RRB NTPC Graduate CBT 1 2024 PYQs

Concept: BODMAS: Brackets → Multiplication → Addition/Subtraction (left to right).

1. $23+19=42$.
2. $6\times 42=252$.
3. $285-252+83=33+83=116$.

Trick: Never subtract before multiplying: $285-6$ first would be wrong.

Answer: (D) 116

Concept: Required number $=\text{HCF}(247-3,189-6)$.

1. $247-3=244=4\times 61$.
2. $189-6=183=3\times 61$.
3. HCF $=61$ (and $61>6$, the larger remainder).

Trick: Subtract the remainders, then take HCF; check the answer exceeds every remainder.

Answer: (C) 61

Concept: Net rate = sum of filling rates − emptying rate (tank per hour).

1. Net rate $=\frac{1}{3}+\frac{1}{36}-\frac{1}{4}=\frac{12+1-9}{36}=\frac{4}{36}=\frac{1}{9}$.
2. Full tank in 9 hours.
3. $\frac{2}{3}$ tank: $\frac{2}{3}\times 9=6$ hours.

Trick: LCM method: capacity 36 units; rates $12+1-9=4$ units/h; $\frac{2}{3}\times 36=24$ units → 6 h.

Answer: (C) 6

Concept: Volume stays the same on melting: $\frac{1}{3}\pi r_1^2{h}_1=\frac{1}{3}\pi r_2^2{h}_2$.

1. ${17.6}^2\times 6.7={8.8}^2\times h$.
2. $h={\left(\frac{17.6}{8.8}\right)}^2\times 6.7=2^2\times 6.7$.
3. $h=26.8\text{cm}$.

Trick: Radius halved → height becomes 4 times.

Answer: (D) $26.8$

Concept: Equal distances: $\text{Avg speed}=\frac{3}{\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}}$.

1. Let each third $=d$. Total time $=d(\frac{1}{67}+\frac{1}{66}+\frac{1}{17})$.
2. $\frac{1}{67}+\frac{1}{66}+\frac{1}{17}\approx 0.014925+0.015152+0.058824=0.088901$.
3. Average $=\frac{3}{0.088901}\approx 33.75\to 33.7\text{km/h}$.

Trick: Average speed is pulled towards the slowest speed (17), so it must be well below the simple mean $50$.

Answer: (C) $33.7$

Concept: $\text{Area of sector}=\frac{\theta }{{360}^{\circ }}\times \pi r^2$.

1. $r=\frac{14}{2}=7\text{cm}$.
2. $\pi r^2=\frac{22}{7}\times 49=154$.
3. Area $=\frac{120}{360}\times 154=\frac{154}{3}=51\frac{1}{3}{\text{cm}}^2$.

Trick: ${120}^{\circ }$ is one-third of the circle: $\frac{154}{3}$.

Answer: (A) $51\frac{1}{3}{\text{cm}}^2$

Concept: Half-yearly compounding: rate is halved and time (periods) doubled.

1. Rate per half-year $=\frac{2}{2}=1\%$, periods $=2$.
2. $A=910000\times (1.01{)}^2=910000\times 1.0201=928291$
3. $CI=928291-910000=18291$

Trick: For 2 periods at $r\%$, effective rate $=2r+\frac{r^2}{100}=2.01\%$; $910000\times 0.0201=18291$.

Answer: (A) ₹$18,291$

Concept: Mode = the value that occurs most often.

1. Count: 21 → 3 times; 25, 27, 29, 31, 32 → 2 times each; others once.
2. Highest frequency is 3, for 21.

Trick: Tally just the repeated values; ignore the singles.

Answer: (D) $21$

Concept: A number is divisible by 11 if (sum of alternate digits) − (sum of other digits) is 0 or a multiple of 11.

1. 2118: $(2+1)-(1+8)=-6$ ✗
2. 1440: $(1+4)-(4+0)=1$ ✗; 2652: $(2+5)-(6+2)=-1$ ✗
3. 1749: $(1+4)-(7+9)=-11$ ✓ ($1749=11\times 159$)

Trick: Difference of alternate-digit sums must be $0,\pm 11,\dots$

Answer: (D) $1749$

Concept: Put $x^2=t$ to turn $x^4-17x^2+60=0$ into a quadratic.

1. $t^2-17t+60=0\Rightarrow (t-12)(t-5)=0$
2. $t=12$ or $t=5$, so $x=\pm \sqrt{12}$ or $x=\pm \sqrt{5}$.
3. Among the options only $-\sqrt{12}$ fits.

Trick: Check options: $x^2=12\Rightarrow 144-204+60=0$ ✓.

Answer: (D) $-\sqrt{12}$

Concept: Combined average $=\frac{n_1{a}_1+n_2{a}_2}{n_1+n_2}$.

1. Total weight $=22\times 40+44\times 37=880+1628=2508$
2. Average $=\frac{2508}{66}=38$ kg

Trick: Ratio $22:44=1:2$, so average $=\frac{40+2\times 37}{3}=\frac{114}{3}=38$.

Answer: (B) $38$

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

1. Distance $=20\times 80+65\times 70=1600+4550=6150$ m
2. Time $=80+70=150$ min
3. Average $=\frac{6150}{150}=41$ m/min

Trick: Never average the two speeds directly ($\frac{20+65}{2}=42.5$ is wrong) – weight them by time.

Answer: (C) $41$

Concept: Successive discounts multiply the remaining fractions.

1. Remaining $=0.76\times 0.82\times 0.80$
2. $0.76\times 0.82=0.6232$; $0.6232\times 0.8=0.49856$
3. Single discount $=1-0.49856=0.50144\approx 50.14\%$

Trick: Two at a time: $a+b-\frac{ab}{100}$: $24,18\to 37.68$; $37.68,20\to 50.144$.

Answer: (D) $50.14\%$

Concept: Time $=\frac{\text{capacity}}{\text{rate}}$. Let filling rate $=x$, emptying rate $=x+10$ (m³/min).

1. $\frac{1800}{x}-\frac{1800}{x+10}=6$
2. $1800\times 10=6x(x+10)\Rightarrow x(x+10)=3000$
3. $x^2+10x-3000=0\Rightarrow (x+60)(x-50)=0\Rightarrow x=50$

Trick: Option check: $\frac{1800}{50}=36$, $\frac{1800}{60}=30$, difference $6$ ✓.

Answer: (C) $50$

Concept: Successive % increases multiply: $P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})$.

1. After year 1: $110000\times 1.05=115500$
2. After year 2: $115500\times 1.30=150150$

Trick: Net increase $=5+30+\frac{5\times 30}{100}=36.5\%$; $110000\times 1.365=150150$.

Answer: (A) $1,50,150$

Concept: Let apple $=a$, mango $=m$ (in ₹).

1. $2a+5m=239$
2. Apple at half price: $2\times \frac{a}{2}+4m=184\Rightarrow a+4m=184$
3. Double it: $2a+8m=368$; subtract: $3m=129\Rightarrow m=43$, $a=184-172=12$
4. $8a+8m=8\times 55=440$

Trick: You only need $a+m=55$, then multiply by 8.

Answer: (B) ₹$440$

Concept: Clear fractions by multiplying by the LCM of denominators.

1. $\frac{1}{14}:\frac{1}{2}:\frac{9}{3}=\frac{1}{14}:\frac{1}{2}:3$
2. Multiply by 14: $1:7:42$, total $=50$ parts
3. Third monkey $=\frac{42}{50}\times 600=504$

Trick: $600÷50=12$ per part; $42\times 12=504$.

Answer: (D) $504$

Concept: Convert each ratio to a decimal (or compare with 1).

1. $\frac{18}{37}\approx 0.49$, $\frac{13}{32}\approx 0.41$ – both below $\frac{1}{2}$.
2. $\frac{23}{25}=0.92$, $\frac{25}{26}\approx 0.96$
3. Greatest: $25:26$

Trick: For fractions just below 1, the one with the smaller gap from 1 is bigger: $\frac{1}{26}<\frac{2}{25}$.

Answer: (C) $25:26$

Concept: A cube has 12 equal edges, so its perimeter (total edge length) $=12a$. Use $x^3+y^3=(x+y{)}^3-3xy(x+y)$.

1. $12x+12y=300\Rightarrow x+y=25$; $xy=156$
2. $x^3+y^3={25}^3-3\times 156\times 25$
3. $=15625-11700=3925{\text{cm}}^3$

Trick: $x+y=25$, $xy=156$ gives $x=13$, $y=12$: $2197+1728=3925$.

Answer: (D) $3925{\text{cm}}^3$

Concept: $(1-a)(1+a)=1-a^2$; ${\sec }^{2}A-1={\tan }^{2}A$, ${\csc }^{2}A-1={\cot }^{2}A$.

1. Expression $=\frac{1-{\sec }^{2}A}{1-{\csc }^{2}A}=\frac{-{\tan }^{2}A}{-{\cot }^{2}A}={\tan }^{4}A$
2. $\tan A=\frac{4}{3}$
3. ${\tan }^{4}A=\frac{256}{81}$

Trick: $\frac{{\tan }^{2}A}{{\cot }^{2}A}={\tan }^{4}A$ – no need to find sec or cosec.

Answer: (B) $\frac{256}{81}$