CBSE Class 7 Mathematics Chapter 1: Integers NCERT Solutions

(a) By observing the number line, we can determine the temperature of each marked place:

• Srinagar: -2°C
• Lahulspiti: -8°C
• Shimla: 5°C
• Ooty: 14°C
• Bangalore: 22°C

(b) The hottest place is Bangalore with a temperature of 22°C. The coldest place is Lahulspiti with a temperature of -8°C. The temperature difference is calculated as:

$$\backslash text\{Difference\}\; =\; \backslash text\{Temperature\; of\; hottest\; place\}\; -\; \backslash text\{Temperature\; of\; coldest\; place\}$$

$$\backslash text\{Difference\}\; =\; 22^\{\backslash circ\}\backslash text\{C\}\; -\; (-8^\{\backslash circ\}\backslash text\{C\})\; =\; 22^\{\backslash circ\}\backslash text\{C\}\; +\; 8^\{\backslash circ\}\backslash text\{C\}\; =\; 30^\{\backslash circ\}\backslash text\{C\}$$

Therefore, the temperature difference is 30°C.

(c) The temperature of Lahulspiti is -8°C and the temperature of Srinagar is -2°C. The temperature difference is:

$$\backslash text\{Difference\}\; =\; \backslash text\{Temperature\; of\; Srinagar\}\; -\; \backslash text\{Temperature\; of\; Lahulspiti\}$$

$$\backslash text\{Difference\}\; =\; (-2^\{\backslash circ\}\backslash text\{C\})\; -\; (-8^\{\backslash circ\}\backslash text\{C\})\; =\; -2^\{\backslash circ\}\backslash text\{C\}\; +\; 8^\{\backslash circ\}\backslash text\{C\}\; =\; 6^\{\backslash circ\}\backslash text\{C\}$$

Therefore, the temperature difference between Lahulspiti and Srinagar is 6°C.

(d) The temperature of Srinagar is -2°C and the temperature of Shimla is 5°C. The sum of their temperatures is:

$$\backslash text\{Sum\}\; =\; \backslash text\{Temperature\; of\; Srinagar\}\; +\; \backslash text\{Temperature\; of\; Shimla\}$$

$$\backslash text\{Sum\}\; =\; (-2^\{\backslash circ\}\backslash text\{C\})\; +\; 5^\{\backslash circ\}\backslash text\{C\}\; =\; 3^\{\backslash circ\}\backslash text\{C\}$$

Now, we compare this sum with the temperature at Shimla (5°C):

So, the temperature of Srinagar and Shimla taken together (3°C) is less than the temperature at Shimla (5°C).

Next, we compare the sum (3°C) with the temperature at Srinagar (-2°C):

$$3^\{\backslash circ\}\backslash text\{C\}\; >\; -2^\{\backslash circ\}\backslash text\{C\}$$

Thus, the combined temperature is not less than the temperature at Srinagar.

Jack's scores in the five successive rounds are 25, -5, -10, 15, and 10.

To find his total score, we need to add all these scores together:

$$\backslash text\{Total\; Score\}\; =\; 25\; +\; (-5)\; +\; (-10)\; +\; 15\; +\; 10$$

First, sum the positive numbers: $25\; +\; 15\; +\; 10\; =\; 50$

Next, sum the negative numbers: $-5\; +\; (-10)\; =\; -15$

Now, combine these sums:

$$\backslash text\{Total\; Score\}\; =\; 50\; +\; (-15)\; =\; 50\; -\; 15\; =\; 35$$

Therefore, Jack's total score at the end of the five rounds was 35 marks.

The temperature on Monday in Srinagar was -5°C.

On Tuesday, the temperature dropped by 2°C. A drop in temperature means we subtract.

$$\backslash text\{Temperature\; on\; Tuesday\}\; =\; \backslash text\{Temperature\; on\; Monday\}\; -\; \backslash text\{Drop\; in\; temperature\}$$

$$\backslash text\{Temperature\; on\; Tuesday\}\; =\; -5^\{\backslash circ\}\backslash text\{C\}\; -\; 2^\{\backslash circ\}\backslash text\{C\}\; =\; -7^\{\backslash circ\}\backslash text\{C\}$$

So, the temperature on Tuesday was -7°C.

On Wednesday, the temperature rose by 4°C from Tuesday's temperature. A rise in temperature means we add.

$$\backslash text\{Temperature\; on\; Wednesday\}\; =\; \backslash text\{Temperature\; on\; Tuesday\}\; +\; \backslash text\{Rise\; in\; temperature\}$$

$$\backslash text\{Temperature\; on\; Wednesday\}\; =\; -7^\{\backslash circ\}\backslash text\{C\}\; +\; 4^\{\backslash circ\}\backslash text\{C\}\; =\; -3^\{\backslash circ\}\backslash text\{C\}$$

Thus, the temperature on Tuesday was -7°C and on Wednesday was -3°C.

The plane is flying at a height of 5000 m above the sea level. We can represent this as +5000 m.

The submarine is floating at a depth of 1200 m below the sea level. We can represent this as -1200 m.

The vertical distance between the plane and the submarine is the sum of the distance of the plane above sea level and the distance of the submarine below sea level.

$$\backslash text\{Vertical\; Distance\}\; =\; \backslash text\{Height\; of\; plane\; above\; sea\; level\}\; +\; \backslash text\{Depth\; of\; submarine\; below\; sea\; level\}$$

$$\backslash text\{Vertical\; Distance\}\; =\; 5000\; \backslash text\{\; m\}\; +\; 1200\; \backslash text\{\; m\}\; =\; 6200\; \backslash text\{\; m\}$$

Alternatively, we can find the difference between their positions:

$$\backslash text\{Vertical\; Distance\}\; =\; (\backslash text\{Position\; of\; plane\})\; -\; (\backslash text\{Position\; of\; submarine\})$$

$$\backslash text\{Vertical\; Distance\}\; =\; 5000\; \backslash text\{\; m\}\; -\; (-1200\; \backslash text\{\; m\})\; =\; 5000\; \backslash text\{\; m\}\; +\; 1200\; \backslash text\{\; m\}\; =\; 6200\; \backslash text\{\; m\}$$

Therefore, the vertical distance between the plane and the submarine is 6200 m.