Home
Class 7
MATHS
An insect is crawling up on a vertical w...

An insect is crawling up on a vertical wall. It crawls up `12 (5)/(7)` inch. Then it slips back `3 (11)/(14)` inch and then again crawls up `5 (1)/(6)` inch. How far is it from the starting point?

A

`11 (6)/(25)` inch

B

`12 (2)/(21)` inch

C

`14 (2)/(21)` inch

D

`13 (5)/(28)` inch

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow these steps: ### Step 1: Convert Mixed Numbers to Improper Fractions 1. **Crawling Up**: The insect crawls up `12 (5)/(7)` inches. - Convert to improper fraction: \[ 12 \frac{5}{7} = \frac{12 \times 7 + 5}{7} = \frac{84 + 5}{7} = \frac{89}{7} \] 2. **Slipping Back**: The insect slips back `3 (11)/(14)` inches. - Convert to improper fraction: \[ 3 \frac{11}{14} = \frac{3 \times 14 + 11}{14} = \frac{42 + 11}{14} = \frac{53}{14} \] 3. **Crawling Up Again**: The insect crawls up `5 (1)/(6)` inches. - Convert to improper fraction: \[ 5 \frac{1}{6} = \frac{5 \times 6 + 1}{6} = \frac{30 + 1}{6} = \frac{31}{6} \] ### Step 2: Calculate the Total Distance from the Starting Point Now we will calculate the total distance the insect has crawled up and slipped back: \[ \text{Total Distance} = \left(\frac{89}{7} - \frac{53}{14} + \frac{31}{6}\right) \] ### Step 3: Find the Least Common Multiple (LCM) To add and subtract these fractions, we need to find the LCM of the denominators (7, 14, and 6): - Prime factors: - 7: \(7\) - 14: \(2 \times 7\) - 6: \(2 \times 3\) - LCM: \(2 \times 3 \times 7 = 42\) ### Step 4: Convert Each Fraction to Have a Common Denominator 1. Convert \(\frac{89}{7}\) to have a denominator of 42: \[ \frac{89}{7} = \frac{89 \times 6}{42} = \frac{534}{42} \] 2. Convert \(\frac{53}{14}\) to have a denominator of 42: \[ \frac{53}{14} = \frac{53 \times 3}{42} = \frac{159}{42} \] 3. Convert \(\frac{31}{6}\) to have a denominator of 42: \[ \frac{31}{6} = \frac{31 \times 7}{42} = \frac{217}{42} \] ### Step 5: Combine the Fractions Now we can combine the fractions: \[ \frac{534}{42} - \frac{159}{42} + \frac{217}{42} = \frac{534 - 159 + 217}{42} \] Calculating the numerator: \[ 534 - 159 + 217 = 534 + 217 - 159 = 592 \] Thus, we have: \[ \frac{592}{42} \] ### Step 6: Simplify the Fraction Now we simplify \(\frac{592}{42}\): - Divide both the numerator and the denominator by 2: \[ \frac{592 \div 2}{42 \div 2} = \frac{296}{21} \] ### Step 7: Convert to Mixed Number Convert \(\frac{296}{21}\) to a mixed number: - Divide 296 by 21: \[ 296 \div 21 = 14 \quad \text{(whole number)} \] - Remainder: \[ 296 - (21 \times 14) = 296 - 294 = 2 \] Thus, we have: \[ 14 \frac{2}{21} \] ### Final Answer The insect is \(14 \frac{2}{21}\) inches from the starting point. ---
Promotional Banner

Topper's Solved these Questions

  • FRACTIONS

    ICSE|Exercise Warm-up Exercise|24 Videos
  • FRACTIONS

    ICSE|Exercise Exercise 2.1|14 Videos
  • FRACTIONS

    ICSE|Exercise Example|52 Videos
  • EXPONENTS AND POWERS

    ICSE|Exercise Unit practice paper -1|7 Videos
  • FUNDAMENTAL GEOMETRICAL CONCEPTS

    ICSE|Exercise Exercise |14 Videos

Similar Questions

Explore conceptually related problems

An insect is at a point A on a wall. It crawls up 3 inches and thenn slides down by 6 inches. The insect then again crawls up 5 inches. How many inches above or below the point A is the insect now?

A centipede was crawling in a straight line. First it crawled for 8 (3)/(7) inch, then it crawled in the reverse direction for 5 (5)/(8) inch. Once again it reversed direction and crawled for 11 (1)/(2) inch in the original direction. How far away was it finally from the starting point?

A person moves towards East for 3 m, then towards North for 4 m and then moves vertically up by 5 m. What is his distance now from the starting point ?

Katie hikes 5 miles north, 7 miles east, and then 3 miles north again. What number of miles, measured in a straight line, is katie from her starting point?

An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom . If the coefficient of friction between the ground and the insect is 0.75 , then h is : (g = 10 ms^(-2))

A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of radius R=20cm as shown in the figure. The refractive index of the material of the sphere is equal to sqrt(2) and the height of the spider from the centre of the sphere is 2R(=40cm) . An insect, initially sitting at the bottom, starts crawling on the sphere along a vertical circular path with the constant speed of pi//4 cm//s . For how long time the insect will be invisible for the spider, assume that it crawls once round the vertical circle.

An insect crawls up a hemispherical surface as shown (see the figure) the coefficient offriction between the insect and the surface is 1/3. If the line joining the centre of the hemisperical surface to the insect makes an angle theta with the vertical, the maximum possible value of theta is given by

A heavy block B is sliding with constant velocity u=5m//s on a horizontal table. The width of the block is L=4m . There is an insect A at distance d=3m from the block as shown in the figure. The insect wants to cross to the oposite side of the table. It begins to crawl at a constant velocity v at the instant shown in the figure. Find the least value of v (in m/s) for which the insect can cross to the other side without getting hit by the block.

An insect of mass m, starts moving on a rough inclined surface from point A. As the surface is vert sticky, the corfficient of fricion between the insect and the incline is mu=1 . Assume that it can move in any direction, up the incline or down the incline then