RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Consecutive natural numbers: $n(n+1)=30$.

1. $5\times 6=30$.
2. Greater number = 6.

Trick: Take $\sqrt{30}\approx 5.5$ — the two numbers lie on either side: 5 and 6.

Answer: (C) 6

Concept: Required speed = remaining distance ÷ remaining time.

1. First part: $39\times 4=156$ km in 4 h.
2. Next 140 km at 35 km/h → $\frac{140}{35}=4$ h.
3. Used: $8$ h, $296$ km. Left: $977-296=681$ km in $3$ h.
4. Speed $=\frac{681}{3}=227\text{km/h}$.

Trick: Track both distance and time used so far in one line: (296 km, 8 h).

Answer: (A) 227 km/hr

Concept: 5, 12, 13 is a Pythagorean triplet, so the triangle is right-angled; area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. $5^2+{12}^2=25+144=169={13}^2$ → right angle between sides 5 and 12.
2. Area $=\frac{1}{2}\times 5\times 12=30{\text{cm}}^2$.

Trick: Remember triplets (3,4,5), (5,12,13), (8,15,17), (7,24,25) — no Heron's formula needed.

Answer: (C) $30{\text{cm}}^2$

Concept: Expand each bracket and combine like terms.

1. $3(\frac{4}{3}{x}^2-25x+12)=4x^2-75x+36$.
2. $-4(x^2+7x-18)=-4x^2-28x+72$.
3. Add: $(4x^2-4x^2)+(-75x-28x)+(36+72)=-103x+108$.

Trick: The $x^2$ terms cancel; just check the constant: $36+72=108$ (positive) → only option D.

Answer: (D) $-103x+108$

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

1. $14.20$ and $0.05$ → $1420$ and $5$ (2 decimal places).
2. LCM$(1420,5)=1420$ (as 1420 is a multiple of 5).
3. LCM $=14.20=14.2$.

Trick: If the larger number is a multiple of the smaller, it is itself the LCM: $14.2÷0.05=284$.

Answer: (D) 14.2

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

1. Discount $=2500-1360=$ ₹1,140.
2. $\frac{1140}{2500}\times 100=45.6\%$.

Trick: Divide by 25: $\frac{1140}{25}=45.6$.

Answer: (C) 45.6%

Concept: Equalise decimal places, take LCM of the integers, restore the decimal.

1. $10.80$ and $0.06$ → $1080$ and $6$.
2. LCM$(1080,6)=1080$.
3. LCM $=10.80=10.8$.

Trick: $10.8÷0.06=180$ (whole number), so the larger number itself is the LCM.

Answer: (C) 10.8

Concept: Overall average = total runs ÷ total matches.

1. First 23 matches: $23\times 44=1012$ runs.
2. Next 10: $10\times 14=140$ runs.
3. Average $=\frac{1012+140}{33}=\frac{1152}{33}=34.909\dots \approx 34.91$.

Trick: Deviation method: $44-\frac{10\times 30}{33}=44-9.09=34.91$.

Answer: (D) 34.91

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. Interior sum $=2\times {360}^{\circ }={720}^{\circ }$.
2. $(n-2)\times 180=720\Rightarrow n-2=4\Rightarrow n=6$.

Trick: Interior sum = $k\times {360}^{\circ }$ gives $n=2k+2$; here $k=2$ → $n=6$.

Answer: (A) 6

Concept: $V=\pi r^{2}h$; new volume factor $=(\text{radius factor}{)}^2\times \text{height factor}$.

1. Radius factor $=0.81$, height factor $=2.22$.
2. ${0.81}^2\times 2.22=0.6561\times 2.22=1.4565$.
3. Increase $\approx 45.65\%\approx 46\%$.

Trick: Successive %: $-19\%$ twice $=-34.39\%$; then $-34.39+122-\frac{34.39\times 122}{100}=45.65\%$.

Answer: (D) 46%

Concept: Form the equation from the past condition, then the future condition.

1. Let present ages be father $F$, son $S$. Four years ago: $F-4=2(S-4)+33\Rightarrow F=2S+29$.
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$.
3. $x=29$.

Trick: For "after how many years will he be twice", answer $=F-2S$ directly.

Answer: (A) $29$ years

Concept: Diagonals of a rhombus bisect each other at right angles; each side is the hypotenuse of a right triangle formed by half-diagonals.

1. Side $=\frac{40}{4}=10\text{cm}$; half of given diagonal $=8\text{cm}$.
2. Other half-diagonal $=\sqrt{{10}^2-8^2}=\sqrt{36}=6\text{cm}$.
3. Other diagonal $=2\times 6=12\text{cm}$.

Trick: $d_1^2+d_2^2=4a^2$: $256+d_2^2=400\Rightarrow d_2=12$.

Answer: (B) $12\text{cm}$

Concept: Average speed = total distance ÷ total time (not the average of the speeds).

1. Times: $\frac{432}{72}=6\text{h}$, $\frac{275}{55}=5\text{h}$, $\frac{680}{85}=8\text{h}$; total $=19\text{h}$.
2. Total distance $=432+275+680=1387\text{km}$.
3. Average speed $=\frac{1387}{19}=73\text{km/h}$.

Trick: Every distance here is an exact multiple of its speed – find hours first, then divide.

Answer: (D) $73$

Concept: Find each square root separately; for decimals, count decimal places in pairs.

1. $\sqrt{169}=13$
2. $\sqrt{0.0324}=0.18$ (since ${18}^2=324$, four decimal places → two).
3. $\sqrt{7.29}=2.7$ (since ${27}^2=729$).
4. $13+0.18-2.7=10.48$.

Trick: Decimal places in the root = half the decimal places in the number.

Answer: (D) $10.48$

Concept: Mode is the value that occurs most often in the data.

1. Frequencies: 10 → 1, 11 → 2, 13 → 4, 14 → 1, 16 → 1, 17 → 1.
2. 13 occurs most (4 times), so mode = 13.

Trick: Sort the data first: 10, 11, 11, 13, 13, 13, 13, 14, 16, 17 – the longest run is the mode.

Answer: (A) $13$

Concept: Use the angle-sum property of a triangle and the linear pair at E.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{63}^{\circ }={27}^{\circ }$.
2. In $△AEC$: $\angle AEC={180}^{\circ }-{76}^{\circ }-{27}^{\circ }={77}^{\circ }$.
3. $\angle AEB$ and $\angle AEC$ form a linear pair: $\angle AEB={180}^{\circ }-{77}^{\circ }={103}^{\circ }$.

Trick: Exterior angle: $\angle AEB=\angle CAE+\angle C={76}^{\circ }+{27}^{\circ }={103}^{\circ }$.

Answer: (B) ${103}^{\circ }$

Concept: Extra expenses (decoration, repair, transport) are added to the cost price.

1. Effective CP = ₹$461+15=476$; SP = ₹$918$.
2. Profit = ₹$918-476=442$.
3. $\text{Profit}\%=\frac{442}{476}\times 100=\frac{13}{14}\times 100=\frac{650}{7}\%$.

Trick: $442=34\times 13$ and $476=34\times 14$ – cancel 34 first.

Answer: (C) $\frac{650}{7}\%$ profit

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; first find the principal, then the new amount.

1. $R\times T=3.6\times \frac{41}{9}=16.4\%$, so $0.164P=5453\Rightarrow P=33250$.
2. New $R\times T=2.7\times \frac{56}{9}=16.8\%$; SI $=0.168\times 33250=5586$.
3. Amount $=33250+5586=38836$.

Trick: Total SI % = rate × time; Amount $=P\times \frac{100+16.8}{100}$.

Answer: (D) ₹38,836

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

1. $P=400$, $R=8\%$, $T=5.5$ years.
2. $\text{SI}=\frac{400\times 8\times 5.5}{100}=4\times 44=176$.

Trick: 1 year SI $=32$; $32\times 5.5=176$.

Answer: (A) ₹176

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

1. $2\times 410=82\times x$
2. $x=\frac{820}{82}=10$.
3. Check: $\text{HCF}(82,10)=2$, $\text{LCM}=410$ ✓.

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

Answer: (C) $10$