trick
(r/10)*(r/10)
(20/10)*(20/10)
4% decrese
TOTAL NUMBER OF BOOKS 40+40=80
ISSUED BOOKS 30
REMAINING BOOKS 80-30=50
RATE=50/80=5/8
bogus, counterfeit, fake, false, illegitimate, sham, unreal
D. Vice President
Correct option is C)
Let be students are consider as child
Let the age of child added later be x years.
average age of 12 children =20 years
∴ Total age of 12 children =20×12=240 years
After one more child is added-
Average of 13 children =20−1=19 years
Average age of 13 children =
13
sum of 12 children+x
19=
13
240+x
⇒240+x=247
⇒x=247−240=7 years
Hence the age of child added later is 7 years.
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
1:2:3 or 1:2:4
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
562.
35 hours
Yes answer is 18 days
Question
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.