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TGT Maths – Practice Set (Day 1)

TGT Mathematics – Previous Year Practice Set (With Answers)

Day 1 – Number Systems, Algebra, Geometry, Trigonometry, Mensuration, Arithmetic

1. Solve the equation:

$$\frac{5x-7}{3}=\frac{2x+11}{5}$$

Answer:
$x = \dfrac{68}{19}$

2. A number is divisible by 9 and 11 and lies between 900 and 1000. Find the number.

Answer:
$\boxed{990}$

3. If $\sqrt{75} - \sqrt{12} = a\sqrt{3}$, find $a$.

Answer:
$a = 3$

4. Solve:

$$x - \frac{9}{x} = 4$$

Answer:
$x = 2 + \sqrt{13}$ or $x = 2 - \sqrt{13}$

5. If $(x-3)(x-4)=7$, find the roots of the quadratic.

Answer:
Equation reduces to:
$x^2 - 7x + 5 = 0$

Roots:

$$x = \frac{7 \pm \sqrt{29}}{2}$$

6. For the polynomial

$$p(x) = 2x^3 - 5x^2 + kx + 4,$$

if $(x - 1)$ is a factor, find $k$.

Answer:
$k = -1$

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7. Find the distance between A(–2, 3) and B(4, –1).

Answer:

$$\sqrt{(4 - (-2))^2 + (-1 - 3)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$$

8. The midpoint of a line segment is (3, –2). One endpoint is (7, 4). Find the other endpoint.

Answer:
Other endpoint $= ( -1, -8 )$

(Worked: If midpoint M = ((x1+x2)/2, (y1+y2)/2) = (3, -2) and one end (7,4), solve for the other.)

9. If $\sin \theta = \dfrac{3}{5}$, find $\cos \theta$ and $\tan \theta$.

Answer:

$$\cos \theta = \frac{4}{5}, \qquad \tan \theta = \frac{3}{4}$$

(First quadrant assumed.)

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10. Evaluate:

$$\sin 30^\circ + \cos 60^\circ + \tan 45^\circ$$

Answer:

$$\frac12 + \frac12 + 1 = 2$$

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11. A triangle has sides 7 cm, 8 cm, 9 cm. Find its area (Heron’s formula).

Solution:
$s = \dfrac{7+8+9}{2} = 12$

Area

$$= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}$$

Answer:

$$\boxed{12\sqrt{5}\ \text{cm}^2}$$

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12. A cone has radius 7 cm and height 24 cm. Find its slant height.

Answer:

$$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\ \text{cm}$$

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13. The average of 7 numbers is 18. If one number, 42, is removed, what is the new average?

Solution:
Total initially $= 7 \times 18 = 126$. After removing 42, new total $= 126 - 42 = 84$.
New average $= 84/6 = 14$.

Answer:
$14$

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14. A class has boys and girls in the ratio 5:3. If there are 40 students, how many girls are there?

Solution:
Total parts $= 5+3 = 8$. Each part $= 40/8 = 5$.
Girls $= 3 \times 5 = 15$.

Answer:
$15$ girls

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15. A bag contains 3 red, 5 blue and 2 green balls. One ball is drawn at random. What is the probability that it is blue?

Solution:
Total balls $= 3+5+2 = 10$. Favorable outcomes (blue) $=5$.
Probability $= 5/10 = 1/2$.

Answer:
$\dfrac{1}{2}$