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QUANTITAITVE APTITUDE CLUBBED MCQs
Total Questions : 1394
| Page 137 of 140 pages
Answer: Option D. ->
20
:
D
If 24! ends in n zeroes, 25! will end in (n+2) zeroes.
Generalising this to all multiples of 25, they satisfy this condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.
However, if 124! ends in n zeroes, 125! will end in (n+3) zeroes.
Generalising this to all multiples of 125, they do not satisfy the condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.
Infact, they end with (n+3) or (n+4) zeroes.
Therefore, the number of favourable cases between 24 to 626 can be represented as:
(Number of multiples of 25 - Number of multiples of 125)
:
D
If 24! ends in n zeroes, 25! will end in (n+2) zeroes.
Generalising this to all multiples of 25, they satisfy this condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.
However, if 124! ends in n zeroes, 125! will end in (n+3) zeroes.
Generalising this to all multiples of 125, they do not satisfy the condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.
Infact, they end with (n+3) or (n+4) zeroes.
Therefore, the number of favourable cases between 24 to 626 can be represented as:
(Number of multiples of 25 - Number of multiples of 125)
25 - 5 = 20.
Answer: Option D. ->
1859C4
:
D
A + B + C + D + E = 2012 (A>11, B>21, C>30, D>40, E>50)
This is a lower limit Similar->Different question.
So, we assign the minimum values that can be taken by each of A, B, C, D and E and reframe the equation:
A + B + C + D + E = 1855 (Assigning 12 to A, 22 to B, 31 to C, 41 to D and 51 to E)
Using the 1-0 method:
:
D
A + B + C + D + E = 2012 (A>11, B>21, C>30, D>40, E>50)
This is a lower limit Similar->Different question.
So, we assign the minimum values that can be taken by each of A, B, C, D and E and reframe the equation:
A + B + C + D + E = 1855 (Assigning 12 to A, 22 to B, 31 to C, 41 to D and 51 to E)
Using the 1-0 method:
Required Answer = 1859C4.Option (d)
Question 1364.
In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same staring point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?
Answer: Option C. ->
10 min
:
C
:
C
The ratio of the speeds of the fastest and the slowest runners is 2 : 1. Hence they should meet at only one point on the circumference i.e. the starting point (As the difference in the ratio in reduced form is 1). For the two of them to meet for the first time, the faster should have completed one complete round over the slower one. Since the two of them meet for the first time after 5 min, the faster one should have completed 2 rounds (i.e. 2000 m) and the slower one should have completed 1 round. (i.e. 1000 m) in this time. Thus, the faster one would complete the race (i.e. 4000 m) in 10 min.
Answer: Option B. ->
49{109(10n−1)−n}
:
B
Reverse Gear Approach
Assume n= 1, then S1 =4
Now put n=1in options.Eliminate those options , where you are not getting 4.
Only option (b) will give 4.
:
B
Reverse Gear Approach
Assume n= 1, then S1 =4
Now put n=1in options.Eliminate those options , where you are not getting 4.
Only option (b) will give 4.
Answer: Option B. ->
42
:
B
:
B
n(n+1)2−1+2+32=900:n(n+1)2=903;
n(n+1) = 1806, n = 42.
Answer: Option A. ->
120
:
A
:
A
If BJP received 60% of the votes this implies that Congress received 40% of the total number of votes. The difference between them, 20% , represents 24 votes. Therefore, the total number of votes cast was 5 × 24 =120.
Question 1368.
The integers 1,2,3,4,......n are written on a blackboard. The following operation is then repeated until we will get only one number: In each repetition, any two numbers say a and b, currently on the blackboard are erased and a new number a + b is written. But one number was missed by mistake. The sum obtained was 1525. Which number was missed?
Answer: Option A. ->
15
:
A
:
A
n(n+1)2=1525; n(n+1)=3050; n=55 (approx.)
55×562=1540; missed number = 1540-1525 = 15.
Answer: Option D. ->
91
:
D
Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.
Among these only 31 and 91 are a part of the answer choices.
Among these, (31)10=(11111)2=(1011)3=(111)5. Thus, all three forms have leading digit 1.
:
D
Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.
Among these only 31 and 91 are a part of the answer choices.
Among these, (31)10=(11111)2=(1011)3=(111)5. Thus, all three forms have leading digit 1.
Hence the answer is 91.
Answer: Option B. ->
6
:
B
:
B
To find the remainder when an expression is divided by 50, we need to find the last two digits of that number.
Last two digits of the given expression are 1 + 91 + 91 + 91 + 91 + 91 = 456 (Ten's digit of powers of 71 follow a cycle of 10).
∴45650, remainder=6.
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