RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: $n(A\cup O)=n(A)+n(O)-n(A\cap O)$; "neither" $=n(U)-n(A\cup O)$.

1. $n(A\cup O)=100+150-75=175$.
2. Neither $=260-175=85$.

Trick: Only apple $=25$, both $=75$, only orange $=75$; total $175$; the rest $85$ take neither.

Answer: (B) 85

Concept: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$; total outcomes $=42$.

1. Multiples of 7 up to 42: $7,14,\dots ,42$ → 6 numbers.
2. Multiples of 8: $8,16,24,32,40$ → 5 numbers.
3. Common multiple (LCM $=56$) → none up to 42.
4. Favourable $=6+5=11$, so $P=\frac{11}{42}$.

Trick: Check the LCM first — if LCM $>$ range, just add the two counts.

Answer: (D) $\frac{11}{42}$

Concept: $D-C$ = elements of $D$ that are not in $C$.

1. $C=\{2,4,6,8,10,12,14,16\}$, $D=\{5,10,15,20\}$.
2. Common element: only $10$.
3. $D-C=\{5,15,20\}$ → 3 elements.

Trick: $n(D-C)=n(D)-n(D\cap C)=4-1=3$.

Answer: (A) 3

Concept: Follow BODMAS — bracket first, then the division.

1. $\frac{3}{4}+\frac{4}{6}=\frac{3}{4}+\frac{2}{3}=\frac{9+8}{12}=\frac{17}{12}$.
2. $\frac{1}{17/12}=\frac{12}{17}$.
3. $\frac{12}{17}÷\frac{2}{15}=\frac{12}{17}\times \frac{15}{2}=\frac{90}{17}=5\frac{5}{17}$.

Trick: $\frac{1}{a/b}=\frac{b}{a}$ — flip the fraction instead of dividing.

Answer: (A) $5\frac{5}{17}$

Concept: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $26.7=3\times 71-2\text{Mean}=213-2\text{Mean}$.
2. $2\text{Mean}=213-26.7=186.3$.
3. $\text{Mean}=93.15\approx 93.2$.

Trick: $\text{Mean}=\frac{3\text{Median}-\text{Mode}}{2}$.

Answer: (B) 93.2

Concept: Shoelace formula: $\text{Area}=\frac{1}{2}|\sum (x_i{y}_{i+1}-x_{i+1}{y}_i)|$.

1. Vertices in order: $(3,0),(4,5),(-1,4),(-2,-1)$.
2. $\sum x_i{y}_{i+1}=3(5)+4(4)+(-1)(-1)+(-2)(0)=32$.
3. $\sum y_i{x}_{i+1}=0(4)+5(-1)+4(-2)+(-1)(3)=-16$.
4. $\text{Area}=\frac{1}{2}|32-(-16)|=24$ sq. units.

Trick: Write the points in a column, repeat the first, cross-multiply down-right minus down-left, halve.

Answer: (C) 24 sq.units

Concept: Area of four walls $=2(l+b)h$; floor area $=l\times b$.

1. Let $b=x$, $l=2x$, $h=13$: $2(2x+x)\times 13=780$.
2. $78x=780\Rightarrow x=10$, so $b=10\text{m}$, $l=20\text{m}$.
3. Floor area $=20\times 10=200{\text{m}}^2$.

Trick: $l+b=\frac{780}{2\times 13}=30$; with $l=2b$ → $b=10$, $l=20$.

Answer: (D) $200{\text{m}}^2$

Concept: For $a{x}^2+bx+c$ with $a>0$, the minimum is $c-\frac{b^2}{4a}$ at $x=-\frac{b}{2a}$.

1. $3x^2+8x+14$: $a=3,b=8,c=14$.
2. Minimum $=14-\frac{64}{12}=14-\frac{16}{3}=\frac{42-16}{3}=\frac{26}{3}$.
3. $\frac{26}{3}=8\frac{2}{3}$ (at $x=-\frac{4}{3}$).

Trick: Minimum of a quadratic $=\frac{4ac-b^2}{4a}=\frac{168-64}{12}=\frac{104}{12}=\frac{26}{3}$.

Answer: (A) $8\frac{2}{3}$

Concept: Marks obtained $=\frac{\text{percentage}}{100}\times \text{maximum marks}$.

1. Mary's percentage in History $=60\%$ and the paper is out of $75$.
2. Marks $=\frac{60}{100}\times 75=0.6\times 75=45$.

Trick: $60\%$ of $75$ = $75-40\%$ of $75$ = $75-30=45$.

Answer: (A) 45

Concept: A radius is perpendicular to the tangent at the point of contact, so $\angle O_{1}PR=\angle O_{2}QR={90}^{\circ }$; and vertically opposite angles at $R$ are equal.

1. In right $△O_{1}PR$: $\angle PR{O}_1={180}^{\circ }-{90}^{\circ }-{60}^{\circ }={30}^{\circ }$.
2. $\angle QR{O}_2=\angle PR{O}_1={30}^{\circ }$ (vertically opposite angles).
3. In right $△O_{2}QR$: $\angle Q{O}_{2}R={180}^{\circ }-{90}^{\circ }-{30}^{\circ }={60}^{\circ }$.

Trick: $△O_{1}PR~△O_{2}QR$ (AA), so the angles at the two centres are equal – the answer must be ${60}^{\circ }$ and ${30}^{\circ }$.

Answer: (D) ${60}^{\circ }$ and ${30}^{\circ }$

Concept: For $a{x}^2+bx+c=0$: $\alpha +\beta =-\frac{b}{a}$, $\alpha \beta =\frac{c}{a}$ and ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$.

1. $\alpha +\beta =11$, $\alpha \beta =24$.
2. ${\alpha }^2+{\beta }^2={11}^2-2\times 24=121-48=73$.

Trick: Roots are $3$ and $8$ ($3+8=11$, $3\times 8=24$): $9+64=73$.

Answer: (D) 73

Concept: When the first term $a$ and last term $l$ are known, $S_n=\frac{n}{2}(a+l)$.

1. $396=\frac{n}{2}(18+48)=\frac{n}{2}\times 66=33n$.
2. $n=\frac{396}{33}=12$.

Trick: Sum = number of terms × average of first and last term: $396÷33=12$.

Answer: (D) 12

Concept: The smallest number divisible by several numbers is their LCM.

1. $24=2^3\times 3$, $6=2\times 3$, $36=2^2\times 3^2$, $13$ is prime.
2. LCM $=2^3\times 3^2\times 13=8\times 9\times 13=936$.

Trick: LCM$(24,36)=72$; $72\times 13=936$ (6 is already a factor of 24).

Answer: (C) 936

Concept: Volume is conserved on melting: $\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$. Surface area of a sphere $=4\pi r^2$.

1. $R^3=64r^3\Rightarrow r=\frac{R}{4}=1\text{cm}$.
2. Original surface area $=4\pi \times 4^2=64\pi$; 5 small spheres: $5\times 4\pi \times 1^2=20\pi$.
3. Ratio $=64\pi :20\pi =16:5$.

Trick: $64=4^3$ ⇒ radius becomes $\frac{1}{4}$, area becomes $\frac{1}{16}$; ratio $=16:5$ for 5 spheres.

Answer: (D) 16 : 5

Concept: In an A.P. $a_n=a_m+(n-m)d$; consecutive terms differ by $d$.

1. $d=a_{24}-a_{23}=312-42=270$.
2. $a_{37}=a_{23}+14d=42+14\times 270=42+3780=3822$.

Trick: Jump directly from the 23rd term: 14 steps of 270 = 3780.

Answer: (D) 3822

Concept: Write each decimal as a perfect square: $\sqrt{0.0324}=\sqrt{\frac{324}{10000}}=\frac{18}{100}$, $\sqrt{6.76}=\sqrt{\frac{676}{100}}=\frac{26}{10}$.

1. $\sqrt{144}=12$, $\sqrt{0.0324}=0.18$, $\sqrt{6.76}=2.6$.
2. $12+0.18-2.6=9.58$.

Trick: Even number of decimal places → take the root of the digits and halve the decimal places ($324\to 18$, 4 places → 2 places).

Answer: (B) 9.58

Concept: For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides.

1. $\frac{\text{ar}(ABC)}{\text{ar}(LMN)}={\left(\frac{BC}{MN}\right)}^2\Rightarrow \frac{25}{36}={\left(\frac{2.5}{MN}\right)}^2$.
2. $\frac{BC}{MN}=\frac{5}{6}\Rightarrow MN=\frac{2.5\times 6}{5}=3\text{cm}$.

Trick: $25:36$ areas → $5:6$ sides; $2.5\times \frac{6}{5}=3$.

Answer: (C) 3

Concept: $\frac{1}{{\sin }^{2}A}={\csc }^{2}A$ and the identity $1+{\cot }^{2}A={\csc }^{2}A$.

1. $\frac{1}{{\sin }^{2}A}-1={\csc }^{2}A-1={\cot }^{2}A$.
2. Directly: $\frac{1-{\sin }^{2}A}{{\sin }^{2}A}=\frac{{\cos }^{2}A}{{\sin }^{2}A}={\cot }^{2}A$.
3. Check with $A={30}^{\circ }$: $\frac{1}{{\sin }^2{30}^{\circ }}-1=4-1=3={\cot }^2{30}^{\circ }$ ✓ (while ${\tan }^2{30}^{\circ }=\frac{1}{3}$, so ${\tan }^{2}A$ is wrong).

Yaad rakhein: $1+{\cot }^{2}A={\csc }^{2}A$ and $1+{\tan }^{2}A={\sec }^{2}A$ – cot pairs with cosec, tan with sec.

Answer: (A) ${\cot }^{2}A$

Concept: The centre of a circle is the midpoint of any diameter: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

1. Let $A=(x,y)$, $B=(1,4)$, centre $=(2,-3)$.
2. $\frac{x+1}{2}=2\Rightarrow x=3$; $\frac{y+4}{2}=-3\Rightarrow y=-10$.

Trick: $A=2\times \text{centre}-B=(4-1,-6-4)=(3,-10)$.

Answer: (C) (3, −10)

Concept: $\tan \theta =\frac{8}{15}$ ⇒ perpendicular 8, base 15, hypotenuse $\sqrt{64+225}=17$, so $\sin \theta =\frac{8}{17}$.

1. $1-\sin \theta =\frac{9}{17}$, $1+\sin \theta =\frac{25}{17}$.
2. $\frac{\sqrt{1-\sin \theta }}{\sqrt{1+\sin \theta }}=\sqrt{\frac{9}{25}}=\frac{3}{5}$.

Trick: 8-15-17 is a Pythagorean triplet; $\frac{17-8}{17+8}=\frac{9}{25}$ – both perfect squares, root $=\frac{3}{5}$.

Answer: (A) $\frac{3}{5}$