RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: $(A+C)=2(A+B+C)-(A+B)-(B+C)$ in terms of per-minute rates.

1. $A+B+C=\frac{1}{1.5}=\frac{2}{3}$, $A+B=\frac{1}{3}$, $B+C=\frac{1}{4.5}=\frac{2}{9}$.
2. $A+C=\frac{4}{3}-\frac{1}{3}-\frac{2}{9}=1-\frac{2}{9}=\frac{7}{9}$ keg/min.
3. Time $=\frac{9}{7}$ minutes.

Trick: Take LCM 9 units: all three drain 6/min, A+B 3/min, B+C 2/min ⇒ A+C $=12-3-2=7$/min ⇒ $\frac{9}{7}$ min.

Answer: (C) $\frac{9}{7}$

Concept: Mean $=\frac{\text{sum of observations}}{n}$.

1. Sum $=12+25+37+48+56+61+73+85+92+104=593$.
2. Mean $=\frac{593}{10}=59.3$.

Trick: Pair for quick adding: $(12+85)+(25+73)+(37+61)+(48+56)+92+104=97+98+98+104+196=593$.

Answer: (A) 59.3

Concept: Take MP = ₹100 per toy (₹900 for 9) and compute the amount paid under each scheme; the least beneficial is the one costing the most.

1. (A) $900\times 0.65\times 0.96=561.6$.
2. (B) $900\times 0.55\times 0.54\times 0.50=133.65$.
3. (C) $3\times 85+6\times 95=255+570=825$.
4. (D) pay for 7 only $=700$.
5. Highest payment is scheme C ⇒ least beneficial.

Trick: "2 free on 9" $=\frac{2}{9}\approx 22\%$ off; scheme C is barely 8.3% off overall.

Answer: (A) C

Concept: TSA of a solid hemisphere $=3\pi r^2$; surface area of sphere $=4\pi r^2$.

1. Two hemispheres: $2\times 3\pi r^2=6\pi r^2$.
2. Sphere formed: $4\pi r^2$; reduction $=2\pi r^2$ (the two flat faces).
3. Percentage $=\frac{2\pi r^2}{6\pi r^2}\times 100=33.33\%$.

Trick: Joining hides $2$ of the $6$ units of area ⇒ $\frac{1}{3}$ reduction.

Answer: (A) 33.33

Concept: If after meeting the trains take $t_1$ and $t_2$ hours to finish, $\frac{S_A}{S_B}=\sqrt{\frac{t_2}{t_1}}$.

1. $\frac{S_A}{S_B}=\sqrt{\frac{9}{16}}=\frac{3}{4}\Rightarrow S_B=\frac{4}{3}\times 60=80\text{km/h}$.
2. $80\text{km/h}=80\times \frac{5}{18}=\frac{200}{9}\text{m/s}$.
3. Length $=\frac{200}{9}\times 18=400\text{m}$.

Yaad rakhein: Speed ratio is the inverse square root of the post-meeting times: $16:9\to 3:4$.

Answer: (B) 400

Concept: Savings = salary − expenditure; write both savings in terms of the old salary $S$.

1. Old savings $=35\%$ of $S=0.35S$.
2. New salary $=1.1S$, savings $=40\%$ of $1.1S=0.44S$.
3. $0.44S-0.35S=0.09S=3150\Rightarrow S=35000$.
4. Present salary $=1.1\times 35000=$ ₹38,500.

Trick: The question asks the present salary – not the old ₹35,000 (option D is the trap).

Answer: (C) 38,500

Concept: Divisible by 45 ⇔ divisible by both 5 (last digit 0 or 5) and 9 (digit sum multiple of 9).

1. $b$ is a positive integer ⇒ $b=5$ (not 0).
2. Digit sum $=7+a+3+5+5=20+a$ ⇒ $a=7$ (sum 27).
3. $a+b=7+5=12$.

Trick: "Positive" rules out $b=0$; then only one $a$ in 0–9 makes the sum a multiple of 9.

Answer: (C) 12

Concept: Assume CP = 100; loss% is always on the (new) cost price.

1. CP $=100$, SP $=125$.
2. New CP $=120$, new SP $=125\times 0.9=112.5$.
3. Loss $=7.5$; loss% $=\frac{7.5}{120}\times 100=6.25\%$.

Trick: New SP/CP $=\frac{1.25\times 0.9}{1.2}=0.9375$ ⇒ $6.25\%$ loss.

Answer: (A) 6.25%

Concept: For an odd count of consecutive integers the average is the middle term.

1. Middle (3rd) integer $=10$ ⇒ numbers are $8,9,10,11,12$.
2. Smallest $=8$.

Trick: Smallest $=\text{average}-\frac{n-1}{2}=10-2=8$.

Answer: (C) 8

Concept: Mean proportional between $p$ and $q$ is $\sqrt{pq}$.

1. $x=\sqrt{(a+b)(a-b)}$.
2. $(a+b)(a-b)=a^2-b^2\Rightarrow x=\sqrt{a^2-b^2}$.

Yaad rakhein: $(a+b)(a-b)=a^2-b^2$ – never $a^2+b^2$ (option B is the trap).

Answer: (D) $\sqrt{a^2-b^2}$

Concept: Net change for successive changes $a\%$ and $b\%$ is $a+b+\frac{ab}{100}$.

1. $a=+15,b=-12$.
2. Net $=15-12+\frac{15\times (-12)}{100}=3-1.8=1.2$.
3. Positive sign means an increase of $1.2\%$.

Trick: Multiply factors: $1.15\times 0.88=1.012$, i.e. $+1.2\%$.

Answer: (B) 1.2% increase

Concept: If $a+b+c=0$ then $a^3+b^3+c^3=3abc$.

1. $21+(-2)+(-19)=0$, so the identity applies.
2. $3abc=3\times 21\times (-2)\times (-19)$.
3. $=3\times 21\times 38=2394$.

Trick: Whenever the three bases add to zero, never cube them – just find $3abc$.

Answer: (C) 2394

Concept: $88=8\times 11$; a number divisible by 88 must be divisible by both 8 and 11.

1. $1232=88\times 14$, $1144=88\times 13$, $968=88\times 11$ – all divisible.
2. 1058: last three digits 058 → $58÷8$ leaves remainder 2, so not divisible by 8.
3. Hence 1058 is not divisible by 88.

Trick: Check divisibility by 8 (last 3 digits) first – it eliminates quickly.

Answer: (D) 1058

Concept: Sum $=$ average $\times$ count; removed number $=$ old sum $-$ new sum.

1. Sum of 5 numbers $=5\times 18.5=92.5$.
2. Sum of remaining 4 $=4\times 16=64$.
3. Removed number $=92.5-64=28.5$.

Trick: Removed number $=$ new average $+$ (drop in average) $\times$ old count $=16+2.5\times 5=28.5$.

Answer: (C) 28.5

Concept: Add the same $x$ years to both ages and form the equation.

1. $36+x=2(12+x)$.
2. $36+x=24+2x\Rightarrow x=12$.
3. Check: after 12 years, mother 48, daughter 24 – exactly double.

Trick: Difference of ages (24) is constant; when mother is twice, difference $=$ daughter's age, so daughter must be 24 → after 12 years.

Answer: (D) 12 years

Concept: A sphere inscribed in a cube of edge $a$ has diameter $a$, so radius $r=\frac{a}{2}$.

1. $a=10\text{cm}\Rightarrow r=5\text{cm}$.
2. Surface area of sphere $=4\pi r^2=4\pi \times 25=100\pi$.
3. Surface area of cube $=6a^2=600$.
4. Ratio $=100\pi :600=\pi :6$.

Trick: For any inscribed sphere, $\frac{4\pi (a/2{)}^2}{6a^2}=\frac{\pi }{6}$ – independent of the edge length.

Answer: (B) $\pi :6$

Concept: Take capacity $=\text{LCM}(12,18)=36$ units. A fills $3$ units/h, B empties $2$ units/h.

1. First 4 hours (only A): $4\times 3=12$ units filled.
2. Remaining $36-12=24$ units; net rate with both $=3-2=1$ unit/h.
3. Time for remaining $=24$ h; total $=4+24=28$ hours.

Trick: Fractions: A alone does $\frac{4}{12}=\frac{1}{3}$; remaining $\frac{2}{3}$ at net $\frac{1}{36}$/h takes 24 h.

Answer: (C) 28 hours

Concept: Convert each ratio to a decimal (or compare with a common benchmark) to find the greatest.

1. $\frac{7}{11}\approx 0.636$, $\frac{9}{14}\approx 0.643$.
2. $\frac{11}{18}\approx 0.611$, $\frac{13}{21}\approx 0.619$.
3. Greatest is $\frac{9}{14}$, i.e. $9:14$.

Trick: Compare with $\frac{2}{3}$: $\frac{9}{14}$ is short of it by only $\frac{1}{42}$, the smallest gap among all four.

Answer: (A) 9 : 14

Concept: ${\left(\frac{6}{14}\right)}^k={\left(\frac{14}{6}\right)}^{-k}$; with equal bases, equate the exponents.

1. LHS $={\left(\frac{14}{6}\right)}^8\times {\left(\frac{14}{6}\right)}^{-3}\times {\left(\frac{14}{6}\right)}^{10}={\left(\frac{14}{6}\right)}^{15}$.
2. RHS $={\left(\frac{6}{14}\right)}^{9m+9}={\left(\frac{14}{6}\right)}^{-(9m+9)}$.
3. $15=-(9m+9)\Rightarrow 9m=-24\Rightarrow m=-\frac{8}{3}$.

Trick: Flip every reciprocal base to a single base first, then just add the powers.

Answer: (C) $-\frac{8}{3}$

Concept: From $\sin A=\frac{5}{7}$: $\cos A=\sqrt{1-\frac{25}{49}}=\frac{2\sqrt{6}}{7}$, $cosecA=\frac{7}{5}$, $\sec A=\frac{7}{2\sqrt{6}}$.

1. Numerator: $5\times \frac{7}{5}-7\times \frac{5}{7}=7-5=2$.
2. Denominator: $2\sqrt{6}\times \frac{7}{2\sqrt{6}}+7\times \frac{24}{49}=7+\frac{24}{7}=\frac{73}{7}$.
3. Value $=2÷\frac{73}{7}=\frac{14}{73}$.

Trick: The coefficients are chosen so that $5cosecA$ and $2\sqrt{6}\sec A$ become whole numbers (7 each) – look for such cancellations.

Answer: (B) $\frac{14}{73}$