Mathematics Part i Solutions Solutions for Class 7 Math Chapter 4 Relation Of Parts are provided here with simple step-by-step explanations. These solutions for Relation Of Parts are extremely popular among class 7 students for Math Relation Of Parts Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 7 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.
Page No 47:
Question 1:
To make a mixture which can dissolve gold, nitric acid and hydrochloric acid are to be mixed in the ratio 3 : 1. With 12 milliliters of hydrochloric acid, how much nitric acid should be mixed?
Answer:
Ratio of nitric acid to hydrochloric acid in the mixture = 3 : 1
Quantity of hydrochloric acid = 12 millilitres
Quantity of nitric acid to be mixed = 
× quantity of hydrochloric acid
= 3 × 12 millilitres
= 36 millilitres
Thus, 36 millilitres of nitric acid should be mixed with 12 millilitres of hydrochloric acid to form the mixture.
Page No 47:
Question 2:
To build the walls of a house, cement and sand are to be mixed in the ratio 1 : 6. How many bowls of sand are to be mixed with three bowls of cement?
Answer:
Ratio of cement to sand = 1 : 6
∴ Ratio of sand to cement = 6 : 1
Number of bowls of cement = 3
Number of bowls of sand to be mixed = × number of bowls of cement = 6 × 3 = 18
Thus, 18 bowls of sand are to be mixed with 3 bowls of cement to build the wall of a house.
Page No 47:
Question 3:
To fuel autorikshaws, petrol and oil are mixed in the ratio 40 : 1. How many liters of oil is needed with 5 liters of petrol? And with 1 liter of petrol?
Answer:
Ratio of petrol to oil in the fuel = 40 : 1
∴ Ratio of oil to petrol in the fuel = 1 : 40
Quantity of petrol in the fuel = 5 litres
Quantity of oil needed for the fuel = × quantity of petrol in the fuel
= × 5 litres
= litre
= 0.125 litre
Thus, 0.125 litres of oil is needed with 5 litres of petrol.
If quantity of petrol in the fuel = 1 litre
Quantity of oil needed in the fuel = × 1 litre = litre = 0.025 litre
Thus, 0.025 litre of oil is needed with 1 litre of petrol.
Page No 50:
Question 1:
We have seen that to make idlis rice and urad must be in the ratio 2 : 1. If the total of rice and urad taken is 9 cups, how many cups of rice was taken? And how many cups of urad?
Answer:
Ratio of rice to urad in the mixture = 2 : 1
So, 3 (= 2 + 1) cups of the mixture contain 2 cups of rice and 1 cup of urad.
Total number of cups of rice-and-urad mixture = 9
Number of cups of rice taken = 9 × = 6
Number of cups of urad taken = 9 × = 3
Page No 50:
Question 2:
The population of Cholathadam Panchayat ward is 6321. The ratio of men to women here is 10 : 11. How many men are there in this Ppanchayat? And women?
Answer:
Ratio of men to women in the Cholathadam Panchayat = 10 : 11
So, if the panchayat had 21 members (= 10 + 11), there would be 10 men and 11 women.
Total number of members of the Cholathadam Panchayat = 6321
Number of men in the panchayat = 6321 × = 3010
Number of women in the panchayat = 6321 × = 3311
Page No 50:
Question 3:
A 30 centimeters long rope is used to make a rectangular plot. If the length to width ratio is 1 : 2, what is the area of the rectangle?
What if this ratio is 2 : 3?
Answer:
(i) Ratio of length to width of the rectangle = 1 : 2
Total length of the rope = 30 cm
⇒ Perimeter of rectangular plot = 30 cm
⇒ 2(l + b) = 30
⇒ l + b = 15
Let the length of the rectangular plot be x cm and breadth be 2x cm.
⇒ x + 2x = 15
⇒ x = 5
∴ Length of the rectangle = 5 cm
Width of the rectangle = 2 × 5 cm = 10 cm
Area of the rectangle = Length × width
= 5 cm × 10 cm
= 50 cm2
(ii) Ratio of length to width of the rectangle = 2 : 3
Total length of the rope = 30 cm
⇒ Perimeter of rectangular plot = 30 cm
⇒ 2(l + b) = 30
⇒ l + b = 15
Let the length of the rectangular plot be 2x cm and breadth be 3x cm.
⇒ 2x + 3x = 15
⇒ x = 3
Length of the rectangle = 2 × 3 cm = 6 cm
Width of the rectangle = 3 × 3 cm = 9 cm
Area of the rectangle = Length × width
= 6 cm × 9 cm
= 54 cm2
Page No 50:
Question 4:
AB is a line 24 centimeters long. A point P is to be marked on it such that AP : PB is 3 : 5. How far away from A should P be marked?
Answer:
Ratio of AP to PB = 3 : 5
So, a line 8 (= 3 + 5) cm long has a point P marked on it such that AP = 3 cm and PB = 5 cm.
Length of AB = 24 cm
Length of AP = 24 × cm = 9 cm
Thus, point P should be marked 9 cm away from point A.
Page No 50:
Question 5:
A point P is to be marked on AB and a line PQ drawn such that the measures of ∠APQ and ∠BPQ are in the ratio 4 : 5. What should be ∠APQ?
Answer:
We know that the sum of the angles in a linear pair is 180°.
∠APQ and ∠BPQ form a linear pair.
⇒ ∠APQ + ∠BPQ = 180°
Ratio of the measures of ∠APQ to ∠ BPQ = 4 : 5
Measure of ∠APQ = 180° × = 80°
Page No 51:
Question 1:
Meena and Mariam are basket weavers. One day Meena made 2 baskets and Mariam, 3 baskets. When these were sold at the market, they got 350 rupees in all. How do they split this money?
Answer:
Number of baskets made by Meena = 2
Number of baskets made by Mariam = 3
Ratio of the number of baskets made by Meena to the number of baskets made by Mariam = 2 : 3
Total amount earned = Rs.350
Amount earned by Meena = = Rs.140
Amount earned by Mariam = = Rs.210
Thus, Meena’s share was Rs.140 and Mariam’s share was Rs.210 in the total earnings.
Page No 51:
Question 2:
Damu works in Nanu’s farm. They have decided to divide the profit in the ratio 4 : 3. One month, they got 2100 rupees. What is the share of each?
Answer:
Ratio of profit-sharing between Damu and Nanu = 4 : 3
Total profit earned = Rs.2100
Damu’s share of the profit = = Rs.1200
Nanu’s share of the profit = = Rs.900
Page No 53:
Question 1:
In a co-operative society; there are 600 men and 400 women. An executive committee of 30 members is to be formed. The ratio of men and women in this committee is to be the same as the ratio in the whole society. How many men and how many women should be members of the committee.
Answer:
Number of men in the society = 600
Number of women in the society = 400
Ratio of men to women in the society =
Total members in the committee to be formed = 30
Number of men in the committee = = 18
Number of women in the committee = = 12
Page No 53:
Question 2:
63000 rupees is set aside for bringing drinking water to the first and second wards of Cheriyakara Panchayat. This amount is to be divided in the ratio of the population of the wards. The population of the first ward is 900 and that of the second ward is 1600. What share of the money does each get?
Answer:
Population of the first ward = 900
Population of the second ward = 1600
Ratio of population of the first ward to that of the second ward =
Total amount set for drinking water supply = Rs.63000
First ward’s share = = Rs.22680
Second ward’s share = = Rs.40320
Page No 53:
Question 3:
Raju and Joseph started a business together investing 60000 rupees together. When the profit for a month was divided, Raju got 1400 rupees and Joseph got 1000 rupees. How much did each invest to start the business?
Answer:
Profit earned by Raju = Rs.1400
Profit earned by Joseph = Rs.1000
Ratio of profit earned by Raju to profit earned by Joseph =
Total amount invested by Raju and Joseph = Rs.60000
Amount invested by Raju = = Rs.35000
Amount invested by Joseph = = Rs.25000
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