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

Answer: (c)

$2^17 × 6^31 × 7^5 × 10^11 × 11^10 × (323)^23$

= $2^17 × 2^31 × 3^31 ×7^5 × 2^11 × 5^11 × 11^10 × 17^23 × 19^23$

= $2^59 × 3^31 × 5^11 × 7^5 × 11^10 × 17^23 × 19^23$

∴ Total number of prime factors

= 59 + 31 + 11 + 5 + 10 + 23 + 23 = 162

Answer: (b)

Let Farah's age at the time of her marriage be x.

Then, $(x + 8) = x × 9/7$

${9x}/7 - x = 8$

$x = {8 × 7}/2 = 28$ years

∴ Farah's present age = 28 + 8 = 36 years

∴ Daughter's age 3 years ago

= $36 × 1/6 - 3$= 3 years

Answer: (d)

As the larger number is written on the left, the larger number is either 54 or 55.

Let the smaller number be x.

Case I: The larger number is 54.

5400 + x = 5481 + 54 – x

2x = 5535 – 5400 = 135

(In this case x will not be a natural number.)

Case II: The larger number is 55.

5500 + x = 5481 + 55 – x

2x = 5536 – 5500 = 36

⇒ x = 18

Hence, the required sum = 73.

Answer: (a)

Suppose, possible, $log_2{7}$ is rational, say p/q where p and q are integers, prime to each other.

Then, $p/q = log_2{7} ⇒ 7 = 2^{p/q} ⇒ 2^p = 7^q$

which is false since L.H.S. is even and R.H.S. is odd.

Obviously $log_2{7}$ is not an integer and hence not a prime number.

Answer: (a)

Let Farah's age at the time of her marriage be x.

Then, $(x + 8) = x × 9/7$

${9x}/7 - x = 8$

$x = {8 × 7}/2 = 28$ years

Farah's present age = 28 + 8 = 36 years

Daughter's age 3 years ago

= $36 × 1/6 - 3$ = 3 years