According to the question:
⇒(x+5)(y−10)=300
⇒xy+5y−10x−50=xy
⇒5×300−10x−50=0
⇒−150+x2+5x=0
⇒x2+15x−10x−150=0
⇒x(x+15)−10(x+15)=0
⇒x=10 or −15
Price cannot be negative
So, As x>0,x=10.
Y R G
B B G
Y G Y
Its 2 times faster than the other train
v1*t=v2
v2*t=4*v1
solving these two,we get
v2/v1=2
4 days
as i assume the time taken to complete the job is not to be
changed while increasing the employees as it does not
clearly states so……….
30 men wrk for 9 hrs
therefore for the work to be done if no of workers is
increases the no of hrs / day to work should decrease so
here s a inverse relation
solving further we get
30 = K /9 ==> K= 270
now
40 = 270/y ==> y= 6.75
therefore ans is d)none
Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .
Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South
Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles
Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles
Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south
68 is the least number when divided by 8, 12, 18 and 24 leaves the remainders 4, 8, 14 and 20
Let the total number of matches to played in the tournament be ‘x’.
Given that A county cricket team has won 10 matches and lost 4.
That means total number of matches played = 14.
So,
= > 70% of x = 14
= > (70/100) * x = 14
= > 70x = 1400
= > x = 20.
So, the number of matches to be played = 20.
They have a record of exactly 75% wins = 20 * 75/100
= > 15.
Therefore, the number of matches should the team win = 15 – 10 = 5.
Hope this helps!
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
Let numbers be 2x, 3x
Given: (2x)^3 + (3x)^3 = 945
8x^3 + 27x^3 = 945
x = 3
Difference between numbers = 3x – 2x = x
Hence difference = 3
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.
444km
Since the speed diff is 5km/h
It takes 12 hrs to cover 60 extra kms
Therefore in 12 hrs the total dist covered is
Relative speed×time=(21+16)×12=444 kms