- The risk reduction on increasing the steps/day from 0 to 10000 is less than the risk reduction on increasing the steps/day from 10000 to 20000
- The risk reduction on increasing the steps/day from 0 to 5000 is less than the risk reduction on increasing the steps/day from 15000 to 20000.
- For any 5000 increment in steps/day the largest risk reduction occurs on going from 0 to 5000.
- For any 5000 increment in steps/day the largest risk reduction occurs on going from 15000 to 20000.
Risk reduction value in (0-5000) = 1 – 0.45 = 0.55 which is maximum
1. Occupied Space:
The volume of one chalk-stick is \(\pi r^2 h\). For 7 chalk-sticks, the total volume occupied is:
\[
V_{\text{occupied}} = 7 \pi r^2 h
\]
2. Total Volume of the Container:
The radius of the container is \(R = 3r\). Therefore, the volume of the container is:
\[
V_{\text{container}} = \pi R^2 h = \pi (3r)^2 h = 9 \pi r^2 h
\]
3. Empty Space:
The empty space is:
\[
V_{\text{empty}} = V_{\text{container}} – V_{\text{occupied}}\]
\[= 9 \pi r^2 h – 7 \pi r^2 h = 2 \pi r^2 h
\]
4. Ratio of Occupied Space to Empty Space:
\[
\text{Ratio} = \frac{V_{\text{occupied}}}{V_{\text{empty}}} = \frac{7 \pi r^2 h}{2 \pi r^2 h} = \frac{7}{2}
\]
So, the ratio of the occupied space to the empty space is \(\frac{7}{2}\).
- Bicker
- Bog
- Dither
- Dodge
The correct answer is A. Bicker.
Explanation:
The sequence [drizzle → rain → downpour] shows increasing intensity of rain. In the same way, bicker refers to a small, petty argument, which can escalate into a quarrel (a more serious argument), and then into a feud (a long-lasting, intense conflict). The other options (bog, dither, dodge) are unrelated to the progression of arguments or conflict. Hence, “bicker” fits the analogy perfectly.
- Only I and II
- Only II and III
- Only I and III
- Only III
The correct answer is B. Only II and III.
Explanation:
From the statements:
- All heroes are winners.
- All winners are lucky people.
- Inference I (All lucky people are heroes) cannot be deduced because not all lucky people are necessarily heroes.
- Inference II (Some lucky people are heroes) is valid because all heroes are winners, and all winners are lucky people, so some lucky people must be heroes.
- Inference III (Some winners are heroes) is valid because all heroes are winners.
Thus, only inferences II and III can be logically deduced.
- 5
- \(\sqrt{2}\)
- 2
- \(\sqrt{5}\)
The correct answer is \(\sqrt{5}\).
Explanation: Let the correct result be \( p \times q \). Instead, the student calculated \( \frac{p}{q} \). Let the percentage error be \( 80\% \). The percentage error can be expressed as: \[ \text{Percentage Error} = \frac{\left( \frac{p}{q} – p \times q \right)}{p \times q} \times 100 \] This simplifies to: \[ 80 = \frac{\left( \frac{p}{q} – p \times q \right)}{p \times q} \times 100 \] Let’s set up the ratio \( \frac{\frac{p}{q}}{p \times q} \), solve it, and use algebra to find that \( q = \sqrt{5} \).
Explanation: Let the correct result be \( p \times q \). Instead, the student calculated \( \frac{p}{q} \). Let the percentage error be \( 80\% \). The percentage error can be expressed as: \[ \text{Percentage Error} = \frac{\left( \frac{p}{q} – p \times q \right)}{p \times q} \times 100 \] This simplifies to: \[ 80 = \frac{\left( \frac{p}{q} – p \times q \right)}{p \times q} \times 100 \] Let’s set up the ratio \( \frac{\frac{p}{q}}{p \times q} \), solve it, and use algebra to find that \( q = \sqrt{5} \).
- 1
- 20
- 2
- 1/2
The correct answer is A. 1.
Explanation: The sum of the first \( n \) consecutive odd numbers is given by the formula: \[ S_n = n^2 \] For the first 20 odd numbers: \[ S_{20} = 20^2 = 400 \] Now, divide the sum by 202: \[ \frac{400}{202} = \frac{200}{101} = 1.98 \approx 1 \] So, the result is 1.
Explanation: The sum of the first \( n \) consecutive odd numbers is given by the formula: \[ S_n = n^2 \] For the first 20 odd numbers: \[ S_{20} = 20^2 = 400 \] Now, divide the sum by 202: \[ \frac{400}{202} = \frac{200}{101} = 1.98 \approx 1 \] So, the result is 1.
- 150
- 200
- 250
- 175
Explanation: Given: – Number of girls in both classes is \( X \). – Total students in 8th class = 450. – Total students in 9th class = 360. The ratio condition is: \[\frac{X}{450 – X} = \frac{360 – X}{X}\] Cross-multiplying and solving: \[X^2 = (450 – X)(360 – X)\] Expanding and solving the quadratic equation: \[X^2 = 162000 – 450X – 360X + X^2\] \[0 = 162000 – 810X\] \[X = \frac{162000}{810} = 200\] So, the number of girls in each class is 200.
- (i) out (ii) down (iii) in (iv) for
- (i) down (ii) out (iii) by (iv) in
- (i) down (ii) out (iii) for (iv) in
- (i) out (ii) down (iii) by (iv) for
The correct answer is D. (i) out (ii) down (iii) by (iv) for.
Explanation:
- (i) out: “Yoko Roi stands out as an author” — to “stand out” means to be noticeable or prominent.
- (ii) down: “for standing down as an honorary fellow” — “standing down” means to resign or step down from a position.
- (iii) by: “after she stood by her writings” — “stood by” means to support or defend.
- (iv) for: “that stand for the freedom of speech” — “stand for” means to represent or support a cause or idea.
Hence, the correct sequence is out, down, by, for.
- Site preparation equipment
- Road building machinery
- Heavy Earth Moving Machinery
- Material Handling machinery
Dozer, motor graders used in site preparation are also deployed for auxiliary operation in road making, excavation and loading and
dumpsite etc.
- Expansion
- Shrinkage
- Contraction
- Rise
a. Expansion
Expansion; Removal of overburden involve fragmentation due to blasting. The fragmented rocks are also vulnerable to more weathering. The previously loaded rocks get relaxed with the release of the forces leading to volume expansion. Creation of new surface area is associated with expansion of volume.
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