RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: A number is divisible by $14=2\times 7$ only if it is divisible by both 2 and 7.

1. 64,36,639 is odd, so it is not divisible by 2 – rejected.
2. Test of 7 (alternate sum of 3-digit groups from right): $6794266\to 266-794+6=-522$ ✗; $7132960\to 960-132+7=835$ ✗.
3. $7048048\to 048-048+7=7$ ✓, and it is even.
4. Check: $7048048÷14=503432$.

Trick: Divisible by a composite number = divisible by each of its co-prime factors ($14=2\times 7$).

Answer: (D) 70,48,048

Concept: Take distance MN $=D$. Speed = distance ÷ time; when two trains move towards each other, relative speed = sum of speeds.

1. A takes 8:20 AM to 3:20 PM $=7$ h → speed $\frac{D}{7}$; B takes 10:20 AM to 3:20 PM $=5$ h → speed $\frac{D}{5}$.
2. By 10:20 AM, A has run 2 h: covered $\frac{2D}{7}$, gap left $=\frac{5D}{7}$.
3. Relative speed $=\frac{D}{7}+\frac{D}{5}=\frac{12D}{35}$.
4. Time to meet $=\frac{5D}{7}÷\frac{12D}{35}=\frac{25}{12}\text{h}=2\text{h}5\text{min}$ → 10:20 + 2 h 5 min = 12:25 PM.

Trick: Take $D=\text{LCM}(7,5)=35$ km: speeds 5 and 7; at 10:20 gap $=25$, time $=\frac{25}{12}$ h.

Answer: (D) 12:25 PM

Concept: $\text{Discount}\%=\frac{\text{Discount}}{\text{MP}}\times 100$.

1. MP = ₹600, discount = ₹50.
2. $\text{Discount}\%=\frac{50}{600}\times 100=\frac{25}{3}=8.33\%$.

Trick: $\frac{50}{600}=\frac{1}{12}$, and $\frac{1}{12}=8\frac{1}{3}\%$.

Answer: (C) $8.33\%$

Concept: Profit share $\propto$ capital $\times$ time.

1. Shares: $A+G=8725$, $A-G=1745$ → $A=5235$, $G=3490$; ratio $=3:2$.
2. Let Gaurav's capital $=x$; Anubhav's $=x+37000$.
3. $\frac{3(x+37000)}{4x}=\frac{3}{2}\Rightarrow 6x+222000=12x\Rightarrow x=37000$.
4. Anubhav's capital $=37000+37000=$ ₹74,000.

Trick: Option check: A = 74000 → G = 37000; $74000\times 3:37000\times 4=3:2$ ✓.

Answer: (C) ₹74,000

Concept: Same time period, so profit is shared in the ratio of capitals.

1. $2410:1170:700=241:117:70$.
2. B's 117 units = ₹1092 → 1 unit $=\frac{1092}{117}=\frac{28}{3}$.
3. A − C $=241-70=171$ units $=171\times \frac{28}{3}=57\times 28=1596$.

Trick: Difference directly: $1092\times \frac{171}{117}=1092\times \frac{19}{13}=84\times 19=1596$.

Answer: (D) ₹1,596

Concept: BODMAS: innermost bracket first, then multiplication, then addition, then division.

1. $(3-7)=-4$.
2. $2\times (-4)=-8$, so $\{12+(-8)\}=4$.
3. $592÷4=148$.

Trick: Solve brackets from inside out: ( ) → { } → [ ].

Answer: (C) 148

Concept: Mode = the value that occurs most often.

1. Frequencies: 10 → 1, 12 → 2, 13 → 1, 14 → 2, 15 → 1, 16 → 3.
2. 16 occurs the maximum number of times (3).

Trick: Arrange data in order: 10, 12, 12, 13, 14, 14, 15, 16, 16, 16 – the longest run is 16.

Answer: (C) 16

Concept: For a fixed distance, speed $\times$ time is constant: $S_1{T}_1=S_2{T}_2$.

1. Distance $=28\times 20=560\text{km}$.
2. Required speed $=\frac{560}{14}=40\text{km/h}$.

Trick: $S_2=28\times \frac{20}{14}=28\times \frac{10}{7}=40$.

Answer: (B) $40\text{km/h}$

Concept: Work step by step on registered voters $R$: polled → valid → margin.

1. Votes polled $=0.95R$; valid $=80\%$ of polled $=0.76R$.
2. Winner 59%, loser 41% of valid → margin $=18\%$ of $0.76R=0.1368R$.
3. $0.1368R=1881\Rightarrow R=\frac{1881}{0.1368}=13750$.

Trick: Margin as % of R $=0.95\times 0.8\times 0.18=13.68\%$; check $13750\times 0.1368=1881$ ✓.

Answer: (B) 13750

Concept: Loss of 15% means selling price $=85\%$ of CP. Take CP $=100$.

1. $\frac{5}{11}$ of actual SP $=85\Rightarrow \text{SP}=\frac{85\times 11}{5}=187$.
2. 90% of SP $=0.9\times 187=168.3$.
3. Profit $\%=\frac{168.3-100}{100}\times 100=68.3\%$.

Trick: New SP/CP $=0.85\times \frac{11}{5}\times 0.9=1.683$ → profit 68.3%.

Answer: (B) $68.3\%$

Concept: Make the decimal places equal, find the LCM of the whole numbers, then put the decimal back.

1. $2.1=\frac{210}{100}$ and $0.35=\frac{35}{100}$
2. $\text{LCM}(210,35)=210$ (since $210=35\times 6$)
3. $\text{LCM}=\frac{210}{100}=2.1$

Trick: If one number is a multiple of the other ($2.1=0.35\times 6$), the bigger number itself is the LCM.

Answer: (D) $2.1$

Concept: For two numbers, $\text{HCF}\times \text{LCM}=\text{first number}\times \text{second number}$.

1. $25\times 250=50\times x$
2. $x=\frac{6250}{50}=125$
3. Check: $\text{HCF}(50,125)=25$, $\text{LCM}=250$ ✓

Trick: Other number $=\frac{\text{HCF}\times \text{LCM}}{\text{given number}}$.

Answer: (D) $125$

Concept: The four angles of a quadrilateral add up to ${360}^{\circ }$.

1. $\angle D={69}^{\circ }-8^{\circ }={61}^{\circ }$
2. $3(x+2)+2(2x-7)+69+61=360\Rightarrow 7x+122=360$
3. $7x=238\Rightarrow x=34$
4. $\angle B=2(2\times 34-7)=2\times 61={122}^{\circ }$

Trick: Check: ${108}^{\circ }+{122}^{\circ }+{69}^{\circ }+{61}^{\circ }={360}^{\circ }$ ✓

Answer: (C) ${122}^{\circ }$

Concept: Selling price = cost price × (1 + markup) × (1 − discount).

1. Let $\text{CP}=100$; $\text{MP}=140$
2. $\text{SP}=140\times \frac{80}{100}=112$
3. $\text{Gain}\%=\frac{112-100}{100}\times 100=12\%$

Trick: Net effect $=a-b-\frac{ab}{100}=40-20-\frac{800}{100}=12\%$ gain.

Answer: (A) $12\%$ gain

Concept: Sum of the first $n$ odd natural numbers $=n^2$, so their average $=\frac{n^2}{n}=n$.

1. Here $n=153$
2. $\text{Sum}={153}^2$
3. $\text{Average}=\frac{{153}^2}{153}=153$

Yaad rakhein: Average of first $n$ odd numbers $=n$; of first $n$ even numbers $=n+1$.

Answer: (A) $153$

Concept: Form two linear equations from the statements and solve them.

1. Let Ram $=R$, Ramya $=r$: $2R=4r+30\Rightarrow R=2r+15$
2. $5r=R-6\Rightarrow R=5r+6$
3. $2r+15=5r+6\Rightarrow r=3$, $R=21$
4. $\text{Difference}=21-3=18$ years

Trick: Equate the two expressions for $R$ directly – one step solves it.

Answer: (B) $18$

Concept: $\text{TSA of cuboid}=2(lb+bh+hl)$.

1. $lb=84\times 45=3780$
2. $bh=45\times 25=1125$, $hl=25\times 84=2100$
3. $\text{TSA}=2(3780+1125+2100)=2\times 7005=14010{\text{cm}}^2$

Trick: Only one option ends in 10 – the unit digit of $2\times 7005$ is 0 and tens digit 1.

Answer: (B) $14,010{\text{cm}}^2$

Concept: Take total work = LCM of the times; add the hourly efficiencies of pipes working together.

1. $\text{LCM}(12,21,16)=336$ units
2. Efficiencies: A $=28$, B $=16$, C $=21$ units/hour
3. Together $=28+16+21=65$ units/hour
4. $\text{Time}=\frac{336}{65}=5\frac{11}{65}$ hours

Trick: All options share $\frac{11}{65}$; only the whole part matters: $65\times 5=325<336<390$, so 5.

Answer: (C) $5\frac{11}{65}$ hours

Concept: Let pen = ₹$p$, pencil = ₹$c$; form two equations and eliminate one variable.

1. $7p+13c=128$
2. $14(p-7)+10(c+2)=98\Rightarrow 14p+10c=176\Rightarrow 7p+5c=88$
3. Subtract: $8c=40\Rightarrow c=5$, $7p=63\Rightarrow p=9$
4. $9p+7c=81+35=116$

Trick: Halve the second equation first so the $7p$ terms cancel at once.

Answer: (D) ₹116

Concept: For a fixed SP of ₹1, the number of articles sold is inversely proportional to the SP-factor $(100\pm \%)$.

1. At $35\%$ loss, factor $=65$; at $30\%$ gain, factor $=130$
2. $\text{New number}=6\times \frac{65}{130}$
3. $=6\times \frac{1}{2}=3$

Trick: $n_2=n_1\times \frac{100-\text{loss}}{100+\text{gain}}$.

Answer: (A) $3$