TET Paper 1 Mathematics Content Bits Previous year question Paper Key with solutions

111) If the angles of a triangle are in the ratio 1:4:5, then the triangle is:
త్రిభుజం యొక్క కోణాలు 1:4:5 నిష్పత్తిలో ఉంటే, ఆ త్రిభుజం:

A) A right-angled triangle
లంబకోణ త్రిభుజం
B) An acute-angled triangle
అల్పకోణ త్రిభుజం
C) An obtuse-angled triangle
అధిక కోణ త్రిభుజం
D) An equal-angled triangle
సమ కోణ త్రిభుజం

View Answer
A) A right-angled triangle

Explanation:We are given that the angles of a triangle are in the ratio (1:4:5).
Let’s use this information to determine the type of triangle.
Step 1: Understand the sum of the angles of a triangle
The sum of the interior angles of any triangle is always (1800).
Step 2: Let the angles be (x), (4x), and (5x)
Since the angles are in the ratio (1:4:5), we can represent the angles as:
– First angle = (x),
– Second angle = (4x),
– Third angle = (5x).
Step 3: Set up the equation
The sum of the angles in a triangle is 1800:
x + 4x + 5x = 180
Step 4: Solve for (x)10x = 180x = \frac{180}{10} = 18
Step 5: Calculate the individual angles
Now that we know (x = 18), we can find the actual angles:
– First angle = (x = 180),
– Second angle = (4x = 4 X 18 = 720),
– Third angle = (5x = 5 X 18 = 900).
Step 6: Identify the type of triangle
Since one of the angles is 900, the triangle is a right triangle.

112) When a Robin bird flies, it flaps its wings 23 times in ten seconds. The number of times it flaps its wings in one and half minute is:
ఒక రాబిన్ పిట్ట ఎగిరేటప్పుడు, రెక్కలను పది సెకన్లలో 23 సార్లు ఆడిస్తుంది. అయితే, అది ఒకటిన్నర నిమిషంలో రెక్కలను ఎన్నిసార్లు ఆడిస్తుంది?

A) 2070
B) 138
C) 207
D) 1380

View Answer
C) 207

Explanation:To determine the number of times the robin flaps its wings in one and a half minutes, we can use the information given:
– The robin flaps its wings 23 times in 10 seconds.
– We need to find the total flaps in 1.5 minutes (which is 90 seconds).
Step 1: Find the number of flaps per second
First, determine how many times the robin flaps its wings per second:
Flaps per second = \frac{23 \, \text{flaps}}{10 \, \text{seconds}} = 2.3 flaps per second
Step 2: Find the number of flaps in 90 seconds
Now, multiply the flaps per second by 90 seconds:
Flaps in 90 seconds} = 2.3flaps per second X 90seconds}
= 207flaps

113) In a triangle, the height is double of its base and its area is 484sq.cm. The length of its height (in cm) is:
ఒక త్రిభుజంలో ఎత్తు దాని భూమికి రెండు రెట్లు ఉంది మరియు దాని వైశాల్యం 484చ.సెం.మీ. అయితే, ఆ త్రిభుజ ఎత్తు (సెం.మీ.లలో):

A) 22
B) 44
C) 34
D) 28

View Answer
B) 44

Explanation:We are given a triangle with the following information:
– The height of the triangle is double the length of its base.
– The area of the triangle is 484 sq.cm
Step 1: Formula for the area of a triangle
The formula for the area ( A ) of a triangle is:
A = \frac{1}{2} \times \text{base} \times \text{height}
Step 2: Let the base of the triangle be (b) cm
Since the height is double the base, the height will be ( 2b ) cm.
Substituting these values into the area formula:
484 = \frac{1}{2} \times b \times 2b
Step 3: Simplify the equation
484 = \frac{1}{2} \times 2b^2484 = b2
Step 4: Solve for ( b )
Taking the square root of both sides:
b = \sqrt{484} = 22 cm
Step 5: Find the height
Since the height is double the base, the height is:
Height = 2b = 2 X 22 = 44 cm

114) (105)2 – (95)2 =
Options:

A) 100
B) 200
C) 1000
D) 2000

View Answer
D) 2000

Explanation:We are asked to simplify the expression:(105)2 – (95)2
Step 1: Use the difference of squares formula
The difference of squares formula states:
a2 – b2 = (a – b)(a + b)
In our case, ( a = 105 ) and ( b = 95 ).
Applying the formula:
(105)2 – (95)2 = (105 – 95)(105 + 95)
Step 2: Simplify the factors
Now, simplify each part:105 – 95 = 10 and 105 + 95 = 200
Thus, the expression becomes:(105)2 – (95)2 = 10 X 200 = 2000

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