In a cube all the diagonal and sides are equal, we can go diagonally.
40
414 Rs
1.2*37.8=45.36=46 sq.feet
46*9=414 RS
let x be lenghth of candles.
thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.
let he lit for n hrs.
so in n hrs thicker one goes n/6.
and thinner one goes n/4.
so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.
a = 2
d = 5 – 2 = 3
Then 12th term is :
This in arithmetic progression
L = a + (12 – 1) d
L = 2 + 11 * 3
L = 2 + 33 = 35
first fill 3quart pail and pour in 5quart pail.fill again 3
quart pail and fill remaining 2 quarts in 5 quart
pail.hence 1 quart remains in 3quart pail.empty 5quart pail
and pour 1 quart from 3 quart pail. also add another 3
quart from 3 quart pail. hence 1 + 3 quarts = 4 quarts.
8, 27, 64, 100, 125, 216, 343
100
500 = Total+50
Total(450) = only one paper(p) + 29+20+35 + all three (g)
285+212+127 = p + 2( 29+20+35 )+ 3g
solve above .. to get g = 45 …
( small corrctn .. i think .. questn shud be 20 read ONLY
hindu and
times of India and 29 read ONLY hindu and Indian express
and 35
read ONLY times of India and Indian express)
0
36,20
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
297
sorry the above ans was typed by mistake the correct ans is…..
1027.05 ? 314.005 + 112.25 = ?
ans:
1027.05 – 314.005 + 112.25 = 825.295
FASHION= FOIHSAN
It is F-ASHIO-N to F-OIHSA-N.
So PROBLEM will be P-ROBLE-M,
the answer is PELBORM.
Calculation:
⇒ If 1000 divided by 112, the remainder is 104. ⇒ 112 – 104 = 8 ⇒ If 8 is added to 1000 it will become the smallest four-digit number and a multiple of 112. ⇒ 1000 + 8 = 1008 ∴ The required result will be 1008.
Reliable Assistance: I promise to provide reliable and accurate assistance in answering questions, providing information, and engaging in meaningful conversations. I strive to deliver helpful and relevant responses to the best of my abilities, drawing from the vast knowledge and information I have been trained on.
Continuous Improvement: I promise to continuously improve my capabilities and expand my knowledge base. As technology evolves and new information becomes available, I aim to stay up-to-date and provide the most accurate and relevant information possible. I strive to enhance my understanding of different topics and improve the quality of my responses over time.
Respectful and Ethical Interaction: I promise to interact with users in a respectful and ethical manner. I am programmed to respect user privacy, confidentiality, and cultural sensitivities, aiming to provide a positive and helpful experience for those who engage with me.
Please note that while I can provide valuable assistance and information, there are certain tasks that may be better suited for human professionals with expertise in specific fields.
10%=394;50%=1970;250%=9850;30%=1182;
250%+30%=9850+1182=11032
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds