MCQ: Ranking & Time sequence test - 1 - SSC CGL MCQ

Given number: 7 8 3 4 3 2 9 5 1 3

After interchanging the digits,
New number: 8 7 4 3 2 3 5 9 3 1

6th digit from right end

Original number - 7 6 5 3 4 2 1 8

Hence, required digit = 6

When the digits of the number '6379452' are arranged in ascending order,
then rearrangement will be 2 3 4 5 6 7 9

Now, fourth digit from the left = 5

Second digit from the right = 7

Sum of fourth and second digit = 5 + 7 = 12

Ascending order:
686,   719,   765,   795,   815,   822,   835

Middle Number

Middle digit of 795 is 9.

Given number 7 5 9 3 8 4 6 2

Arrange the digits in ascending order,

Therefore, such pair = six

i.e. (7,9), (5,8), (3,6), (8,6), (5,3) and (4,2).

Given number -        8 3 7 9 1 2

Descending order -  9 8 7 3 2 1

Hence, the position of only one digit, i.e. 7 remains unchanged.

Subtract 2 from odd digits in given number and add 3 in even digits.

So, in new number there are two digits of five and two digits of seven.

There are five such 5's as shown below

3 5 9 5 4 5 5 3 5 8 4 5 6 7 3 5 7 5 5 4 5 2 3 5 1 0

8965359683496526973729413794173498
453976153195742968532957489451

In the given series, there are six 9's followed by and preceded by number divisible by 2.

184381483287848568784186

There are four such 8's which are followed by even number and preceded by an odd number.

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C
Descending order of weights of the articles will be
A, B, E, D, C

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C

So A is the heaviest in the weight.

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C
E is heavier than D and C.

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C
So, E is lighter in weight than A and B.

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C
So lightest in weight is C.