Quant by Arun Sharma

MBA CAT 2024
6026
@Raghu5jan said:
Facing prob in Averages chapt lod 2 question 1 With an average speed of 40km/h a train reaches its destination in time.. If it goes with an average speed of 35km/h it is late by 15mint..The length of the total journey is?????
D = 40 *t
D = 35 ( t + 1/4)

so
40t = 35t + 35/4
t = 7/4
D = 70km ?
6027

dude can u explain me the D=35(t+1/4) part how u got it??

6028

Yes dude i hav got it..thanks yes ur answer is right

6029

Ratio of speed will be inversly proportional to time take. therefor
40/35 = (t+1/4)/t
t= 7/4
distance is 40*7/4 = 70

6030

Average chap arun sharma lod 2 sum number 5

One-fourth of a certain journey is covered at the rate of 25km/h,one third at the rate of 30km/h and the rest at 50km/h..Find the average speed of the whole journey.

6031

Average speed = total distance/total time
Assume the distance be 12. Since 1/4, 1/3 and 5/12 are distances to be covered.
now 12/(3/25 + 4/30 + 5/50)
Solving this will get 1800/53.

6032

sonic why did u assume distance as 12?

6033
@Raghu5jan said:
sonic why did u assume distance as 12?
Since the distances covered are 1/4, 1/3 and remaining 5/12. So if u take the LCM of denominators then u will get 12. And the final distance come out to be integers. Distances are 12/4 = 3, 12/3=4, and 5*12/12=5. Now the calculation will be easier.

6034

dude i din get ths equation 12/(3/25+4/30+5/50)

6035
@Raghu5jan said:
dude i din get ths equation 12/(3/25+4/30+5/50)
this equation is avg speed = total distance/total time
total distance is 12
time taken when traveling at 25kmph is distance/speed = (1*12/4)/25 = 3/25
likewise when traveling at 30kmph, time is (1*12/4)/3 = 4/30
similarly 5/50
adding all the time gives total time = 3/25 + 4/30 + 5/50
6036

plz help
Q ) The cost of an article(which is composed of raw materials and wages) was 3 times the value of the raw materials used. The cost of materials increased in the ratio 3:7 and wages increased in the ratio 4:9. Find the present cost of the article if its original cost was Rs. 18?

a) Rs. 41 b) Rs. 30 c) Rs.40 d) Rs. 46

6037
@ramirez said:
plz help Q ) The cost of an article(which is composed of raw materials and wages) was 3 times the value of the raw materials used. The cost of materials increased in the ratio 3:7 and wages increased in the ratio 4:9. Find the present cost of the article if its original cost was Rs. 18?a) Rs. 41 b) Rs. 30 c) Rs.40 d) Rs. 46
Is the answer 41 ?
6038
@Raghu5jan 40*t = 35*(t+15) logic is the distance travelled in two cases is equal and t+15 is the time because in the second case it is late by 15 mins.
solving we will get t=105 mins
now d= 40*t
=40*105/60 (converting 105 mins in hrs since speed 40 is in km/hrs)
=70km
6039
@sonic_boom explain the method plz : )
6040
@ramirez said:
@sonic_boom explain the method plz : )
Initial cost is 18 which is 3 times of raw material, that means raw material is 6 and wages is 12.
now initial raw material : final raw material = 3:7
=>final raw material is 14
like wise final wages is 27
so final value of article is 14+27 = 41
I am not sure if the ans is correct.
6041
@sonic_boom thank you : )
6042
@sonic_boom your right the ans is correct thanks : )
6043

In two a alloys, aluminium and iron are in the ratios of 4:1 and 1:3. After alloying together 10 kg of the first alloy, 16 kg of the second and several kilograms of aluminum, an alloy was obtained in which the ratio of aluminium Anne iron was 3:2. Find the weight of the new alloy.

a) 15 b) 35 c) 65 d) 95. Ans is 35. Plz explain.. : )
6044

OK guys got the answer in the first mixture amounts of Al =4/5*10=8kg and amount of Fe = 1/5 *10=2kg. Similarly in the second mix Al =1/4*16=4 kg and Fe=3/4*16=12kg..

Now after mixing them new ratio of Al/Fe=3/2 (given)
Therefore this ratio is nothing but ratio of sum of the two quantities
3/2=sum of aluminum/ sum of iron
3/2=8+4+x/2+12
x=9 now
Total amount of mixture = sum of Al + sum of Fe
=8+4+9+2+12
=35
6045
@ramirez said:
In two a alloys, aluminium and iron are in the ratios of 4:1 and 1:3. After alloying together 10 kg of the first alloy, 16 kg of the second and several kilograms of aluminum, an alloy was obtained in which the ratio of aluminium Anne iron was 3:2. Find the weight of the new alloy.
a) 15 b) 35 c) 65 d) 95. Ans is 35. Plz explain.. : )
Aluminium in 10 kg Alloy1 = 10*4/5 = 8, so iron = 2 kg
Aluminium in 16kg of Alloy2 = 16*1/4 = 4, so iron = 12kg
let x kg of aluminium is mixed, so total aluminium = 8+4+x=12+x
final ratio of alumin to iron = 3/2
=>(12+x)/14 = 3/2
x = 9
final weight = 10+16+9=35kg