number systems Model Questions & Answers, Practice Test for ssc mts paper 1

Answer: (d)

$1^{st}$ time seen

= 1 hour 54 min 50 sec

= 3600 sec + 54 × 60 sec + 50 sec

= 6890 sec

$2^{nd}$ time seen

= 3 hour + 17 min + 49 sec

= 11869 sec.

Interval between light seen

= 11869 – 6890 = 4979

Number of times light was seen

= $4979/13$ + 1 = 384

Answer: (a)

Given x + y + z = 0

$x^3 + y^3 + z^3 + 3xyz – 3xyz + 3xyz$

= $x^3 + y^3 + z^3 –3xyz + 6xyz$

= $(x + y + z) (x^2 + y^2 + z^2 – xy – yz –zx) + 6xyz$

= $(0) (x^2 + y^2 + z^2 – xy – yz – zx) + 6xyz$

= 6xyz

Answer: (b)

Let there be w wide runs.

Byes = w + 8

Runs scored by batsmen = 26 w

Total run = 232

or w + w + 8 + 26W = 323

⇒ w = $224/28 = 8$

∴ Run scored by Ram = $6/13 × 208 = 96$

Answer: (a)

The expression can be written as a + $1/{b + x}$,

where x lies in the interval (0, 1).

Since $(1.25)^3 = 1.953125$ and $(1.3)^3 = 2.197$, it can be concluded that $2^{1/3}$ belongs to the interval (1.25, 1.3).

Hence, a = 1.

This implies that $1/{b + x}$ lies in the interval (0.25, 0.3).

The only possible value of b = 3.

Answer: (a)

Given, numbers are 50, 35 and 35.

Now, place value of 3 is 30 and 30 in the numbers 35 and 35 respectively.

∴ Sum of the place values = 30 + 30 = 60