Question

SECTION - B
Questions 11 to 16 carry 2 marks each.
11. The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until
midnight, at what time would the temperature be 8°C below zero? What would be the
temperature at mid-night?
12. Evaluate each of the following:
[(-36) = 12] = 3 (b) [(-6) + 5)] = [(-2) + 1]
13. Find the values of the angles x, y, and z in the given figure:
14. PQR is a triangle right angled at P. If PQ - 10 cm and PR - 24 cmn, find QR​

Answer 1

Answer:

11. At 12 noon, the temperature =10°C above zero

The temperature is decreasing at 2°C per hour

8°C below zero =−8°C

So, total fall in temperature from 10°C to 0°C and from 0°C to −8°C=18°C

Since the temperature falls 2°C in every one hour.

Therefore, to decrease 18°C, time taken = 18/2

= 9 hours

Present time is 12 noon so, the time when the temperature is −8°C

=12noon+9hours

So, The time at which the temperature will be −8°C=9PM

At 9 PM, the temperature is −8°C, the temperature at 12 midnight will be =

(−8°C)+(−2°C×3hours)

=(−8°C)+(−6°C)

=−14°C

So, at 12 midnight the temperature will be−14°C.

12.a.[(-36) ÷ 12] = 3

=[-3] ÷ 3

=-1

b.[(-6) + 5)] ÷ [(-2) + 1]

=(-1) ÷ (-1)

=1

13. We know that y=58° (alternate angles)

By supplementary angles property:

x° + y° = 180°

58° + x° = 180°

x° = 122°

z = 122° (alternate angles)

14. Given: PQ=10cm, PR=24cm

Let QR be x cm.

In right angled triangle QPR,

(Hypotenuse)

2

=(Base)

2

+(Perpendicular)

2

[By Pythagoras theorem]

⇒(QR)

2

=(PQ)

2

+(PR)

2

⇒x

2

=(10)

2

+(24)

2

⇒x

2

=100+576=676

⇒x=

676

=26cm

Thus, the length of QR is 26cm.

solution