How to do quick user assign in kanban in Odoo 17 ERP
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1. Number System
1. The greatest among 1.5, 1.05, 1.05, 1.5 is
(1) 1.05 (2) 1.05
(3) 1.5 (4) 1.5
2. If a number is as much greater than 31 as it is less than 75,
then the number is
(1) 106 (2) 44
(3) 74 (4) 53
3. The greatest among the following numbers
(3)1/3, (2)1/2, 1, (6)1/6 is :
(1) 1/ 2
(2) (2) 1
(3) (6)1/6 (4) (3)1/3
4. The largest among the numbers ( )3
0.1 , 0.0121,0.12 and
0.0004 is
(1) ( )2
0.1 (2) 0.0121
(3) 0.12 (4) 0.0004
5. Which of the following numbers is the greatest of all ?
0.9,0.9,0.09,0.09
(1) 0.9 (2) 0.9
(3) 0.09 (4) 0.09
6. Anumber x when divided by289 leaves 18 as the remainder.
The same number when divided by 17 leaves y as a
remainder. The value of y is
( 1) 5 (2) 2
(3) 3 (4) 1
7. In a division sum, the divisor is 3 times the quotient and
6 times the remainder. If the remainder is 2, then the
dividend is
(1) 50 (2) 48
(3) 36 (4) 28
8. Two numbers 11284 and 7655, when divided by a certain
number ofthree digits, leaves the same remainder. The sum
of digits of such a three-digit number is
(1) 8 (2) 9
(3) 10 (4) 11
9. The remainder when 321 is divided by 5 is
(1) 1 (2) 2
(3) 3 (4) 4
10. A number when divided by 119 leaves remainder 19. If the
same number is divided by 17, the remainder will be
(1) 12 (2) 10
(3) 7 (4) 2
11. A nmber consists of two digits. If the number formed by
interchanging the digits is added tothe original number, the
resulting number (i.e. the sum) must be divisible by
(1) 11 (2) 9
(3) 5 (4) 3
12. In a question on division, the division, the divisor is 7 times
the quotient and 3 times the remainder. If the remainder is
28, then the dividend is
(1) 588 (2) 784
(3) 823 (4) 1036
13. Which of the following numbers is NOT divisible by 18 ?
(1) 54036 (2) 50436
(3) 34056 (4) 65043
14. If
3
4
of the difference of
1
2
4
and
2
1
3
is subtracted from
2
3
of
1
3
4
the result is
(1)
48
83
-
(2)
48
83
(3)
83
48
-
(4)
83
48
15. A number when divided by899 gives a remainder 63. Ifthe
same number is divided by 29, the remainder will be :
(1) 10 (2) 5
(3) 4 (4) 2
16. The numerator ofa fraction is 4 less than its denominator. If
the numerator is decreased by 2 and the denominator is
increased by 1, then the denominator becomes eight times
the numerator. Find the fraction.
(1)
3
8
(2)
3
7
(3)
4
8
(4)
2
7
17. A fraction becomes
9
11
, if 2 is added to both the numerator
and the denominator. If 3 is added to both the numerator
and the denominator it becomes
5
6
. What is the fraction?
(1)
7
9
(2)
3
7
(3)
5
9
(4)
7
10
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2. 2
CAREER POWER
CAREER POWER, G-7, Roots Tower, Laxmi Nagar, District Centre, Delhi-92, www.careerpower.in
Chapterwise SSC Quantitative Aptitude
18. A, B, C and D purchase a gift worth ` 60. A pays
1
2
of
what others are paying, B pays
1
3
rd of what others are
paying and C pays
1
4
th of what others are paying. What is
the amount paid by D ?
(1) 16 (2) 13
(3) 14 (4) 15
19. In an office, there are 108 tables and 132 chairs. If
1
6
of
the tables and
1
4
ofthe chairs are broken, how manypeople
can work in the office if each person requires one table and
one chair ?
(1) 86 (2) 90
(3) 92 (4) 99
20. Aperson gives
1
4
of his property to his daughter,
1
2
tohis
sons and
1
5
for charity. How much has he given away ?
(1)
1
20
(2)
19
20
(3)
1
10
(4)
9
10
21. By how much does
6
7
8
exceed
6
7
8
?
(1)
3
6
4
(2)
1
6
8
(3)
3
6
28
(4) 3
7
4
22. The product of two fractions is
14
15
and their quotient is
35
24
. The greater of the fraction is
(1)
7
4
(2)
7
6
(3)
7
3
(4)
4
5
23. In an examination, a student was asked to find
3
14
of a
certain number. By mistake, he found
3
4
of it. His answer
was 150 more than the correct answer. The given number is
(1) 500 (2) 280
(3) 240 (4) 180
24. Arrange
4 7 6 5
, , ,
5 8 7 6
in the ascending order :
(1)
4 7 6 5
, , ,
5 8 7 6
(2)
5 6 7 4
, , ,
6 7 8 5
(3)
4 5 6 7
, , ,
5 6 7 8
(4)
7 6 5 4
, , ,
8 7 6 5
25. If the sum of five consecutive integers is S, then the largest
of those integers in terms of S is
(1)
10
5
S -
(2)
4
4
S +
(3)
5
5
S +
(4)
10
5
S +
26. The sum of three consecutive odd natural numbers is 147.
Then, the middle number is :
(1) 47 (2) 48
(3) 49 (4) 51
27. If we write 45 as sum of four numbers so that when 2 is
added to first number, 2 subtracted from second number,
third multiplied by 2 and fourth divided by 2, we get the
same result, then the four numbers are :
(1) 1, 8, 15, 21 (2) 8, 12, 5, 20
(3) 8, 12, 10, 15 (4) 2, 12, 5, 26
28. A number consists of two digits and the digit in the ten’s
place exceeds that in the unit’s place by 5. If 5 times the
sum of the digits be subtracted from the number, the digits
of the number are reversed. Then the sum of digits of the
number is
(1) 11 (2) 7
(3) 9 (4) 13
29. 0.393939....... is equal to
(1)
39
100
(2)
13
33
(3)
93
100
(4)
39
990
30. The number, which is to be added to 0.01 to get 1.1, is
(1) 1.11 (2) 1.09
(3) 1 (4) 0.10
31. If *
a
a b a b
b
= + + , then the value of 12 * 4 is :
(1) 20 (2) 21
(3) 48 (4) 19
32. If sum of twonumbers be a and their product be b, then the
sum of their reciprocals is
(1)
1 1
a b
+ (2)
b
a
(3)
a
b
(4)
1
ab
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3. 3
CAREER POWER
CAREER POWER, G-7, Roots Tower, Laxmi Nagar, District Centre, Delhi-92, www.careerpower.in
Chapterwise SSC Quantitative Aptitude
1. (4) 1.05 < 1.05 < 1.5 < 1.5
Remember :
1.05 = 1.0555.......
1.5 = 1.555.........
2. (4) If the number be x, then x – 31 = 75 – x
Þ 2x = 75 + 31 = 106 Þ x = 53
3. (4) LCM of 3, 2 and 6 = 6
( )
1 1
3 6
1
2 63 (3 ) (9)= =
1 11
6 62
2 (23) (9)= =
1 1 1
6 6 6
(1) 1;(6) (6)= =
4. (3) (0.1)2 = 0.01
0.0121 = 0.11 0.11´ = 0.111
0.0004 = 0.02
5. (2) 0.9 =
9
10
; 0.9 =
9
9
= 1,
0.09 =
9
90
=
1
10
; 0.09 =
9
99
=
1
11
6. (4) Here, the first divisor (289) is a multiple of second
divisor (17).
Required remainder = Remainder obtained on
dividing 18 by 17 = 1
7. (1) Divisor = 6 × 2 = 12
Again, Divisor = 3 × quotient
Quotient =
12
3
= 4
Dividend = 12 × 4 + 2 = 48 + 2 = 50
8. (4) If the remainder be x, then (11284 – x) and (7655 – x)
are divisible by three digit number.
i.e. (11284 – x) – (7655 – x)
= 3629 is divisible by that number.
3629 = 19 × 191
Hence, required number = 191
Sum of digits = 1 + 9 + 1 = 11
9. (4) 31 = 3; 32 = 9; 33 = 27; 34 = 81; 35 = 243
i.e. unit’s digit is repeated after index 4.
Remainder after dividing 21 by 4 = 1
Unit’s digit in the expansion of (3)21 = 3
Remainder after dividing by 5 = 3
10. (4) When we divide the number by 119, remainder = 19
Q 119 is exactly divisible by 17
the required remainder = remainder obtained by
dividing
19 by 17 = 2
11. (1) Let the number be 10x + y
After interchanging the digits, the number obtained
= 10y + x
According to the question
Resulting number
= 10x + y + 10y + x = 11x + 11y
= 11 (x + y) which is exactly divisible by 11.
12. (4) Let the quotient be Q and the remainder be R. Then
Divisor = 7Q = 3R
Q =
3
7
R =
3
7
× 28 = 12
Divisor = 7Q = 7 × 12 = 84
Dividend = Divisor × Quotient + Remainder
= 84 × 12 + 28 = 1008 + 28 = 1036
13. (4) A number will be exactlydivisible by18 ifit isdivisible
by 2 and 9 both. Clearly 65043 is not divisible by 2.
Required number = 65043
14. (3)
13 2 9 5 3 13 27 20 3
4 3 4 3 4 6 12 4
-æ ö æ ö
´ - - ´ = - ´ç ÷ ç ÷è ø è ø
13 7 3 13 7
6 12 4 6 16
= - ´ = -
104 21 83
48 48
-
= =
15. (2) Required remainder = remainder got when 63 is
divided by 29 = 5
HINTS & SOLUTIONS
.ANSWER KEY.
1. (4) 2. (4) 3. (4) 4. (3) 5. (2) 6. (4)
7. (1) 8. (4) 9. (4) 10. (4) 11. (1) 12. (4)
13. (4) 14. (3) 15. (2) 16. (2) 17. (1) 18. (2)
19. (2) 20. (2) 21. (1) 22. (2) 23. (2) 24. (3)
25. (4) 26. (3) 27. (2) 28. (3) 29. (2) 30. (2)
31. (4) 32. (3)
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4. 4
CAREER POWER
CAREER POWER, G-7, Roots Tower, Laxmi Nagar, District Centre, Delhi-92, www.careerpower.in
Chapterwise SSC Quantitative Aptitude
16. (2) Original fraction =
4x
x
-
In case II, 8 (x – 4 – 2) = x + 1
Þ 8x – 48 = x + 1 Þ 7x = 49 Þ x = 7
Original fraction =
7 4 3
7 7
-
=
17. (1) Solve this question by options.
Original fraction =
7
9
Adding 2 to numerator and denominator, fraction
=
9
11
Adding 3 to numerator and denominator, fraction
=
10
12
=
5
6
18. (2) A + B + C + D = 60
A =
B C D
2
+ +
Þ 3A = 60 Þ A = ` 20
B =
A C D
3
+ +
Þ 4B = 60 Þ B = ` 15
C =
A B D
4
+ +
Þ 5C = 60 Þ B = ` 12
D = 60 – (20 + 15 + 12) = ` 13
19. (2) Unbroken table =
5
108 90
6
´ =
Unbroken chairs = 132 99
4
3
´ =
Unbroken pairs = 90
20. (2) Part of the property given away
1 1 1
4 2 5
= + +
5 10 4 19
20 20
+ +
= =
21. (1) First number = 6 ×
8
7
=
48
7
Second number
6 1 3
7 8 28
= ´ =
Required difference
48 3
7 28
= -
192 3 189 27 3
6
28 28 4 4
-
= = = =
22. (2) Let the fractions be x and y, where x > y
14
15
xy = and
35
24
x
y
=
14 35
15 24
x
xy
y
´ = ´
Þ x2 =
49
36
Þ x =
7
6
23. (2) Let the number be x.
Then,
3 3
150
4 14
x x- = Þ
21 6
150
28
x x-
=
Þ 15 x = 28 × 150 Þ x =
28 150
15
´
= 280
24. (3) Firstly, we express every fraction in decimal form.
4 7
0.8; 0.875
5 8
= = ;
6
0.857
7
= ;
5
0.833 0.83
6
= =
25. (4) Sum of five consecutive integers = S
Third integer =
5
S
Largest integer =
5
S
+ 2 =
10
5
S +
26. (3) x + x + 2 + x + 4 = 147
Þ 3x + 6 = 147 Þ 3x = 141
Þ x =
141
3
47
Middle Number = x + 2 = 47 + 2 = 49
27. (2) The number are : 8, 12, 5, 20
28. (3) Let unit's digit be x
Ten's digit = x + 5
Number = 10 (x + 5) + x = 11x + 50
Again, 11x + 50 – 5 (2x + 5)
Þ 10x + x + 5
Þ 11x + 50 – 10x – 25
Þ 10x = 20 Þ x = 2
Required sum = 2x + 5 = 2 × 2 + 5 = 9
29. (2) 0.393939.......
39 13
0.39
99 33
= = =
g g
30. (2) Required number = 1.1 – 0.01 = 1.09
31. (4) a * b = a + b +
a
b
12 * 4 = 12 + 4 +
12
4
= 16 + 3 = 19
32. (3) If the numbers be x and y then
x + y = a and xy = b
1 1x y a
xy y x b
+
= + =
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