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Class 12 Applied Maths Chapter 1 Solutions
Number, Quantification and Numerical Applications
EXERCISE- 1.2
Q.1 In what ratio must a grocer mix two varieties of pulses worth 85 per kg and 100 per kg respectively so as to get a mixture worth 92 per kg?
Ans. Cost of the first variety of pulses = ₹85 per kg
Cost of the second variety of pulses = ₹100 per kg
Cost of the mixture = ₹92 per kg
Using the allegation method, the formula is:
Ratio of the two varieties = (Price of second variety−Price of mixture)/(Price of mixture – Price of First variety)
Substitute the values:
Ratio = (100 – 92)/(92 – 85)
= 8/7
So, the ratio of the two varieties of pulses to be mixed is 8:7.
Q.2 The cost price of type A apples is 120 per kg and that of type B apples is 180 per kg. If both types of apples are mixed in the ratio 2 : 3 respectively, then find the price per kg of mixed apples.
Ans. Cost price of type A apples = ₹120 per kg
Cost price of type B apples = ₹180 per kg
Mixing ratio of type A to type B = 2:3
Ratio of the two varieties = (Price of second variety−Price of mixture)/(Price of mixture – Price of First variety)
Substitute the values:
2/3 = (180 – m)/(m – 120)
540 – 3m = 2m – 240
780 = 5m
m = 780/5
m = 156
The price per kg of the mixed apples is ₹156.
Q.3 In what ratio must rice at 45 per kg is mixed with rice at 60 per kg so that the mixture be worth 54 per kg?
Ans. Cost of the first variety of rice = ₹45 per kg
Cost of the second variety of rice = ₹60 per kg
Cost of the mixture = ₹54 per kg
Using the allegation method, the formula is:
Ratio of the two varieties = (Price of second variety−Price of mixture)/(Price of mixture – Price of First variety)
Substitute the values:
Ratio = (60 – 54)/(54 – 45)
= 2/3
So, the ratio of the two varieties of pulses to be mixed is 2:3.
Q.4 In what ratio must a grocer mix two varieties of tea worth 180 per kg and Rs.200 per kg respectively so that selling the mixture at 216.20, he may gain 15%?
Ans. The selling price of the mixture is ₹216.20, and the grocer gains 15% on the cost price.
Using Formula :
CP = (SP x 100)/(100 + 15)
= (216.20 x 100)/(100 + 15)
= 21620/115
= 188
Price of the first variety of tea = ₹180 per kg
Price of the second variety of tea = ₹200 per kg
Price of the mixture = ₹188 per kg
Using the allegation method, the formula is:
Ratio of the two varieties = (Price of second variety−Price of mixture)/(Price of mixture – Price of First variety)
Substitute the values:
Ratio = (200 – 188)/(188 – 180)
= 3/2
So, the ratio of the two varieties of pulses to be mixed is 3:2.
Q.5 A shopkeeper mixes two varieties of rice one costs ₹ 161 per kg and other costs Rs.179 per kg in the ratio 5: 4 respectively. He sells the mixed varieties at Rs.202.80 per kg. Find his profit percentage.
Ans. Cost price of the first variety of rice = ₹161 per kg
Cost price of the second variety of rice = ₹179 per kg
Ratio of mixing = 5:4
Ratio of the two varieties = (Price of second variety−Price of mixture)/(Price of mixture – Price of First variety)
Substitute the values:
5/4 = (179 – m)/(m – 161)
5m – 805 = 716 – 4m
9m = 1521
m = 1521/9
m = 169
So, the cost price of the mixture is ₹169 per kg.
Now, Cost price of mixture = Rs.169
Selling price of mixture = Rs.202.80
Profit = 202.80 – 169
= Rs.33.80
Profit% = 33.80/169 x 100
= 20%
Q.6 How many kg of sugar costing Rs.45 per kg must be mixed with 30 kg sugar costing Rs.35 per kg so that there may be a gain of 12% by selling the mixture at Rs.47.04 per kg ?
Ans. S.P of 1kg of mixture = Rs.47.04
Gain = 12%
Therefore,
C.P of 1kg of mixture = Rs. 100
= 47.04/100 + 12
= 4704/112
= Rs.42
Now, By allegation method
= (45 – 42)/(42 – 35) = 30 – x
= 3/7 = 30/x
= x = 70
Q.7 Three varieties A, B and C of rice are mixed together in the ratio 4:1:1 respectively. The cost price of rice B is 82 per kg and that of rice C is 90 per kg. If the price of the mixture is ₹ 94 per kg, then find the price per kg of rice A.
Ans. n1 = 4, m1 = Rs.x
n2 = 1, m2 = Rs.82
n3 = 1, m3 = Rs.90
m = 94
m = (m1n1 + m2n2 + m3n3)/(n1 + n2 + n3)
94 = (4x + 82 + 90)/(4 + 1 + 1)
(4x + 172)/6 = 94
4x + 172 = 94 x 6
4x + 172 = 564
4x = 564 – 172
4x = 392
x = 392/4
x = 98
Q.8 A vessel contains a mixture of two liquids X and Y in the ratio 3: 5. 8 litres of mixture are drawn off from the vessel and 8 litres of liquid X is filled in the vessel. If the ratio of liquids X and Y is now becomes 7: 10, how many litres of liquids X and Y were contained by the vessel initially?
Ans. Let,
3x be initial liquid of X
5x be initial liquid of Y
8 L of mixture drawn off
In Vessel x = 3x – 3 x 8/8
= 3x – 3
but we add 8 l in x = 3x – 3 + 8
= 3x + 5
In Vessel Y = 5x – 5 x 8/8
= 5x – 5
Now,
7/10 = (3x + 5)/(5x – 5)
7 (5x – 5) = 10 (3x + 5)
35x – 35 = 30x + 50
35x – 30x = 50 + 35
5x = 85
X = 17
Therefore,
Liquid of X = 3 x 17 = 51L
Liquid of Y = 5 x 17 = 85L
Q.9 Two vessels P and Q contain milk and water in the ratio 5: 3 and 13: 3 respectively. In what ratio mixtures from two vessels should be mixed to get a new mixture containing milk and water in the ratio 3: 1 respectively?
Ans. Quantity of milk in vessel P = 5/8
Quantity of milk in vessel Q = 13/16
Quantity of mixture of milk = 3/4
L.C.M = 8, 16, 4 = 16
Therefore,
Quantity of milk in vessel P = 10/16 ( C )
Quantity of milk in vessel Q= 13/16 ( d )
Quantity of milk in mixture = 12/16 ( m )
(d – m)/(m – c) = [(13/16) – (12/16)]/[(12/16) – (10/16)]
= (1/16)/(2/16) = 1/2
SO,
Required Ratio = 1: 2
Q.10 Two vessels A and B contain milk and water in the ratio 3:2 and 7: 13 respectively. They are mixed in the ratio 2:3. Find the ratio of milk and water in the final mixture.
Ans. Quantity of milk in vessel A = 3/5
Quantity of milk in vessel B = 7/20
Ratio of milk and water = 2/3
By using allegation method,
2/3 = [(7/20 – x)/(x – 3/5)]
2/3 = [(7 – 20x)/20]/[(5x – 3/5)]
2/3 = (7 – 20x)/(4(5x-3)
40x – 24 = 21 – 60x
100x = 45
X = 45/100
Quantity of milk (x) = 9/20
Therefore,
Milk/Water = 9:11
Q.11 A milkman has two cans. He first contained 75% milk and the rest water, whereas the second contained 50% milk and the rest of the water. How much mixture should he mix from each can to get 20 liters of the mixture, such that the ratio of milk and water is 5: 3 respectively?
Ans. Milk in first can (c) = 75% = 3/4
Milk in second can (d)= 50% = 1/2
Ratio of final mixture (m) = 5/8
(d – m)/(m – c) = [(1/2) – (5/8)]/[(5/8) – (3/4)]
= [(4-5)/8]/[(5-8)/8]
= (-1/8)/(-1/8)
Ratio = 1 : 1
Therefore, quantity of milk = 10L
quantity of water = 10L
Q.12 In what ratio must a person mix two sugar solutions of 30% and 50% concentration respectively so as to get a solution of 45% concentration?
Ans. Concentration of the first solution = 30%
Concentration of the second solution = 50%
Desired concentration of the mixture = 45%
Using the allegation rule :
Ratio of two solutions = Concentration of second solution−Concentration of mixture/Concentration of mixture-Concentration of first solution
Ratio = (50 – 45)/(45 – 30)
= 5/15
= 1/3
Therefore, the person must mix the two sugar solutions in a ratio of 1:3
Q.13 A shopkeeper has 1 quintal of wheat, part of which she sells at 18% gain and the rest at 28% gain. In total she gains 24%. Find the quantity of wheat sold at 18% and 28%.
Ans. Let, x be the sugar sold at 18% profit and
Y be the sugar sold at 28% profit
M = 24%
Now,
According to the question, by alligation method
= [(24/100) – (18/100)]/[(28/100) – (24/100)] = x/y
= (6/100)/(4/100) = x/y
= 3/2 = x/y
SO, Ratio = 3:2
Therefore,
Quantity of wheat sold at 18% profit = 2/5 x 100 = 40kg
Quantity of wheat sold at 28% profit = 3/6 x 100 = 60kg
Q.14 The average salary per worker of the entire staff of a polythene bag manufacturing unit including machine operator and labours is Rs.7500. The average salary per head of the machine operator is Rs.25000 and that of labours is Rs.6000. Find the number of workers in the unit if there are 3 machine operators.
Ans. Let, No. of labours be x
Given,
Average salary of labour ( c ) = Rs.6000
Average salary of machine operator ( d ) = Rs. 25000
Average mean ( m ) = Rs. 7500
Therefore,
By allegation method
= (25000 – 7500)/(7500 – 600) = x/3
= 17500/1500 = x/3
= 52500/1500 = x
= x = 35
No. of labours = 35
No. of total worker = 35 + 3 = 38
Q.15 900 g of sugar solution has 40% sugar in it. How much sugar should be added to make it 55% in the solution?
Ans. Let, total of sugar solution = 100%
Given,
In 900g of solution
Percentage of sugar = 40%
Mean ( m ) = 55%
According to the question
Let, x gm of sugar be added using allegation method
= (100 – 55)/(55 – 40) = 900/x
= 45/15 = 900/x
= x = 900 x 15/45
= x = 300g
Q.16 In what ratio water must be added in milk costing Rs.60 per litre, so that resulting mixture would be worth Rs.50 per litre?
Ans. C.P of 1 litre of water = 0
C.P of 1 litre of milk = 60
Mean Price = 50
By using allegation method,
d – m = 0 – 50 = -50
m – c = 50 – 60 = -10
Therefore,
Ratio of milk and water = 5 : 1
Q.17 A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated two more times. How much milk is now left the container?
Ans. X = 40L , Y = 4L , n = 3
Therefore,
X(1 – x/y)^n = 40 (1 – 4/40)^3
= 40 (40 – 4/40)^3
= 40 (36/40)^3
= 40 x 729/1000
= 40 x 7.29
= 29.16 L
Q.18 From a container containing 50 litres of milk, 8 litres of milk was taken out and replaced by water. This process is repeated two more times. How much milk is now left in the container ?
Ans. X = 50L , Y = 8L , n = 3
Therefore,
X (1 – y/x)^n = 50 (1 – 8/50)^3
= 50 (50 – 8/50)^3
= 50( 42/50)^3
= 50 x 9.261/15625
= 29.635 L
Q.19 From a container full of alcohol 6 litres of alcohol was drawn out and replaced by water. This process is repeated one more time. The ratio of the quantity of alcohol and water left in the container is 9:16. How many litres of alcohol did the container hold originally ?
Ans. Let, quantity of alcohol be x
Y = 6 , n = 2 , Ratio = 9:16
Final quantity of alcohol = 9/25
Therefore,
X (1 – y/x)^n = 9/25
= (1 – 6/x) = 3/5
= x (1 – 6/x) = 3/5x
= x – 6/x = 3/5
= 5x – 30 = 3x
= 5x – 3x = 30
= x = 15 L
FAQ’s related to Class 12 Applied Maths Chapter 12 on Number, Quantification and Numerical Applications:
Q.1 What is the primary focus of Chapter 12 in Applied Mathematics for Class 12?
Ans. The chapter primarily deals with numerical applications like time and work, pipes and cisterns, clocks, calendars, and quantitative estimation. It also includes topics such as numbers and numerical problems in real-life situations.
Q.2 What are the types of numbers covered in this chapter?
Ans. This chapter covers different types of numbers, including natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers.
Q.3 What is Quantification?
Ans. Quantification refers to the process of expressing a numerical value or a quantity. It is often used in mathematical modeling, calculations, and real-life applications like estimating cost, time, or distances.
Q.4 What is the importance of Pipes and Cisterns problems?
Ans. Pipes and cisterns problems are analogous to time and work problems. Here, pipes filling or emptying a cistern at different rates are analyzed. It helps in understanding flow rates and capacities.
These are a few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 13
In Class 12 Applied Maths chapter 1, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 12 Applied Maths Chapter 1 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 1, we delve deep into advanced mathematical concepts that are crucial for understanding.
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