Latest TCS Aptitude Question SOLUTION: How many parallelograms are formed by a set of 4 parallel lines intersecting on other set of 7 parallel lines ? ans126 7. If A works total in 10 days and B works total in 15 Days.

7) A=10 days
B= 15 days
A:B = 10:15 =2:3
So 2x+3x=5000=> x= 1000
A’s share = 2000
B’s share = 3000
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8) 2 as a^p – a is dividable by p when p is prime
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9) TOGETHER
Starting with ‘O’ = 6!/2! =360
Starting with ‘E’ the last letter is either ‘O’ or’E’ = 2*(6!/2!) = 720
360+720=1080
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10) Take last 2 digit of 1941 and take last 2 digits of 1961
41^1=41 r=1 61^1=61 r=1
41^2=81(take last 2 digits of ans)r=2 61^2=21 r=2
41^3=21 r=3 61^3=81 r=3
41^4=61 r=4 61^4=41 r=4
41^5=01 r=0 61^5=01 r=0
41^6=41 61^6=61
hence power cycle is repeating and it is 5 hence for 1961 it is 5
r-remainder for 1961 power cycle-5
for 1941 power cycle-5, 4181 mod 5=rem 1=61
3843 mod 5=rem 3=21
therefore 21+61=82
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11) let base be x
convert all the num to base x
(1*x+2*x^0)*(2*x+5*x^0)=(3*x^0+3*x^1+3*x^2)
solving X=7
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12) Words starting with A……..is 120
we have to find the 25th word in dictionary order.
so obviously words start with A.
the there are 24 words starting with AD…
there are again 24 words starting with AE…..
so 25th word will be started with AE..
now we have to find the 25th word
that will be AEDKNR
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13) Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 —(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
So B can complete the work in 75/2 days = 37 (1/2) days
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14) ans Wednesday
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15)n*n!+(n+1)*(n+1)!+------+(n+m)*(n+m)! = (n+m+1)!-1=2012+1=2013!-1
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16) the tank can never be filled because the emptying pipe (B) takes lesser time than filling pipe (A)
=> 1/25 - 1/20 = -1/100
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17) grandmother =x ;; rohit =x/2
1994-x +1944- x/2 = 3844
3988-3844=x+x/2
144=3x/2
X=96
Rohit = x/2 = 96/2=48
Rohit @1994 =48
@1999 =48+5=53
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18) Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 —(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
So B can complete the work in 75/2 days = 37 (1/2) days
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19) Factors of 1260 are 2*2*3*3*7*5
Now minimum possible two numbers A and B formed from these factors are
7*5=35=A
2*2*3*3=36=B
so minimum value of A+B is 71
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20) n(n*n-1)=n^3-n=(n-1)(n)(n+1)
any 3 consecutive natural numbers product is divisible by 6
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21) sqrt(3)/4)*(a*a)=1/2 * 12(base) * altitude
so altitude is=6sqrt(3)
now the small triangle below that altitude having the altitude is
(sqrt(3)/4)*(a*a) where a=4 because all 12 triangle is equilateral = 1/2 * 4(base) * altitude
sp altitude=2sqrt(3)
so diameter of circle=6sqrt(3) + 2sqrt(3) = 8sqrt(3)
so radius is = 4sqrt(3)
remaining area= 22/7 * r * r - 12* (sqrt(3)/4) * a * a where a=4
so ans is=67.718
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22) a^2 + 2ab + b^2 = (a+b)^2
Here, 2^74 + 2^2058 + 2^2 = (2^37)^2 + 2*2^37*2^2020 + (2^1)^2
If the number is a perfect square, then 2^1 should equal 2^2020.
Hence, 1=2020
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23) 43+45+………499 are in a.p.
n/2(2a+(n-1)d)
last digit of this sum is 9.
Similarly 92+94+………..9100.
Last digit of this sum is 0.
So 9+4+0 = 13.
3 is last digit.
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24) a^2 + 2ab + b^2 = (a+b)^2
Here, 2^74 + 2^2058 + 2^2n = (2^37)^2 + 2*2^37*2^2020 + (2^n)^2
If the number is a perfect square, then 2^n should equal 2^2020.
Hence, n=2020
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25) milk:water=2:3 =40:60
take out 50 %(50 ltr means milk=20 ltr water=30 ltr) remaining
milk:water=20:30
add 50 ltr of water
ratio
milk:water=20:(30+50) =20:80
1:4
so initially milk= 40%
after milk=20%
so reduction in concentration= 20%
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26) 4kg of a metal contains 1/5 ‘parts of ‘ copper and rest is zinc. another 5kg of metal contains 1/6 ‘parts of ‘of copper and rest is zinc.
Then amt of copper in 4 kg mixture = 4/5 kg.
Then amt of zinc in 4 kg mixture = 16/5 kg
Then amt of copper in 5 kg mixture = 5/6 kg
Then amt of zinc in 5 kg mixture = 25/6 kg
Ratio of copper to zinc in final mixture = (4/5 +5/6) : (16/5+25/6)
= 49/30 : 221/30 = 49:221
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27)if the total quantity=100
Actual quantity of spirit to be sold to get 12.5% gain=100*100/112.5=800/9
So the water =100-(800/9)=100/9
Proportion of water:spirit =100/9 : 800/9= 1:8
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28) Number of factors = either (2a)^n or (2a-1)^n
Based on following conditions
If 2(a-1) > 3(b-1) , then (2a-1)^n
If 2(a-1) < 3(b-1) , then (2a)^n
If 2(a-1) = 3(b-1) , then both are possible.
Here 2(a-1) > 3(b-1)
=> no of factors = (2a-1)^n = 9^3= 729
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29) last 2 digits can be 12, 24, 32, 44, 52 .
so 5^(8-1) = 5^7
=78125
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30) 40*80 /(40+80-104) = 3200/16 =200m
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parallelogram qn => 4C2*7C2 = 6*21=126

30. 40*80 /(40+80-104) = 3200/16 =200m

10 sol : Remember 1 raised to any power will give 1 as unit digit.
To find the digit in the 10th place, we have to multiply, 10th digit in the base and unit digit in the power.
So the Last two digits of the given expression = 21 + 61 = 82

30.answer is 200

23.answer will be 0.

distance=s*t.
so, 18*(t+40*60)=28*t.
solving we get 72 mins =1 hr 12 mins. so 7.12 pm