**1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?**

**Sol:** log 0.317=0.3332 and log 0.318=0.3364, then

log 0.319=log0.318+(log0.318-log0.317)** = 0.3396**

**2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?**

**Sol:** x= 2 kg Packs

y= 1 kg packs

x + y = 150 .......... Eqn 1

2x + y = 264 .......... Eqn 2

Solve the Simultaneous equation; x = 114

so, y = 36

**ANS : Number of 2 kg Packs = 114.**

3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs
later at a place with coordinates 36N70W. What is the local time when my
plane landed?

**Options:** a)6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00

**Sol:** The destination place is 80 degree west to the starting place.
Hence the time difference between these two places is 5 hour 20 min.
(=24hr*80/360).

When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).

**Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM**

**4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am
and takes 8 hours to reach the destination, find the local arrival time ?**

**Sol:** Since it is moving from east to west longitide we need to add both

ie,40+40=80

multiply the ans by 4

=>80*4=320min

convert this min to hours ie, 5hrs 33min

It takes 8hrs totally . So 8-5hr 30 min=2hr 30min

So the ans is 10am+2hr 30 min

=>**ans is 12:30 it will reach**

**5) The size of the bucket is N kb. The bucket fills at the rate of 0.1
kb per millisecond. A programmer sends a program to receiver. There it
waits for 10 milliseconds. And response will be back to programmer in 20
milliseconds. How much time the program takes to get a response back to
the programmer, after it is sent? answer with
explanation.**

**Sol:** see it doesn't matter that wat the time is being taken to fill the
bucket.after reaching program it waits there for 10ms and back to the
programmer in 20 ms.then total time to get the response is 20ms +10
ms=30ms...it's so simple....

6**) A fisherman's day is rated as good if he catches 9 fishes, fair if 7
fishes and bad if 5 fishes. He catches 53 fishes in a week n had all
good, fair n bad days in the week. So how many good, fair n bad days did
the fisher man had in the week**

**Ans: **4 good, 1 fair n 2 bad days

**Sol:** Go to river catch fish

4*9=36

7*1=7

2*5=10

36+7+10=53...

take what is given 53

good days means --- 9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days.

fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.

bad days means -------5 fishes so remaining 10---10/5=2days.

**Ans:** 4 good, 1 fair, 2bad. ==== total 7 days.

x+y+z=7--------- eq1

9*x+7*y+5*z=53 -------eq2

multiply eq 1 by 9,

9*x+9*y+9*z=35 -------------eq3

from eq2 and eq3

2*y+4*z=10-----eq4

since all x,y and z are integer i sud put a integer value of y such that
z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z
= 2 or 1 from eq 4

for first y=1,z=2 then from eq1 x= 4

so 9*4+1*7+2*5=53.... satisfied

now for second y=3 z=1 then from eq1 x=3

so 9*3+3*7+1*5=53 ......satisfied

so finally there are two solution of this question

(x,y,z)=(4,1,2) and (3,3,1)...

**7) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?**

**Sol:** Let no. of fish x catches=p

no. caught by y =r

r=5p.

r+p=42

then p=7,r=35

**8) Three companies are working independently and receiving the savings
20%, 30%, 40%. If the companies work combinely, what will be their net
savings?**

**Sol: **suppose total income is 100

so amount x is getting is 80

y is 70

z =60

total=210

but total money is 300

300-210=90

so they are getting 90 rs less

90 is 30% of 300 so they r getting 30% discount

**9) The ratio of incomes of C and D is 3:4.the ratio of their
expenditures is 4:5. Find the ratio of their savings if the savings of C
is one fourths of his income?**

**Sol:** incomes:3:4

expenditures:4:5

3x-4y=1/4(3x)

12x-16y=3x

9x=16y

y=9x/16

(3x-4(9x/16))/((4x-5(9x/16)))

**ans:**12/19

**10) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?**

**Ans:** -1

bcoz g(2)=g(1)-g(0)=1+1=2

g(3)=1

g(4)=-1

g(5)=-2

g(6)=-1

Read more ...