## 3 May 2016

### SBI Clerk Prelims 2016- Practice Aptitude Questions (Number System)

SBI Clerk Prelims 2016- Practice Aptitude Questions (Number System) Set-28:
Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

1).A no. when divided by 115 leaves remainder 69. What is the remainder, if the no. is divided by 23?
a)   21
b)   zero
c)   6
d)   18
e)   None of these

2).On dividing a certain number by 4, 7 and 9 successively the remainders are 3, 5 and 2 respectively. What would be the remainders if the order of division is reversed ?
a)   1, 7, 1
b)   7, 1 , 1
c)   7, 1, 2
d)   None of these
e)   2, 7, 1

3). Find the largest 3 – digit no. which when divided by 4 leaves remainder 3 and when divided by 3, leaves remainder 2.
a)   983
b)   95
c)   971
d)   None of these
e)   Cannot be determined

4).Find the largest 5 – digit number which when added to 12345 makes the number exactly divisible by 15, 21, 23 & 25.
a)   84255
b)   96330
c)   99999
d)    None of these
e)   85435

5). The LCM of two numbers is 72 and their HCF is 9. If one of the numbers is 12, what is the other number ?
a)   48
b)   52
c)   54
d)   45
e)   None of these

6).when a positive no. is divided by 6, the remainder is 2 and when the same no. is divided by 7, the remainder is 3. If the difference between the two quotients so obtained is 3, then find the number.
a)    38
b)   206
c)   122
d)   None of these
e)   44

7).6 bells ring at intervals of 2, 4, 6, 8, 10  and 12 min. respectively. If they rang at 1 : 00 p.m., together for the first time then how many times in all, would they ring till 11 : 00 p.m(excluding 1:00 p.m.)
a)   4
b)   3
c)   2
d)   5
e)   6

8).A no. when divided by 243 leaves a remainder 26, another number when divided by 351 leaves a remainder 13. What is the remainder when the sum of these two numbers is divided by 27 ?
a)   15
b)   12
c)   13
d)   14
e)   None of these

9).Arjun wanted to make a number of groups of balls, such that each group had equal no.of balls. When he made groups of 3, 5, 7 and 9 balls, exactly one ball was left behind. If total no. of balls with Arjun is a 3 – digit number, then how many different possible solutions are there for total no. of students?
a)   1
b)   2
c)   3
d)   More than three
e)   None of these

10). A no. ‘x’ is reduced by 5 and the result is multiplied by 5. The operation is done for a total of 4 times. If the answer after the 4th operation is 5, then find ‘x’.
a)   781 / 125
b)   781 / 25
c)   81 / 25
d)   None of these
e)   718 / 25

1). b) 2). b) 3). d) 4). b) 5). c) 6). c) 7). d) 8). b) 9). c) 10). a)

Solution:

1). The given number is of the form 115k + 69,
Now, (115k + 69) / 23   = remainder is zero [because 115 is a multiple of 23 and 69 is also a multiple of 23.]

2). Divisors 4        7        9
Remainder   3        5        2

The lowest such number is 2 × 7 + 5 = 19, 19 × 4 + 3 = 79.
When 79 is divided in reverse order (i.e.) 9, 7 and 4 successively

Respective remainders are 7, 1 & 1.

3). The required no. is of the form , (4k + 3) and (3m + 2)
3m + 2 = 4k + 3
3m = 4k + 1
When k = 2, m = 3
Smallest such number is (4k + 3) = 4 × 2 + 3 = 11
General form of such numbers is L.C.M (4,3) + 11
(i.e) 12n + 11 when n = 0, 1 , 2 , 3…
We know largest 3 digit number is 999
12 × 80 + 11 = 960 + 11 = 971
12 × 81 + 11 = 983
12 × 82 + 11 = 995
12 × 83 + 11 = 1007
Hence, 995 is the greatest 3 digit no. which when divided by 4 leaves remainder 3 and when divided by 3 leaves remainder 2.

4). The lowest number divisible by 15, 21, 23 & 25 is
L.C.M (15, 21, 23, 25)
15 = 3 × 5
21 = 3 × 7
23 = 23 × 1
25 = 5 × 5
L.C.M = 3 × 52 × 7 × 23 = 12075
We have to go closest to largest 5 digit no. which is 99999
12075 × 7 = 84525, 12075 × 8 = 96600
Next multiple of 12075 would be 12075 × 9 = 108675
Which is a 6 – digit no & 12075 × 10 = 120750
Required no. would be 108675 – 12345 = 96330, (if we consider 120750, then 120750 – 12345 = 108405, Which will be a 6 – digit no )
You can check through options also.
Start from highest 5 digit no. let it be 99999,
99999 + 12345 = 112344, which is not a common multiple of 15, 21, 23 & 25. Clearly, it is not divisible by 25 or 15.
Next option is 96330
96330 + 12345 = 108675 which is common multiple of 15, 21, 23 & 25.
Still you will have to check no higher than 5 – digit no, (which satisfies the given condition exits) so as to rule out option d.

5). H.C.F × L.C.M = Product of two numbers
72 × 9 = 12x
x = (72 × 9) / 12 = 54

6). Check through options → option a
38 / 6 → remainder is 2, quotient is 6
38 / 7 → remainder is 3, quotient is 5
Difference between quotient is not equal to 3.
Option b
206 / 6 → remainder is 2, quotient is 34
206 / 7 → remainder is 3, quotient is 29
34 – 29 ≠3
Option c
122 / 6 → remainder is 2, quotient is 20
122 / 7 → remainder is 3, quotient is 17
20-17 = 3

7). L.C.M (2, 4, 6, 8, 10, 12) = 120

All the six bells would ring together at an interval of 120 min or 2 hours, hence in 10 hours (i.e from 1 : 00 – 11: 00 P.M) they will ring together for 10 / 2 or 5 times

8). One no. is(243k + 26), another no. is (351m + 13). Sum of the numbers  (243k + 26 + 35 l m + 13) = 243k + 35 lm + 39
When this no. is divided by 27, the remainder is
Remainder of (243k / 27) + Remainder (35lm / 27) + Remainder (39 / 27) = 0 + 0 + 12
Remainder = 12

9). When Arjun made groups of 3, 5, 7 and 9, exactly one ball was left behind. Hence the total no. of balls was left behind. Hence the total no. of balls is (a common multiple of 3, 5, 7, 9) + 1;
L.C.M of 3, 5, 7, 9 = 315.
Minimum no.of total balls = 315 + 1 = 316
General form → 315 m + 1, where m = 1, 2, 3, 4, 5..
m = 1, no. of balls = 315 + 1 = 316
m = 2, no. of balls = (315) 2 + 1 = 631
m = 3, no. of balls = (315) 3 + 1 = 946
m = 4, no. of balls = (315) 4 + 1 = 1261
→ 4 digit no
No. of possible solutions are 3.