RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Find the part filled by A alone, then fill the rest at the net rate.

1. In 3 min A fills $\frac{3}{9}=\frac{1}{3}$; remaining $=\frac{2}{3}$
2. Net rate $=\frac{1}{9}-\frac{1}{10}=\frac{1}{90}$ per min
3. Time $=\frac{2}{3}÷\frac{1}{90}=\frac{2}{3}\times 90=60$ min

Trick: Tank $=90$ units: A $=10$, B $=-9$; after 3 min, $60$ units left at $1$ unit/min → $60$ min.

Answer: (B) $60$

Concept: Angle sum of a triangle $={180}^{\circ }$; an exterior angle equals the sum of the two interior opposite angles.

1. In right $△BDC$: $\angle C={90}^{\circ }-{22}^{\circ }={68}^{\circ }$
2. In $△AEC$: $\angle AEC={180}^{\circ }-{36}^{\circ }-{68}^{\circ }={76}^{\circ }$
3. $\angle AEB={180}^{\circ }-{76}^{\circ }={104}^{\circ }$ (linear pair on BC)

Trick: $\angle AEB$ is exterior to $△AEC$: ${36}^{\circ }+{68}^{\circ }={104}^{\circ }$.

Answer: (D) ${104}^{\circ }$

Concept: Water flowing per minute forms a cuboid: depth × width × distance moved in 1 minute.

1. Speed $=2.7\text{km/h}=\frac{2700}{60}=45\text{m/min}$.
2. Volume $=20\times 36\times 45$.
3. $=720\times 45=32400{\text{m}}^3$.

Trick: Convert km/h to m/min by multiplying by $\frac{50}{3}$: $2.7\times \frac{50}{3}=45$.

Answer: (B) 32,400

Concept: If $a:b::c:d$ then $ad=bc$; and if $a:y::y:b$ then $y^2=ab$ (mean proportional).

1. $\frac{20+x}{34+x}=\frac{11+x}{16+x}$ ⇒ $(20+x)(16+x)=(34+x)(11+x)$.
2. $320+36x=374+45x$ ⇒ $9x=-54$ ⇒ $x=-6$.
3. $3x=-18$ and $2x-6=-18$.
4. $y^2=(-18)(-18)=324$ ⇒ $y=18$ (since $y>0$).

Trick: After expanding, $x^2$ cancels, so it is just a linear equation.

Answer: (C) $18$

Concept: For $a{x}^3+b{x}^2+cx+d=0$: sum of roots $=-\frac{b}{a}$, product of roots $=-\frac{d}{a}$.

1. Here $a=4,b=-8,c=-13,d=19$.
2. Sum $=-\frac{-8}{4}=2$.
3. Product $=-\frac{19}{4}$.

Trick: Signs alternate for a cubic: sum $-\frac{b}{a}$, pair-sum $+\frac{c}{a}$, product $-\frac{d}{a}$.

Answer: (D) $2,\frac{-19}{4}$

Concept: If $m$ is the mean proportional between $a$ and $b$, then $m^2=ab$.

1. ${276}^2=368x$.
2. $x=\frac{276\times 276}{368}$.
3. $\frac{276}{368}=\frac{3}{4}$, so $x=\frac{3}{4}\times 276=207$.

Trick: Cancel first: $276=92\times 3$, $368=92\times 4$ – no need to square 276.

Answer: (A) $207$

Concept: In similar triangles, the ratio of perimeters equals the ratio of corresponding sides.

1. Perimeter of XYZ $=8+12+16=36\text{cm}$.
2. $\frac{AB}{XY}=\frac{6}{8}=\frac{3}{4}$.
3. Perimeter of ABC $=\frac{3}{4}\times 36=27\text{cm}$.

Trick: No need to find BC and CA separately – scale the whole perimeter.

Answer: (C) 27 cm

Concept: Open brackets from the innermost: ( ) → { } → [ ], then divide.

1. $(1+8)=9$; $4+7-9=2$.
2. $\{9+2-2\}=9$.
3. $[\frac{4}{2}\times 9]=18$.
4. $72÷18=4$.

Trick: Work inside-out and write each bracket's value before moving outward.

Answer: (A) 4

Concept: Half-yearly compounding: rate is halved and periods are doubled: $A=P{(1+\frac{r}{200})}^{2n}$.

1. Rate per half-year $=10\%$, periods $=2$.
2. $A=9900\times (1.1{)}^2=9900\times 1.21$.
3. $A=11979$, i.e. ₹11,979.

Trick: $10\%$ twice = $21\%$ total: $9900+21\%$ of $9900=9900+2079=11979$.

Answer: (D) ₹11,979

Concept: $\text{CP}=\frac{\text{SP}}{1+\frac{\text{gain}\%}{100}}$.

1. SP of 5 chocolates = ₹1, so SP of 1 $=\frac{1}{5}$.
2. CP of 1 $=\frac{1/5}{1.2}=\frac{1}{6}$.
3. So for ₹1 he bought $6$ chocolates.

Trick: With 20% gain, number bought per ₹1 $=5\times 1.2=6$.

Answer: (D) 6

Concept: If every group has the same average, the overall average is also that value.

1. Total $=12\times 60+18\times 60=30\times 60$.
2. Average $=\frac{30\times 60}{30}=60$.

Trick: Equal group averages → class average is the same, no calculation needed.

Answer: (A) 60

Concept: If the tank is still 10% empty, the 50 litres fill 90% of it. Filling the empty tank completely needs its full capacity.

1. $90\%$ of capacity $=50$ litres.
2. Capacity $=\frac{50\times 100}{90}=55.555\dots$
3. $\approx 55.56$ litres in all to fill the tank.

Trick: $\frac{50}{0.9}$: divide by 9 and multiply by 10 → $5.556\times 10$.

Answer: (A) 55.56

Concept: Net effect of $+a\%$ and $-b\%$: $a-b-\frac{ab}{100}$.

1. $40-15-\frac{40\times 15}{100}=25-6=19\%$.
2. Check: MP $=8100\times 1.4=11340$; SP $=11340\times 0.85=9639$.
3. Profit $=1539$; $\frac{1539}{8100}\times 100=19\%$.

Trick: $1.4\times 0.85=1.19$ → 19% profit directly.

Answer: (D) 19%

Concept: Savings = 100% minus total spending.

1. Spending $=30+25+15+16=86\%$.
2. Savings $=14\%=42000$.
3. Salary $=\frac{42000}{14}\times 100=300000$, i.e. ₹3,00,000.

Trick: $1\%=3000$, so $100\%=3,00,000$.

Answer: (B) 3,00,000

Concept: Write all terms with the same base: ${\left(\frac{10}{11}\right)}^n={\left(\frac{11}{10}\right)}^{-n}$, then add powers.

1. LHS $={\left(\frac{11}{10}\right)}^{7-10+9}={\left(\frac{11}{10}\right)}^6={\left(\frac{10}{11}\right)}^{-6}$.
2. So $3m+17=-6$.
3. $3m=-23$ ⇒ $m=-\frac{23}{3}$.

Trick: Flipping the fraction just changes the sign of the power.

Answer: (B) $-\frac{23}{3}$

Concept: $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$. From $\cot \theta =\frac{3}{4}$ build a 3-4-5 right triangle.

1. Base $=3$, perpendicular $=4$, hypotenuse $=5$ ⇒ $\sin \theta =\frac{4}{5}$.
2. $\sin 3\theta =3\times \frac{4}{5}-4\times \frac{64}{125}$.
3. $=\frac{300}{125}-\frac{256}{125}=\frac{44}{125}$.

Trick: 3-4-5 triple: $\cot =\frac{\text{base}}{\text{perp}}$, so perp $=4$.

Answer: (A) $\frac{44}{125}$

Concept: Product of two numbers $=\text{LCM}\times \text{HCF}$. With ratio $17:7$ the numbers are $17k$ and $7k$, HCF $=k$.

1. $17k\times 7k=119$ ⇒ $k^2=1$ ⇒ $k=1$.
2. HCF $=1$, LCM $=119$.
3. $\frac{1}{119}+\frac{1}{1}=\frac{120}{119}$.

Trick: Product equals $17\times 7$ itself, so the numbers are 17 and 7 (co-prime).

Answer: (A) $\frac{120}{119}$

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. Sum $=308+127+90+45+125+103=798$.
2. Average $=\frac{798}{6}=133$.

Trick: Check unit digits: $8+7+0+5+5+3=28$ → sum ends in 8; $133\times 6=798$ ✓.

Answer: (D) 133

Concept: $\sigma =\sqrt{\frac{\sum (x-\overline{x}{)}^2}{n}}$.

1. Mean $\overline{x}=\frac{225}{6}=37.5$.
2. Deviations $\pm 2.5,\pm 7.5,\pm 12.5$; squares $6.25,56.25,156.25$ (each twice).
3. $\sum (x-\overline{x}{)}^2=2\times 218.75=437.5$.
4. $\sigma =\sqrt{\frac{437.5}{6}}=\sqrt{72.92}\approx 8.54$.

Trick: Equally spaced data with gap $d$: $\sigma =d\sqrt{\frac{n^2-1}{12}}=5\sqrt{\frac{35}{12}}\approx 8.54$.

Answer: (C) 8.54

Concept: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

1. Take $a=2x,b=3y,c=-4z$.
2. Numerator: $8x^3+27y^3-64z^3-3(2x)(3y)(-4z)=8x^3+27y^3-64z^3+72xyz$ ✓.
3. Denominator: $4x^2+9y^2+16z^2-6xy+12yz+8zx$ ✓.
4. Quotient $=a+b+c=2x+3y-4z$.

Trick: The sign of $z$ in $-64z^3$ fixes $c=-4z$; answer must contain $-4z$ with $+3y$.

Answer: (B) $(2x+3y-4z)$