Practice Aptitude Questions (Mixed Problems) With Solutions Set-50

Practice Aptitude Questions (Mixed Problems)
Practice Aptitude Questions (Mixed Problems) With Solutions Set-50:
Dear Readers, Important Aptitude Questions for SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.



1). Find the sum of 18²+20²+22²+…50².
a)    53582
b)    73892
c)    633552
d)    21284

2). Find the unit’s digit of (496)496×(524)524×(1016)260
a)    4
b)    2
c)    6
d)    8

3). If cosθ=-1/2 and п< θ<3п/2, find the value of 4 tan²θ-3cosec²θ
a)    12
b)    8
c)    4
d)    2

4). Find the value of
(cos60°/ sin30°) + [(cos65°.cosec25°) / (tan10°.tan30°.tan45°.tan60°.tan80°]
a)    1
b)    2
c)    0
d)    -1

5). Find the least number which when divided by 15, 16, 18, 20 and 24 leaves 3 as remainder in every case, but when it is divided by 13 leaves no remainder.
a)    2886
b)    1443
c)    732
d)    728

6). The LCM of two numbers is 33times of their HCF. The sum of the LCM and their HCF is 952. If one number is 42, the other number is
a)    588
b)    616
c)    634
d)    226

7). If tan θ+ cot θ=16, then find the ratio of tan²θ + cot²θ to tan²θ+cot²θ+20 tanθ.cotθ
a)    64:65
b)    129:137
c)    27:29
d)    127:137

8). Find the value of (1/2) + (1/4) + (1/12) + (1/20) + (1/30) + (1/42) + (1/56) + (1/72)
a)    1/5840
b)    15/36
c)    1/18
d)    35/36

9). Find the value of [1+ (1/x)] [1+ (1/x+1)] [1+ (1/x+2)] [1+ (1/x+3)]
[1+ (1/x+4)] [1+ (1/x+5)]
a)    1+ (1/(x+6))
b)    x+6
c)    1/x
d)    (x+6)/x

10). The simplified form of 1038(1/13) +1038(2/13) +1038(3/13)
+1038(4/13) +1038(7/13) +1038(9/13)
a)    6130
b)    6230
c)    12456
d)    6320

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

Solutions:


1). The sum of the square of even numbers from 2 to n= [n(n+1) (n+2)] / 6
2²+4²+6²+8²+…+50²
= [50(50+1) (50+2)] / 6
= (50×51×52) / 6 =22100
And 2²+4²+6²+8²+…+16²= [16(16+1) (16+2)] / 6
= (16×17×18) / 6 =816
Now, 18²+20²+22²+24²+…+50²
=22100-816 =21284
= (Sum of squares of 2 to 50 –Sum of squares of 2 to 16)
Answer: (d)

2). If the last digit of a number is 6, that number raised to any power will give a number with its last digit as 6.
∴Last digit of (496)496=6
Similarly, digit of (1016)260=6
For a number with 4 as last digit,  42n with have 6 as its last digit.
Last digit of (524)524=6
Now, 6×6×6 will have its last digit as 6.
Hence, Required answer=6
Answer: (c)

3). Since θ lies in the third quadrant, sin θ is negative and tan θ is positive.
Now, sin θ=± √1- cosθ² θ
or, sin θ=√(1-1/4)=-3/2
or, cosec θ=-2/3     and tan θ= sinθ/cos θ =√3
Now, 4tan²θ-3cosec² θ=4×3-3×4÷3=8
Answer: (b)




5). LCM of 15, 16, 18, 20 and 24=720
Let the number to be divided by 13 be 720k+3.
Now, 720k+3=13×55k+5k+3
Then, 720k+3 is divisible by 13.
Put k=2
So, 5k+3=10+3=13which is divisible by13
∴Required number is 720k+3=720×2+3=1440+3=1443
Answer: (b)

6). Let the HCF be x. Then LCM=33x
Now, x+33x=952 or, 34x=952
∴ x=28
We know LCM×HCF=First number× Second number,
Then 33x×x=42×Seond number or, 33×28×28=42×Second number
∴ Second number= (33×28×28) ÷42=616
Answer: (b)

7). Given tanθ+ cotθ=16
Squaring both sides, we get tan²θ+2tanθ.cotθ+cot²θ=256
Or, tan²θ+2+cot²θ=256 or, tan²θ+cot²θ=256-2
∴tan²θ+cot²θ=254
Now, tan²θ+cot²θ+20tanθ.cotθ
= (tan²θ+cot²θ) +20tanθ.1÷tanθ
=254+20=274
∴ Required ratio=254÷274 =127÷137 =127:137
Answer: (d)





10). 1038(1/13) +1038(2/13) +1038(3/13) +1038(4/13) +1038(7/13) +1038(9/13)
=6×1038+ [(1/13) + (2/13) + (3/13) + (4/13) + (7/13) + (9/13)]
=6×1038+ [(1+2+3+4+7+9) /13]
=6×1038+ (26/13)
=6×1038+2=6230
Answer: (b)

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