### SSC CHSL- Practice Aptitude Questions (Algebra) With Solutions Set-46

SSC CHSL- Practice Aptitude Questions (Algebra) With Solutions Set-46:
Dear Readers, Important Aptitude Questions for SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1). When n is divided by 6, the remainder is 4. When 2n is divided by 6, the remainder is
a)    1
b)    2
c)    0
d)    4
2). The value of
1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 is
a)    7/20
b)    1/10
c)    3/5
d)    3/20
3). Let a = [ 1/(2-√3)] + [ 1/(3-√8)] + [ 1/(4-√15)] . Then we have
a)    a=9
b)    a<18 but a≠9
c)    a>18
d)    a = 18
4). The smallest among the numbers 2250, 3150, 5100 and 4200 is
a)    2250
b)        4200
c)    5100
d)        3150
5). The greatest of the following numbers 0.16, √0.16, (0.16)2, 0.04 is
a)        (0.16)2
b)    0.16
c)    √0.16
d)    0.04
6). If A : B = 3 : 4 and B : C = 6 : 5, then C : A is
a)    9 : 8
b)    10 : 9
c)    9 : 10
d)    8 : 9
7). If a, b are rationals and a√2 + b√3 = √98 + √108 - √48 - √72, then the values of a, b are respectively
a)    2, 3
b)    1, 2
c)    1, 3
d)    2, 1
8). If 1/a – 1/b = 1/(a-b) , then the value of a3 + b3 is
a)    2
b)    0
c)    -1
d)    1
9). If a2 + b2 + c2 = ab + bc + ca, then (a+c)/b is equal to
a)    4
b)    1
c)    2
d)    3
10). If the graphs of the equations x+y = 0 and 5y + 7x = 24 interest at (m,n), then the value of m+n is
a)    -1
b)    2
c)    1
d)    0

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

Solutions:

1). When n us divided by 6, remainder is 4
n = 6k + 4
2n = 12k + 8
= 12k +6 + 2
= 6(2k+1)+2
When 2n is divided by 6, Remainder = 2
Example:
Take n = 10
When 10 is divided by 6, remainder = 4
2n = 2 × 10 = 20
When 20 is divided by 6, Remainder = 2.
2).

L.C.M of (20, 30, 42, 56, 72, 90 ) = 2×2×2×3×3×5×7
= 2520
Now,  1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90
= (126+84+60+15+35+28)/2520
= 378/2520 = 3/20
3).
Formula : (a+b)(a-b)=a2-b2
1/(2-√3)  = [ 1/(2-√3) ] × [ (2+√3)/(2+√3) ]
= (2+√3) / (22 – (√3)2)
= (2 + √3) / (4-3) = 2+√3 / 1
= 2 +√3
Similarly,
1 / (3-√8) = [1/(3-√8)] × [ (3+√8)/(3+√8) ] = 3+√8
1/(4-√15) = [ 1/(4-√15) ] × [ (4+√15)/(4+√15) ]
= 4 + √15
a= 1/(2-√13) + 1/(3-√8) + 1/(4-√15)
=  2+ √3 + 3 + √8 + 4 + √15
= 9 + √3 + √8 + √15
Clearly
√3 < √4 = 2
√8 > √9  = 3
√15 < √16 = 4
Therefore
a= 9 +√3 +√8 + √15
< 9 + 2 + 3 4
= 18
a < 18
4). GCD of (250, 150, 100, 200) = 50
2250 = 25×50 = (25)50 = 3250
3150 = (35)50 = 2750
5100 = (52)50 = 2550
4200 = (44)50 = 25650
2250 = 3250
3150 = 2750
5100 = 2550
4200 = 25650
Smallest number = 5100
5).
0.16
√0.16 = 0.4
(0.16)2 = 0.0256 = 0.04
Greatest number = √0.16
6). A : B = 3 : 4 = 9 : 12
B : C = 6 : 5 = 12 : 10
A : B : C = 9 : 12 : 10
C : A = 10 : 9
7).
√98 = √(2×49) = 7√2
√108 =  √(3×36) = 6√3
√48 = √(3×16) = 4√3
√72 = √(2×36) = 6√2
Now,
a √2 + b√3 = √98 + √108 - √48 - √72
= 7√2 + 6√3 - 4√3 - 6√2
= √2+2√3
a = 1
b = 2
8).
1/a – 1/b = 1/(a-b)
(b-a)/ab = 1/(a-b)
-(a-b)/ab = 1/(a-b)
(a-b)2 = -ab
a2+b2-2ab = -ab
now, a2+b2 = (a+b)(a2-ab+b2) (formula)
= (a+b)( a2 + b2 -ab)
=  (a+b)(ab-ab)
= (a+b) × 0 = 0

10). Solve x+y=0 and 5y+7x=24
X+y = 0 à x=-y
5y + 7x = 24
- 2y =24
y = -12
x = -y = 12
m=12; n= -12
now, m+n = 12-12 = 0

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