RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Increasing a number by $100\%$ doubles it.

1. Let the number be $x$.
2. $x+100\%\text{ of }x=2x=2760$
3. $x=\frac{2760}{2}=1380$

Trick: Increase by $100\%$ → divide the result by 2.

Answer: (D) 1380

Concept: Find each pipe's full-tank time, take the LCM as tank capacity and add rates (emptying pipe negative).

1. P: $\frac{2}{9}$ in 20 h → full tank in $20\times \frac{9}{2}=90$ h. Q: $\frac{5}{9}$ in 10 h → $10\times \frac{9}{5}=18$ h.
2. In 3 h, P and Q fill $\frac{3}{90}+\frac{3}{18}=\frac{1}{30}+\frac{5}{30}=\frac{1}{5}$ of the tank.
3. R empties $\frac{1}{5}$ in 8 h → full tank in 40 h.
4. Tank $=\text{LCM}(90,18,40)=360$ units; rates $4+20-9=15$ units/h → time $=\frac{360}{15}=24$ h.

Trick: Net rate $=\frac{1}{90}+\frac{1}{18}-\frac{1}{40}=\frac{1}{24}$ directly.

Answer: (C) 24 hours

Concept: Profit share ∝ capital × time.

1. M + N shares $=5977$ and M − N $=139$ → M $=\frac{5977+139}{2}=3058$, N $=2919$.
2. Let N's capital $=x$; M's $=x+31000$. Ratio: $\frac{2(x+31000)}{7x}=\frac{3058}{2919}$.
3. $5838x+180978000=21406x\Rightarrow 15568x=180978000\Rightarrow x=11625$.
4. M's capital $=11625+31000=₹42625$.

Trick: Check options: M $=42625$ → N $=11625$; $\frac{2\times 42625}{7\times 11625}=\frac{85250}{81375}=\frac{3058}{2919}$ ✓.

Answer: (B) ₹42,625

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with R and T in the same time unit.

1. As printed (4% per month for 6 months): $\text{SI}=\frac{1200\times 4\times 6}{100}=₹288$.
2. (If the rate were 4% per annum, 6 months $=\frac{1}{2}$ year would give $\frac{1200\times 4}{100}\times \frac{1}{2}=₹24$; but the question states per month.)

Trick: Match units first: monthly rate × number of months.

Answer: (D) ₹288

Concept: Form two equations in pen price $p$ and pencil price $c$ and eliminate.

1. $9p+14c=95$ … (i)
2. New prices: $9(p-3)+3(c+7)=78\Rightarrow 9p+3c=84$ … (ii)
3. (i) − (ii): $11c=11\Rightarrow c=1$, then $9p=81\Rightarrow p=9$.
4. $17p+13c=153+13=₹166$

Trick: Same coefficient of $p$ (9) in both → subtract straight away.

Answer: (B) ₹166

Concept: Expand each bracket and combine like terms.

1. $3(\frac{4}{3}{x}^2-21x+12)=4x^2-63x+36$
2. $-4(x^2+4x-14)=-4x^2-16x+56$
3. Sum: $(4-4)x^2+(-63-16)x+(36+56)=-79x+92$

Trick: $x^2$ cancels, so look only at $x$ and constant: $-63-16=-79$, $36+56=92$.

Answer: (C) $-79x+92$

Concept: In a right triangle the largest side $(x+4)$ is the hypotenuse; use Pythagoras.

1. $x^2+(x+2{)}^2=(x+4{)}^2\Rightarrow x^2-4x-12=0$
2. $(x-6)(x+2)=0\Rightarrow x=6$
3. Sides 6, 8, 10; area $=\frac{1}{2}\times 6\times 8=24$ ✓
4. Perimeter $=6+8+10=24\text{cm}$

Trick: Sides in steps of 2 → think of the triplet 6-8-10.

Answer: (B) 24

Concept: Use the triangle angle sum, alternate angles (BC ∥ AD) and the isosceles triangle ACD.

1. In △ABC: $\angle BCA={180}^{\circ }-{70}^{\circ }-{62}^{\circ }={48}^{\circ }$.
2. BC ∥ AD, AC is a transversal → $\angle CAD=\angle BCA={48}^{\circ }$ (alternate angles).
3. AC = CD → $\angle CDA=\angle CAD={48}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{48}^{\circ }-{48}^{\circ }={84}^{\circ }$

Trick: Answer $={180}^{\circ }-2({180}^{\circ }-{70}^{\circ }-{62}^{\circ })$.

Answer: (D) 84

Concept: Evaluate powers and roots first (BODMAS).

1. $3^2=9$, $\sqrt{4^2}=4$, $5\sqrt{81}=5\times 9=45$
2. $9+4-45-4=-36$

Trick: $\sqrt{a^2}=a$ for positive $a$.

Answer: (D) $-36$

Concept: Mode is the value that occurs most often.

1. Frequencies: 20 → 4 times; 19, 14, 18, 12, 11, 17 → once each.
2. Highest frequency is for 20.

Yaad rakhein: Just count repeats — the most repeated value is the mode.

Answer: (A) 20

Concept: If SP of $a$ articles = CP of $b$ articles and $a>b$, there is a loss: $\text{Loss}\%=\frac{a-b}{a}\times 100$.

1. Let CP of 1 book = ₹1, so SP of 40 books = CP of 30 books = ₹30.
2. CP of 40 books = ₹40, SP = ₹30 → loss = ₹10.
3. $\text{Loss}\%=\frac{10}{40}\times 100=25\%$ (note $\frac{100}{10}\%=10\%$, which is wrong).

Trick: Goods count more than money count → loss; divide difference by the larger count.

Answer: (A) 25% loss

Concept: $SI=\frac{P\times R\times T}{100}$, with $T$ in years; count days excluding the deposit day (2024 is a leap year, 366 days).

1. Days: February $29-24=5$, March 31, April 25 → $5+31+25=61$ days.
2. $T=\frac{61}{366}=\frac{1}{6}$ year.
3. $SI=\frac{4000\times 6.75\times 1}{100\times 6}=\frac{270}{6}=45$.

Trick: 61 days of a 366-day year is exactly $\frac{1}{6}$ year — the numbers are chosen to cancel.

Answer: (D) ₹45

Concept: $(a^m{)}^n=a^{mn}$; equal bases ⇒ equal exponents.

1. $x^z=({95}^{0.53}{)}^z={95}^{0.53z}$.
2. $y^8=({95}^{0.83}{)}^8={95}^{6.64}$.
3. $0.53z=6.64\Rightarrow z=\frac{6.64}{0.53}\approx 12.53$.

Trick: Just compare powers of 95: $z=\frac{8\times 0.83}{0.53}$.

Answer: (A) 12.53

Concept: Literate females = total literate − literate males; then express as a percentage of total females.

1. Females $=342000-228000=114000$.
2. Total literate $=44\%\times 342000=150480$; literate males $=59\%\times 228000=134520$.
3. Literate females $=150480-134520=15960$.
4. $\frac{15960}{114000}\times 100=14\%$.

Trick: Weighted average: males are $\frac{2}{3}$ and females $\frac{1}{3}$ of population, so $44=\frac{2\times 59+f}{3}\Rightarrow f=14$.

Answer: (C) 14%

Concept: Take distance $MN=D$. Speed = $\frac{D}{\text{time}}$; after both trains are moving, time to meet = remaining gap ÷ sum of speeds.

1. A takes 7 h → speed $\frac{D}{7}$; B takes 5 h → speed $\frac{D}{5}$.
2. By 9:55 a.m., A has covered $2\times \frac{D}{7}=\frac{2D}{7}$; gap $=\frac{5D}{7}$.
3. Time $=\frac{5D}{7}÷(\frac{D}{7}+\frac{D}{5})=\frac{5}{7}\times \frac{35}{12}=\frac{25}{12}$ h = 2 h 5 min.
4. 9:55 a.m. + 2 h 5 min = 12:00 p.m.

Trick: Take $D=35$ (LCM of 7 and 5): speeds 5 and 7, gap at 9:55 is 25, time $=\frac{25}{12}$ h.

Answer: (D) 12:00 p.m.

Concept: Write the condition as an equation and solve for $\frac{a}{b}$.

1. $\frac{79}{100}a=\frac{4}{9}b$.
2. $\frac{a}{b}=\frac{4}{9}\times \frac{100}{79}=\frac{400}{711}$.
3. So $a:b=400:711$.

Trick: Cross-multiply directly: $a:b=(4\times 100):(9\times 79)$.

Answer: (A) 400 : 711

Concept: $SP=MP\times \frac{100-d}{100}$, so $MP=\frac{SP\times 100}{100-d}$.

1. $MP=\frac{34340\times 100}{85}$.
2. $\frac{34340}{85}=404$, so $MP=40400$.
3. Check: 15% of 40400 = 6060; $40400-6060=34340$ ✓.

Trick: SP is 85% of MP — divide by 0.85.

Answer: (B) ₹40,400

Concept: Test the options: the pair must add to 52 and have LCM 651.

1. $651=3\times 7\times 31$.
2. 31 and 21: sum $=52$, and since they are co-prime, $\text{LCM}=31\times 21=651$ ✓.
3. Other pairs: (24, 28) LCM 168; (29, 23) LCM 667; (36, 16) LCM 144.

Trick: Factorise 651 and look for a pair of its factors adding to 52: $31+21$.

Answer: (D) 31, 21

Concept: Simplify the bracket first, then take LCM.

1. $1+\frac{5}{7}=\frac{12}{7}$, so $\frac{1}{12/7}=\frac{7}{12}$.
2. Expression $=\frac{5}{4}+\frac{7}{12}-\frac{3}{5}$.
3. LCM 60: $\frac{75+35-36}{60}=\frac{74}{60}=\frac{37}{30}$.

Trick: Reciprocal of $1+\frac{5}{7}$ is $\frac{7}{12}$ — do this before anything else.

Answer: (D) $\frac{37}{30}$

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=\frac{64}{2}=32$ cm.
2. $4\pi r^2=4\pi \times 1024=4096\pi {\text{cm}}^2$.

Trick: With diameter $d$: area $=\pi d^2=\pi \times {64}^2=4096\pi$.

Answer: (A) $4096\pi {\text{cm}}^2$