RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Simplify the complex fraction first, then add/subtract using the LCM of denominators.

1. $1+\frac{5}{9}=\frac{14}{9}$, so $\frac{1}{14/9}=\frac{9}{14}$
2. Expression $=\frac{3}{4}+\frac{9}{14}-\frac{2}{3}$; LCM of 4, 14, 3 is 84.
3. $=\frac{63+54-56}{84}=\frac{61}{84}$

Trick: Denominator 84 appears in only one option – just check numerator $63+54-56=61$.

Answer: (A) $\frac{61}{84}$

Concept: Take CP = 100 units; mark-up and discount act one after the other.

1. List price $=134$; after 25% discount, SP $=134\times \frac{3}{4}=100.5$
2. Profit $=100.5-100=0.5$ unit $=₹177$ ⇒ 1 unit $=₹354$
3. List price $=134\times 354=₹47,436$ (CP $=₹35,400$, SP $=₹35,577$).

Trick: Net effect $=34-25-\frac{34\times 25}{100}=0.5\%$ profit ⇒ CP $=\frac{177}{0.005}=35400$.

Answer: (C) 47,436

Concept: HCF divides every number, so it also divides their differences.

1. $120=2^3\times 3\times 5$, $144=2^4\times 3^2$, $168=2^3\times 3\times 7$
2. Common factors with lowest powers: $2^3\times 3=24$
3. Check: differences are 24 and 24, and 24 divides 120, 144, 168.

Trick: Difference trick: $144-120=24$, $168-144=24$ ⇒ HCF is 24 or its factor; 24 divides all.

Answer: (D) 24

Concept: Increasing a number by 100% means adding the whole number again, i.e. doubling it.

1. Let the number be $x$; then $x\times \frac{200}{100}=2750$
2. $2x=2750$ ⇒ $x=1375$

Trick: +100% = ×2, so just halve: $2750÷2=1375$.

Answer: (C) 1375

Concept: Form two linear equations from the two conditions and eliminate $8P$.

1. Now: $8P-Q=46$ ⇒ $8P=Q+46$
2. After 10 years: $3(Q+10)=8(P+10)-12$ ⇒ $3Q+30=8P+68$ ⇒ $3Q=8P+38$
3. Substitute: $3Q=Q+46+38$ ⇒ $2Q=84$ ⇒ $Q=42$
4. Check: $8P=88$, $P=11$; after 10 years $3\times 52=156=8\times 21-12$ ✓

Trick: Option check: $Q+46$ must be a multiple of 8 – $42+46=88$ ✓, while 49, 37, 43 give 95, 83, 89 ✗.

Answer: (C) 42

Concept: A number is prime if no prime up to its square root divides it; here $\sqrt{322}<18$, so test 2, 3, 5, 7, 11, 13, 17.

1. Even numbers 318, 320, 322 are not prime; $319=11\times 29$; $321=3\times 107$.
2. 317 is not divisible by 2, 3, 5, 7, 11, 13 or 17 ⇒ prime.
3. So only one prime (317) lies in the range.

Trick: Quick tests: digit sum for 3 (3+2+1=6), alternate-sum for 11 (3−1+9=11).

Answer: (D) 1

Concept: Bring CP and SP to the same unit (per pencil), then use $\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times 100$.

1. CP of 1 pencil $=\frac{180}{12}=₹15$; SP $=₹16$
2. Profit $=₹1$
3. Profit% $=\frac{1}{15}\times 100=6.67\%$

Trick: $\frac{1}{15}=6\frac{2}{3}\%$ – remember the fraction table.

Answer: (A) 6.67

Concept: Follow BODMAS: brackets/powers first, then division and multiplication, then addition and subtraction.

1. $3^2=9$, $5\times 9=45$, $1\times 45=45$
2. $45÷17=\frac{45}{17}$
3. $\frac{45}{17}+44-78=\frac{45}{17}-34=\frac{45-578}{17}=-\frac{533}{17}$

Trick: All options are over 17: numerator $=45-34\times 17=45-578=-533$.

Answer: (A) $-\frac{533}{17}$

Concept: Selling price after discount $=\text{Marked Price}\times \frac{100-d}{100}$.

1. Discount $=10\%$ of ₹15,000 $=₹1,500$
2. Amount paid $=15000-1500=₹13,500$

Trick: 10% off ⇒ pay 90%: $15000\times 0.9=13500$.

Answer: (B) ₹13,500

Concept: Take tank capacity = LCM of the times; each pipe's efficiency = capacity ÷ its time.

1. LCM of 12, 25, 10 $=300$ units.
2. A $=\frac{300}{12}=25$, B $=\frac{300}{25}=12$, C $=\frac{300}{10}=30$ units per hour; together $=67$
3. Time $=\frac{300}{67}=4\frac{32}{67}$ hours

Trick: All options end in $\frac{32}{67}$ – just find the whole part: $300÷67\approx 4.5$ ⇒ 4.

Answer: (C) $4\frac{32}{67}$ hours

Concept: Expand the brackets; the $x^2$ terms cancel and a linear equation remains.

1. $4(4x^2-5)=16x^2-20$ and $4(4x^2+6x-4)=16x^2+24x-16$
2. Subtract: $16x^2-20-16x^2-24x+16=-24x-4$
3. $-24x-4=15\Rightarrow -24x=19\Rightarrow x=-\frac{19}{24}$

Trick: Check: $-24\times (-\frac{19}{24})-4=19-4=15$ ✓

Answer: (C) $-\frac{19}{24}$

Concept: For equal distances at speeds $a$ and $b$, average speed $=\frac{2ab}{a+b}$ (harmonic mean), not $\frac{a+b}{2}$.

1. $a=20,b=30$
2. Average speed $=\frac{2\times 20\times 30}{20+30}=\frac{1200}{50}=24\text{km/h}$

Trick: Average speed for a round trip is always a little less than the simple average ($25$).

Answer: (A) $24$

Concept: Under compound interest, if a sum becomes $k$ times in $T$ years, it becomes $k^n$ times in $nT$ years.

1. $3\times$ in $15$ years
2. $81=3^4$, so time $=4\times 15=60$ years

Trick: CI multiples go by powers: $3\to 9\to 27\to 81$ at $15,30,45,60$ years.

Answer: (B) $60$ years

Concept: Order: Division and Multiplication first (left to right), then Addition and Subtraction.

1. $12÷6=2$, $2\times 3=6$
2. $16+2-6=12$

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

Answer: (A) $12$

Concept: If the three adjacent face areas of a cuboid are $lh,bh,lb$, then $V=\sqrt{(lh)(bh)(lb)}=lbh$.

1. Product of areas $=45\times 24\times 30=32400$
2. $(lbh{)}^2=32400\Rightarrow V=\sqrt{32400}=180{\text{cm}}^3$
3. (e.g. $l=7.5,b=4,h=6$ fits: $7.5\times 6=45$, $4\times 6=24$, $7.5\times 4=30$)

Trick: Volume $=\sqrt{\text{product of three face areas}}$.

Answer: (D) $180{\text{cm}}^3$

Concept: Successive percentages of a salary are applied on the remaining amount, so multiply the fractions left.

1. Let salary $=S$. After loan payment ($20\%$), left $=0.8S$
2. He spends $25\%$ of this, so savings $=0.75\times 0.8S=0.6S$
3. $0.6S=780\Rightarrow S=\frac{780}{0.6}=1300$

Trick: Savings fraction $=\frac{4}{5}\times \frac{3}{4}=\frac{3}{5}$; $780÷3\times 5=1300$.

Answer: (B) ₹$1300$

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$. Doubling the speed halves the time.

1. Let Amit’s time $=t_A=\frac{198}{a}$, Vinay’s time $=t_V$. Then $t_A=t_V+4$
2. With double speed Amit takes $\frac{t_A}{2}=t_V-5$
3. Subtract: $t_A-\frac{t_A}{2}=9\Rightarrow t_A=18$ h
4. $a=\frac{198}{18}=11\text{km/h}$

Trick: Check: Vinay $=14$ h; Amit doubled $=9$ h $=14-5$ ✓

Answer: (C) $11\text{km/h}$

Concept: Successive discounts multiply: $SP=MP\times (1-d_1)(1-d_2)(1-d_3)$.

1. After $25\%$: $2500\times 0.75=1875$
2. After $20\%$: $1875\times 0.8=1500$
3. $1500\times (1-\frac{K}{100})=1380\Rightarrow 1-\frac{K}{100}=0.92\Rightarrow K=8$

Trick: Last step: discount $=1500-1380=120$, and $\frac{120}{1500}\times 100=8\%$.

Answer: (D) $8$

Concept: The string is the hypotenuse: $\sin \theta =\frac{\text{height}}{\text{string}}$.

1. $\sin {45}^{\circ }=\frac{353.55}{L}$
2. $L=353.55\times \sqrt{2}=353.55\times 1.4142\approx 500\text{m}$

Trick: $353.55=\frac{500}{\sqrt{2}}$ – at ${45}^{\circ }$, hypotenuse $=\text{side}\times \sqrt{2}$.

Answer: (C) $500\text{m}$

Concept: Arrange the data in order; for odd $n$ the median is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: $23,30,36,44,46,\mathbf{49},54,55,59,96,97$
2. $n=11$, median $=6^{\text{th}}$ value $=49$

Trick: Median ignores extreme values like $96,97$ – only the middle position matters.

Answer: (A) $49$