Answer: (a)Take options and check. If 10 are
correct, his score is 10 x 8 = 80. But 16 are
wrong. So total negative marking is 16 x 5 = 80.
So final score is zero.
Answer : (c)
Step 1: (3 x 2) - 1 = 5 ( 2^2 + 1)
Step 2: (5 x 2) - 1 = 9 (2^3 + 1)
Step 3: (9 x 2) - 1 = 17 (2^4 + 1)
Step 4: (17 x 2) - 1 = 33 (2^5 + 1)
.
.
.
So After 30 steps we have 2^31 + 1
Answer : (c) Given that, pq=28 and p, q are integers.
The factors of 28 are 1, 2, 4, 7, 14 and 28.
Then the possibilities of pq = 28 are (p,q) = (1, 28), (2,
14), (4, 7), (7,4), (14, 2) and (28,1)
We would have p > q since the given options are
positive.
(That is, if p < q then (p2 - q2) will be negative).
Therefore, the possibilities of (p, q) reduced to (7,4),
(14, 2) and (28,1).
If (p,q) = (28,1) then (p2 - q2) = 282 - 12 = 784 - 1 = 783.
If (p,q) = (14, 2) then (p2 - q2) = 142 - 22 = 196 - 4 =
192
If (p,q) = (7,4) = (p2 - q2) = 72 - 42 = 49 - 16 = 33
From the given options, required answer is option c.
Answer : (a)
A's share = 1/2
Then remaining = 1 - 1/2 = 1/2 ...(A)
2/3 of this remaining 1/2 (as in eq A) goes to B,C and D.
Total share of B, C and D = 2/3 of 1/2 = 1/3
Therefore, now remaining property =
Total share of E,F,G and H
= value of (A) - total share of B,C and D
= 1/2 - 1/3 = 1/6
Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x
1/4 = 1/24
Given 1/24 = Rs.30,000
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000
Total share of B, C and D = 1/3 x 7,20,000
Then their individual share amount = (1/3 x 7,20,000) / 3 =
Rs.80,000.
Hence D's share is Rs.80000.
First digit (100’s place) of a 3-digit number cannot be
zero, let us start with 1.If the first digit is 1, then the
possible last two digits to get the product 1 is
(1,1)Therefore, the number is 111.
If the first digit is 2, then the possible last two digits to
get the product 2 are (1,2) and (2,1).
Therefore, the numbers are 212 and 221.
Proceeding like this, we get,
First digit
Corresponding
3-digit numbers
1
111
2
212, 221
3
313, 331
4
414, 441,422
5
515, 551
6
616, 661,623,632
7
717, 771
8
818, 881,842,824
9
919, 991,933
Thus, the required number of such 3-digit numbers is 23
[(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) = ? / [(74 /
49)1/6]
a) 1 b) 724 c) 76 d) 74
Given expression is, [(74)/(493/2)](1/6) x
[(73 /(3434/3)](1/2) = ? / [(74 / 49)1/6]
= [(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) x [(74 /
49)1/6] = ?
= [(74)/(72)(3/2)](1/6) x [(73/(73)(4/3)](1/2) x
[(74/72)1/6]
= (74)/(73)](1/6) x [(73)/(74)](1/2) x [(74/72)1/6]
= (74-3)(1/6) x [(73-4)](1/2) x [(74-2)1/6
= (71)(1/6) x [(7-1)](1/2) x [(72)1/6
= (7)(1/6) x (7)(-1/2) x (7)1/3
= (7)(1/6 -1/2 + 1/3)
= (7)(2-6+4/12)
= (70 )(1/12) = 1
Jake can dig a well in 16 days. Paul can dig
the same well in 24 days. Jake, Paul and Hari
together dig the well in 8 days. Hari alone
can dig the well in
a. 96 days
b. 48 days
c. 32 days
d. 24 days
Answer : (b)
Simple one. Let the total work to be done be
48 meters. Now Jake can dig 3 mts, Paul can
dig 2 mts a day. Now all of them combined
dug in 8 days so per day they dug 48/8 = 6
mts. So Of these 8 mts, Hari capacity is 1 mt.
So he takes 48 /1 = 48 days to complete the
digging job.
Answer : (b)
Let the money with Mark and John are M
and J respectively.
Now M + J/2 = 75 & M/3 + J = 75
Solving we get M = 45, and J = 60.
Answer: (B)
Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 36
9(x - y) = 36
x - y = 4.
Answer : (c)
Lets say his age is X.
So years spent as student = 5X / 16 + 2,
as Husband = X/4 + 1,
as Politician = X/4,
and as Father = 3X/40.
Remaining 6 years (as grandfather) = X - { 5X/16 +
2 + X/4 + 1 + X/4 + 3X/40}
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X /
40}
6 = X - [13X / 16 + 3 + 3X / 40]
6 = X - [71X / 80 + 3]
9X/80 = 9
X = 80.
Hence the age of the man is 80 years.
Ans: D
40,000 x (3/4)^3=16875
Answer : (b)
(a) Since loga a = 1, so log10 10 = 1.
(b) log (2 + 3) = log 5 and log (2 x 3) = log 6
= log 2 + log 3
Hence log (2 + 3)≠ log (2 x 3)
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) =
log 1 + log 2 + log 3.
So, (b) is incorrect.
Answer: (B)
Using logarithm base switch rule
logb(c) = 1 / logc(b)
Given expression
= log60 3 + log60 4 + log60 5
= log60 (3 x 4 x 5)
= log60 60
=1
Answer: (C)
log 27 = 1.431
log (3^3 ) = 1.431
3 log 3 = 1.431
log 3 = 1.431 / 3 = 0.477
log 9 = log(3^2 ) = 2 log 3 = (2 x 0.477) = 0.954.
Answer: (a)
Required run rate = ( 282 - (3.2 x 10) ) / 40
= 250 / 40
= 6.25
Answer: (b)
Required average
= (67 x 2 + 35 x 2 + 6 x 3 )/(2 + 2 + 3)
= (134 + 70 + 187 )/ 7
= 222/7
= 31 5/7
Answer : (b)
Let there be x pupils in the class.
Total increase in marks =X/2
X / 2= (83 - 63)
X /2 = 20
X = 40.
Answer: (b)
Required average
= (55 x 50 + 60 x 55 + 45 x 60) / (55 + 60 + 45)
= (2750 + 3300 + 2700) / 160
= 8750 / 160
= 54.68
CI = X (1+ 4/100) ^ 2 – X = (676 X / 675 ) – X
= 51x /625
SI = X * 4 * 2 /100 = 2X/25
51X /625 – 2X/25 = 1
51X/ 625 – 50X/625 = 1
X = 625.
Answer : (b)
Given that, the ratio of variety1 and variety2 in each
container is 2:3 and 5:9
Quantity of variety1 in 1st container = 2/(2 + 3) = 2/5
(that is, 2 parts of 5 parts is variety1)
Quantity of variety1 in 2nd container = 5/(5 + 9) = 5/14.
Therefore, total quantity of variety1 in new container =
2/5 + 5/14 = (28 + 25)/70 = 53/70
Similarly, quantity of variety2 in two containers are
3/5 and 9/14 respectively.
And, the total quantity of variety2 in new container =
3/5 + 9/14 = (42 + 45)/70 = 87/70
Now, the ratio of variety1 and variety2 in new
container = (53/70) : (87/70) = 53:87
i.e., 53 parts of (53 + 87) parts are variety1 => 53 parts of
140 parts are variety1.
Since the total quantity in new container is 280 kg
then the kg of variety1 is (53/140) x 280 = 2x53 = 106.
Hence, the answer is 106kg
Answer : (a)
Given that, total number of students in A, B, and C is 30,
40 and 60 respectively.
And the respective ratio of boys and girls is 2:3, 3:7 and
4:11.
Number of boys in A = [[2/(3+2)] x 30 = 2/5 x 30 = 12
So, the remaining = number of girls in A = 30-12 = 18.
Number of boys in B = [[3/(3+7)] x 40 = 3/10 x 40 = 12
So, the remaining = number of girls in B = 40-12 = 28.
Number of boys in C = [[4/(4+11)] x 60 = 4/15 x 60 = 16
So, the remaining = number of girls in C = 60-16 = 44.
Therefore, total number of boys = 12 + 12 + 16 = 40
And, total number of girls = 18 + 28 + 44 = 90
Now, the required ratio = number of boys : number of
girls = 40:90 = 20:45
Answer (a) 3 4/9 km/hr
Let the speed of upstream be X km/hr.
Then, speed in downstream = 2X km/hr (since boat takes
twice as much as time to cover up than as to cover
down the same distance in running water).
Speed in still water = (2X+X)/2 km/hr. (formula 3)
= 3X/2 km/hr.
Given that, boat covers 10 1/3 km in 1 hour in still water.
Therefore, 3X/2 = 10 1/3
X = 62/9
So, speed in upstream = 62/9 km/hr.
And, speed in downstream = 2 x 62/9 = 124/9 km/hr
Hence, speed of the current = [(124/9 - 62/9)]/2 km/hr
= 62/9x2 = 34/9 = 3 4/9 km/hr
x----------------------------x