Question 1. Solve the inequality 27(7x-2) > 3
a) x > 1/7
b) x > 1/3
c) x > 2/7
d) x > 1
Answer : b) x > 1/3
Solution : Given inequality is 27(7x-2) > 3.
we can re-write 27 with powers of 3 as
[33](7x-2) > 3
33(7x - 2) > 31
Let us clear out base 3 from both sides,
3(7x - 2) > 1
21x - 6 > 1
21x > 7
x > 7/21 = 1/3
x > 1/3
Question 2. Solve the inequality 6(6x-2) < 1/36
a) x < 0
b) x > 0
c) x = 1/3
Answer : a) x < 0
Solution :
Given inequality is 6(6x-2) < 1/36
Multiply by 36 on both sides, we get
36[6(6x-2)] < 1
Make each term as the power of 6
(62)[6(6x-2)] < 60
Now, using the formula (ax)(ay)= a(x+y)
d) x = 6
6(6x-2+2) < 60
6(6x) < 60
Cancelling the base 6 on both sides,
6x < 0
x<0
Hence the answer is option a
Question 3. Solve 1/(25) < 2[(6-2x)/3] < 4
a) x < 0 `
b) x > 31/3
c) x < 21/2
d)none of these
Answer : c) x < 21/2
Solution :
The given inequality is 1/(25) < 2[(6-2x)/3] < 4
Rewrite the given equation with powers of 2 as
2(-5) < 2[(6-2x)/3] < 2^2
Let us clear out the base 2 from each term
-5 < (6-2x)/3 < 2
Now, let us clear out "/3" by multiplying each term by 3:
Since we are multiplying by a positive number, the inequalities will not change.
Then the inequality becomes -15 < 6-2x < 6
Now subtract 6 from each part then -21 < -2x < 0
Now multiplying each part by -1/2.
Since, we are multiplying by a negative number, the inequalities change direction.
i.e., 21/2 > x > 0
Arrange the smaller number left, larger number right then we have 0 < x < 21/2
Hence the answer is x < 21/2
Question 1. The code for creating stack in DOS is _____ .
a) STACKS = (number),(size)
b) STACKS = number
c) STACKS = (size),(number)
d) STACKS = size
Answer : a) STACKS = (number),(size)
Question 2. Unix Operating System is an example of ______ .
a) Macro or Monolithic Kernel
b) Micro Kernel
c) Hybrid Kernel
d) None of these
Answer : a) Macro or Monolithic Kernel
Question 3. In Linux, the data structure of file system is called as ______ .
a) inode
b) struct inode
c) vnode
d) lnode
Answer : b) struct inode
Question 4. Interrupt Latency means _______ .
a) time gap between two interrupts
b) time gap between the first interrupt and service of the second interrupt
c) time gap between interrupt and service of the interrupt
d) None of these
Answer : c) time gap between interrupt and service of the interrupt
Question 5. Among the following options which is true about Remote IPC ?
a) Allows more than two processes to communicate.
b) Allows more than two processes to communicate in different systems.
c) Allows more than two processes to communicate in same systems.
d) Allows two processes to communicate in different systems
Answer : b) Allows more than two processes to communicate in different systems.
Question 1. A man lent out a certain amount of money to A on simple interest (S.I) at the rate 20% per
annum for 2 years and the same amount of money at same rate to B on compound interest (C.I) which is
compounded half yearly for 2 years. At the end of 2 years, he got Rs.1282 as the difference amount of
S.I and C.I then what was the amount he lent out?
a) Rs.20,000
b) Rs.21,000
Answer : a) Rs.20,000
Solution : Let the required amount be Rs. P.
c) Rs.22,000
d) Rs.23,000
S.I on P for 2 years at 20% per annum = P x R x T /100 = P x 2 x 20 / 100 = 2P/5.
C.I on P for 2 years at 20% (compounded half yearly) = P {1 + (R/2)/100}2n
= P {1 + (20/2)/100}2(2) – P
= P {1 + 10/100}4 – P
= P {1 + 1/10}4 – P
= P {11/10}4 – P
Given that, the difference is Rs.1282.
i.e., P{11/10}4 – P - 2P/5
= P {11/10}4– 7P/5
= P{14641/10000 – 7/5} = Rs.1282
= P{14641- 14000/10000 } = Rs.1282
= P{641/10,000} = Rs.1282
= P = Rs. 20,000.
Hence, the answer is Rs.20,000.
Question 2. A man invested Rs. A in bank X which offers simple interest at 10% per annum and he
invested Rs. B in bank Y which offers compound interest at 6% per annum. If the total amount of
interest (together bank X and Y) in 2 years was Rs.8017 and the total amount invested by him was Rs.
52500 then which of the following is equal to A?
a) Rs.20,000`
b) Rs.21,500
c) Rs.22,000
Answer : a) Rs.20,000.
Solution :We have to find Rs. A
Total amount invested by him is Rs. 52500
i.e., Rs. (A + B) = Rs. 52,500
Then B = Rs.52500 - A
He invests Rs. Rs.52500 - A in bank Y on C.I at 6% per annum for 2 years.
d) Rs.23,500
Total S.I from bank X = PRT/100 = A x 10 x 2 / 100 = Rs. A / 5 …..(1)
Total C.I from bank Y = P {1 + R/100}n – P = P [{1 + R/100}2 – 1]
= (52500 - A)[{1+ 6/100}2 – 1]
= (52500 - A)[{1+ 3/50}2 - 1]
= (52500 - A)[{53/50}2 - 1]
= (52500 - A)[{2809/2500} -1 ]
= (52500 - A){309/2500} ……(2)
Total interest amount of S.I and C.I = Rs.8017.
From (1) and (2), A/5 + (52500 – A)(309 / 2500) = 8017
A/5 + 52500 x (309/2500) – A (309/2500) = 8017
191A/2500 + 21x309 = 8017
191A/2500 = 8017 – 6489
191A/2500 = 1528
A = 8 x 2500 = 20,000
Hence the required answer is Rs.20,000.
Question 3. A man borrowed a certain amount on S.I and C.I at same rate of interest. At the end of two
years, he paid back Rs.1600 and Rs.1664 as S.I and C.I respectively. If he took 3 years to return the
amount then what was the difference of C.I and S.I ?
a) Rs.197.12
b) Rs.128.12
c) Rs.293.12
Answer : a) Rs.197.12
Solution : Let the amount he borrowed be Rs.P
Let C.I on P for 2 years be X and S.I on P for 2 years be Y.
S.I on P for 1st year = Y/2 …..(1)
d) Rs.179.12
C.I for 2nd year = Fixed Interest + Interest on interest
Therefore, X = Y + S.I on Y/2 for 1st year (by eqn(1))
X - Y = S.I on Y/2 for 1st year ….(2)
He paid S.I at the end of 2nd year (Y) = Rs.1600 then S.I for 1st year (Y/2) = Rs.800
S.I for 3rd year = S.I for 2nd years + S.I for 1st year = Rs.1600 + Rs.800 = Rs.2400 ….(3)
He paid C.I at the end of 2nd year = Rs. 1664
Difference S.I – C.I = (X - Y) = Rs.1664 – 1600 = Rs. 64.
sub. X - Y in eqn(2), we get
Rs.64 = S.I on Y/2 for 1st year.
S.I on Rs.800 for 1st year = Rs.64
Therefore, principal P1 = Rs.800, S.I = 64, T = 1 year.
Then R = 100 x S.I / P x T
= 100 x 64 / 800 x 1 = 8%
Now, R = 8%, t = 2 years and S.I = 1600
Then, principal P = 100 x S.I / R x T
= 100 x 1600 / 8 x 2 = Rs.10,000
Now, we have to calculate the C.I on Rs.10,000 for 3 years at 8% per annum.
C.I = P [{1 + R/100}n – 1]
= 10000[{1+ 8/100}3 – 1]
= 10000[{27/25}3 – 1]
= 10000[{273 – 253}/ 253]
= Rs.2597.12
C.I for 3 years = Rs.2597.12
Required difference = C.I for 3 years – S.I for 3 years = Rs.( 2597.12 – 2400) = Rs.197.12