1. If x is even, y is odd, then which of the following are false?
a. $x(x + y)$ is even
b. ${x^n} \times {y^n}$ is even
c. $3x + 5y$ is odd
d. $\left( {{x^2} + x - 1} \right)\left( {{y^2} + y - 1} \right)$ is even
Answer: D
Explanation:
Option A: $E(E + O)$ $= E\times O $ $= E$
Option B: $E^n \times O^n$ $=E \times O$ $=E$
Option C: $3\times E + 5 \times O$ $=E + O = O$
Option D: $\left( {E + E - 1} \right) \times \left( {{O} + O - 1} \right) = $ $O \times O = O$
2. Let \(x,y\) are even numbers and \(z\) is an odd number. Then which of the following is false?
a. \({\left( {x - z} \right)^2}\) is odd number
b. \(\left( {x - y} \right)z\) is even number
c. \(\left( {z - x} \right){y^2}\) is even number
d. \({\left( {x - y} \right)^2} + z\) is even number
Answer: D
Explanation:
Recap the rules related to odd and even numbers.
\(x - z\) = even - odd = odd and odd2 = odd. So option a is correct.
\(\left( {x - y} \right)z\) = (even - even) × odd = even × odd = even. So option b is correct
\(\left( {z - x} \right){y^2}\). We know that y is even. So \({y^2}\) is even. So anything multiplied by even is even. So option c also correct.
\({\left( {x - y} \right)^2} + z\) = even2 + odd = even + odd = odd. So option d is false.
Shortcut:
To solve these type of questions, assume numbers and check options.
Take x = 2, y = 4, z = 3
Clearly, the fourth option = \({\left( {2 - 4} \right)^2} + 3 = 7\) is odd.
3. Which of the following is a prime number?
a. 221
b. 323
c. 629
d. 727
Answer: D
Explanation:
If a number $N$ has to be prime, it should not be divisible by all primes below $\sqrt N $.
Let us check the options.
Option A: $\sqrt {221} $ < $15^2 = 225$. So we need to check the primes below 15. 221 is divisible by 13. So not a prime number.
Option B: $\sqrt {323} $ < $18^2 = 324$. So we need to check the primes below 18. 323 is divisible by 17. So not a prime number.
Option C: $\sqrt {629} $ < $26^2 = 676$. So we need to check the primes below 26. 629 is divisible by 17. So not a prime number.
Option D: $\sqrt {727} $ < $27^2 = 729$. So we need to check the primes below 27. 629 is not divisible by any prime below 27. So it is a prime number.
4. Let \(x\) be prime and \(y\) be composite. Then
a. \(xy\) is always even
b. \(y - x\) is never even
c. \(\dfrac{{x + y}}{x}\) is never even
d. None of the above
Answer: D
Explanation:
Interesting question. We try to give counter examples to eliminate options one by one.
Take \(x = 3,y = 9\). Then \(xy = 27\) is odd. So option a ruled out.
Take \(y = 4,x = 2\). Then \(y - x = 2\) is even. So option b ruled out.
Take \(x = 2,y = 6\). Then \(\dfrac{{x + y}}{x} = \dfrac{{2 + 6}}{2} = 4\) is even. So option c also ruled out.
So Answer option d.
5. Prime factorization of 12600 is
a. \({2^3} \times {3^2} \times {5^2} \times 7\)
b. \({2^2} \times {3^2} \times {5^2} \times 7\)
c. \({2^2} \times {3^2} \times {5^3} \times 7\)
d. \({2^1} \times {3^1} \times {5^2} \times 7\)
Answer: A
Explanation:
To find the prime factorization of a number, we continuously divide the given number by prime numbers until the we get prime number.
$\qquad \begin{array}{|l}
\llap{2~~~~} 12600 \\[4px] \hline
\llap{2~~~~} 6300 \\[4px] \hline
\llap{2~~~~} 3150 \\[4px] \hline
\llap{5~~~~} 1575 \\[4px] \hline
\llap{5~~~~} 315 \\[4px] \hline
\llap{7~~~~} 63 \\[4px] \hline
\llap{3~~~~} 9 \\[4px] \hline
3
\end{array}$
So \(12600 = {2^3} \times {3^2} \times {5^2} \times 7\)
Important Tip:
The clue in solving this question is, when you observe the number, it is an even number ending with '0'. So it must be divisible by both 2 and 5. So start with 2. After 3 divisions we get 1575 which is clearly divisible by 5. So after 2 divisions we got 63 which is divisible by 7 and 9.
6. What least number must be subtracted from 12702 to get number exactly 99 ?
a. 49
b. 30
c. 29
d. 31
Answer: B
Explanation:
Divide the given number by 99 and find the remainder. If you subtract the remainder from the given number then it is exactly divisible by 99.
$\quad\require{enclose}
\begin{array}{rll}
128\,\,\,\, &\\[2px]
99 \enclose{longdiv}{12702}\kern-.2ex \\[2px]
\underline{99}{\phantom{00}} \\[2px]
280\phantom{0} \\[2px]
\underline {198}{\phantom{0}} \\[2px]
822\phantom{} \\[2px]
\underline{792}\phantom{} \\[2px]
30\phantom{}
\end{array}$
Required number is 30.
