Quantitative Aptitude: Mixture and Alligation Questions Set 11
- Varsha covered 180 km distance in 10 hours. The first part of his journey she covered by bike, then he hired a auto. The speed of the bike and auto is 25 kmph and 15 kmph respectively. The ratio of the distances covered by the bike and the auto is which of the following?
7:51:53:109:11None of theseOption A
Average Speed of bike and auto both= 180/10 = 18kmph
15………………….25
…………18…………..
7…………………….3
The ratio of time taken by auto to bike = 7:3
The ratio of distances covered by auto to bike = 7 * 15 : 3 * 25 =7:5 - A mixture of rice is sold at Rs.3 per Kg. This mixture is formed by mixing the rice of Rs.2.10 per kg and Rs.2.52 per kg. What is the ratio of price of cheaper to costlier quality in the mixture if profit of 25% is being earned?
1:82:53:45:9None of theseOption B
S.P= x + 25*x/100 = 3;
x = 2.4
210………………….252
…………240…………..
12…………………….30
=2:5 - From the 50 l of a solution, 5 l of solution is taken out and after it, 5 l of water is added to the rest amount of solution. Again 5 l of solution and water is drawn out and it was replaced by 5 l of water. If this process is continued similarly for the third time, the amount of solution left after the third replacement is what?
38.50 l36.45 l40.50 l35.85 lNone of theseOption B
=50*45/50 *45/50 *45/50
=36.45 l - Wheat worth Rs. 126 per kg and Rs. 134 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 177 per kg, the price of third variety of whaet per kg will be which of the following?
224230245265None of theseOption A
126*1+134*1+2y=177*4
260+2y=708
2y=708-260=448
y=448/2 = 224 - One quantity of boron priced at Rs 9.30 per Kg is mixed with another quality at a certain rate in the ratio 8:7. If the mixture so formed be worth Rs 10 per Kg, what is the rate per kg of the second quality of aluminium?
9.5010.6010.8011.50None of theseOption C
Let the rate of second quality be Rs p per Kg.
9.30…..…………p
…………10………….
8………………………7
7p– 70 = 5.6
p = 10.80 - A bottle contains 50 l of milk. 10 l of this milk is taken out and replaced with water. This process is repeated twice. Find the amount of remaining milk in mixture?
128/5 l75 l125 6/3 l91 lNone of theseOption A
50 [1 – (10/50)]3 =128/5 l - A 144 l of mixture contains HCL and H2SO4 in the ratio 5 : 7. How much HCL need to be added to this mixture so that the new ratio is 23 : 21 respectively?
32 l25 l27 l29 lNone of theseOption A
=144 =5:7
=60 : 84
Now == 21 = 84
=23 = 92
=92-60 = 32 l - A bottle contains 30 l mixture of HCL and H2SO4 in the ratio of p:q respectively. When 10 l of the mixture is taken out and replaced it H2SO4, then the ratio becomes 2:3. Then what is the initial quantity of HCL in bottle?
30 l25 l18 l15 lNone of theseOption C
p+q =30 l
(p-10*p/p+q)/ (q-10*q/(p+q) + 10) = 2/3
2p-4/3q = 20
p =18 l - The ratio of Solution “X” and Solution “Y” in the jar is 3:2 when 10 l of the mixture is taken out and is replaced by the Solution “Y”, the ratio become 2:3. The total quantity of the mixture in the bar is ?
30 l25 l35 l40 l45 lOption A
Initial =3:2 ;
After replacement its = 2:3
2/3 = (1 – 10/p)
p=30 l - ‘P’ Liters of the mixture contains HCL and H2SO4 in the ratio 4:3. If 13 l of H2SO4 is added then the ratio becomes 1:1. Then what is the final quantity of H2SO4 in the mixture?
50 l52 l55 l58 l69 lOption B
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