a), when n is even natural number.

Question :2

At a college football game, 4/5 of the seats in the lower deck of the stadium were sold. If 1/4 of all the seating in the stadium is located in the lower deck, and if 2/3 of all the seats in the stadium were sold, then what fraction of the unsold seats in the stadium was in the lower deck ?

Answer: (a)

Let total number of seats in the stadium be p; number of seats in the lower deck be x and number of seats in upper deck be y.

∴ p = x + y, x = p/4, y = 3p/4

Now in the lower deck, 4x/5 seats were sold and x/5 seats were unsold.

No. of total seats sold in the stadium = 2p/3.

No. of unsold seats in the lower deck = x/5 = p/20

No. of unsold seats in the stadium = p/3

∴ Required fraction = ${p/20}/{p/3} = 3/20$

Question :3

If $(12)^3$ is subtracted from the square of a number the answer so obtained is 976. What is the number?

Answer: (d)

Let the number be x.

∵ $x^2 – (12)^3 = 976$

∴ $x^2 = 976 + 1728 = 2704$

∴ $x = √{2704} = 52$

Question :4

A number which when divided by 32 leaves a remainder of 29. If this number is divided by 8 the remainder will be

Answer: (a)

Let this number be N then

N = 32 × $Q_1$ + 29 ...(1)

Again N = 8 × $Q_2$ + R …(2)

From (1) and (2)

$32Q_1 + 29 = 8Q_2$ + R (where R is the remainder)

$8Q_2 – 32Q_1$ = 29 – R

$8(Q_2 – 4Q_1)$ = 29 – R

or $(Q_2 – 4Q_1 ) = {29 - R}/8$

Since $Q_1 , Q_2$ , R are integers also $Q_2 – 4Q_1$ is an integer.

Therefore 29 – R must be divisible by 8.

Question :5

The sum of the digits of a 3 digit number is subtracted from the number. The resulting number is always :

Answer: (a)

Let the hundred's, ten's and unit's digit of the

required number be x, y and z respectively.

Then the number = 100x + 10y + z ...(1)

And sum of digits = x + y + z ...(2)

According to the question,

(1) – (2) gives 99x + 9y = 9 (11x + y)

which is always divisible by 9.

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