RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Digit sum of 253118139: $2+5+3+1+1+8+1+3+9=33$.
2. $33÷3=11$, so the number is divisible by 3.
3. It is odd, so it is not divisible by 8 or 12; it does not end in 0 or 5, so not by 5.

Trick: Odd number → straight away cut 8 and 12 (both even); last digit 9 → cut 5.

Answer: (C) 3

Concept: With successive transactions, final price $=\text{CP of A}\times (1-\frac{60}{100})\times (1+\frac{25}{100})$.

1. Let CP of A $=x$. A sells to B at 60% loss: $0.4x$.
2. B sells to C at 25% profit: $0.4x\times 1.25=0.5x$.
3. $0.5x=47236\Rightarrow x=94472$.

Trick: Net multiplier $=\frac{2}{5}\times \frac{5}{4}=\frac{1}{2}$, so A’s cost is just double: $2\times 47236$.

Answer: (A) ₹94,472

Concept: Sum of first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. Here $n=170$.
2. Sum $=170\times 171$.
3. Average $=\frac{170\times 171}{170}=171$.

Yaad rakhein: Average of first $n$ even numbers $=n+1$; of first $n$ odd numbers $=n$.

Answer: (A) 171

Concept: If $4A=9B=3C=k$, then $A:B:C=\frac{1}{4}:\frac{1}{9}:\frac{1}{3}$.

1. Multiply by LCM 36: $A:B:C=9:4:12$.
2. Total parts $=9+4+12=25$.
3. Share of A $=\frac{9}{25}\times 58300=9\times 2332=20988$.

Trick: In $pA=qB=rC$, A’s part is $\frac{1}{p}$ — just take reciprocals and clear fractions.

Answer: (D) ₹20,988

Concept: $\text{SP}=\text{MP}\times (1-\frac{\text{Discount}\%}{100})$.

1. Discount $=10\%$ of 900 $=90$.
2. SP $=900-90=810$.

Trick: 10% discount = multiply by 0.9: $900\times 0.9=810$.

Answer: (D) ₹810

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (work left to right).

1. ${50}^{-13}÷{50}^{15}={50}^{-13-15}={50}^{-28}$.
2. ${50}^{-28}\times {50}^{-11}={50}^{-28-11}={50}^{-39}$.

Trick: Just add/subtract the powers: $-13-15-11=-39$.

Answer: (B) ${50}^{-39}$

Concept: A number increased by $r\%$ becomes $x\times \frac{100+r}{100}$.

1. $x\times 1.7=3910$.
2. $x=\frac{3910}{1.7}=2300$.

Trick: Check: $2300+70\%\text{ of }2300=2300+1610=3910$.

Answer: (D) 2300

Concept: Write each statement as a linear equation in M (mother) and D (daughter).

1. $2M=4D+26$ and $3D=M-2\Rightarrow M=3D+2$.
2. Substitute: $2(3D+2)=4D+26\Rightarrow 6D+4=4D+26\Rightarrow D=11$.
3. $M=3\times 11+2=35$; difference $=35-11=24$.

Trick: Check: $2\times 35=70=44+26$ ✓ and $3\times 11=33=35-2$ ✓.

Answer: (A) 24

Concept: Angle sum of a triangle is ${180}^{\circ }$; alternate angles between parallel lines are equal; angles opposite equal sides are equal.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{18}^{\circ }-{93}^{\circ }={69}^{\circ }$.
2. $BC\parallel AD$ with transversal AC: $\angle CAD=\angle BCA={69}^{\circ }$ (alternate angles).
3. $AC=CD\Rightarrow \angle CDA=\angle CAD={69}^{\circ }$.
4. In $△ACD$: $\angle ACD={180}^{\circ }-{69}^{\circ }-{69}^{\circ }={42}^{\circ }$.

Trick: $\angle ACD={180}^{\circ }-2({180}^{\circ }-{18}^{\circ }-{93}^{\circ })={180}^{\circ }-{138}^{\circ }={42}^{\circ }$.

Answer: (B) ${42}^{\circ }$

Concept: With the same base, $(a^m{)}^n=a^{mn}$; equate the powers.

1. $x^z=(4^{0.72}{)}^z=4^{0.72z}$.
2. $y^8=(4^{0.86}{)}^8=4^{6.88}$.
3. $0.72z=6.88\Rightarrow z=\frac{6.88}{0.72}\approx 9.56$.

Trick: $\frac{688}{72}=\frac{86}{9}=9.555\dots \approx 9.56$.

Answer: (A) 9.56

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=40+49+56+75+39=259$.
2. Average $=\frac{259}{5}=51.8$ kg.

Trick: Assume 50: deviations $-10,-1,+6,+25,-11$ sum to $+9$; $50+\frac{9}{5}=51.8$.

Answer: (B) 51.8

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$.

1. After 10%: $1200\times 0.9=1080$.
2. After 5%: $1080\times 0.95=1026$.

Trick: Equivalent single discount $=10+5-\frac{10\times 5}{100}=14.5\%$; $1200\times 0.855=1026$.

Answer: (C) ₹1,026

Concept: Perimeter of an equilateral triangle $=3\times \text{side}$.

1. Side $=13$ cm.
2. Perimeter $=3\times 13=39$ cm.

Yaad rakhein: Equilateral triangle: perimeter $=3a$, area $=\frac{\sqrt{3}}{4}{a}^2$.

Answer: (B) 39 cm

Concept: One-day work $=\frac{1}{\text{days}}$; subtract combined rates to get an individual rate.

1. Anjali + Ayushi $=\frac{1}{18}$; all three $=\frac{1}{9}$.
2. Ankita $=\frac{1}{9}-\frac{1}{18}=\frac{1}{18}$.
3. Anjali + Ankita $=\frac{1}{30}+\frac{1}{18}=\frac{3+5}{90}=\frac{8}{90}$.
4. Days $=\frac{90}{8}=\frac{45}{4}$.

Trick: Take total work $=90$ units: Anjali 3/day, Ankita $10-5=5$/day; together $\frac{90}{8}=\frac{45}{4}$ days.

Answer: (A) $\frac{45}{4}$ days

Concept: C’s one-day work $=(\text{A+B+C})-\text{A}-\text{B}$.

1. A+B+C $=\frac{1}{5}$, A $=\frac{1}{19}$, B $=\frac{1}{16}$.
2. C $=\frac{1}{5}-\frac{1}{19}-\frac{1}{16}=\frac{304-80-95}{1520}=\frac{129}{1520}$.
3. C alone $=\frac{1520}{129}=11\frac{101}{129}$ days.

Trick: C must take more than 5 days but less than A or B alone (16 days); $1520÷129\approx 11.8$, so option A.

Answer: (A) $11\frac{101}{129}$ days

Concept: Profit is shared in the ratio of investments (same time period).

1. $63:35:42=9:5:6$ (divide by 7); total $=20$ parts.
2. One part $=\frac{5120}{20}=256$.
3. B − C difference $=6-5=1$ part $=256$.

Trick: Simplify the ratio first — a difference of 1 part means the answer is just one part.

Answer: (D) ₹256

Concept: Divisible by 5 if the last digit is 0 or 5; by 15 only if divisible by both 3 and 5.

1. 245015 ends in 5 → divisible by 5.
2. Digit sum $=2+4+5+0+1+5=17$, not divisible by 3 → not by 15.
3. Odd → not by 2; for 11: alternate digit sums $2+5+1=8$ and $4+0+5=9$ differ by 1 → not by 11.

Trick: 15 is the tempting trap — always check the 3-rule too.

Answer: (C) 5

Concept: The largest circle inside a rectangle has diameter equal to the shorter side.

1. Shorter side $=126$ m, so $d=126$ m.
2. Circumference $=\pi d=\frac{22}{7}\times 126$.
3. $=22\times 18=396$ m.

Trick: 126 is a multiple of 7 → $126÷7=18$, then $\times 22$.

Answer: (B) 396

Concept: For equal distances at speeds $a$ and $b$: $\text{Average speed}=\frac{2ab}{a+b}$.

1. $a=72$, $b=90$.
2. Average $=\frac{2\times 72\times 90}{72+90}=\frac{12960}{162}$.
3. $=80\text{km/h}$.

Trick: Average speed is always a little less than the simple mean $\frac{72+90}{2}=81$ → 80.

Answer: (A) 80

Concept: ${\text{cosec}}^2\beta -1={\cot }^2\beta$ and ${\sec }^2\beta -{\tan }^2\beta =1$.

1. $\frac{1}{{\text{cosec}}^2\beta -1}=\frac{1}{{\cot }^2\beta }={\tan }^2\beta$.
2. Expression $={\sec }^2\beta -{\tan }^2\beta$.
3. $=1$.

Trick: Put $\beta ={45}^{\circ }$: $2-\frac{1}{2-1}=1$.

Answer: (C) 1