RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: The distance of a point $(x,y)$ from the x-axis is $|y|$ and from the y-axis is $|x|$.

1. For $P(2,3)$: $x=2$, $y=3$.
2. Distance from the x-axis $=|y|=3$.
3. Option D (2) is the distance from the y-axis – the usual trap.

Trick: "From x-axis → take y"; "from y-axis → take x" (the opposite coordinate).

Answer: (B) 3

Concept: An open drum has a curved inner wall and a circular base (no lid). Area to paint $=\text{CSA}+\text{base}=2\pi rh+\pi r^2=\pi r(2h+r)$.

1. $r=\frac{78}{2}=39\text{m}$, $h=79\text{m}$.
2. $2h+r=158+39=197$.
3. Area $=\frac{22}{7}\times 39\times 197=\frac{169026}{7}\approx 24146.57\approx 24146.6{\text{m}}^2$.

Trick: "Open" → add only one circle; "closed" → add two. Factor out $\pi r$ to cut the arithmetic.

Answer: (D) 24146.6

Concept: In a percentage pie chart, amount of a sector $=\frac{\text{sector }\%}{100}\times \text{total}$.

1. From the chart, Food $=25\%$ (Education 27%, Rent 16%, Bills 32%).
2. Expenditure on Food $=\frac{25}{100}\times 10000=$ ₹$2500$.
3. ₹2000 would be 20% and ₹1700 is 17% – neither is the Food sector.

Trick: $25\%=\frac{1}{4}$ → simply divide the total by 4.

Answer: (C) 2500

Concept: Evaluate each square root of a perfect square separately, then add/subtract.

1. $\sqrt{144}=12$, $\sqrt{0.04}=\sqrt{\frac{4}{100}}=0.2$, $\sqrt{9}=3$.
2. $12+0.2-3=9.2$.

Trick: For decimals, $\sqrt{0.04}=0.2$ (half the decimal places), not $0.02$.

Answer: (C) 9.2

Concept: Two consecutive positive integers are $x$ and $x+1$; translate "product is 277" into an equation.

1. $x(x+1)=277$.
2. $x^2+x=277\Rightarrow x^2+x-277=0$.
3. Option C would come from $x(x-1)$ and A/B have the wrong sign of 277.

Trick: "Product = k" → move $k$ to the left as $-k$.

Answer: (D) $x^2+x-277=0$

Concept: In an A.P., $a_n=a+(n-1)d$; the difference between two terms is $(\text{gap in positions})\times d$.

1. $a_{20}-a_{16}=4d=441-385=56\Rightarrow d=14$.
2. $a_{90}=a_{20}+70d=441+70\times 14=441+980=1421$.

Trick: No need to find $a$: jump directly from a known term, $a_m=a_n+(m-n)d$.

Answer: (C) 1421

Concept: For two numbers, $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. Let the numbers be $\frac{4}{9}b$ and $b$. Then $\frac{4}{9}b\times b=12\times 300=3600$.
2. $b^2=3600\times \frac{9}{4}=8100\Rightarrow b=90$.
3. Smaller number $=\frac{4}{9}\times 90=40$. The pair satisfying the ratio and the product is $40$ and $90$, whose smaller member is $40$.

Trick: Ratio $4:9$ → numbers $4k,9k$; $36k^2=3600\Rightarrow k=10$ → $40,90$.

Answer: (C) 40

Concept: The tower, its shadow and the line to the sun form a right triangle; $\tan \theta =\frac{\text{height}}{\text{shadow}}$.

1. Let shadow $=x$; then height $=\sqrt{3}x$.
2. $\tan \theta =\frac{\sqrt{3}x}{x}=\sqrt{3}$.
3. $\tan {60}^{\circ }=\sqrt{3}$, so $\theta ={60}^{\circ }$.

Trick: height $=\sqrt{3}\times$ shadow → ${60}^{\circ }$; shadow $=\sqrt{3}\times$ height → ${30}^{\circ }$; equal → ${45}^{\circ }$.

Answer: (C) ${60}^{\circ }$

Concept: ${\sigma }^2=\frac{\sum x^2}{n}-{\overline{x}}^2$. Recover $\sum x$ and $\sum x^2$, replace the wrong value, and recompute.

1. Wrong data: $\sum x=25\times 42=1050$; $\sum x^2=n({\sigma }^2+{\overline{x}}^2)=25(25+1764)=44725$.
2. Corrected: $\sum x=1050-10+210=1250\Rightarrow \overline{x}=50$; $\sum x^2=44725-{10}^2+{210}^2=44725-100+44100=88725$.
3. ${\sigma }^2=\frac{88725}{25}-{50}^2=3549-2500=1049\Rightarrow \sigma =\sqrt{1049}$.

Trick: Only $\sum x$ and $\sum x^2$ change: subtract the wrong value (and its square), add the correct one.

Answer: (A) $\sqrt{1049}$

Concept: A quadratic with roots $\alpha ,\beta$ is $x^2-(\alpha +\beta )x+\alpha \beta =0$.

1. Sum of roots $=9+11=20$; product $=9\times 11=99$.
2. Equation: $x^2-20x+99=0$.
3. Check: $(x-9)(x-11)=x^2-20x+99$ ✓.

Trick: "minus sum, plus product" – signs are the usual trap (options B, D).

Answer: (A) $x^2-20x+99=0$

Concept: Simplify the denominator first, then divide by a fraction by multiplying with its reciprocal.

1. $\frac{5}{6}+\frac{2}{4}=\frac{5}{6}+\frac{1}{2}=\frac{5+3}{6}=\frac{8}{6}=\frac{4}{3}$
2. $\frac{1}{4/3}=\frac{3}{4}$
3. $\frac{3}{4}÷\frac{4}{29}=\frac{3}{4}\times \frac{29}{4}=\frac{87}{16}=5\frac{7}{16}$

Trick: $\frac{1}{a/b}=\frac{b}{a}$ — flipping a fraction is just swapping numerator and denominator.

Answer: (B) $5\frac{7}{16}$

Concept: Three points are collinear if the slope between any two pairs is equal.

1. Slope of $(-5,1)$ and $(4,-2)$: $\frac{-2-1}{4-(-5)}=\frac{-3}{9}=-\frac{1}{3}$
2. Slope of $(-5,1)$ and $(1,k)$: $\frac{k-1}{1-(-5)}=\frac{k-1}{6}$
3. $\frac{k-1}{6}=-\frac{1}{3}\Rightarrow k-1=-2\Rightarrow k=-1$

Trick: Or use area $=0$: $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$.

Answer: (A) $-1$

Concept: $B^{\prime }=U-B$ = elements of $U$ that are not in $B$; $n(B^{\prime })=n(U)-n(B)$.

1. $n(U)=9$, $n(B)=4$
2. $B^{\prime }=\{1,3,5,7,9\}$, so $n(B^{\prime })=9-4=5$

Trick: Complement count = total − given.

Answer: (A) 5

Concept: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and for independent events $P(A\cap B)=P(A)\times P(B)$.

1. $P(A\cap B)=0.4\times 0.6=0.24$
2. $P(A\text{ or }B)=0.4+0.6-0.24=0.76$

Trick: $P(A\cup B)=1-P(A^{\prime })P(B^{\prime })=1-0.6\times 0.4=0.76$.

Answer: (D) 0.76

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

1. $236=\frac{n}{2}(28+31)=\frac{59n}{2}$
2. $n=\frac{236\times 2}{59}=\frac{472}{59}=8$

Trick: Sum = number of terms × average of first and last term: $\frac{236}{29.5}=8$.

Answer: (A) 8

Concept: $n(M\cup P)=n(M)+n(P)-n(M\cap P)$.

1. $26=14+n(P)-6\Rightarrow n(P)=18$ teach physics in all.
2. Teachers who teach only physics $=26-14=12$ (everyone outside the mathematics circle).
3. Only mathematics $=14-6=8$; check: $8+6+12=26$ ✓

Trick: Draw the Venn diagram: 8 | 6 | 12. 'Physics' (total) = 18, 'only physics' = 12 — read the question carefully.

Answer: (B) 12 — the number who teach physics but not mathematics (the intended reading of the question).

Concept: In similar triangles, corresponding sides and perimeters are in the same ratio.

1. $YZ$ corresponds to $BC$: ratio $=\frac{YZ}{BC}=\frac{8}{4}=2$
2. Perimeter of $△ABC=5+4+4.5=13.5\text{cm}$
3. Perimeter of $△XYZ=2\times 13.5=27\text{cm}$

Trick: $\frac{{\text{Perimeter}}_1}{{\text{Perimeter}}_2}=\frac{{\text{side}}_1}{{\text{side}}_2}$ — no need to find each side.

Answer: (B) 27 cm