RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Form two linear equations in the price of a pen $p$ and a pencil $c$ and solve them.

1. $2p+19c=118$ ...(i)
2. New prices: $5(p-5)+8(c+4)=144\Rightarrow 5p-25+8c+32=144\Rightarrow 5p+8c=137$ ...(ii)
3. (i)$\times 5$ − (ii)$\times 2$: $95c-16c=590-274\Rightarrow 79c=316\Rightarrow c=4$, so $p=\frac{118-76}{2}=21$
4. Required cost $=14\times 21+6\times 4=294+24=318$

Trick: Since $118-19c$ must be even, $c$ is even; try $c=4$: $p=21$ and $5(16)+8(8)=144$ ✔.

Answer: (A) ₹318

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $182\times x=91\times 546$
2. $x=\frac{91\times 546}{182}=\frac{546}{2}=273$
3. Check: $182=2\times 7\times 13$, $273=3\times 7\times 13$ → HCF $=91$, LCM $=546$ ✔

Trick: Both numbers are multiples of the HCF: $182=91\times 2$, so other $=91\times 3=273$ (since $\frac{546}{91}=6=2\times 3$).

Answer: (A) 273

Concept: If a number $x$ is increased by $r\%$, new value $=x\times \frac{100+r}{100}$.

1. $x\times \frac{160}{100}=3550$
2. $x=\frac{3550\times 100}{160}=\frac{3550\times 5}{8}=\frac{17750}{8}=2218.75$

Trick: $60\%$ increase = multiply by $\frac{8}{5}$; reverse by multiplying with $\frac{5}{8}$.

Answer: (A) 2218.75

Concept: A dishonest seller gains twice: on the marked-up price and on the short weight. $\text{Gain}\%=(\frac{\text{price received}}{\text{cost of goods given}}-1)\times 100$.

1. Let cost of 1 unit = ₹1. He claims 100 units and charges $100\times 1.39=139$.
2. He actually gives $100-17=83$ units, costing ₹83.
3. $\text{Gain}\%=\frac{139-83}{83}\times 100=\frac{5600}{83}\approx 67.47\%\approx 67\%$

Trick: $\frac{1+\frac{39}{100}}{1-\frac{17}{100}}=\frac{139}{83}\approx 1.675$ → about $67\%$.

Answer: (C) $67\%$

Concept: Take the square root of each term separately, then add/subtract.

1. $\sqrt{400}=20$
2. $\sqrt{0.0225}=\sqrt{\frac{225}{10000}}=\frac{15}{100}=0.15$
3. $\sqrt{9}=3$
4. $20+0.15-3=17.15$

Trick: Decimal places in a square root = half of those in the number: $0.0225$ has 4 places → root $0.15$ has 2.

Answer: (D) 17.15

Concept: If a number $N$ leaves remainders $r_1,r_2$ on dividing $a,b$, then $N=\text{HCF}(a-r_1,b-r_2)$.

1. $527-7=520$ and $632-8=624$
2. $520=2^3\times 5\times 13$, $624=2^4\times 3\times 13$
3. $\text{HCF}=2^3\times 13=104$
4. Check: $527=5\times 104+7$, $632=6\times 104+8$ ✔ (and $104>8$)

Trick: $624-520=104$; the HCF must divide this difference, and 104 itself divides both → answer 104.

Answer: (B) 104

Concept: If $9A=3B=7C=k$, then $A:B:C=\frac{1}{9}:\frac{1}{3}:\frac{1}{7}$; multiply by LCM of $9,3,7=63$.

1. $A:B:C=\frac{63}{9}:\frac{63}{3}:\frac{63}{7}=7:21:9$
2. Total parts $=7+21+9=37$
3. 1 part $=\frac{74000}{37}=2000$
4. $A=7\times 2000=14000$

Trick: The share is inversely proportional to the multiplier: smallest multiplier 3 (B) gets the largest share.

Answer: (C) ₹14,000

Concept: Expand each bracket, then combine like terms.

1. $x(6x-3)=6x^2-3x$
2. $5(x^2-4)=5x^2-20$
3. Sum: $6x^2-3x+5x^2-20+18=11x^2-3x-2$

Trick: Put $x=1$: expression $=3+(-15)+18=6$; option (B) gives $11-3-2=6$ ✔.

Answer: (B) $11x^2-3x-2$

Concept: To test $n$ for primality, check divisibility by primes up to $\sqrt{n}$. Here $\sqrt{476}<22$, so test 2, 3, 5, 7, 11, 13, 17, 19.

1. Even numbers 466, 468, 470, 472, 474, 476 are out; 475 ends in 5.
2. $469=7\times 67$, $471=3\times 157$, $473=11\times 43$
3. 467 is not divisible by 2, 3, 5, 7, 11, 13, 17 or 19 → prime.
4. So only one prime: 467.

Trick: Between consecutive odd numbers, quickly test 3 (digit sum), 7 and 11 (alternating sum) first – they remove most candidates.

Answer: (A) 1

Concept: $\text{Profit}=\text{SP of plates sold}-\text{CP of all plates}$; broken plates add to cost but give no revenue.

1. CP of 79 plates = ₹79
2. Plates left $=79-13=66$
3. SP $=66\times 5.90=389.40$
4. Profit $=389.40-79=310.40$

Trick: $66\times 5.9=66\times 6-66\times 0.1=396-6.6=389.4$.

Answer: (D) ₹310.40

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

1. Total distance $=282+120+496=898\text{km}$.
2. Times: $\frac{282}{47}=6$, $\frac{120}{20}=6$, $\frac{496}{62}=8$ hours; total $=20$ h.
3. Average speed $=\frac{898}{20}=44.9\text{km/h}$.

Trick: Each distance is an exact multiple of its speed – divide first, then add the hours.

Answer: (B) $44.9$

Concept: Volume of a sphere $=\frac{4}{3}\pi r^3$.

1. Diameter $=30\text{cm}$, so $r=15\text{cm}$ and $r^3=3375$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 3375=\frac{99000}{7}\approx 14142.9{\text{cm}}^3$.
3. The nearest option is $14130{\text{cm}}^3$ (this exact value comes from $\pi =3.14$: $\frac{4}{3}\times 3.14\times 3375=14130$).

Trick: $\frac{4}{3}\pi r^3=4\pi \times \frac{r^3}{3}$: $\frac{3375}{3}=1125$, then $4\times 1125=4500$, $4500\pi$.

Answer: (A) $14,130{\text{cm}}^3$

Concept: Parallel lines give equal alternate angles; equal sides of a triangle face equal angles.

1. In $△ABC$: $\angle ACB={180}^{\circ }-{54}^{\circ }-{60}^{\circ }={66}^{\circ }$.
2. $BC\parallel AD$, so $\angle CAD=\angle ACB={66}^{\circ }$ (alternate angles).
3. $AC=CD\Rightarrow \angle CDA=\angle CAD={66}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{66}^{\circ }-{66}^{\circ }={48}^{\circ }$.

Trick: Once you get ${66}^{\circ }$, the answer is simply ${180}^{\circ }-2\times {66}^{\circ }$.

Answer: (B) ${48}^{\circ }$

Concept: A number is divisible by 5 if its last digit is 0 or 5.

1. $740705$ ends in $5$ → divisible by $5$.
2. It is odd, so not divisible by $4$ or $12$.
3. Digit sum $=7+4+0+7+0+5=23$, not a multiple of $9$ (or $3$).

Trick: Check the last digit first – it settles 2, 4, 5 and 10 at a glance.

Answer: (C) $5$

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

1. Ratio $1350:1470:1380=135:147:138=45:49:46$ (sum $140$).
2. Isha's share $=\frac{46}{140}\times 1330=46\times 9.5=437$.

Trick: Divide total profit by the ratio sum first: $\frac{1330}{140}=9.5$ per part.

Answer: (C) ₹$437$

Concept: Combined average $=\frac{\text{total runs}}{\text{total matches}}$.

1. First 23 matches: $23\times 41=943$ runs.
2. Next 10 matches: $10\times 14=140$ runs.
3. Average $=\frac{943+140}{33}=\frac{1083}{33}\approx 32.82$.

Trick: Answer must lie between $14$ and $41$, closer to $41$ because 23 matches carry more weight.

Answer: (D) $32.82$

Concept: $SI=P\times r\times t$, with rate and time in the same unit (here per month).

1. $P=1200$, $r=4\%$ per month, $t=3$ months.
2. $SI=1200\times \frac{4}{100}\times 3=144$.

Trick: $4\%$ per month for 3 months $=12\%$; $12\%$ of $1200=144$.

Answer: (D) ₹$144$

Concept: If $p\%$ of $A$ = $k$ times $B$, then $\frac{A}{B}=\frac{k}{p/100}$.

1. $\frac{39}{100}A=\frac{8}{9}B$.
2. $\frac{A}{B}=\frac{8}{9}\times \frac{100}{39}=\frac{800}{351}$.

Trick: Cross-multiply: $A:B=(8\times 100):(9\times 39)$.

Answer: (B) $800:351$

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$; use $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$, $\tan {60}^{\circ }=\sqrt{3}$.

1. Roof: $h_1=30\tan {30}^{\circ }=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{m}$.
2. Top of pole: $h_2=30\tan {60}^{\circ }=30\sqrt{3}\text{m}$.
3. Flagstaff $=h_2-h_1=20\sqrt{3}\text{m}$.

Trick: With ${30}^{\circ }$ and ${60}^{\circ }$ from the same point, the heights are in ratio $1:3$, so pole $=\frac{2}{3}$ of $30\sqrt{3}$.

Answer: (B) $20\sqrt{3}$

Concept: BODMAS: division and multiplication before addition and subtraction.

1. $12÷6=2$ and $2\times 4=8$.
2. $16+2-8=10$.

Trick: Never add $16+12$ first – solve $÷$ and $\times$ before $+$, $-$.

Answer: (D) $10$