RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: Drawing one by one without replacement – multiply the successive probabilities; a pack has 4 jacks in 52 cards.

1. $P=\frac{4}{52}\times \frac{3}{51}\times \frac{2}{50}$.
2. $=\frac{24}{132600}=\frac{1}{5525}$.

Trick: Same as choosing 3 jacks together: $\frac{{}^4{C}_3}{{}^{52}{C}_3}=\frac{4}{22100}=\frac{1}{5525}$.

Answer: (B) $\frac{1}{5525}$

Concept: In an isosceles triangle the angles opposite equal sides are equal; the three angles add to ${180}^{\circ }$.

1. $AC=BC$ ⇒ $\angle A=\angle B=\frac{2}{5}\angle C-9$.
2. Sum: $2(\frac{2}{5}C-9)+C=180$ ⇒ $\frac{9}{5}C=198$ ⇒ $\angle C={110}^{\circ }$.
3. $\angle A=\angle B=44-9={35}^{\circ }$.
4. $\frac{2(110)-3(35)+35}{2}=\frac{220-105+35}{2}=\frac{150}{2}={75}^{\circ }$.

Trick: Since $\angle A=\angle B$, the expression simplifies to $\frac{2C-2A}{2}=C-A=110-35=75$.

Answer: (B) ${75}^{\circ }$

Concept: Simplify the bracket first, take its reciprocal, then divide by a fraction = multiply by its reciprocal.

1. $\frac{5}{6}+\frac{7}{9}=\frac{15+14}{18}=\frac{29}{18}$, so $\frac{1}{29/18}=\frac{18}{29}$.
2. $\frac{18}{29}÷\frac{5}{23}=\frac{18}{29}\times \frac{23}{5}=\frac{414}{145}=2\frac{124}{145}\approx 2.855$.
3. Compare with options: $2\frac{59}{69}\approx 2.855$, $2\frac{5}{6}\approx 2.833$, $2\frac{53}{61}\approx 2.869$, $4\frac{1}{27}\approx 4.04$. The value matches option A.

Trick: When exact fractions in options look odd, convert everything to decimals and pick the closest – it saves time in the exam.

Answer: (A) $2\frac{59}{69}$

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

1. Side ratio $=6:2$.
2. Area ratio $=6^2:2^2=36:4$ (i.e. $9:1$).
3. Option C gives exactly $36:4$ in the order mentioned.

Trick: Sides → square → areas; areas → square root → sides. For volumes of similar solids, cube the ratio.

Answer: (C) 36 : 4

Concept: Shadow length $=\frac{h}{\tan \theta }$; a higher sun gives a shorter shadow.

1. At ${30}^{\circ }$: shadow $=h\cot {30}^{\circ }=\sqrt{3}h$; at ${45}^{\circ }$: shadow $=h\cot {45}^{\circ }=h$.
2. Difference: $\sqrt{3}h-h=80$ ⇒ $h(\sqrt{3}-1)=80$.
3. $h=\frac{80}{\sqrt{3}-1}=\frac{80(\sqrt{3}+1)}{3-1}=40(\sqrt{3}+1)$ (units as given in the options).

Trick: For ${30}^{\circ }\to {45}^{\circ }$, $h=\frac{\text{decrease}}{\sqrt{3}-1}=\frac{\text{decrease}(\sqrt{3}+1)}{2}$ – rationalise directly.

Answer: (A) $40(\sqrt{3}+1)$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$.

1. Here $a=7$, $b=-18$, $c=-11$.
2. $D=(-18{)}^2-4(7)(-11)=324+308=632$.

Trick: When $c$ is negative, $-4ac$ becomes positive – add, don't subtract. $D>0$ ⇒ two real distinct roots.

Answer: (C) 632

Concept: Distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.

1. $d=\sqrt{(-3-0{)}^2+(0-3{)}^2}=\sqrt{9+9}$.
2. $=\sqrt{18}=3\sqrt{2}$ units.

Trick: The points are the ends of the legs of a right isosceles triangle with legs 3 and 3, so the hypotenuse is $3\sqrt{2}$.

Answer: (B) $3\sqrt{2}$ units

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

1. $574=\frac{n}{2}(36+46)=\frac{n}{2}\times 82=41n$.
2. $n=\frac{574}{41}=14$.

Trick: Number of terms = Sum ÷ (average of first and last term) = $574÷41=14$.

Answer: (B) 14

Concept: For $n$ independent tosses, $P(\text{exactly }r\text{ tails})=\left(\genfrac{}{}{0ex}{}{n}{r}\right){\left(\frac{1}{2}\right)}^n$.

1. $n=8$, $r=5$: $\left(\genfrac{}{}{0ex}{}{8}{5}\right)=\left(\genfrac{}{}{0ex}{}{8}{3}\right)=\frac{8\times 7\times 6}{6}=56$.
2. Total outcomes $=2^8=256$.
3. $P=\frac{56}{256}=0.21875\approx 0.22$.

Trick: $\left(\genfrac{}{}{0ex}{}{8}{5}\right)=\left(\genfrac{}{}{0ex}{}{8}{3}\right)$ – always use the smaller index to compute faster.

Answer: (A) 0.22

Concept: Total surface area of a cone $=\pi r(l+r)$, where slant height $l=\sqrt{r^2+h^2}$.

1. $l=\sqrt{{2.1}^2+{2.8}^2}=\sqrt{4.41+7.84}=\sqrt{12.25}=3.5\text{cm}$.
2. $\text{TSA}=\frac{22}{7}\times 2.1\times (3.5+2.1)=6.6\times 5.6=36.96{\text{cm}}^2$.

Trick: 2.1, 2.8, 3.5 is the 3-4-5 triplet × 0.7 – spot it to get $l$ instantly. Only CSA would be $\pi rl=23.1$.

Answer: (D) 36.96 cm²

Concept: $X\cap Y$ (intersection) contains only the elements common to both sets.

1. $X=\{a,b,c,d\}$, $Y=\{f,b,d,g\}$.
2. Elements present in both: $b$ and $d$ ⇒ $X\cap Y=\{b,d\}$.
3. $\{a,b,c,d,g,f\}$ would be the union $X\cup Y$.

Trick: ∩ looks like a cap – "common" elements only; ∪ looks like a cup – "collect everything" (union).

Answer: (D) { b, d }

Concept: Two dice give $6\times 6=36$ equally likely outcomes; probability = favourable ÷ total.

1. Pairs with sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) – 5 outcomes.
2. $P=\frac{5}{36}$.

Trick: For two dice, number of ways for sum $s$ is $s-1$ for $s\le 7$ and $13-s$ for $s\ge 7$; sum 8 → $13-8=5$.

Answer: (A) $\frac{5}{36}$

Concept: Identify the figure from side lengths and angles of the given vertices.

1. Take the points in order: $A(9,0)$, $B(9,6)$, $C(-9,6)$, $D(-9,0)$.
2. $AB=6$ (vertical), $BC=18$ (horizontal), $CD=6$, $DA=18$.
3. Adjacent sides are horizontal and vertical, so every angle is ${90}^{\circ }$; opposite sides equal but adjacent sides unequal ($6\ne 18$) → a rectangle, not a square.

Trick: Same $x$ for two points → vertical side; same $y$ → horizontal side. All right angles with unequal adjacent sides = rectangle.

Answer: (B) rectangle

Concept: $T_n=a+(n-1)d$; the difference of two terms gives $d$ directly.

1. $T_{12}-T_8=4d=478-82=396\Rightarrow d=99$.
2. $T_{50}=T_{12}+38d=478+38\times 99=478+3762=4240$.
3. (Check: $a=82-7\times 99=-611$; $-611+49\times 99=4240$ ✓)

Trick: $T_m-T_n=(m-n)d$ – no need to find $a$.

Answer: (D) 4240

Concept: Read the value from the graph and convert to a percentage of the maximum marks (50).

1. From the graph, Devi's marks in Punjabi $=45$ (out of 50).
2. Percentage $=\frac{45}{50}\times 100=90\%$.
3. Other points: Math 40 (80%), English 25 (50%), History 35 (70%).

Trick: Out of 50 → simply double the marks to get the percentage.

Answer: (D) 90%

Concept: Division and multiplication are done left to right before subtraction.

1. $9÷6=1.5$.
2. $1.5\times 6=9$.
3. $38-9=29$.

Trick: $÷6\times 6$ cancels out, leaving $38-9$.

Answer: (A) 29

Concept: $\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.

1. $A(-5,7)$, $B(-4,-5)$, $C(4,5)$.
2. $=\frac{1}{2}|(-5)(-5-5)+(-4)(5-7)+4(7+5)|$
3. $=\frac{1}{2}|50+8+48|=\frac{106}{2}=53$ sq. units.

Trick: Write the coordinates in a column, repeat the first row, and cross-multiply (shoelace method).

Answer: (A) 53 sq. units

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

1. $19.5=3\times 85-2\text{Mean}=255-2\text{Mean}$.
2. $2\text{Mean}=255-19.5=235.5$.
3. $\text{Mean}=117.75\approx 117.8$.

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

Answer: (C) 117.8

Concept: If adjacent faces have areas $lb$, $bh$, $hl$, then $(lbh{)}^2=lb\times bh\times hl$, so $V=\sqrt{A_1{A}_2{A}_3}$.

1. $V=\sqrt{260\times 104\times 10}=\sqrt{270400}$.
2. $270400=2704\times 100={52}^2\times {10}^2$.
3. $V=52\times 10=520{\text{cm}}^3$.

Trick: Volume = square root of the product of the three face areas.

Answer: (B) 520

Concept: The building (vertical), its shadow (horizontal) and the line from the top of the building to the shadow's end form a right triangle.

1. Required distance = hypotenuse $=\sqrt{8^2+6^2}$.
2. $=\sqrt{64+36}=\sqrt{100}=10\text{m}$.

Trick: 6-8-10 is the 3-4-5 triplet doubled.

Answer: (D) 10 m