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Averages, Mixtures and Alligation questions for IPMAT, with solutions

10 free Averages, Mixtures and Alligation practice questions for IPMAT Indore, Rohtak and JIPMAT, 3 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Averages, Mixtures and Alligation on Learn.

1. Question 1Multiple choiceAsked in IPMAT Rohtak 2020Averages

   The height of nineteen people of comic book is in Arithmetic progression. The average height of them is $19$ feet. If the tallest is $37$ feet. Then what is the height of the shortest?

   1. A.$2$
   2. B.$1$
   3. C.$3$
   4. D.$4$
   Show the answer and solution

   Answer: B. $1$

   In an AP the average equals the middle term, and the terms are symmetric about it. The tallest is $37-19=18$ feet above the average. So the shortest is $18$ feet below it: $19-18=1$ foot. Equivalently, shortest $=2\times19-37=1$.

2. Question 2Multiple choiceAsked in IPMAT Rohtak 2020Averages

   The average spending of Lata Mangeshkar for the part of week from Monday to Saturday is Rs. $4200$. She spent Rs. $1200$ on Monday and Rs. $1500$ on Sunday that week. The average spending for the part of week from Tuesday to Sunday that week is?

   1. A.$4100$
   2. B.$4200$
   3. C.$4250$
   4. D.$4300$
   Show the answer and solution

   Answer: C. $4250$

   Monday–Saturday total $=6\times4200=25200$. Tuesday–Sunday total drops Monday's $1200$ and adds Sunday's $1500$: $25200-1200+1500=25500$. Average over these $6$ days $=\frac{25500}{6}=4250$.

3. Question 3Type the answerWritten by IPMATwayMixtures and Alligation

   How many litres of a $20\%$ acid solution must be added to $30$ litres of a $50\%$ acid solution to obtain a $30\%$ acid solution?

   Show the answer and solution

   Answer: 60

   Let $x$ litres of the $20\%$ solution be added. Acid: $0.2x+0.5\times30=0.3(x+30)$. So $0.2x+15=0.3x+9$. Then $0.1x=6$ and $x=60$ litres. Check: $12+15=27$ litres of acid in $90$ litres is $30\%$.

4. Question 4Type the answerWritten by IPMATwayMixtures and Alligation

   Alloy X has gold and silver in the ratio $2:3$; alloy Y has gold and silver in the ratio $3:7$. How many kg of Y must be melted with $10$ kg of X to obtain an alloy that is $35\%$ gold?

   Show the answer and solution

   Answer: 10

   X is $\frac25=40\%$ gold and Y is $\frac3{10}=30\%$ gold. Let $y$ kg of Y be used with $10$ kg of X. Gold: $0.4\times10+0.3y=0.35(10+y)$. So $4+0.3y=3.5+0.35y$, $0.05y=0.5$ and $y=10$ kg.

5. Question 5Type the answerWritten by IPMATwayMixtures and Alligation

   A tank holds $50$ litres of pure alcohol. $10$ litres are taken out and replaced with water, and the process is repeated once more. How many litres of water does the tank now contain?

   Show the answer and solution

   Answer: 18

   Each replacement removes $\frac{10}{50}=\frac15$ of the liquid, so $\frac45$ of the alcohol stays. After the first: $50\times\frac45=40$ litres of alcohol. After the second: $40\times\frac45=32$ litres. The tank still holds $50$ litres, so water $=50-32=18$ litres.

6. Question 6Type the answerWritten by IPMATwayMax and Min under an Average

   The sum of $10$ distinct positive integers is $100$. What is the largest possible value of the greatest of them?

   Show the answer and solution

   Answer: 55

   To make the greatest integer as large as possible, make the other nine as small as possible. Nine distinct positive integers are at least $1,2,\dots,9$, whose sum is $45$. The greatest is then $100-45=55$.

7. Question 7Multiple choiceAsked in IPMAT Rohtak 2020Averages

   Imagine that there are four hundred and fifty nine managers in a hotel. If the number of managers is enlarged by $36$, the expenses of the food increased by Rs. eighty one per day while the average spending per head decreased by one. Find the original spending of the hotel on food?

   1. A.$7344$
   2. B.$7244$
   3. C.$7044$
   4. D.$7444$
   Show the answer and solution

   Answer: A. $7344$

   Let the original spending per head be $y$ rupees, so the original total is $459y$. After adding $36$ managers there are $495$ heads, each costing $y-1$, and the total is $81$ more. So $459y+81=495(y-1)=495y-495$. Then $36y=576$, so $y=16$. The original spending is $459\times16=7344$.

8. Question 8Type the answerWritten by IPMATwayMax and Min under an Average

   The average age of $20$ students is $24$ years. No student is younger than $18$ or older than $30$. At most how many students can be $30$ years old?

   Show the answer and solution

   Answer: 10

   Total age $480$. Making $k$ students 30 and the rest as young as allowed (18) uses $30k+18(20-k)$ years, which must not exceed 480: $k\le10$. With $k=10$ the other ten are exactly 18.

9. Question 9Type the answerWritten by IPMATwayMax and Min under an Average

   A set of $5$ distinct positive integers has mean $20$ and median $22$. What is the largest possible value of the greatest integer in the set?

   Show the answer and solution

   Answer: 52

   The middle number is 22. Keep the two below it as small as possible (1 and 2) and the one above it just above (23). The largest is $100-(1+2+22+23)=52$.

10. Question 10Multiple choiceWritten by IPMATwayMixtures and Alligation

    A $60$-litre mixture of milk and water is in the ratio $7:3$. $20$ litres of the mixture are removed and replaced with $20$ litres of water. What is the new ratio of milk to water?

    1. A.$21:19$
    2. B.$7:8$
    3. C.$11:9$
    4. D.$8:7$
    Show the answer and solution

    Answer: B. $7:8$

    Initially there are $42$ L milk and $18$ L water. The removed $20$ L has the same ratio $7:3$, so it takes out $14$ L milk and $6$ L water, leaving $28$ L milk and $12$ L water. Adding $20$ L water gives $28$ L milk and $32$ L water, a ratio of $28:32=7:8$. Adding water without removing gives $42:38=21:19$, and removing milk only gives $22:18=11:9$. Treating the removed 20 L as half milk and half water leaves $32:8$ and, after adding water, $32:28=8:7$.