Q25 . Mini and Vinay are quiz masters preparing for a quiz. In x min, Mini makes y questions more than Vinay. If it were possible to reduce the time needed by each to make a question by two min, then in x min. Mini would makes 2y questions more than Vinay. How many questions does Mini makes in x min?
a. 1/4[2(x+y)-(2x^2 + 4y^2)^2]
b. 1/4[2(x-y)-(2x^2 + 4y^2)^2]
c. Either a or b
d. 1/4[2(x+y)-(2x^2 -4y^2)^2]
@hemant89 said:Q2 . The duration of a railway journey varies as the distance and inversely as the velocity, the velocity varies directly as the square root of the quantity of coal used per kilometer , and inversely as the number of carriages in the train. In a journey of 50 km in half an hour with 18 carriages, 100 kg of coal is required . How much coal will be consumed in a journey of 42km in 28 min with 16 carriages.a. 53.76 b. 179.33 c. 47.51 d. 48.43 e. None of these
is it 53.76??
t duration, d distance, v velocity and k,k1 constant
t=d*k/v....(1)
and v=k1*rtcoal/n
substituting v in (1)
t=d*k*n/k1*rtcoal.... (2)
now 100kg is used for 50km.
hence coal used per km is 2.
substitute the given data in (2)..
u'll get k/k1 as 1/900rt2
then substitute this k/k1 also substitute 42km,28min and 16 and find coal. it comes out 1.28.
multiply this 1.28 with 42 to get coal used for the journey.
it comes out 53.76.
t duration, d distance, v velocity and k,k1 constant
t=d*k/v....(1)
and v=k1*rtcoal/n
substituting v in (1)
t=d*k*n/k1*rtcoal.... (2)
now 100kg is used for 50km.
hence coal used per km is 2.
substitute the given data in (2)..
u'll get k/k1 as 1/900rt2
then substitute this k/k1 also substitute 42km,28min and 16 and find coal. it comes out 1.28.
multiply this 1.28 with 42 to get coal used for the journey.
it comes out 53.76.
@hemant89 said:Q28 . A tank of 425 litres capacity has been filled with water through two pipes, the first pipe having been opened 5 hours longer than the secong. If the first pipe were open as long as the second pipe; if the two pipes were open simultaneously, the tank would be filled up in 17 hours. How long was the second pipe open? a. 10 b. 12 c. 15 d. 18
is it 15?
let me know if it's right, i'll post the solution then. :)
let me know if it's right, i'll post the solution then. :)
@hemant89 said:Q28 . A tank of 425 litres capacity has been filled with water through two pipes, the first pipe having been opened 5 hours longer than the secong. If the first pipe were open as long as the second pipe; if the two pipes were open simultaneously, the tank would be filled up in 17 hours. How long was the second pipe open? a. 10 b. 12 c. 15 d. 18
c. 15
@hemant89 said:Q10 . Brass is an alloy of copper and zinc. Bronze is an alloy containing 80% of copper, 4% ofzinc and 16% of tin. A fused mass of brass and bronze is found to contain 74% of copper,16% of zinc and 10% of tin. The ratio of copper to zinc in brass is:64 and 3633 and 6750 and 7535 and 65None of these
brass=x(cu)+y(zn)+0(sb)
where cu=copper, zn=zinc and sb=tin
x and y are ratio of cu and zn
bronze= 80(cu)+4(zn)+16(sb)
now let brass and bronze is mized in l:m ratio.
so,
for cu=> x.l+80.m=74........(1)
for zn => y.l+ 4.m=16........(2)
and for tin=> m.16=10.......(3)
from eq (3) we get ,m= 10/16
substitute in (1) and (2)
u'll get,
x.l=24
y.l=13.5
divide them.
x:y = 16:9
multiply by 4 to scale it to 100.
u'll get 64:36 :)
where cu=copper, zn=zinc and sb=tin
x and y are ratio of cu and zn
bronze= 80(cu)+4(zn)+16(sb)
now let brass and bronze is mized in l:m ratio.
so,
for cu=> x.l+80.m=74........(1)
for zn => y.l+ 4.m=16........(2)
and for tin=> m.16=10.......(3)
from eq (3) we get ,m= 10/16
substitute in (1) and (2)
u'll get,
x.l=24
y.l=13.5
divide them.
x:y = 16:9
multiply by 4 to scale it to 100.
u'll get 64:36 :)
@ChirpiBird said:is it 15? let me know if it's right, i'll post the solution then.
yes it is. Solution please.:-)
@hemant89 said:yes it is. Solution please.
i did it by trial method.
first case it was give A takes 5 hours more than B.
and in second case they exchanged.
so B takes 5 more than A
the eq forms to be .. 17(A+B)=425
or A+B=25.
by options, B=15 and A=10
although for original case... A=20 and B=15.
if someone has a better explanation, plz update. :)
first case it was give A takes 5 hours more than B.
and in second case they exchanged.
so B takes 5 more than A
the eq forms to be .. 17(A+B)=425
or A+B=25.
by options, B=15 and A=10
although for original case... A=20 and B=15.
if someone has a better explanation, plz update. :)
Q60 . Without stoppage , a train travels at an average speed of 75kmph and with stoppage it covers the same distance at an average speed of 60kmph. How many min per hour does the train stop?
a. 10 min
b. 12 min
c. 15 min
d. 18 min
a. 10 min
b. 12 min
c. 15 min
d. 18 min
Can someone guide me with one question in Arun Sharma. The question goes :51^203 divided by 7. Find remainder.??
Please guide with whatever comes to ur mind.
Thanks in advance
@maanas.vyas said:Can someone guide me with one question in Arun Sharma. The question goes :51^203 divided by 7. Find remainder.??Please guide with whatever comes to ur mind.Thanks in advance
51/7 = 2
so it becomes 2^203
now 2^3/7 gives you remainder 1
so 2^201 gives 1
the remainder = 1*2*2 = 4
@hemant89 said:Q60 . Without stoppage , a train travels at an average speed of 75kmph and with stoppage it covers the same distance at an average speed of 60kmph. How many min per hour does the train stop? a. 10 min b. 12 min c. 15 min d. 18 min
take total distance = 300kms
without stoppage it will take 4hrs
with stoppage it will take 5hrs
so in 5 hrs it will stop for 1 hr
so 1/5
in 1 hr it will stop for 1/5*60 = 12mins
@hemant89
is the answer to question 60 - B
Solution (as i think..:P)
in 1 hour train covers (without stoppage):75 kms
to cover same distance with stoppages it takes: 75/60 i.e. 1.25 hrs.
i.e. 15 mins extra
now ,
in 1.25 (75mins) hrs it stops for 15 mins
so in 1hr (60 mins) it stops for: 15/75*60= 12 mins.
is the answer to question 60 - B
Solution (as i think..:P)
in 1 hour train covers (without stoppage):75 kms
to cover same distance with stoppages it takes: 75/60 i.e. 1.25 hrs.
i.e. 15 mins extra
now ,
in 1.25 (75mins) hrs it stops for 15 mins
so in 1hr (60 mins) it stops for: 15/75*60= 12 mins.
Q34 . A and B start a two-length swimming race at the same moment but from opposite ends of the pool. They swim in lane and at uniform speeds, but A is faster than B. They first pass at a point 18.5 m from the deep end and having completed one length each one is allowed to rest on the edge for exactly 45 sec. After setting off on the return length , the swimmer pass for the second time just 10.5 m from the shallow end. What is the length of pool?
55.5
45
66
49
55.5
45
66
49
sol34 :
Deep__18.5___.___d-18.5________Shallow
assumption, the slower person B started from the deep end as all the options show almost double of 18.5, which means the faster person must have covered more distance hence is at the shallow end.
speed of B be b
speed of A be a
18.5/ b=(d-18.5)/a
or a/b=(d-18.5)/18.5....eq 1
now for the second round
Deep__d-10.5___.___10.5________Shallow
d/a + 45 + (d-10.5)/a = d/b + 45 + 10.5/b
or 2d - 10.5= a/b (d+10.5)
put the value of a/b ffrom eq1
(2d - 10.5)*18.5= (d-18.5 )*(d+10.5)
or 37d - 10.5*18.5= d^2-8d -18.5*10.5
or d= 45.
Q15 . A supply of water last for 150 days if 12 gallons leak of every day, but only for 100 days if 15 gallons leak of f daily. What is the total quantity of water in the supply?
a. 900
b. 1125
c. 3350
d. 1250
a. 900
b. 1125
c. 3350
d. 1250
Q19 . 6 technicians working at the same rate completely work of one server in 2.5 hours. If they start at 11:00 am and one additional technician per hour being added beginning at 5:00 PM, at what time the server will be complete?
6:40 PM
7:00 PM
7:20 PM
8:00 PM
6:40 PM
7:00 PM
7:20 PM
8:00 PM
@lostanand For Q89:The super computer at Ram Mohan Roy Seminary takes an i/p of a no. N & a X where X is a factor of N. In a particular case N is equal to 83p796161q & X is equal to 11 where 0A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
ANSWER:
ANSWER:
X is a factor of N.
N=83p796161q. X=11
For N to be divisible by 11:
(q+6+6+7+3)-(1+1+9+p+8) should be 0 or a multiple of 11.
(q+6+6+7+3)-(1+1+9+p+8)=(22+q)-(19+p)
The following values satisfy the conditions:
p=1 and q=9
p+q=10 and p=1
Dividing 8317961619 by p+q and p
gives remainders: 9 and 0
9+0=9(Ans)