Elitmus previous questions - 1


1. If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?
Explanation:
When you convert a decimal number into octal form, think which numbers end with zero.  The multiples of 8. For example ${\left( 8 \right)_{10}} = {\left( {10} \right)_8}$, ${\left( {16} \right)_{10}} = {\left( {20} \right)_8}$, ${\left( {24} \right)_{10}} = {\left( {30} \right)_8}$, ${\left( {32} \right)_{10}} = {\left( {40} \right)_8}$, ${\left( {40} \right)_{10}} = {\left( {50} \right)_8}$, . . .
So out of five numbers end with zero in octal form, only one ${\left( {50} \right)_8} = {\left( {40} \right)_{10}}$ has zero in decimal format.  So the required probability is 1/5

My initial thinking:
Let the number be "ab0" in octal form.  Remember here a, b < 8.  So total possibilities are 64.
If we convert the given number into decimal format, then (ab0)8 = 64a  + 8b.
We will find for how many values of 64a + 8b ends in zero.
64a + 8b = 60a + (4a + 8b)
For (0,0), (1, 2), (1, 7), (2, 4), (3, 1), (3, 6), (4, 3), (5, 5), (6, 2), (6, 7), (7, 4), (0, 0) the above expression gives unit digit 0.
So required probability = 11/64
(Imp: Still I am not comfortable with the solution I have written as it is not specified whether the given number is 3 digit or not.   Just learn how I tried to solve it. )

2. log xy= a; log x2y = b then then value of log y /log x
Explanation:
log xy3 = logx + 3logy = a - - - (1)
log x2y = 2logx + logy = b - - - (2)
Multiplying (1) by 2 and subtracting (2) from it
5logy = 2a-b
logy = $\dfrac{{2a - b}}{5}$
Put the value of log y in equation (2),
2logx + $\dfrac{{2a - b}}{5}$ = b
logx = $\dfrac{{3b - a}}{5}$
$\dfrac{{\log y}}{{\log x}} = \dfrac{{\dfrac{{2a - b}}{5}}}{{\dfrac{{3b - a}}{5}}}$ = $\dfrac{{2a - b}}{{3b - a}}$

3. Einstein walks on an escalator at a rate of 5steps per second and reaches the other end in 10 sec. while coming back, walking at the same speed he reaches the starting point in 40secs. What is the number of steps on the escalator?
a) 40
b) 60
c) 120
d) 80
e) data insufficient
Explanation:
Escalator problems are similar to boats and streams problems.  If we assume man's speed as 'a m/s' and escalator speed as 'b m/sec' then while going up man's speed becomes 'a -b' and while coming down 'a + b'.
In this question, Let the speed of escalator be b steps per sec.  And length of escalator be L.   Einstein's speed = 5 steps/ sec
While going down,  $\dfrac{L}{{5 + x}} = 10$ $ \Rightarrow $ L = 50 + 10x
While coming up, $\dfrac{L}{{5 - x}} = 40$ $ \Rightarrow $ L = 200 - 40x
Multiply the first equation by 4, and add to the second, we get L = 80

4. Photography changed drastically _________ introduced in the market.
a) when polariod
b) as polaroid
c) since polaroid
d) that polaroid
Explanation:
There is no adequation information is given about details of subsequent changes.  So "when" is the right word. Option a. "As" inditates reason.

5. Data sufficiency question:
Is x has only '3' factors?
a) x^2 has only '5' factors.
b) one of its factor must be prime number
Explanation:
Only squares of primes has 3 factors.  For example, 4 has three factors. 1, 2, 4.
Now when we square 4, we get 16, which has 5 factors. 1, 2, 4, 8, 16.  So Statement 1 is sufficient.
Statement two says that one of the factor must be prime number.  10 has 4 factors.  i.e., 1, 2, 5, 10 and one of which is prime. but we cannot say whether numbers of this format always have three factors or not. So statement 2 is insufficient.

6.  Data sufficiency question:
If solution (x) with alcohol (14%) and solution (y) with alcohol (20%) mixed together to form 24 liters of mixture.  Then the amount of solution(x) is?
a) The alcohol in solution(mixture) is 15%
b) If we inter changed the amounts of solutions 'x' and 'y' we get mixture of alcohol %,44%
Explanation:
Let a liters of first solution, and 'b' liters of second solution has been taken.
So using weighted average formula = $\dfrac{{{n_1} \times {a_1} + {n_2} \times {a_2}}}{{{n_1} + {n_2}}}$ = $\dfrac{{{n_1} \times 14 + {n_2} \times 20}}{{{n_1} + {n_2}}} = 15$
$\Rightarrow {n_1} = 5{n_2} \Rightarrow \frac{{{n_1}}}{{{n_2}}} = \frac{5}{1}$
As total volume was given as 24 liters, first solution = $\dfrac{1}{6} \times 24 = 4$
So statement 1 is sufficient.

Let us take volume of the first solution is 'a' liters. Then volume of the second = 24 - a.
Again by using the question statement, and statement 2, $\dfrac{{(24 - a) \times 14 + a \times 20}}{{24}} = 44$
From this also, value of 'a' can be calculated.  So Statement 2 also sufficient.

7. How many solution  does the expression $\sqrt {x + 3}  = x\sqrt {x + 3} $ has?
Explanation:
Squaring on the both sides, $x + 3 = {x^2}(x + 3)$
Bring x+3 to one side and take common, $\left( {{x^2} - 1} \right)\left( {x + 3} \right)$
So roots are, 1, -1, 3. But -1 is not a root. So 1, -3 are the roots.

($\because$ when you square the equation additional roots may be generated. So finally you have to check manually, whether the given roots satisfy the original equation or not)

8. In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag-
(i) the least chance of drawing a red ball ?
(ii) the greatest chance of drawing a red ball ?
Explanation:
(1) Put 1 white in first bag, and all the remaining ball in second bag.  Probability = $\dfrac{1}{2} \times 0 + \dfrac{1}{2} \times \dfrac{5}{{16}}$ = $\dfrac{5}{{32}}$
(2) To maximize, we put 1 red in the first bag, and all the remaining in the second. Probability = $\dfrac{1}{2} \times 1 + \dfrac{1}{2} \times \dfrac{4}{{32}}$ = $\dfrac{9}{{16}}$

9. The dairy condensed milk has only 20% water while the cow milk has 90% water. How many litres of milk maid will be obtained from 72 litres of milk?
a) 7.2l
b) 8l
c) 9l
d) 14.4l
Answer: 9
Explanation:
In this question, Pure milk in condensed milk = Pure milk in cow milk
80% (x) = 10% (72)
Therefore, x = 9 Liters.

10. How many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number? for example if we take 98 then 9*8=72, 72=7*2=14, 14=1*4=4. Consider only 4 prime numbers (2,3,5,7)
Explanation:
2 = 12 or 21 So 1×2, 2×1, 3×4, 4×3, 2×6, 6×2, 3×7, 7×3
3 = 13 or 31 So 1×3, 3×1
5 = 15, 51 So 1×5, 5×1, 3×5, 5×3, 7×5, 5×7
7 = 17 or 71 So 1×7, 7×1
15 = 3×5 = 5×3
So total 18 numbers = 12,13,15,17,21,26,31,34,35,37,43,51,53,57,62,71,73,75