Amcat previous questions set - 5


1. Which is greatest among 5^1/2 ,11^1/3, 123^1/6 ?
Explanation: Raising each term to their 6th power as the LCM of the denominators of the powers = 6, we get,
${\left( {{5^{{\textstyle{1 \over 2}}}}} \right)^6} = {5^3} = 125$
${\left( {{{11}^{{\textstyle{1 \over 3}}}}} \right)^6} = {11^2} = 121$
${\left( {{{123}^{{\textstyle{1 \over 6}}}}} \right)^6} = 123$
Hence,  ${5^{{\textstyle{1 \over 2}}}}$ is greatest.

2.  A fair is coin is tossed repeatedly.  If head appears on first four tosses then what is the probability of appearance of tail in the fifth toss?
Explanation:
As the given coin is a fair coin, the probability of getting tail on the fifth toss is independent of the out comes in the previous 4 tosses.  So probability = 1/2

3.  Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 15 times as heavy as water ?
Explanation:
Let us say, water weighs 1kg/1 liter.  Then Gold weighs 19 kgs and copper weighs 9 kgs.
Assume x kgs of gold and y kgs of copper required to get the required alloy.
We use weighted average formula = ${A_x} = \dfrac{{{n_1}.{A_1} + {n_2}.{A_2}}}{{{n_1} + {n_2}}}$
Therefore, $15 = \dfrac{{x.19 + y.9}}{{x + y}}$
15x + 15y = 19x + 9y
4x = 6y ⇒ x/y = 3 : 2
So Gold and Copper should be mixed in the ratio 3 : 2.

4. 7:11::31: ? 
1. 33 
2. 37 
3. 39 
4. 42 
Explanation:
The prime number after 7 is 11, therefore the prime number after 31 is 37

5. From the given choices select the odd man out
1. 11, 3, 3, 17
2. 41, 5, 3, 47
3. 71, 7, 3, 17
4. 37, 14, 19, 7
5. 67, 71, 3, 5
Explanation:
All the numbers in those options are prime numbers except 14 in option 4.

6. Find the number of zero's in the expression 1^1 * 2^2 * 3^3 * 4^4 * 5^5 * ……..10^10.
Explanation:
To find the number of zero's we have to find the powers of 2's and 5's.
Powers of 2's = 2 + 4 + 6 + 8 + 10 = 26
But 4, 8 contains more 2's which are equal to 4 + 8  = 12
8 contains another eight 2's.
So total 2's = 46
Number of 5's = 5 + 10 = 15
So trailing zeroes = 15.

7. What are the unit's digits of 369, 6864, 4725 respectively ?
1. 9,6 and 6
2. 6, 6 and 6
3. 3,6 and 4
4. None of these
Explanation:
Cyclicity of 3 is 4.  369  =  ${\left( {{3^4}} \right)^k}{.3^1}$  = ${\left( 1 \right)^k}{.3^1}$ = 3
Power of 6 always gives units digit 6 only.
If power of 4 is odd, then we will get 4 as units digit, if power of 4 is even, then we get 6 as units digit.
Hence, Option 3 is correct.

8. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: 
1. 1 
2. 2 
3. 3 
4. 4
Answer: 2
Explanation:
Let the numbers be 13a and 13b.
Then, 13a × 13b = 2028
⇒ ab = 12
Here a and b are coprimes.  Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13× 1, 13×12) and (13×3, 13×4).
Hence, there are 2 such pairs.

9. If 72X23Y is divisible by 72, then what is the value of X and Y?
1. X=0;Y=1
2. X=2;Y=3
3. X=7;Y=2
4. X=2;Y=2
Correct Option : 4
Explanation:
If a number has to be divisible by 72, then it has to be divisible by both 8 and 9.
Divisibility for 8 is last three digits should be divisible by 8. So Y takes only 2.
Now divisibility for 9 is sum of the digits should be divisible by 9. So by substituting Y = 2, the sum is divisible by 9 if X = 2.
So X = 2 and Y = 2

10.  (((1/8)-1)-4)-1) = 
1. 4096
2. - 4096
3. 4092
4. 4090
Correct Option:
Explanation:
We hae to use BODMAS rule. First simplify the terms in the bracket.
(((1/8)-1)-4)-1) = $\left[ {\left( {\dfrac{1}{8} - 1} \right) - 4} \right] - 1$ = $\left[ { - \dfrac{7}{8} - 4} \right] - 1$ = $ - \dfrac{{39}}{8} - 1$ = $ - \dfrac{{47}}{8}$