RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Selling price after discount $=\text{MP}\times \frac{100-\text{Discount}\%}{100}$.

1. After a $28\%$ discount, SP $=72\%$ of MP.
2. $72\%$ of MP $=$ ₹$3240$
3. MP $=\frac{3240\times 100}{72}=45\times 100=$ ₹$4500$

Trick: Check: $10\%$ of 4500 = 450; $28\%$ = 1260; $4500-1260=3240$ ✓.

Answer: (D) ₹4,500

Concept: A number is divisible by 11 if (sum of digits at odd places) $-$ (sum of digits at even places) is $0$ or a multiple of 11.

1. Odd places: $N+4+6=N+10$; even places: $8+M+9=M+17$.
2. Difference: $(N+10)-(M+17)=N-M-7$.
3. $N-M-7=0\Rightarrow N-M=7$, i.e. $M-N=-7$ (e.g. $N=9,M=2$: 984269 $=11\times 89479$).
4. $N-M-7=-11$ gives $M-N=4$, not an option; $+11$ is impossible for digits.

Trick: Write the digits with alternate $+$ and $-$ signs: $N-8+4-M+6-9=N-M-7$.

Answer: (D) $M-N=-7$

Concept: Take tank capacity $=$ LCM of times; each pipe's efficiency $=\frac{\text{capacity}}{\text{time}}$.

1. LCM$(12,22,8)=264$ units.
2. A $=\frac{264}{12}=22$, B $=\frac{264}{22}=12$, C $=\frac{264}{8}=33$ units/h.
3. Together $=22+12+33=67$ units/h.
4. Time $=\frac{264}{67}=3\frac{63}{67}$ hours.

Trick: All options end in $\frac{63}{67}$; only the whole part matters: $67\times 3=201<264<268$, so it is 3.

Answer: (B) $3\frac{63}{67}$ hours

Concept: Same direction: relative speed $=$ difference; opposite direction: relative speed $=$ sum. $1\text{km/h}=\frac{5}{18}\text{m/s}$.

1. Overtaking the man: relative speed $=\frac{220}{44}=5\text{m/s}=5\times \frac{18}{5}=18\text{km/h}$.
2. Train speed $=18+10=28\text{km/h}$.
3. Opposite train: relative speed $=28+79=107\text{km/h}=\frac{107\times 5}{18}\text{m/s}$; distance $=220+315=535\text{m}$.
4. Time $=\frac{535\times 18}{107\times 5}=\frac{5\times 18}{5}=18\text{s}$.

Trick: $535=5\times 107$ – the numbers are chosen to cancel neatly.

Answer: (D) 18

Concept: $A=P{(1+\frac{R}{100})}^n$.

1. $\frac{24389}{15625}={(1+\frac{R}{100})}^3$
2. $15625={25}^3$ and $24389={29}^3$, so ${\left(\frac{29}{25}\right)}^3={(1+\frac{R}{100})}^3$.
3. $1+\frac{R}{100}=\frac{29}{25}\Rightarrow \frac{R}{100}=\frac{4}{25}\Rightarrow R=16\%$.

Trick: Know cubes up to 30: ${25}^3=15625$, ${29}^3=24389$. Rate $=\frac{29-25}{25}\times 100=16\%$.

Answer: (B) $16\%$

Concept: BODMAS: solve innermost brackets first, then multiplication/division, then addition/subtraction.

1. $(3-7)=-4$
2. $6\times (-4)=-24$
3. $\{26-24\}=2$
4. $868÷2=434$

Trick: Watch the negative sign: $6\times (-4)$ is $-24$, not $+24$.

Answer: (D) 434

Concept: $\text{SI}=\frac{P\times R\times T}{100}$. Equal SI for the same time $\Rightarrow$ principal $\propto \frac{1}{\text{rate}}$.

1. Let $x$ at $3\%$ and $y$ at $12\%$: $\frac{x\times 3\times 7}{100}=\frac{y\times 12\times 7}{100}\Rightarrow 3x=12y$.
2. $x:y=4:1$ (5 parts in total).
3. 1 part $=\frac{21700}{5}=4340$, so $x=4\times 4340=17360$.

Trick: Equal interest: ratio of sums $=$ inverse ratio of rates $=12:3=4:1$.

Answer: (A) 17,360

Concept: If a number leaves the same remainder with $a$, $b$, $c$, it divides their differences. Greatest such number $=$ HCF of the differences.

1. Differences: $183-127=56$, $211-183=28$, $211-127=84$.
2. HCF$(56,28,84)=28$.
3. Check: $127=4\times 28+15$, $183=6\times 28+15$, $211=7\times 28+15$ – remainder 15 each time.

Trick: HCF of $|a-b|$ and $|b-c|$ is enough: HCF$(56,28)=28$.

Answer: (B) 28

Concept: Use $\tan A=\frac{\sin A}{\cos A}$ and ${\sin }^{2}A=1-{\cos }^{2}A=(1-\cos A)(1+\cos A)$.

1. $\frac{\sin A\tan A}{1-\cos A}=\frac{{\sin }^{2}A}{\cos A(1-\cos A)}$
2. $=\frac{(1-\cos A)(1+\cos A)}{\cos A(1-\cos A)}=\frac{1+\cos A}{\cos A}$
3. $=\sec A+1=1+\sec A$

Trick: Put $A={60}^{\circ }$: $\frac{\frac{\sqrt{3}}{2}\times \sqrt{3}}{\frac{1}{2}}=3$, and $1+\sec {60}^{\circ }=1+2=3$ ✓.

Answer: (A) $1+\sec A$

Concept: Product of consecutive numbers $n(n+1)$ lies close to $n^2$, so look near $\sqrt{182}$.

1. $\sqrt{182}\approx 13.5$, so try 13 and 14.
2. $13\times 14=182$ ✓
3. The greater number is 14.

Trick: Unit digit 2 comes from $3\times 4$ – so the numbers end in 3 and 4: 13, 14.

Answer: (A) 14

Concept: If $x$ is the third proportional of $a$ and $b$, then $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $25:35=35:x$
2. $x=\frac{{35}^2}{25}=\frac{1225}{25}=49$

Trick: Third proportional $=\frac{(\text{second}{)}^2}{\text{first}}$.

Answer: (C) 49

Concept: Check for a Pythagorean triplet first; then area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. $33:44:55=3:4:5$ (each $\times 11$), so the triangle is right-angled.
2. ${33}^2+{44}^2=1089+1936=3025={55}^2$
3. Area $=\frac{1}{2}\times 33\times 44=726{\text{cm}}^2$

Trick: Spot $3:4:5$ multiples – no need for Heron's formula.

Answer: (B) 726

Concept: After a 60% increase, the number becomes $160\%$ of itself.

1. $1.6\times N=3570$
2. $N=\frac{3570}{1.6}=2231.25$

Trick: Divide by the multiplying factor $1+\frac{r}{100}$.

Answer: (C) 2231.25

Concept: Test each number from 306 to 313 for divisibility by primes up to $\sqrt{313}\approx 17$.

1. 306, 308, 310, 312 are even; $309=3\times 103$.
2. 307, 311 and 313 have no prime factor up to 17, so they are prime.
3. Count $=3$

Trick: Only odd numbers not ending in 5 need checking.

Answer: (C) 3

Concept: Successive discounts multiply, so their order does not change the final price.

1. Store A: $12000\times 0.85\times 0.90=9180$
2. Store B: $12000\times 0.90\times 0.85=9180$
3. Both prices are ₹9,180.

Trick: Equivalent discount of $a\%$ and $b\%$ $=a+b-\frac{ab}{100}=23.5\%$ either way.

Answer: (D) Both stores A and B offer the same price

Concept: Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$.

1. $a=\frac{22.2}{3}=7.4\text{cm}$
2. $a^2=54.76$
3. Area $=\frac{\sqrt{3}}{4}\times 54.76=13.69\sqrt{3}{\text{cm}}^2$

Trick: $\frac{a^2}{4}={\left(\frac{a}{2}\right)}^2={3.7}^2=13.69$.

Answer: (A) $13.69\sqrt{3}$

Concept: Use the angle sum of a triangle, alternate angles of parallel lines and equal base angles of an isosceles triangle.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{74}^{\circ }-{54}^{\circ }={52}^{\circ }$
2. $BC\parallel AD$ ⇒ $\angle CAD=\angle BCA={52}^{\circ }$ (alternate angles)
3. $AC=CD$ ⇒ $\angle CDA=\angle CAD={52}^{\circ }$
4. $\angle ACD={180}^{\circ }-{52}^{\circ }-{52}^{\circ }={76}^{\circ }$

Trick: Diagonal + parallel sides ⇒ look for alternate angles first.

Answer: (A) ${76}^{\circ }$

Concept: Form two linear equations from the statements and solve.

1. $3M=5D+18$ and $2D=M-3$, i.e. $M=2D+3$
2. $3(2D+3)=5D+18\Rightarrow 6D+9=5D+18\Rightarrow D=9$
3. $M=2\times 9+3=21$
4. Difference $=21-9=12$ years

Trick: Substitute the simpler equation into the other one.

Answer: (A) 12

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

1. $25x^2-30-25x^2-10x+20=20$
2. $-10x-10=20$
3. $-10x=30\Rightarrow x=-3$

Trick: Check: $5(45-6)-5(45-6-4)=195-175=20$ ✓

Answer: (C) $-3$

Concept: Total = average × count; the common person is counted in both pair sums.

1. $M+K+T=3\times 45=135$
2. $M+K=88$, $K+T=90$
3. $K=(M+K)+(K+T)-(M+K+T)=88+90-135=43\text{kg}$

Trick: Sum of the two pairs minus the total gives the common member.

Answer: (C) 43