TS TET Mathematics 8 Jan 2025 Shift 1 Solved Paper

D) no predecessor
ముందు సంఖ్య ఉండదు
Explanation:Whole numbers are 0, 1, 2, 3, …
0 is the smallest whole number, so it has no predecessor.
Correct Answer: (no predecessor)

A) 4
Explanation:Let's solve step-by-step.
Step 1: Given data
Rectangle: length \( l = 32 \, \text{cm} \), breadth \( b = 28 \, \text{cm} \)
Perimeter of rectangle:
\( P_{\text{rect}} = 2 \times (l + b) = 2 \times (32 + 28)
= 2 \times 60 = 120 \, \text{cm} \)
Step 2: Perimeter of square
Given \( P_{\text{rect}} = P_{\text{square}} = 120 \, \text{cm} \)
Perimeter of square \( = 4 \times \text{side} \)
So, \( 4 \times s = 120 \)
\( s = 30 \, \text{cm} \)
Step 3: Areas
Area of rectangle:
\( A_{\text{rect}} = l \times b = 32 \times 28 = 896 \, \text{cm}^2 \)
Area of square:
\( A_{\text{sq}} = s^2 = 30 \times 30 = 900 \, \text{cm}^2 \)
Step 4: Difference
\( A_{\text{sq}} – A_{\text{rect}} = 900 – 896 = 4 \, \text{cm}^2 \)
Final answer:
\( \boxed{1} \)

D) 5
Explanation:A regular polygon has as many lines of symmetry as its number of sides.
A regular pentagon has 5 sides.

B) 6
Explanation:Let's understand the problem.
The number is: \( 6 ? 5 ? 4 \).
Let the first “?” digit (thousands place) be \( a \) and the second “?” digit (tens place) be \( b \).
So the number is:
6 a 5 b 4
That means digits (from left to right) are:
– Ten-thousands place: 6
– Thousands place: a
– Hundreds place: 5
– Tens place: b
– Units place: 4
Step 1: Interpret “difference between its place value”
The phrase likely means the difference between the place value of digit \(a\) and the place value of digit \(b\).
Place value of a = \( a \times 1000 \)
Place value of b = \( b \times 10 \)
The problem says: difference between these two place values = 5940.
Which is larger? Possibly \(a\)'s place value minus \(b\)'s place value = 5940, because \(1000a > 10b\) for most \(a, b\) digits (0–9).
\( 1000a – 10b = 5940 \)
Step 2: Solve equation
\( 1000a – 10b = 5940 \)
Divide by 10:
\( 100a – b = 594 \)
So:
\( b = 100a – 594 \)
Step 3: Digit constraints
\(a\) and \(b\) are digits 0–9.
\(b \ge 0\) ⇒ \(100a – 594 \ge 0\) ⇒ \(100a \ge 594 \) ⇒ \(a \ge 5.94 \) ⇒ \(a \ge 6\)
\(b \le 9\) ⇒ \(100a – 594 \le 9\) ⇒ \(100a \le 603\) ⇒ \(a \le 6.03\) ⇒ \(a \le 6\)
So \(a = 6\) only.
Then \(b = 100 \times 6 – 594 = 600 – 594 = 6\).
Thus \(a=6, b=6\).
Step 4: Check the other way?
Could it be \(10b – 1000a = 5940\)?
That would mean \(10b – 1000a = 5940\) ⇒ \(10b = 1000a + 5940\) ⇒ \(b = 100a + 594\)
That's >594 for any \(a\), impossible since \(b\) is a digit.So not possible.
So \(a=6, b=6\) fits.
Step 5: Match with options
They ask: “find the digit at '?' place” — possibly they mean the same digit for both placeholders? The given options are single digits (7, 6, 8, 5).
Since \(a=6\), the digit at “?” place is \(6\).
\( \boxed{2} \)

C) ₹21700
Explanation:Oil required per day = 7 litres
Cost per litre = ₹100
Cost per day = 7 × 100 = ₹700
For 31 days = 700 × 31 = ₹21700

C) 50
Explanation:To find the median of the runs scored:
Data: 60, 15, 120, 50, 48, 4, 56, 100, 80, 8, 15
Step 1: Arrange in ascending order:
4, 8, 15, 15, 48, 50, 56, 60, 80, 100, 120
Step 2: Since there are \( n = 11 \) players (odd number), median is the middle value, which is the \( \frac{n+1}{2} = \frac{12}{2} = 6 \)th term in ascending order.
Step 3: The 6th term in the sorted list is \( 50 \).
Thus, median = \( 50 \).
\( \boxed{50} \)

D) \( \frac{15}{17} \times (2^{0} – 1) \)
Explanation:Evaluate each option:
1) \( \frac{15}{16} – \frac{3}{4} = \frac{15}{16} – \frac{12}{16} = \frac{3}{16} \) (not whole)
2) \( \frac{2}{5} \div 1 = \frac{2}{5} \) (not whole)
3) \( \frac{11}{12} + \left( \frac{1}{2} \times \frac{5}{6} \right)
= \frac{11}{12} + \frac{5}{12} = \frac{16}{12} = \frac{4}{3} \) (not whole)
4) \( \frac{15}{17} \times (2^{0} – 1)
= \frac{15}{17} \times (1 – 1)
= \frac{15}{17} \times 0
= 0 \) (whole number)

A) \( \pi r^{2} h \)
Explanation:Volume of a cylinder
= Base area × Height
= \( \pi r^{2} h \)

B) 8
Explanation:Let's solve step-by-step.
Let the number be \( x \).
Step 1: Multiply by 8: \( 8x \)
Subtract 16 from the product: \( 8x – 16 \)
Step 2: This gives six times the number itself:
8x – 16 = 6x
Step 3: Solve for x:
8x – 6x = 16
2x = 16
x = 8
Step 4: Check:
\( 8 \times 8 = 64 \)
\( 64 – 16 = 48 \)
\( 6 \times 8 = 48 \) ✅
\( \boxed{8} \)

D) right angle
లంబ కోణం
Explanation:Supplementary angles sum to 180°.
Let angle = x
Supplementary angle = \( 180 – x \)
Given:
\( 180 – x = x \)
\( 2x = 180 \Rightarrow x = 90° \)
90° is a right angle.