CBSE Class 7 Mathematics Rational Numbers Worksheet Set 06

SECTION-A

Question 1. The value of 0.03784x1000 is
(a) 3.784
(b) 37.84
(c) 378.4
(d) 3784
Answer: (b) 37.84
In simple words: When we multiply a decimal number by 1000, we move the decimal point three places to the right.

Exam Tip: Count the number of zeros in the multiplier to know how many places to shift the decimal point.

Question 2. 3,470 g = ___________ kg.
(a) 34.7
(b) 347
(c) 3.47
(d) 0.347
Answer: (c) 3.47
In simple words: There are 1000 grams in a kilogram. To convert 3470 grams, divide it by 1000 to get 3.47 kg.

Exam Tip: Always divide by 1000 when you convert smaller mass units like grams to larger ones like kilograms.

Question 3. On subtracting 4.81 from 43.17 , we get
(a) 38.36
(b) 38.81
(c) 38.17
(d) 47.98
Answer: (a) 38.36
In simple words: Subtract 4.81 from 43.17 by borrowing carefully to get 38.36.

Exam Tip: Align the decimal points vertically before you subtract or add decimal numbers to keep calculations correct.

Question 4. The reciprocal of rational number \( \frac{2}{14} \) is
(a) \( \frac{1}{7} \)
(b) 7
(c) \( \frac{4}{28} \)
(d) none of these
Answer: (b) 7
In simple words: To find the reciprocal, flip the fraction to get 14/2, which simplifies to 7.

Exam Tip: Flip the numerator and the denominator upside down to find the reciprocal of any fraction.

Question 5. The additive inverse of \( \frac{-3}{7} \) is
(a) \( \frac{7}{-3} \)
(b) \( -\frac{7}{3} \)
(c) \( \frac{3}{7} \)
(d) \( \frac{5}{7} \)
Answer: (c) \( \frac{3}{7} \)
In simple words: The additive inverse of a negative number is the same number with a positive sign.

Exam Tip: Simply change the sign of the given rational number to find its additive inverse.

Question 6. Standard form of \( \frac{-45}{75} \) is
(a) \( \frac{9}{15} \)
(b) \( \frac{-9}{15} \)
(c) \( \frac{-3}{5} \)
(d) \( \frac{-3}{-5} \)
Answer: (c) \( \frac{-3}{5} \)
In simple words: Divide both 45 and 75 by their biggest common divisor, which is 15, to get -3/5.

Exam Tip: Simplify fractions completely by dividing both parts by their highest common factor to get the standard form.

SECTION- B

Question 1. Express a) 235 paise in rupees b) 1 kg 5 g as kg
Answer:
a) To convert paise to rupees, divide by 100:
\( 235\text{ paise} = \text{Rs. } \frac{235}{100} = \text{Rs. } 2.35 \)

b) To convert grams to kilograms, divide by 1000:
\( 5\text{ g} = \frac{5}{1000}\text{ kg} = 0.005\text{ kg} \)
\( 1\text{ kg } 5\text{ g} = 1\text{ kg} + 0.005\text{ kg} = 1.005\text{ kg} \)
In simple words: 235 paise is Rs. 2.35, and 1 kg 5 g is 1.005 kg.

Exam Tip: Write single digit grams as 0.00x kg to avoid decimal alignment mistakes.

Question 2. A car covers 120.55 km in 10 litres of petrol.How much distance will it cover in 1 litre of petrol ?
Answer:
Distance covered in 10 litres = \( 120.55\text{ km} \)
Distance covered in 1 litre = \( \frac{120.55}{10} = 12.055\text{ km} \)
The car will cover 12.055 km in 1 litre of petrol.
In simple words: Divide the total distance of 120.55 km by 10 to find the distance for 1 litre.

Exam Tip: Dividing a decimal by 10 shifts the decimal point exactly one place to the left.

Page 2

Question 3. Represent \( \frac{-4}{7} \) on a number line.
Answer:
The rational number \( -\frac{4}{7} \) lies between 0 and -1.
Divide the space between 0 and -1 into 7 equal sections. Starting from 0 and moving to the left, the fourth point is \( -\frac{4}{7} \).

In simple words: To show -4/7, split the space between 0 and -1 into 7 equal parts. Count 4 steps to the left from 0.

Exam Tip: Clearly label 0 and -1 first, then mark the divisions neatly on the negative side.

Question 4. Subtract \( \frac{5}{9} \) from \( \frac{-9}{10} \).
Answer:
Subtract \( \frac{5}{9} \) from \( \frac{-9}{10} \):
\[ \frac{-9}{10} - \frac{5}{9} \]
The common denominator of 10 and 9 is 90. Convert both fractions:
\[ \frac{-9 \times 9}{10 \times 9} = \frac{-81}{90} \]
\[ \frac{5 \times 10}{9 \times 10} = \frac{50}{90} \]
Subtract the numerators:
\[ \frac{-81 - 50}{90} = \frac{-131}{90} = -1\frac{41}{90} \]
In simple words: Change both denominators to 90. Subtract the top numbers to get -131/90.

Exam Tip: Keep the negative sign with the first numerator when subtracting rational numbers.

SECTION- C

Question 1. Find a rational number between \( \frac{-3}{2} \) and \( \frac{2}{3} \).
Answer:
A rational number between two values is their mean (average):
\[ \text{Rational number} = \frac{1}{2} \left( \frac{-3}{2} + \frac{2}{3} \right) \]
The common denominator of 2 and 3 is 6:
\[ \frac{-3}{2} = \frac{-9}{6} \]
\[ \frac{2}{3} = \frac{4}{6} \]
Add the fractions:
\[ \frac{-9}{6} + \frac{4}{6} = \frac{-5}{6} \]
Now multiply by \( \frac{1}{2} \):
\[ \frac{1}{2} \times \frac{-5}{6} = \frac{-5}{12} \]
In simple words: Add both fractions together, then divide the answer by 2 to get -5/12.

Exam Tip: The easiest way to find a fraction in between is to take the average of the two numbers.

Question 2. Find the area of a square each of whose side measures 23.45 m.
Answer:
Area of a square = \( \text{side} \times \text{side} \)
Given side length = \( 23.45\text{ m} \)
\[ \text{Area} = 23.45 \times 23.45 = 549.9025\text{ sq m} \]
The area of the square is 549.9025 square meters.
In simple words: Multiply 23.45 by itself to find the total square area.

Exam Tip: Remember to write the unit as square meters (sq m or \( \text{m}^2 \)) for area values.

Question 3. A retailer sells Basmati rice at the rate of 71.45 per kg. Find what Seema had to pay for 50 kg of rice
Answer:
Cost of 1 kg of rice = Rs. 71.45
Cost of 50 kg of rice = \( 71.45 \times 50 = \text{Rs. } 3572.50 \)
Seema had to pay Rs. 3572.50.
In simple words: Multiply the price of 1 kg by 50 to get the total cost of Rs. 3572.50.

Exam Tip: When multiplying by 50, multiply by 5 first and then move the decimal point one place to the right.

Question 4. Write in ascending order: \( \frac{2}{3}, \frac{1}{9}, \frac{-3}{6}, \frac{1}{3} \)
Answer:
Find the LCM of denominators 3, 9, 6, and 3, which is 18. Convert each fraction:
\[ \frac{2}{3} = \frac{12}{18} \]
\[ \frac{1}{9} = \frac{2}{18} \]
\[ \frac{-3}{6} = \frac{-9}{18} \]
\[ \frac{1}{3} = \frac{6}{18} \]
Comparing the numerators: \( -9 < 2 < 6 < 12 \).
So, the ascending order of fractions is:
\[ \frac{-3}{6}, \frac{1}{9}, \frac{1}{3}, \frac{2}{3} \]
In simple words: Change all denominators to 18. Sort the top numbers from smallest to largest.

Exam Tip: Negative numbers are always the smallest, so place them first in ascending order.

SECTION- D

Question 1. Show that \( \frac{9}{4} \times (\frac{-7}{5} \times \frac{1}{2}) = (\frac{9}{4} \times \frac{-7}{5}) \times \frac{1}{2} \)
Answer:
Let us calculate both sides of the equation.
Left Hand Side (LHS):
\[ \text{LHS} = \frac{9}{4} \times \left( \frac{-7}{5} \times \frac{1}{2} \right) = \frac{9}{4} \times \frac{-7}{10} = \frac{-63}{40} \]
Right Hand Side (RHS):
\[ \text{RHS} = \left( \frac{9}{4} \times \frac{-7}{5} \right) \times \frac{1}{2} = \frac{-63}{20} \times \frac{1}{2} = \frac{-63}{40} \]
Since LHS = RHS = \( \frac{-63}{40} \), the equation is shown to be true.
In simple words: Work out both sides separately. Both answers equal -63/40, so they are the same.

Exam Tip: Solve bracket terms first and write out each side clearly to show they are equal.

Question 2. In a super market, the cost of a table lamp is 870 , on which \( \frac{1}{5} \)th is off. The same table lamp is available at an electric shop for 920 with a discount of \( \frac{1}{10} \)th. From where should one buy the lamp? What is the difference in prices?
Answer:
Calculate the final price for both choices:

Supermarket:
Marked price = Rs. 870
Discount = \( \frac{1}{5} \times 870 = \text{Rs. } 174 \)
Final Price = \( 870 - 174 = \text{Rs. } 696 \)

Electric Shop:
Marked price = Rs. 920
Discount = \( \frac{1}{10} \times 920 = \text{Rs. } 92 \)
Final Price = \( 920 - 92 = \text{Rs. } 828 \)

Comparing prices, Rs. 696 is cheaper than Rs. 828. Therefore, one should buy the lamp from the supermarket.
Difference in price = \( 828 - 696 = \text{Rs. } 132 \).
In simple words: The lamp costs Rs. 696 at the supermarket and Rs. 828 at the electric shop. Buy it from the supermarket to save Rs. 132.

Exam Tip: Find the actual discount amount in rupees first, then subtract it to find the final price.

Question 3. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarla bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Answer:
Calculate the total fruit weight for both buyers:

Shyama:
Apples = 5 kg 300 g = 5.300 kg
Mangoes = 3 kg 250 g = 3.250 kg
Total = \( 5.300 + 3.250 = 8.550\text{ kg} \)

Sarla:
Oranges = 4 kg 800 g = 4.800 kg
Bananas = 4 kg 150 g = 4.150 kg
Total = \( 4.800 + 4.150 = 8.950\text{ kg} \)

Comparing weights, \( 8.950\text{ kg} > 8.550\text{ kg} \). Sarla bought more fruits.
In simple words: Shyama bought 8.55 kg of fruit, and Sarla bought 8.95 kg. Sarla bought more.

Exam Tip: Convert weights into decimals to make adding and comparing different quantities simpler.

Question 4. Simplify : 7.5 - [ 2.25 + { 1.25 - 0.5 ( 1.5 - 0.75 - 0.5 ) }]
Answer:
Solve step-by-step using BODMAS rules:
1. Solve the innermost brackets:
\( 1.5 - 0.75 - 0.5 = 0.25 \)

2. Solve the braces next:
\( 1.25 - 0.5(0.25) = 1.25 - 0.125 = 1.125 \)

3. Solve the square brackets next:
\( 2.25 + 1.125 = 3.375 \)

4. Compute the final value:
\( 7.5 - 3.375 = 4.125 \)
The simplified result is 4.125.
In simple words: Solve innermost brackets first to get 0.25. Multiply by 0.5 to get 0.125, subtract from 1.25, add 2.25, and subtract the total from 7.5 to get 4.125.

Exam Tip: Work carefully from the innermost brackets to the outermost brackets to avoid errors.

Question 5. Simplify : \( \frac{2}{5} + [ \frac{8}{3} \text{ of } \frac{-12}{4} + \{ \frac{4}{5} - \frac{2}{3} \} ] \)
Answer:
Solve step-by-step using BODMAS rules:
1. Solve the innermost braces first:
\( \frac{4}{5} - \frac{2}{3} = \frac{12}{15} - \frac{10}{15} = \frac{2}{15} \)

2. Solve the square brackets next:
\( \frac{8}{3} \text{ of } \frac{-12}{4} + \frac{2}{15} \)
Remember that 'of' means multiply:
\( \frac{8}{3} \times \frac{-12}{4} = \frac{8 \times -12}{3 \times 4} = \frac{-96}{12} = -8 \)
Now add the values:
\( -8 + \frac{2}{15} = \frac{-120 + 2}{15} = \frac{-118}{15} \)

3. Final addition:
\( \frac{2}{5} + \left(\frac{-118}{15}\right) = \frac{6}{15} - \frac{118}{15} = \frac{-112}{15} = -7\frac{7}{15} \)
The simplified result is \( \frac{-112}{15} \).
In simple words: First solve the braces to get 2/15. Multiply 8/3 by -12/4 to get -8. Add these to get -118/15, and finally add 2/5 to get -112/15.

Exam Tip: Convert the term 'of' into multiplication first before performing any other operation inside square brackets.