RRB NTPC Graduate CBT 1 2024 PYQs

Concept: $\text{SI}=\frac{P\times R\times T}{100}$ and $A=P+\text{SI}$.

1. SI for 2 years at $12.5\%$ = $25\%$ of $P=\frac{P}{4}$.
2. $A=P+\frac{P}{4}=\frac{5P}{4}=7500\Rightarrow P=6000$
3. $\text{SI}=7500-6000=1500$

Trick: $12.5\%=\frac{1}{8}$; 2 years → $\frac{1}{4}$ of P. So A = 5 parts = 7500, SI = 1 part = 1500.

Answer: (A) ₹$1500$

Concept: If rent is $28\%$, the remaining $72\%$ covers all other expenses and savings.

1. Other spending + saving $=687+7483+218=8388$
2. $72\%$ of salary $=8388$
3. Salary $=\frac{8388\times 100}{72}=11650$

Trick: Check the unit digit: $0.72\times 11650=8388$ exactly – only this option fits.

Answer: (B) ₹$11650$

Concept: Area of four walls of a cube $=4a^2$ (top and bottom faces excluded).

1. $a=3x+8y+5$
2. $a^2=9x^2+64y^2+25+48xy+80y+30x$
3. $4a^2=36x^2+256y^2+100+192xy+320y+120x$

Trick: Use $(p+q+r{)}^2=p^2+q^2+r^2+2pq+2qr+2rp$; the $y$-term $4\times 2\times 8y\times 5=320y$ separates the options.

Answer: (C) $(36x^2+256y^2+100+192xy+320y+120x){\text{m}}^2$

Concept: Median of grouped data $=L+\frac{\frac{N}{2}-cf}{f}\times h$.

1. $N=18+34+17+13+22=104$, $\frac{N}{2}=52$
2. Cumulative frequencies: $18,52,69,82,104$ → median class $170-220$.
3. $L=170$, $cf=18$, $f=34$, $h=50$
4. Median $=170+\frac{52-18}{34}\times 50=170+50=220$

Trick: Here $\frac{N}{2}=52$ equals the cumulative frequency up to 220, so the median is the upper limit 220.

Answer: (B) $220$

Concept: A wire is a cylinder: $V=\pi r^{2}h$ (keep all lengths in cm).

1. $r=\frac{6}{2}=3\text{cm}$, $h=56\text{m}=5600\text{cm}$
2. $V=\frac{22}{7}\times 3^2\times 5600$
3. $=22\times 9\times 800=158400{\text{cm}}^3$

Trick: Cancel 7 into 5600 first ($5600÷7=800$) – the rest is easy.

Answer: (A) $158400$

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$.

1. Let distances be $d$ and $2d$: onward time $=\frac{d}{52}+\frac{2d}{39}=\frac{11d}{156}$.
2. Total distance $=6d$; total time $=\frac{6d}{56}=\frac{3d}{28}$.
3. Return time $=\frac{3d}{28}-\frac{11d}{156}=\frac{117d-77d}{1092}=\frac{40d}{1092}$
4. Return speed $=\frac{3d\times 1092}{40d}=81.9\text{km/h}$

Trick: Take $d=156$: onward time 11 h, distance 468; total time $\frac{936}{56}$; return time $\frac{936}{56}-11$ → speed 81.9.

Answer: (B) $81.9$

Concept: Required number $=\text{HCF}(a-r_1,b-r_2)$.

1. $48-9=39$ and $46-7=39$
2. $\text{HCF}(39,39)=39$
3. Check: $48=39+9$, $46=39+7$ ✔

Trick: Subtract the remainders; when both results are equal, that number itself is the answer.

Answer: (D) $39$

Concept: Successive discounts multiply: price left $=(1-d_1)(1-d_2)(1-d_3)$.

1. $0.74\times 0.71=0.5254$
2. $0.5254\times 0.98=0.514892$
3. Single discount $=(1-0.514892)\times 100=48.5108\%\approx 48.51\%$

Trick: Combine two at a time: $26+29-\frac{26\times 29}{100}=47.46$, then $47.46+2-\frac{47.46\times 2}{100}=48.51$.

Answer: (C) $48.51\%$

Concept: Area of a quadrilateral $=\frac{1}{2}\times d\times (h_1+h_2)$, where $h_1,h_2$ are perpendiculars on diagonal $d$.

1. $\frac{1}{2}\times AC\times (11+12)=299$
2. $AC\times 23=598$
3. $AC=26\text{cm}$

Trick: Diagonal splits it into two triangles on the same base AC – just add their heights.

Answer: (A) $26$

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

1. $44=P\times {\left(\frac{4}{100}\right)}^2=\frac{16P}{10000}$
2. $P=\frac{44\times 10000}{16}=27500$

Trick: $R=4\%$ → $R^2=16$; $P=\frac{\text{diff}\times 10000}{16}$.

Answer: (C) ₹$27500$

Concept: A ratio needs both quantities in the same unit.

1. $17\text{ days}=17\times 24=408\text{ hours}$
2. Ratio $=\frac{39}{408}$
3. Divide by 3: $\frac{13}{136}$

Trick: Convert the bigger unit to the smaller one first, then cancel the HCF.

Answer: (D) $13:136$

Concept: $SP=MP\times (1-\frac{\text{discount}}{100})$

1. Let $CP=100$, so $MP=145$.
2. Discount $60\%$: $SP=145\times 0.40=58$
3. Loss $=100-58=42$, i.e. $42\%$ loss.

Trick: Net effect $=45-60-\frac{45\times 60}{100}=-42\%$ (minus = loss).

Answer: (B) $42\%$ loss

Concept: $22=2\times 11$, so the number must be even and divisible by 11 (alternate-digit difference is 0 or a multiple of 11).

1. 7019 and 8371 are odd – rejected.
2. 7986: $(7+8)-(9+6)=0$ ✓ and even ✓.
3. 7664: $(7+6)-(6+4)=3$ ✗.

Trick: For a composite divisor, test its co-prime factors separately.

Answer: (A) 7986

Concept: $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$

1. $\frac{0.038025}{0.004225}=\frac{38025}{4225}=9$
2. $\sqrt{9}=3$
3. Check: $\sqrt{0.038025}=0.195$, $\sqrt{0.004225}=0.065$, and $\frac{0.195}{0.065}=3$.

Trick: Same number of decimal places → just divide the whole numbers inside the root.

Answer: (D) 3

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

1. $(a-b{)}^2=169-20=149$
2. $(a+b{)}^2=169+20=189$
3. Since $a>b$, $\frac{a-b}{a+b}=\sqrt{\frac{149}{189}}$

Trick: Square the required ratio: ${\left(\frac{a-b}{a+b}\right)}^2=\frac{a^2+b^2-2ab}{a^2+b^2+2ab}$.

Answer: (C) $\sqrt{\frac{149}{189}}$

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

1. Net rate $=\frac{1}{3}+\frac{1}{57}-\frac{1}{3}=\frac{1}{57}$ per hour.
2. Full cistern takes 57 hours.
3. One-third takes $\frac{57}{3}=19$ hours.

Trick: The 3-hour filler and 3-hour emptier cancel each other – only the 57-hour tap works.

Answer: (A) 19

Concept: Equal roots when discriminant $D=b^2-4ac=0$.

1. Here $a=7$, $b=-4$, $c=k$.
2. $(-4{)}^2-4\times 7\times k=0$
3. $16=28k\Rightarrow k=\frac{16}{28}=\frac{4}{7}$

Trick: For equal roots, $k=\frac{b^2}{4a}=\frac{16}{28}$.

Answer: (A) $\frac{4}{7}$

Concept: Income = Saving + Expenditure.

1. Saving $=30\%$, so expenditure $=70\%$ of income.
2. $70\%=560\Rightarrow 1\%=8$
3. Income $=100\times 8=800$

Trick: $560\times \frac{10}{7}=800$.

Answer: (B) 800

Concept: Use the 3-4-5 right triangle.

1. $\sin A=\frac{4}{5}\Rightarrow \cos A=\frac{3}{5}$, $\tan A=\frac{4}{3}$
2. $2+\tan A=\frac{10}{3}$, $4+\cos A=\frac{23}{5}$
3. Product $=\frac{10}{3}\times \frac{23}{5}=\frac{46}{3}$

Trick: Value is clearly greater than $2\times 4=8$, so only $\frac{46}{3}$ or $\frac{208}{15}$ fit – compute once.

Answer: (C) $\frac{46}{3}$

Concept: Form two linear equations in present ages P and Q.

1. $4P-Q=35$ …(i)
2. After 8 years: $2(Q+8)=6(P+8)-40\Rightarrow 2Q=6P-8\Rightarrow Q=3P-4$ …(ii)
3. Put (ii) in (i): $4P-3P+4=35\Rightarrow P=31$
4. $Q=3\times 31-4=89$

Trick: Check options: $Q=89\Rightarrow P=\frac{89+35}{4}=31$; then $2\times 97=194=6\times 39-40$ ✓.

Answer: (C) 89