RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Sum = average $\times$ number of items; overlapping sums count the common person twice.

1. $G+A+T=3\times 45=135$
2. $G+A=2\times 44=88$ and $A+T=2\times 46=92$
3. Adding: $G+2A+T=88+92=180$
4. $A=180-135=45$ kg

Trick: Common member = (sum of the two pair-totals) $-$ total of all three.

Answer: (D) $45$

Concept: Smallest number to be added = divisor $-$ remainder.

1. $24\times 3124=74976$
2. $74978-74976=2$, so the remainder is $2$.
3. Number to add $=24-2=22$; check: $74978+22=75000=24\times 3125$.

Trick: Divisible by 24 ⇔ divisible by 8 and 3: last three digits of $75000$ give $000$ (÷8) and digit sum $7+5=12$ (÷3).

Answer: (A) $22$

Concept: Apply BODMAS: powers and roots first, then multiplication, then addition/subtraction.

1. $8^2=64$
2. $\sqrt{9^2}=9$
3. $4\sqrt{25}=4\times 5=20$
4. $64+9-20-14=39$

Trick: $\sqrt{a^2}=a$ for positive $a$ – no need to compute $9^2$.

Answer: (A) $39$

Concept: Increasing by $80\%$ means multiplying by $\frac{180}{100}$.

1. $x\times \frac{180}{100}=2520$
2. $x=2520\times \frac{100}{180}=14\times 100=1400$
3. Check: $1400+80\%\text{ of }1400=1400+1120=2520$

Trick: $180\%\to 2520$ ⇒ $1\%\to 14$ ⇒ $100\%\to 1400$.

Answer: (B) $1400$

Concept: Write each number in prime factors; LCM = product of every prime with its highest power.

1. $4^3\times 6^2\times 15=2^6·2^2{3}^2·3·5=2^8\times 3^3\times 5$
2. $4^2\times {15}^2\times 19=2^4\times 3^2\times 5^2\times 19$
3. $6^3\times {15}^2\times {19}^2=2^3\times 3^5\times 5^2\times {19}^2$
4. Highest powers: $2^8,3^5,5^2,{19}^2$

Trick: Only compare the highest power of each prime – no full multiplication needed.

Answer: (C) $2^8\times 3^5\times 5^2\times {19}^2$

Concept: At each vertex of a polygon, interior angle + exterior angle $={180}^{\circ }$; exterior angle of a regular $n$-gon $=\frac{{360}^{\circ }}{n}$.

1. $I+E={180}^{\circ }$ and $I-E={144}^{\circ }$
2. Subtracting: $2E={36}^{\circ }$, so $E={18}^{\circ }$ (and $I={162}^{\circ }$)
3. $n=\frac{{360}^{\circ }}{{18}^{\circ }}=20$

Trick: $E=\frac{{180}^{\circ }-\text{difference}}{2}$, then $n=\frac{{360}^{\circ }}{E}$.

Answer: (D) $20$

Concept: Find each pipe's full-tank time, then use rates (filling $+$, emptying $-$).

1. P: $\frac{5}{7}$ tank in 30 h ⇒ full in $30\times \frac{7}{5}=42$ h. Q: $\frac{4}{7}$ in 12 h ⇒ full in $12\times \frac{7}{4}=21$ h.
2. In 7 h, P and Q fill $7(\frac{1}{42}+\frac{1}{21})=7\times \frac{3}{42}=\frac{1}{2}$ tank.
3. R empties $\frac{1}{2}$ tank in 9 h ⇒ full tank in 18 h.
4. Net rate $=\frac{1}{42}+\frac{1}{21}-\frac{1}{18}=\frac{3+6-7}{126}=\frac{2}{126}=\frac{1}{63}$ ⇒ 63 h.

Trick: Take tank $=\text{LCM}(42,21,18)=126$ units: P $=3$, Q $=6$, R $=-7$ per hour ⇒ $126÷2=63$.

Answer: (D) $63$ hours

Concept: For the same time period, profit is shared in the ratio of investments.

1. Anshuman's profit $=14000-6000=8000$
2. Profit ratio $8000:6000=4:3$ = investment ratio
3. $7$ parts $=48300$ ⇒ $1$ part $=6900$
4. Investments: $27600$ and $20700$; difference $=1$ part $=6900$

Trick: Difference $=\frac{4-3}{4+3}\times 48300=6900$.

Answer: (D) ₹$6900$

Concept: Expand the brackets, then combine like terms.

1. $x(8x-7)=8x^2-7x$
2. $7(x^2-5)=7x^2-35$
3. Sum: $8x^2-7x+7x^2-35+14=15x^2-7x-21$

Trick: Check with $x=1$: $1+(-28)+14=-13$ and $15-7-21=-13$. ✓

Answer: (B) $15x^2-7x-21$

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

1. Times: $\frac{152}{38}=4$ h, $\frac{189}{63}=3$ h, $\frac{375}{75}=5$ h
2. Total time $=12$ h; total distance $=152+189+375=716$ km
3. Average speed $=\frac{716}{12}=59.666\dots \approx 59.67\text{km/h}$

Trick: Each distance divides exactly by its speed – find the times first.

Answer: (A) $59.67$

Concept: Form two equations in pen price $p$ and pencil price $c$, then solve.

1. $9p+9c=252\Rightarrow p+c=28$
2. New prices: $13(p-1)+3(c+3)=100\Rightarrow 13p+3c=104$
3. Subtract $3(p+c)=84$: $10p=20\Rightarrow p=2,c=26$
4. $10p+19c=20+494=514$

Trick: Multiply the simple equation by the coefficient of $c$ and subtract – one variable disappears at once.

Answer: (C) ₹514

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

1. $(5-7)=-2$
2. $8\times (-2)=-16$
3. $17+(-16)=1$
4. $428÷1=428$

Trick: When a bracket simplifies to 1, the answer is the dividend itself.

Answer: (A) 428

Concept: Both numbers must divide the LCM 264; check the options for sum 121 and LCM 264.

1. All options give sum 121, so test the LCM.
2. $88=8\times 11$, $33=3\times 11$
3. $\text{LCM}=8\times 3\times 11=264$ ✔
4. Other pairs (e.g. 81, 40) do not divide 264.

Trick: Pick the pair whose both numbers are factors of the LCM: 264 ÷ 88 = 3 and 264 ÷ 33 = 8.

Answer: (C) 88, 33

Concept: Under annual compounding, interest of the 2nd year $=P\times \frac{R}{100}\times (1+\frac{R}{100})$.

1. $P\times 0.10\times 1.10=1760$
2. $0.11P=1760$
3. $P=\frac{1760}{0.11}=16000$

Trick: At 10%, 2nd-year interest = 11% of the sum, so sum = $1760\times \frac{100}{11}$.

Answer: (C) ₹16,000

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{P}\%}{100}$ and $\text{SP}=\text{MP}\times \frac{100-\text{D}\%}{100}$.

1. $\text{SP}=1575\times 1.2=1890$
2. $\text{MP}\times 0.9=1890$
3. $\text{MP}=\frac{1890}{0.9}=2100$

Trick: $\text{MP}=\text{CP}\times \frac{120}{90}=1575\times \frac{4}{3}=2100$ in one line.

Answer: (A) ₹2,100

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

1. $P=400,R=20,T=4.5$
2. $\text{SI}=\frac{400\times 20\times 4.5}{100}$
3. $=4\times 90=360$

Trick: 20% per year for 4.5 years = 90% of the principal: $0.9\times 400=360$.

Answer: (B) 360

Concept: Let present ages be father $F$, son $S$; write both conditions.

1. $F-6=2(S-6)+34\Rightarrow F=2S+28$
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$
3. $x=28$

Trick: When the father becomes exactly twice the son, years needed $=F-2S$ – here directly 28.

Answer: (B) 28

Concept: ${\sin }^{4}A-{\cos }^{4}A=({\sin }^{2}A-{\cos }^{2}A)({\sin }^{2}A+{\cos }^{2}A)$ and $1+{\cot }^{2}A={\csc }^{2}A$.

1. Numerator: ${\sin }^{2}A-{\cos }^{2}A+1={\sin }^{2}A+{\sin }^{2}A=2{\sin }^{2}A$
2. Denominator: $1+\frac{1}{{\tan }^{2}A}=1+{\cot }^{2}A={\csc }^{2}A=\frac{1}{{\sin }^{2}A}$
3. Value $=2{\sin }^{2}A\times {\sin }^{2}A=2{\sin }^{4}A$

Trick: Put $A={90}^{\circ }$: expression $=\frac{1-0+1}{1+0}=2$; only $2{\sin }^{4}A$ gives 2.

Answer: (A) $2{\sin }^{4}A$

Concept: Together, work per hour = sum of individual rates; take LCM of 16, 21, 15 as tank capacity.

1. Capacity $=\text{LCM}(16,21,15)=1680$ units
2. Rates: $A=105,B=80,C=112$ units/h; total $=297$
3. Time $=\frac{1680}{297}=\frac{560}{99}=5\frac{65}{99}$ hours

Trick: Three pipes of about 15–21 h together take roughly a third of 17 h $\approx 5.6$ h; only $5\frac{65}{99}$ fits.

Answer: (B) $5\frac{65}{99}$ hours

Concept: Lateral surface area of a cuboid $=2h(l+b)$.

1. Side of each cube $=\sqrt[3]{512000}=80\text{cm}$
2. Cuboid: $l=5\times 80=400,b=h=80$
3. $\text{LSA}=2\times 80\times (400+80)=160\times 480=76800{\text{cm}}^2$

Trick: LSA of $n$ cubes in a row $=4\times n\times a^2=4\times 5\times 6400=76800$.

Answer: (C) 76800