RRB NTPC Graduate CBT 1 2024 PYQs

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$ (not the average of the two speeds).

1. Total distance $=5+5=10\text{km}$
2. Time: morning $1\text{h}$, evening $1+0.5=1.5\text{h}$; total $2.5\text{h}$
3. Average speed $=\frac{10}{2.5}=4\text{km/h}$

Trick: Speeds are $5$ and $\frac{10}{3}$; for equal distances use $\frac{2xy}{x+y}=\frac{2\times 5\times \frac{10}{3}}{5+\frac{10}{3}}=4$.

Answer: (D) 4

Concept: Successive change: $\text{SP}=\text{CP}\times (1+\frac{m}{100})(1-\frac{d}{100})$.

1. Let $\text{CP}=100$; $\text{MP}=128$
2. $\text{SP}=128\times \frac{75}{100}=96$
3. Loss $=100-96=4$, i.e. $4\%$ loss

Trick: Net % $=28-25-\frac{28\times 25}{100}=3-7=-4\%$ (minus means loss).

Answer: (C) $4\%$ loss

Concept: Angle at the centre $=2\times$ angle at the circumference, and tangent ⟂ radius, so $\angle ADB={180}^{\circ }-\angle AQB$.

1. $\angle AQB=2\times \angle APB=2\times {39}^{\circ }={78}^{\circ }$
2. $\angle QAD=\angle QBD={90}^{\circ }$ (radius ⟂ tangent)
3. In quadrilateral QADB: $\angle ADB={360}^{\circ }-{90}^{\circ }-{90}^{\circ }-{78}^{\circ }={102}^{\circ }$

Trick: $\angle ADB={180}^{\circ }-2\times \angle APB={180}^{\circ }-{78}^{\circ }={102}^{\circ }$.

Answer: (C) ${102}^{\circ }$

Concept: Under SI the interest is equal every year, so the difference of two amounts gives interest for the extra years.

1. SI for 2 years $=2100-1300=₹800$, so SI per year $=₹400$
2. Principal $=1300-2\times 400=₹500$
3. Rate $=\frac{400}{500}\times 100=80\%$

Trick: $R=\frac{\text{yearly SI}}{P}\times 100$; find P by removing interest from the first amount.

Answer: (A) $80\%$

Concept: $\text{Average price}=\frac{\text{Total cost}}{\text{Total books}}$.

1. $51\times 1120=57120$
2. $49\times 905=44345$
3. Average $=\frac{57120+44345}{100}=\frac{101465}{100}=1014.65$

Trick: With 100 books in total, just add the two costs and shift the decimal two places.

Answer: (B) ₹1,014.65

Concept: Use $(a\pm b{)}^2=a^2+b^2\pm 2ab$ and $a^2-b^2=(a+b)(a-b)$.

1. Given $a^2+b^2=7ab$.
2. $(a+b{)}^2=7ab+2ab=9ab\Rightarrow a+b=3\sqrt{ab}$.
3. $(a-b{)}^2=7ab-2ab=5ab\Rightarrow a-b=\sqrt{5ab}$ (as $a>b$).
4. $a^2-b^2=3\sqrt{ab}\times \sqrt{5ab}=3\sqrt{5}ab$.
5. Option (C) is $3\sqrt{5ab}=3\sqrt{5}\sqrt{ab}$, which equals $3\sqrt{5}ab$ only when $ab=1$; options (B) and (D) are the same expression $5\sqrt{3ab}$ and (A) is $\sqrt{15ab}$. No option equals $3\sqrt{5}ab$.
6. If option (C) had been written $3\sqrt{5}ab$ (radical over 5 only), it would be correct.

Note: Note where the square-root sign ends: $3\sqrt{5}ab$ and $3\sqrt{5ab}$ are different. In the exam, every candidate gets full marks (bonus) for such a question.

Answer: Bonus – no option is correct as printed; full marks to all.

Concept: Form two equations and solve by elimination.

1. $4p+3n=277$ … (i); $16n-12p=144\Rightarrow 4n-3p=36$ … (ii)
2. (i)$\times 3$ + (ii)$\times 4$: $12p+9n+16n-12p=831+144\Rightarrow 25n=975\Rightarrow n=39$
3. $4p=277-117=160\Rightarrow p=40$
4. $19p+15n=760+585=1345$

Trick: Check units digit: $19\times 40$ ends in 0, $15\times 39$ ends in 5 → total ends in 5; only ₹1,345 fits.

Answer: (D) ₹1,345

Concept: If SP of $x$ articles = CP of $y$ articles, then gain/loss $\%=\frac{y-x}{x}\times 100$.

1. SP of 34 = CP of 17 ⇒ $\frac{\text{SP}}{\text{CP}}=\frac{17}{34}=\frac{1}{2}$
2. Let CP $=2$, SP $=1$ ⇒ loss $=1$
3. Loss $\%=\frac{1}{2}\times 100=50\%$

Trick: $\frac{17-34}{34}\times 100=-50\%$ (negative = loss).

Answer: (A) $50\%$ loss

Concept: Total time = Total distance ÷ Average speed.

1. Total distance $=1980+1306+540=3826\text{km}$; total time $=\frac{3826}{14}$ h
2. Known times: $\frac{1980}{12}=165$ h, $\frac{540}{36}=15$ h
3. $\frac{1306}{x}=\frac{3826}{14}-180=\frac{3826-2520}{14}=\frac{1306}{14}$
4. So $x=14$

Trick: Keep the time as a fraction $\frac{3826}{14}$ – the 1306 appears on both sides and $x$ pops out directly.

Answer: (D) 14

Concept: Profit is shared in the ratio of investments (same time period).

1. Amount to be shared $=71\%$ of $49200=34932$
2. Ratio $41:39:84$; sum $=164$
3. Ashok $=\frac{84}{164}\times 34932=84\times 213=17892$

Trick: First divide by the ratio sum ($34932÷164=213$), then multiply by Ashok's part.

Answer: (D) ₹17,892

Concept: Number of diagonals of an $n$-sided polygon $=\frac{n(n-3)}{2}$.

1. From each vertex we can join $n-3=5$ vertices (not itself and not its two neighbours).
2. Each diagonal is counted twice, so total $=\frac{8\times 5}{2}$
3. $=20$

Trick: Octagon: $\frac{8\times 5}{2}=20$; hexagon: 9; pentagon: 5.

Answer: (A) 20

Concept: Equivalent discount $=100-100\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})(1-\frac{d_3}{100})$.

1. $0.87\times 0.80=0.696$
2. $0.696\times 0.96=0.66816$
3. Single discount $=100-66.816=33.184\%\approx 33.18\%$

Trick: Two at a time: $13+20-2.6=30.4\%$; then $30.4+4-\frac{30.4\times 4}{100}=33.184\%$.

Answer: (C) $33.18\%$

Concept: Net work per hour = sum of filling rates − emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{32}-\frac{1}{4}=\frac{1}{32}$ of the cistern per hour
2. Full cistern takes $32$ hours
3. One-fourth takes $\frac{32}{4}=8$ hours

Trick: The first tap and the emptying tap (both 4 h) cancel; only the 32-h tap works.

Answer: (C) 8

Concept: Area $=l\times b$, Perimeter $=2(l+b)$.

1. Let longer side $=l$, shorter $=l-15$
2. $l(l-15)=5\times 2(2l-15)\Rightarrow l^2-15l=20l-150$
3. $l^2-35l+150=0\Rightarrow (l-30)(l-5)=0$
4. $l=5$ gives a negative breadth, so $l=30\text{cm}$

Trick: Option check: $l=30$, $b=15$: area $450$, perimeter $90$, and $5\times 90=450$ ✓

Answer: (A) 30

Concept: Arrange the data in order; for odd $n$, median = ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 14, 29, 31, 36, 40, 44, 46, 47, 74, 87, 97
2. $n=11$, so median is the $\frac{11+1}{2}=6^{\text{th}}$ value
3. 6th value $=44$

Trick: Count 5 from each end – the one left in the middle is the median.

Answer: (C) 44

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $43=P\times {\left(\frac{4}{100}\right)}^2=P\times \frac{16}{10000}$
2. $P=\frac{43\times 10000}{16}=\frac{430000}{16}$
3. $P=26875$

Trick: ${\left(\frac{100}{R}\right)}^2=625$, so $P=43\times 625=26875$.

Answer: (D) ₹26,875

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$, i.e. $\text{Mode}-\text{Median}=2(\text{Median}-\text{Mean})$.

1. Given $\text{Mode}-\text{Median}=80.32$
2. $2(\text{Median}-\text{Mean})=80.32$
3. $\text{Median}-\text{Mean}=40.16$

Trick: Median lies between mean and mode, 2 parts from mode and 1 part from mean → just halve 80.32.

Answer: (C) 40.16

Concept: $h=\sqrt{l^2-r^2}$ and $V=\frac{1}{3}\pi r^{2}h$.

1. $r=\frac{30}{2}=15\text{cm}$, $l=113\text{cm}$
2. $h=\sqrt{{113}^2-{15}^2}=\sqrt{12769-225}=\sqrt{12544}=112\text{cm}$
3. $V=\frac{1}{3}\times \frac{22}{7}\times 225\times 112=22\times 75\times 16=26400{\text{cm}}^3$

Trick: $(15,112,113)$ is a Pythagorean triplet – recognise it to get $h$ instantly.

Answer: (C) 26,400

Concept: A yearly increase of $r\%$ for $n$ years gives $A=P{(1+\frac{r}{100})}^n$.

1. From 2019 to 2021 there are $n=2$ increments.
2. $A=22000\times {\left(\frac{11}{10}\right)}^2=22000\times \frac{121}{100}$
3. $A=220\times 121=26620$

Trick: Two successive 10% increases = $10+10+\frac{10\times 10}{100}=21\%$ total; $22000\times 1.21=26620$.

Answer: (D) ₹26,620

Concept: If $a:b::b:c$, then $b$ is the mean proportional: $b^2=ac$.

1. $b^2=a\times 4a=4a^2\Rightarrow b=2a$
2. $10=2a\Rightarrow a=5$
3. $c=4a=20$

Trick: $c=4a$ means $\frac{c}{a}=4$, so $\frac{b}{a}=\sqrt{4}=2$; $a=5,c=20$.

Answer: (B) 20