Brass is an alloy of Copper and Zinc . Bronze is an alloy containing 80% copper , 4% Zinc and 16% 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:
a. 64% and 36% b. 33% and 67% c. 50% and 75% d. None of these
given answer is a
2)if sum of the roots and the product of the roots of a quadratic equation S are in the ratio of 2:1 then which of the following is true
a) f(S)the answer is e)
pls explain in detail with steps
my take is option (d) what is the correct answer...
best possible way to do such kind of question is assume 2 values...
a=10 and b=20...and try to take bigger value but easy one...or i would say the one's which end with zeros because they will make calculation easy........put them in different options and you will get your answer...
correct me if i am wrong
the given answer is c
Brass is an alloy of Copper and Zinc . Bronze is an alloy containing 80% copper , 4% Zinc and 16% 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:
a. 64% and 36% b. 33% and 67% c. 50% and 75% d. None of these
given answer is a
answer is correct actually i think you also did the same mistake which i did in my earlier attempt to solve this question....
here main catch is the fact....16% of tin
now assume bronze as 100 gram/kg...
this means we have 80g-cu,4g-zn,16g-tin
and don't assume brass quantity....
10% of X=16...(here X=mixture of brass and bronze)
this will come X=160 gram...
this means we have 100g of bronze and just 60 gram of brass....
and then solve it furthur you will get your answer...
hope this will help...
and good going....don't stop and keep the hard work going
Brass is an alloy of Copper and Zinc . Bronze is an alloy containing 80% copper , 4% Zinc and 16% 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:
a. 64% and 36% b. 33% and 67% c. 50% and 75% d. None of these
given answer is a
2)if sum of the roots and the product of the roots of a quadratic equation S are in the ratio of 2:1 then which of the following is true
a) f(S)the answer is e)
pls explain in detail with steps
1. Let brass and bronze be mixed as x:y
=> 10gm of tin must have come from bronze which means amount of bronze in the mixture is:
10 = 16*y
y= (10/16)
=> (10/16)*4 + x*z = 16
=> x*z=13.5
=> (10/16)*80 + x*c = 74
=> x*c=24
=> c:z=240:135=48:27=16:9
=> which is 64:36
2. ax^2 + bx + c= 0 is the quadratic eq say
=> Sum of roots = -b/a and product of roots = c/a
Given,
-b/c = 2/1
=> b = -2c
We cant determine anything about a..so none of these are true.
if a and b are positive integers then 2 always lies between
a) (a+b)/(a-b) and ab
b) a/b and (a+2b)/(a+b)
c) a and b
d)ab/(a+b) and (a-b)/ab
e)a+b and ab
pls explain wit steps
kittu12 Saysthe given answer is c
answer to this question (C) is not possible from my side...i think there must be some mistake....
A contractor employees 200 men to build a bund.they finish 5/6th of the work in 10 weeks.then rain sets in and not only does the work remain suspended for 4 weeks but also half of the work already done is washed away.after the rain when the work is resumed,only 140 men turned up.the total time in which the contractor is able to complete the work assuming that there are no further disruptions in the schedule is: a)25weeks b)26 weeks c) 24 weeks d)20 weeks e)none
the answer is : c
pls explain in detail with stepsss!!!!!!!!!
A contractor employees 200 men to build a bund.they finish 5/6th of the work in 10 weeks.then rain sets in and not only does the work remain suspended for 4 weeks but also half of the work already done is washed away.after the rain when the work is resumed,only 140 men turned up.the total time in which the contractor is able to complete the work assuming that there are no further disruptions in the schedule is: a)25weeks b)26 weeks c) 24 weeks d)20 weeks e)none
the answer is : c
pls explain in detail with stepsss!!!!!!!!!
work = w
work done in 14 weeks = 5/12
7/12 work left
now 200 men do 5/6 work in 10 days
140 men -- 10*200/140*7/12*6/5 = 10 days
24 total days
A contractor employees 200 men to build a bund.they finish 5/6th of the work in 10 weeks.then rain sets in and not only does the work remain suspended for 4 weeks but also half of the work already done is washed away.after the rain when the work is resumed,only 140 men turned up.the total time in which the contractor is able to complete the work assuming that there are no further disruptions in the schedule is: a)25weeks b)26 weeks c) 24 weeks d)20 weeks e)none
the answer is : c
pls explain in detail with stepsss!!!!!!!!!
look...let there was W work is been there...
then in 10 weeks
200*10=(5/6)W
this much amount of work was done,....
now as half the completed work has been washed away this means they have to complete=(7/12)W
{half of 5/6=5/12....hence from whole work only 5/12 was completed....}
now we have 140 men...time taken be T weeks
140*T=7/12 and 200*10 weeks=5/6
now you can solve this...you will get T=10..
add 4weeks to that of rain...and 10weeks of earlier time...
A contractor employees 200 men to build a bund.they finish 5/6th of the work in 10 weeks.then rain sets in and not only does the work remain suspended for 4 weeks but also half of the work already done is washed away.after the rain when the work is resumed,only 140 men turned up.the total time in which the contractor is able to complete the work assuming that there are no further disruptions in the schedule is: a)25weeks b)26 weeks c) 24 weeks d)20 weeks e)none
the answer is : c
pls explain in detail with stepsss!!!!!!!!!
200 men need 10 weeks for 5 units of work.
=> For 1 unit they will need 2 weeks.
Now, 1 unit of work is left and 2.5 units of work is wasted.
So, 3.5 units of work to be done.
=> 3.5 * (200*2) men-weeks needed
=> 3.5 * (200 * 2) / 140 = 10 weeks needed
So, total 10+4+10 = 24 weeks needed
1. Let brass and bronze be mixed as x:y
=> 10gm of tin must have come from bronze which means amount of bronze in the mixture is:
10 = 16*y
y= (10/16)
=> (10/16)*4 + x*z = 16
=> x*z=13.5
=> (10/16)*80 + x*c = 74
=> x*c=24
=> c:z=240:135=48:27=16:9
=> which is 64:36
hai..im unable to understand these steps:
(10/16)*4 + x*z = 16
=> x*z=13.5
=> (10/16)*80 + x*c = 74
=> x*c=24
can u pls elaborate it..thanks in advancee
200 men need 10 weeks for 5 units of work.
=> For 1 unit they will need 2 weeks.
Now, 1 unit of work is left and 2.5 units of work is wasted.
So, 3.5 units of work to be done.
=> 3.5 * (200*2) men-weeks needed
=> 3.5 * (200 * 2) / 140 = 10 weeks needed
So, total 10+4+10 = 24 weeks needed
hai dude im unable to understand tis step.can u pls elaborate
3.5 * (200*2) men-weeks needed
=> 3.5 * (200 * 2) / 140 = 10 weeks needed
hai dude im unable to understand tis step.can u pls elaborate
3.5 * (200*2) men-weeks needed
=> 3.5 * (200 * 2) / 140 = 10 weeks needed
See.. for 3.5 units of work we will need 200 men for 3.5*2 weeks
=> 200*3.5*2 is total amont of work being done
Now, we do not have 200 men and have 140 men.
So, divide by 140
=> 200*3.5*2 / 140 = 10
See.. for 3.5 units of work we will need 200 men for 3.5*2 weeks
=> 200*3.5*2 is total amont of work being done
Now, we do not have 200 men and have 140 men.
So, divide by 140
=> 200*3.5*2 / 140 = 10
thanks alot yaar got it!!!!!!!
if a and b are positive integers then 2 always lies between
a) (a+b)/(a-b) and ab
b) a/b and (a+2b)/(a+b)
c) a and b
d)ab/(a+b) and (a-b)/ab
e)a+b and ab
pls explain wit steps
Let a = 2, b=1
option A. 3 and 2 _/2 =1.41 Ruled out
B. 2 and 4/3 = 1.33
c. 2 and 1
d. 2/3 & 1/2 ruled out
e. 3 &2 ruled out
So now option B &C;
Let a =60 b = 10
in this case option C ruled out So my take option B
Brass is an alloy of Copper and Zinc . Bronze is an alloy containing 80% copper , 4% Zinc and 16% 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:
a. 64% and 36% b. 33% and 67% c. 50% and 75% d. None of these
given answer is a
For Tin.....
16 .....................0
.........10
10.................... 3
5:3 => they are added in 5:3 ratio
80................x
...........74
74-x...........6
74-x/6 = 5/3
solving 64%
similarily other one 36% so my take option A
A contractor employees 200 men to build a bund.they finish 5/6th of the work in 10 weeks.then rain sets in and not only does the work remain suspended for 4 weeks but also half of the work already done is washed away.after the rain when the work is resumed,only 140 men turned up.the total time in which the contractor is able to complete the work assuming that there are no further disruptions in the schedule is: a)25weeks b)26 weeks c) 24 weeks d)20 weeks e)none
the answer is : c
pls explain in detail with stepsss!!!!!!!!!
total work-week required : 2000/5*6 =2400 man-week
5/6*1/2 = 5/12 work has been completed in 10 week +4 week
rest work 7/12 work to be done
man =140
2000*7/12 = 1*140*W
W =10
So total time =10+10+4 =24
tuliram runs in a triathlon consisting of 3 phases in the following manner.running 12 km,cycling 24 kkm and swimming 5 km.his speeds in the 3 phases are in the ratio 2:6:1.he completes the race in n minutes.later he changes his stratgey so that the distances he covers in each phase are constant but his speeds are now in the ratio 3:8:1.the end result is that he completes the race taking 20 mins more than the earlier speed.it is also known that he has not changed his running speed when he changes his strategy.
1)what is his intial speed? ans:0.5km/min
2)if his speeds are in the ratio 1:3:1,with the running time remaining unchanged what is his finishing time ans:250/3 min
3)what is tuliram's original speed of running? 18kmph
pls explainin detail with steps..thanks in advance
A number is such that when divided by 4,5,6, or 7 it leaves the remainder 2,3,4, or 5 resp. Which is the largets number below 4000 that satisfies this property.
1. 3352
2. 3378
3. 3898
4. 2938
A number is such that when divided by 4,5,6, or 7 it leaves the remainder 2,3,4, or 5 resp. Which is the largets number below 4000 that satisfies this property.
1. 3352
2. 3378
3. 3898
4. 2938
Ans : LCM of 4 5 6 7 = 420...SO largest multiple of 420 below 4000 = 420 x 9 =3780. Since the difference between all the divisors and remainders are same which is equal to 2 final answer is 3780 - 2 =3778. You sure about the options dude?....
A number is such that when divided by 4,5,6, or 7 it leaves the remainder 2,3,4, or 5 resp. Which is the largets number below 4000 that satisfies this property.
1. 3352
2. 3378
3. 3898
4. 2938
dude best possible way to do this kind of questions is by using options...they will give you sure shot answer and that also in quick time.....
p.s.-condition apply