RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Overhead (decoration) is added to cost price. $\text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times 100$.

1. CP $=849+6=855$; SP $=441$ → loss.
2. Loss $=855-441=414$.
3. $\text{Loss}\%=\frac{414}{855}\times 100=\frac{920}{19}\%\approx 48.42\%$.

Trick: SP < CP means loss – options with 'profit' are out at once.

Answer: (A) $\frac{920}{19}\%$ loss

Concept: Square roots of decimals: take the root of the digits and halve the decimal places.

1. $\sqrt{169}=13$.
2. $\sqrt{0.0196}=0.14$ (since ${14}^2=196$).
3. $\sqrt{4.41}=2.1$.
4. $13+0.14-2.1=11.04$.

Trick: 4 decimal places under the root → 2 decimal places in the answer.

Answer: (C) 11.04

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

1. Mohsin's share $=\frac{2600}{14400}=\frac{13}{72}$; Abdul's $=\frac{59}{72}$.
2. Mohsin's investment $=\frac{13}{72}\times 79200=14300$; Abdul's $=64900$.
3. Difference $=64900-14300=50600$.

Trick: Difference $=\frac{59-13}{72}\times 79200=46\times 1100=50600$.

Answer: (B) ₹50,600

Concept: Savings = Income − Expenditure; apply each % change separately.

1. Savings $=25\%$ of 91700 $=22925$; expenditure $=68775$.
2. New income $=1.27\times 91700=116459$.
3. New expenditure $=1.40\times 68775=96285$.
4. New savings $=116459-96285=20174$; change $=20174-22925=-2751$ → decrease by ₹2,751.

Trick: Change in savings $=0.27\times 91700-0.40\times 68775=24759-27510=-2751$.

Answer: (A) decrease by ₹2,751

Concept: Take one day's work as a fraction of the whole work and subtract group rates.

1. Shifa $=\frac{1}{30}$, Shifa + Afiya $=\frac{1}{18}$, all three $=\frac{1}{9}$.
2. Mariyam $=\frac{1}{9}-\frac{1}{18}=\frac{1}{18}$.
3. Shifa + Mariyam $=\frac{1}{30}+\frac{1}{18}=\frac{3+5}{90}=\frac{4}{45}$.
4. Days $=\frac{45}{4}=11.25$.

Trick: LCM method: total work 90 units → S = 3, S+A = 5, S+A+M = 10 → M = 5; S+M = 8 → $\frac{90}{8}=\frac{45}{4}$.

Answer: (D) $\frac{45}{4}$

Concept: If $x+\frac{1}{x}=2$ (with $x>0$), then $x=1$.

1. $\text{cosec}\theta =\frac{1}{\sin \theta }$, so $\sin \theta +\frac{1}{\sin \theta }=2\Rightarrow \sin \theta =1$.
2. Then $\text{cosec}\theta =1$.
3. ${\text{cosec}}^6\theta +{\sin }^6\theta =1+1=2$.

Trick: $x+\frac{1}{x}=2\Rightarrow x^n+\frac{1}{x^n}=2$ for every $n$.

Answer: (A) 2

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

1. $2^3=8$, $3^2=9$, ${\left(\frac{2}{2}\right)}^2=1$, $4\times 7=28$.
2. $8-9+1-2+28=26$.

Trick: Any number divided by itself is 1, and $1^2=1$.

Answer: (D) 26

Concept: Buyer's cost = list price after successive discounts; required SP = cost × 1.15 = 90% of new list price.

1. Cost $=500\times 0.90\times 0.86=387$.
2. Required SP $=387\times 1.15=445.05$.
3. New list price $=\frac{445.05}{0.90}=494.50$.

Trick: List price $=\text{CP}\times \frac{100+\text{P}\%}{100-\text{D}\%}=387\times \frac{115}{90}$.

Answer: (C) 494.50

Concept: Profit from a false weight: compare the price received with the cost of the quantity actually given.

1. For 1 kg marked, he gives 0.88 kg (cost 88) and charges $1.48\times 100=148$.
2. Profit% $=\frac{148-88}{88}\times 100=\frac{6000}{88}\approx 68.18\%$.
3. Rounded: 68%.

Trick: Total gain $=\frac{(100+\text{claimed})}{(100-\text{short})}-1=\frac{148}{88}-1$.

Answer: (C) $68\%$

Concept: Sum = average × count; subtract the known parts.

1. First two numbers sum $=2\times 80=160$.
2. Remaining two $=655-160-108=387$.
3. Average $=\frac{387}{2}=193.5$.

Trick: Total − known parts, then divide by how many are left.

Answer: (B) 193.5

Concept: Check each option by division or divisibility rules.

1. $20$: number must end in $0$ – $726179$ ends in $9$, so no. $28$: an odd number cannot be divisible by $28$.
2. $726179÷23$: $23\times 31573=631460+94719=726179$, remainder $0$.
3. $726179÷29$ leaves remainder $29\times 25040=726160$, $726179-726160=19\ne 0$.

Trick: Odd number → strike out every even divisor first ($28$, $20$), then test only $23$ and $29$.

Answer: (A) $23$

Concept: A quadrilateral is cyclic if its opposite angles are supplementary (sum ${180}^{\circ }$).

1. $\angle A+\angle C={47}^{\circ }+{133}^{\circ }={180}^{\circ }$ and $\angle B+\angle D={66}^{\circ }+{114}^{\circ }={180}^{\circ }$.
2. Opposite angles are supplementary, so A, B, C, D lie on a circle – cyclic quadrilateral.
3. Not a parallelogram/rhombus (opposite angles not equal); not a trapezium (adjacent pairs ${47}^{\circ }+{66}^{\circ }$, ${66}^{\circ }+{133}^{\circ }$ etc. are not ${180}^{\circ }$, so no sides parallel).

Trick: Add opposite angles: both pairs $={180}^{\circ }$ → cyclic.

Answer: (C) Cyclic quadrilateral

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

1. $r=24\text{cm}$, so $r^2=576$.
2. $S=4\pi \times 576=2304\pi {\text{cm}}^2$.

Trick: $4r^2=(2r{)}^2={48}^2=2304$ – square the diameter.

Answer: (C) $2304\pi {\text{cm}}^2$

Concept: Let mother's age $=M$ and daughter's age $=D$; form two linear equations.

1. $3M=5D+29$ and $2D=M-2\Rightarrow M=2D+2$.
2. Substitute: $3(2D+2)=5D+29\Rightarrow 6D+6=5D+29\Rightarrow D=23$.
3. $M=2\times 23+2=48$; difference $=48-23=25$ years.

Trick: Check: $3\times 48=144$ and $5\times 23+29=144$ ✓.

Answer: (A) $25$

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

1. $3^3\times 9^2\times 13=3^3\times 3^4\times 13=3^7\times 13$.
2. $3^2\times {13}^2\times 18=3^2\times {13}^2\times 2\times 3^2=2\times 3^4\times {13}^2$.
3. $9^3\times {13}^2\times {18}^2=3^6\times {13}^2\times 2^2\times 3^4=2^2\times 3^{10}\times {13}^2$.
4. LCM $=2^2\times 3^{10}\times {13}^2$.

Trick: Here the third number already contains the other two, so it is itself the LCM.

Answer: (B) $2^2\times 3^{10}\times {13}^2$

Concept: Use $(a-b)(a+b)=a^2-b^2$ to pair the brackets in the denominator.

1. $(x-\frac{1}{3})(x+\frac{1}{3})=x^2-\frac{1}{9}=\frac{9x^2-1}{9}$.
2. $(1-\frac{x}{3})(1+\frac{x}{3})=1-\frac{x^2}{9}=\frac{9-x^2}{9}$.
3. Expression $=\frac{(x^2-9)(9x^2-1)\times 9\times 81}{(9x^2-1)(9-x^2)}$.
4. Since $\frac{x^2-9}{9-x^2}=-1$, value $=-729$.

Trick: Put $x=0$: $\frac{(-9)(-1)}{(-\frac{1}{3})(1)}\times \frac{9}{\frac{1}{3}\times 1}=(-27)\times 27=-729$.

Answer: (A) $-729$

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

1. $168\times x=8\times 4200=33600$.
2. $x=\frac{33600}{168}=200$.

Trick: Check: HCF$(168,200)=8$ ✓.

Answer: (B) $200$

Concept: The car is treated as a point, so length of bridge = distance covered = speed × time (same units).

1. Time $=4.6\text{min}=\frac{4.6}{60}\text{h}$.
2. Distance $=48\times \frac{4.6}{60}=0.8\times 4.6=3.68\text{km}$.

Trick: $48\text{km/h}=0.8\text{km}$ per minute; $0.8\times 4.6=3.68$.

Answer: (D) $3.68\text{km}$

Concept: Let the second number be $y$; then the first is $5y$.

1. $5y\times y=2000\Rightarrow y^2=400\Rightarrow y=20$.
2. First number $=100$; sum $=100+20=120$.

Trick: Sum $=6y$, so it must be a multiple of $6$ – only $120$ fits.

Answer: (A) $120$

Concept: Find each pipe's full-tank time, then add filling rates and subtract the emptying rate.

1. P: $\frac{2}{7}$ in $18\text{h}$ → full in $63\text{h}$. Q: $\frac{5}{9}$ in $25\text{h}$ → full in $45\text{h}$.
2. In $3\text{h}$: $\frac{3}{63}+\frac{3}{45}=\frac{1}{21}+\frac{1}{15}=\frac{4}{35}$ of tank filled.
3. R empties $\frac{4}{35}$ in $8\text{h}$ → full tank in $70\text{h}$.
4. Together: $\frac{1}{63}+\frac{1}{45}-\frac{1}{70}=\frac{10+14-9}{630}=\frac{15}{630}=\frac{1}{42}$ → $42$ hours.

Trick: Use LCM $630$ as total work: P $=10$, Q $=14$, R $=-9$ units/h → net $15$; $\frac{630}{15}=42$.

Answer: (B) $42$ hours