The work done by A in 8 days is = 8/ 12 = 2/3
Means A alone completes 2/3 part of work.
Remaining work which is (1–2/3) = 1/3 is completed by B in 8–2 = 6 days
So the complete work done by B in 6/(1/3) = 18 days.
B alone can complete the work in 18 days.
First write equations from info:
(A) (Mon + Tue + Wed)/3 = 111 Rearrange as ——–> Tue + Wed = 111 – Mon
(B) (Tue + Wed + Thu)/3 =102 Rearrange as ——–> Tue + Wed = 102 – Thu
(C) Thu = 0.8(Mon)
Substitute equation C into B:
(B) Tue + Wed = 102 – 0.8(Mon)
At this point I changed the values for clearer algebra:
Mon = x
Tue + Wed = y
Re-write equations A & B with new values:
(A) y = 111 – x
(B) y = 102 – 0.8x
Solve simultaneous equations:
111 – x = 102 – 0.8x
111 – 102 = x – 0.8x (Re-arraged)
9 = 0.2x
x = 45
Thus, Mon = 45C
Thu = 0.8(45)
Thu = 36C
So the answer is it was 36C on Thursday
40
R
T,P,R,Q,S
(3/4)*(L/Sp)=30
L/Sq=75
(3/4)*(Sq/Sp)=30/75
Sq/Sp=8/15
Sp/Sq=15/8
6,36
let the amount of money be x
cloths 1/3 X x=rs.x/3
balance = x- x/3 = 2x/3
food = 1/5 X 2x/3 = 2x/15
balance = 2x/3 – 2x/15= 8x/15
travel = 1/4 X 8x/15 = 2x/15
now he has 100 rupees
2x/5 = 100
2x= 500
x = 500/2
x = 250
let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
the ratio of present age of x:y is 5:6
And,we have to find the present age of x.
so,the present age of X is definitely will be multiple of 5.
X=35
so,35 is the answer.
1 : 2
speed = 72 Kmph
Speed= 72 * 1000 / (60*60) m/s
speed= 72 * 5 / 18
speed= 20 m/s
time = 30 s
distance = 600 m
here the 30% of 100 is 30
and 10% of the 30 = 3
so 20% = 6%
so ans = 6%
20
(60/100)*60 + (40/100)*40
=52%