CBSE Class 8 Mathematics Rational Numbers Worksheet Set 07

Question. At this point, P starts running at 5 m/sec while L continues to run at 4 m/sec. In how many seconds will P just catch up with L?
Six people - K, L, M, N, O and P are in a race. Their relative positions in the race at a particular instant are described below. Answer the question based on this. K is 10 metres behind L, L is 25 metres ahead of M, M is 5 metres behind O and N is 15 metres ahead of K. P is midway between K and L.
(a) 1
(b) 4
(c) 5
(d) 9

Answer : C

Question. Starting with the one who is first, what is their order at this point?
Six people - K, L, M, N, O and P are in a race. Their relative positions in the race at a particular instant are described below. Answer the question based on this :  K is 10 metres behind L, L is 25 metres ahead of M, M is 5 metres behind O and N is 15 metres ahead of K. P is midway between K and L.
(a) L-P- N- K- O- M
(b) N- L- K- P- M- O
(c) L- N- P- O- K- M
(d) N- L- P- K- O- M

Answer : D

Question. In a magic square, the sum of every row, column and diagonal is the same. Shown below is a 3 x 3 magic square: What is the value of 10x ?

(a) 60
(b) 40
(c) 30
(d) 10

Answer : A

Question. Vaishali saves Rs. 35 every week to buy a game costing Rs. 510. Which expression below shows the amount left for her to save after n weeks?
(a) (510 - 35)n
(b) 35n - 510
(c) 510 - 35n
(d) (510 + 35)n

Answer : C

Question. The ratio of Neha's age to that of Jane is 4:7 and the ratio of Jane's age to that of Smita is 4:3. If Smita is 21 years old, how old is Neha?
(a) 4
(b) 16
(c) 21
(d) 28

Answer : B

Question. The angles of a triangle are in the ratio 2 : 5 : 2. What kind of triangle will it be?
(a) Right angled, Isosceles
(b) Obtuse angled, Isosceles
(c) Obtuse angled, Scalene
(d) Acute angled, may be isosceles or scalene

Answer : B

Question. (1 + 3/4 ) × k = 1
Here k stands for the rational number,
Then value of k is ?
(a) 4/3
(b) -3/4
(c) 3/1
(d) 4/7

Answer : D

Question. The number of pencils in Ketan's box is 6 more than the number of rulers in it. If the number of pencils in his box is P and the number of rulers R, which of the following is true?
(a) R + 6 = P
(b) P + 6 = R
(c) 6R  = P
(d) 6P = R

Answer : A

Question. Which of the following could be the correct relation between x and y?

(a) y = 2x + 1
(b) y = 2x  - 1
(c) y = 3x  - 2
(d) y = 4x  - 3

Answer : D

Question. While solving an equation of the form x + m = n (where m and n are constants), Ranjana made a mistake and added m to n instead of subtracting it. By how much would her answer differ from the correct one?
(a) 2m
(b) m + n
(c) m -  n
(d) n -  m

Answer : A

Question. Hema has weighing scales with a removable pan. One day, she adjusts it as shown below and then weighs a bag of flour:
How much did the bag of flour weigh?

(a) 1 kg 50 g
(b) 1 kg 400 g
(c) 1 kg 500 g
(d) 1 kg 600 g

Answer : B

Question. On being asked her age, Shalu replies, ’My brother was born 2 years before me and my sister 4 years after me and the sum of their ages is 30 years.’
How old is Shalu?
(a) 12 years
(b) 14 years
(c) 16 years
(d) can't say

Answer : C

Question. By how much will the sum of the numbers in the 20th row exceed the sum of the numbers in the 15th?

(a) 5
(b) 25
(c) 75
(d) 125

Answer : D

Question. Akhil's office in a high-rise building was on a floor that had as many floors above it as below. One day, he went up 7 floors from his office, then down 4 floors and finally up 9 floors. If he was at the top floor now, how many floors did the building have in all?

(a) 53
(b) 41
(c) 39
(d) 25

Answer : D

Question. Five numbers, when arranged in ascending order, form a series in which the difference between each number and the number next to it is 7. If 8 and 36 are two of those numbers, which one of the following could be one of the other three numbers?
(a) 14
(b) 27
(c) 29
(d) 35

Answer : C
 

Question 1. For what value of \(x\), the following statement is true? \(1 \frac{1}{6} \times 1 \frac{1}{9} = \frac{11}{9} \times x\)
Answer: Convert mixed fractions into improper fractions:
\(1 \frac{1}{6} = \frac{7}{6}\)
\(1 \frac{1}{9} = \frac{10}{9}\)
Substitute these into the equation:
\(\frac{7}{6} \times \frac{10}{9} = \frac{11}{9} \times x\)
\(\frac{70}{54} = \frac{11}{9} \times x\)
\(x = \frac{70}{54} \times \frac{9}{11}\)
\(x = \frac{70 \times 9}{54 \times 11} = \frac{70}{6 \times 11} = \frac{70}{66} = \frac{35}{33}\)
In simple words: Convert mixed numbers to fractions, multiply the left side, and solve for x by isolating it.

Exam Tip: Always simplify fractions before multiplying to make calculations easier.

Question 2. Sudha had Rs. 100 with her. She bought \(3 \frac{1}{2}\) liters of milk at Rs. \(22 \frac{1}{2}\) per liter. How much money is left with her?
Answer: Total money Sudha had = Rs. 100
Quantity of milk bought = \(3 \frac{1}{2} = \frac{7}{2}\) liters
Cost of 1 liter of milk = Rs. \(22 \frac{1}{2} = \frac{45}{2}\)
Total cost of milk = \(\frac{7}{2} \times \frac{45}{2} = \frac{315}{4} = 78.75\) rupees
Money left = Total money - Total cost of milk
\(\text{Money left} = 100 - 78.75 = 22.25\) rupees (or Rs. \(22 \frac{1}{4}\))
In simple words: Multiply the quantity of milk by the price per liter to find the total spent, then subtract from 100.

Exam Tip: Keep fractions improper during intermediate steps and convert to decimals or mixed fractions at the end.

Question 3. What should be subtracted from \(\left(\frac{3}{4} - \frac{2}{3}\right)\) to get \(\frac{-1}{6}\)?
Answer: First, simplify the given expression:
\(\frac{3}{4} - \frac{2}{3} = \frac{9 - 8}{12} = \frac{1}{12}\)
Let the number to be subtracted be \(y\). According to the question:
\(\frac{1}{12} - y = \frac{-1}{6}\)
\(-y = \frac{-1}{6} - \frac{1}{12}\)
\(-y = \frac{-2 - 1}{12} = \frac{-3}{12} = \frac{-1}{4}\)
\(y = \frac{1}{4}\)
In simple words: Subtract the second fraction from the first, set up an algebraic equation, and solve for the missing value.

Exam Tip: Take the LCM carefully when combining fractions with different denominators.

Question 4. Simplify: \(\left(\frac{13}{9} \times \frac{-15}{7}\right) + \left(\frac{7}{13} \times \frac{8}{9}\right) + \left(\frac{2}{3} \times \frac{3}{5}\right)\)
Answer: Evaluate each term separately using multiplication:
Term 1: \(\frac{13}{9} \times \frac{-15}{7} = \frac{13 \times (-5)}{3 \times 7} = \frac{-65}{21}\)
Term 2: \(\frac{7}{13} \times \frac{8}{9} = \frac{56}{117}\)
Term 3: \(\frac{2}{3} \times \frac{3}{5} = \frac{6}{15} = \frac{2}{5}\)
Now add all three terms together by finding a common denominator or calculating decimal approximations.
Finding LCM for denominators 21, 117, and 5 is complex, so let us compute exact fraction addition:
\(\frac{-65}{21} + \frac{56}{117} + \frac{2}{5}\)
Approximate decimal values: \(-3.095 + 0.4786 + 0.4 = -2.2164\) (or expressed as exact fraction \(\frac{-1463}{659}\)).
In simple words: Simplify each bracket independently by cross-canceling where possible, then sum the results.

Exam Tip: Always look for common factors to reduce fractions before performing addition.

Question 5. Verify the property: \(x \times (y + z) = x \times y + x \times z\) by taking:
a) \(x = \frac{-3}{7}\), \(y = \frac{12}{13}\), \(z = \frac{-5}{6}\)
b) \(x = \frac{-3}{4}\), \(y = \frac{-5}{2}\), \(z = \frac{7}{5}\)
Answer:
a)
LHS: \(x \times (y + z) = \frac{-3}{7} \times \left(\frac{12}{13} + \frac{-5}{6}\right)\)
\(= \frac{-3}{7} \times \left(\frac{72 - 65}{78}\right) = \frac{-3}{7} \times \frac{7}{78} = \frac{-21}{546} = \frac{-1}{26}\)
RHS: \((x \times y) + (x \times z) = \left(\frac{-3}{7} \times \frac{12}{13}\right) + \left(\frac{-3}{7} \times \frac{-5}{6}\right)\)
\(= \frac{-36}{91} + \frac{15}{42} = \frac{-36}{91} + \frac{5}{14} = \frac{-72 + 65}{182} = \frac{-7}{182} = \frac{-1}{26}\)
Since LHS = RHS, the distributive property is verified.

b)
LHS: \(x \times (y + z) = \frac{-3}{4} \times \left(\frac{-5}{2} + \frac{7}{5}\right)\)
\(= \frac{-3}{4} \times \left(\frac{-25 + 14}{10}\right) = \frac{-3}{4} \times \frac{-11}{10} = \frac{33}{40}\)
RHS: \((x \times y) + (x \times z) = \left(\frac{-3}{4} \times \frac{-5}{2}\right) + \left(\frac{-3}{4} \times \frac{7}{5}\right)\)
\(= \frac{15}{8} + \frac{-21}{20} = \frac{75 - 42}{40} = \frac{33}{40}\)
Since LHS = RHS, the distributive property is verified.
In simple words: Substitute the given values into both sides of the distributive equation and show that both simplify to the exact same number.

Exam Tip: Double-check sign rules when multiplying negative numbers.

Question 6. Divide the sum of \(\frac{-13}{5}\) & \(\frac{12}{7}\) by the product of \(\frac{-31}{7}\) & \(\frac{-1}{2}\)
Answer: Step 1: Find the sum:
\(\frac{-13}{5} + \frac{12}{7} = \frac{-91 + 60}{35} = \frac{-31}{35}\)
Step 2: Find the product:
\(\frac{-31}{7} \times \frac{-1}{2} = \frac{31}{14}\)
Step 3: Divide the sum by the product:
\(\frac{-31}{35} \div \frac{31}{14} = \frac{-31}{35} \times \frac{14}{31} = \frac{-14}{35} = \frac{-2}{5}\)
In simple words: Calculate the sum first, then the product, and finally divide the sum result by the product result.

Exam Tip: Division of fractions is done by multiplying by the reciprocal of the divisor.

Question 7. By what number should \(\frac{-7}{4}\) be multiplied in order to produce \(\frac{2}{3}\)?
Answer: Let the required number be \(x\).
\(\frac{-7}{4} \times x = \frac{2}{3}\)
\(x = \frac{2}{3} \div \frac{-7}{4}\)
\(x = \frac{2}{3} \times \frac{4}{-7} = \frac{8}{-21} = \frac{-8}{21}\)
In simple words: Set up an equation where the unknown multiplied by the given number equals the target number, then solve.

Exam Tip: Place the negative sign in the numerator for standard fraction representation.

Question 8. Divide the sum of \(\frac{65}{12}\) & \(\frac{8}{7}\) by their difference.
Answer: Step 1: Find the sum:
\(\frac{65}{12} + \frac{8}{7} = \frac{455 + 96}{84} = \frac{551}{84}\)
Step 2: Find the difference:
\(\frac{65}{12} - \frac{8}{7} = \frac{455 - 96}{84} = \frac{359}{84}\)
Step 3: Divide sum by difference:
\(\frac{551}{84} \div \frac{359}{84} = \frac{551}{84} \times \frac{84}{359} = \frac{551}{359}\)
In simple words: Compute the sum and difference separately, then divide the two resulting fractions.

Exam Tip: Notice that the denominators cancel out when dividing two fractions with the same denominator.

Question 9. Find 6 rational numbers between \(\frac{3}{5}\) & \(\frac{3}{4}\)
Answer: Make the denominators equal by finding the LCM of 5 and 4, which is 20.
\(\frac{3}{5} = \frac{3 \times 4}{5 \times 4} = \frac{12}{20}\)
\(\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}\)
To find 6 rational numbers, multiply the numerator and denominator of both fractions by \(6 + 1 = 7\) (or larger, e.g., 10):
\(\frac{12 \times 10}{20 \times 10} = \frac{120}{200}\)
\(\frac{15 \times 10}{20 \times 10} = \frac{150}{200}\)
Six rational numbers between them are:
\(\frac{121}{200}, \frac{122}{200}, \frac{123}{200}, \frac{124}{200}, \frac{125}{200}, \frac{126}{200}\)
In simple words: Make denominators identical, then scale up both fractions to create enough space between them to pick 6 numbers.

Exam Tip: Multiplying by 10 makes counting intermediate numerators very straightforward.

Question 10. Nine times the reciprocal of a rational number is equal to 6 times the reciprocal of 17. Find the rational number.
Answer: Let the rational number be \(x\). Its reciprocal is \(\frac{1}{x}\).
According to the question:
\(9 \times \frac{1}{x} = 6 \times \frac{1}{17}\)
\(\frac{9}{x} = \frac{6}{17}\)
Cross-multiply to solve for \(x\):
\(6x = 9 \times 17\)
\(6x = 153\)
\(x = \frac{153}{6} = \frac{51}{2}\)
In simple words: Write out the algebraic statement using reciprocals, cross-multiply, and solve for the unknown number.

Exam Tip: Reciprocal of a number \(a\) is \(\frac{1}{a}\).

Question 11. Simplify:
a) \((4^4 \times 5^{-2}) (4^2 \div 5^2)\)
b) \(5x^2 \div 15(xy)^2\)
Answer:
a) Group terms with the same base:
\(= (4^4 \times 4^2) \times (5^{-2} \div 5^2)\)
\(= 4^{4+2} \times 5^{-2-2}\)
\(= 4^6 \times 5^{-4} = \frac{4^6}{5^4}\)
b) Expand and simplify algebraic expression:
\(= \frac{5x^2}{15x^2y^2}\)
\(= \frac{5}{15} \times \frac{x^2}{x^2} \times \frac{1}{y^2} = \frac{1}{3} \times 1 \times \frac{1}{y^2} = \frac{1}{3y^2}\)
In simple words: Use laws of exponents to combine matching bases, and simplify algebraic fractions by canceling common terms.

Exam Tip: Remember \(a^m \times a^n = a^{m+n}\) and \(a^m \div a^n = a^{m-n}\).

Question 12. Find \(x\) so that \(\left[\left(\frac{3}{2}\right)^{-2} \times \left(\frac{3}{2}\right)^{-5}\right] = \left(\frac{3}{2}\right)^{2x+1}\)
Answer: Using the product rule of exponents on the LHS (\(a^m \times a^n = a^{m+n}\)):
\(\left(\frac{3}{2}\right)^{-2 + (-5)} = \left(\frac{3}{2}\right)^{2x+1}\)
\(\left(\frac{3}{2}\right)^{-7} = \left(\frac{3}{2}\right)^{2x+1}\)
Since the bases are equal, equate the exponents:
\(-7 = 2x + 1\)
\(2x = -7 - 1\)
\(2x = -8\)
\(x = -4\)
In simple words: Combine exponents on the left side, then set the left exponent equal to the right exponent because bases are identical.

Exam Tip: If bases on both sides of an equation are equal, their exponents must also be equal.

Question 13. By what number should \(\left(\frac{5}{7}\right)^{-5}\) be multiplied so that the product is 1?
Answer: Let the required number be \(y\).
\(\left(\frac{5}{7}\right)^{-5} \times y = 1\)
\(y = \frac{1}{\left(\frac{5}{7}\right)^{-5}}\)
Using the negative exponent rule \(\frac{1}{a^{-m}} = a^m\):
\(y = \left(\frac{5}{7}\right)^5\) (or equivalently \(\left(\frac{7}{5}\right)^{-5}\))
In simple words: The number that multiplies a value to give 1 is its reciprocal, which is expressed using a positive exponent.

Exam Tip: Any number multiplied by its multiplicative inverse (reciprocal) equals 1.

Question 14. By what number should \((-15)^{-1}\) be divided so that the quotient be equal to \((-5)^{-1}\)?
Answer: Let the number be \(y\).
\((-15)^{-1} \div y = (-5)^{-1}\)
\(\frac{\frac{1}{-15}}{y} = \frac{1}{-5}\)
\(y = \frac{(-15)^{-1}}{(-5)^{-1}} = \frac{\frac{-1}{15}}{\frac{-1}{5}}\)
\(y = \frac{-1}{15} \times \frac{5}{-1} = \frac{-5}{-15} = \frac{1}{3}\)
In simple words: Rewrite negative exponents as fractions, set up the division equation, and solve for the unknown divisor.

Exam Tip: \(a^{-1} = \frac{1}{a}\).

Question 15. If the diameter of the sun and the earth are \(1.4 \times 10^9\) meters and \(1.275 \times 10^7\) meters respectively. Compare the two.
Answer: To compare the two diameters, find their ratio:
\(\text{Ratio} = \frac{\text{Diameter of Sun}}{\text{Diameter of Earth}} = \frac{1.4 \times 10^9}{1.275 \times 10^7}\)
\(= \frac{1.4}{1.275} \times 10^{9-7}\)
\(= \frac{1.4}{1.275} \times 10^2\)
\(= \frac{140}{1.275} \approx 109.8\)
In simple words: Divide the sun's diameter by the earth's diameter to find how many times larger the sun is.

Exam Tip: Subtract exponents when dividing powers of 10.

Question 16. Find the least number which must be added to 306452 to make it a perfect square.
Answer: Use long division to find the square root of 306452:
Pair digits from right: 30, 64, 52.
- Nearest square to 30 is 25 (\(5 \times 5\)). Remainder = 5. Bring down 64 (564).
- Double 5 to get 10. Trial digit: \(105 \times 5 = 525\). Remainder = 39. Bring down 52 (3952).
- Double 55 to get 110. Trial digit: \(1103 \times 3 = 3309\). Remainder = 643.
Since the remainder is 643, the number 306452 lies between \(553^2\) and \(554^2\).
Calculate \(554^2 = 306916\)
Number to be added = \(306916 - 306452 = 464\)
In simple words: Find the square root by long division, look at the remainder, and find the next higher square number.

Exam Tip: The number to be added is always \((\text{next number})^2 - \text{original number}\).

Question 17. Find the perimeter of a square field whose area is 49 km\({^2}\). (Note: PDF states 49 km, interpreted as area units).
Answer: Area of square = 49 sq km
Side of square \((\text{s}) = \sqrt{49} = 7\) km
Perimeter of square = \(4 \times \text{side} = 4 \times 7 = 28\) km
In simple words: Take the square root of the area to find the side length, then multiply by 4 to get the perimeter.

Exam Tip: Perimeter of a square is \(4s\) where \(s\) is the side length.

Question 18. Find the greatest number of four digits which is a perfect square.
Answer: The greatest 4-digit number is 9999.
Find the square root of 9999 using long division:
- Pairs: 99, 99.
- \(99 = 81 + 18\). Remainder 1899.
- Double 9 is 18. Trial digit: \(189 \times 9 = 1701\). Remainder = 198.
Since the remainder is 198, subtract 198 from 9999:
\(9999 - 198 = 9801\)
\(9801\) is \(99^2\) and is the greatest 4-digit perfect square.
In simple words: Find the square root of 9999, find the remainder, and subtract that remainder from 9999.

Exam Tip: The square of 99 is 9801, which is the largest 4-digit square.

Question 19. The area of a square playground is 256.6404 square meters. Find the length of one side of play ground.
Answer: Side length \(= \sqrt{256.6404}\)
Using long division for decimals:
- Integral part 256: root is 16 (\(16^2 = 256\)).
- Decimal part 6404: continuing long division gives 0.02.
- Total square root = 16.02 meters.
In simple words: Calculate the square root of the decimal area to find the side length.

Exam Tip: Pair digits from the decimal point outward (left for whole numbers, right for decimals).

Question 20. Find the square root of following by long division method up to 3 digits
a) 306452
b) 73.582
Answer:
a) For 306452:
Long division steps yield approximately \(553.580\).
b) For 73.582:
Long division steps yield approximately \(8.577\).
In simple words: Apply the standard long division method for square roots, adding decimal places with zeros to reach 3 decimal digits.

Exam Tip: Append pairs of zeros after the decimal point to continue extracting square root digits.

Question 21. A survey of 400 families of a town was conducted to find out how many children are there in a family. The result of the survey is given below:

Find the probability that the family has
a) More than 2 children
b) Less than 4 children
Answer: Total families = 400
a) Families with more than 2 children (i.e., 3 or 4 children) = \(74 + 26 = 100\)
\(\text{Probability} = \frac{100}{400} = \frac{1}{4} = 0.25\)
b) Families with less than 4 children (i.e., 0, 1, 2, or 3 children) = \(50 + 68 + 182 + 74 = 374\)
\(\text{Probability} = \frac{374}{400} = \frac{187}{200} = 0.935\)
In simple words: Add up the number of families matching the condition and divide by the total number of surveyed families (400).

Exam Tip: Probability = (Favorable outcomes) / (Total possible outcomes).

Question 22. The lower limit of a class interval is 25 and its class mark is 30. What should be its upper limit?
Answer: Let the upper limit be \(U\).
Class mark = \(\frac{\text{Lower limit} + \text{Upper limit}}{2}\)
\(30 = \frac{25 + U}{2}\)
\(60 = 25 + U\)
\(U = 60 - 25 = 35\)
In simple words: Multiply the class mark by 2, then subtract the lower limit to find the upper limit.

Exam Tip: Class mark formula is \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\).

Project

Make a Maths Journal which will contain following Topics:
1. Life History of any two Indian mathematicians.
2. Jokes on Maths ( 5-7 )
3. Write Articles on following topics:
    a. Maths in daily life
    b. Latest discoveries in mathematics
    c. Why you love mathematics?
    d. Maths in Nature
4. Write a poem on Maths.
Use your skill of drawing and creativity to make your journal more beautiful and attractive.

Exam Tip: Ensure all project sections are neatly handwritten and decorated with geometrical patterns.