RRB NTPC Undergraduate CBT 1 2024 PYQs

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

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ for one sharpener.

1. CP of one $=\frac{150}{12}=$ ₹12.50
2. SP of one $=$ ₹41, profit $=$ ₹28.50
3. $\text{Profit}\%=\frac{28.5}{12.5}\times 100=228\%$

Trick: $\frac{41}{12.5}=3.28$ ⇒ profit $=3.28-1=2.28=228\%$.

Answer: (D) 228

Concept: Mode is the value that occurs most often.

1. Frequencies: 14 → 4 times, 17 → 2 times, the rest once.
2. Highest frequency is for 14.

Yaad rakhein: Mode = most frequent observation.

Answer: (B) 14

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

1. ${55}^{-13}÷{55}^{14}={55}^{-13-14}={55}^{-27}$
2. ${55}^{-27}\times {55}^{-11}={55}^{-38}$

Trick: Same base: just add/subtract the exponents: $-13-14-11=-38$.

Answer: (C) ${55}^{-38}$

Concept: Express the actual SP in terms of CP using the loss case.

1. $\frac{6}{15}\text{SP}=0.4\text{SP}=0.86\text{CP}$
2. $\text{SP}=\frac{0.86}{0.4}\text{CP}=2.15\text{CP}$
3. $54\%$ of SP $=0.54\times 2.15\text{CP}=1.161\text{CP}$
4. $\text{Profit}\%=16.1\%$

Trick: Take CP $=100$: $0.4\text{SP}=86\Rightarrow \text{SP}=215$; $0.54\times 215=116.1$.

Answer: (C) 16.1%

Concept: A number is divisible by 3 if its digit sum is divisible by 3.

1. Digit sum $=3+2+7+0+8+7=27$, divisible by 3 ✓
2. It is odd, so not divisible by 8 or 12; last digit 7, so not by 5.

Trick: Odd numbers can never be divisible by 8 or 12.

Answer: (A) 3

Concept: Convert everything to man-days / woman-days.

1. Work $=4\times 18=72$ man-days $=9\times 8=72$ woman-days
2. So 1 man = 1 woman in efficiency.
3. $3$ women $+6$ men $=9$ men
4. Days $=\frac{72}{9}=8$

Trick: $M_1{D}_1=M_2{D}_2$ once the units are made equal.

Answer: (C) 8 days

Concept: Distance = speed × time (time in hours).

1. $2.4\text{min}=\frac{2.4}{60}=0.04\text{h}$
2. Length $=41\times 0.04=1.64\text{km}$

Trick: km/h × minutes ÷ 60 gives km directly.

Answer: (D) 1.64 km

Concept: Find the value of one ratio part from the given difference.

1. Largest − smallest $=16-10=6$ parts $=$ ₹29,547
2. 1 part $=\frac{29547}{6}=4924.5$
3. Total $=40$ parts $=40\times 4924.5=$ ₹1,96,980

Trick: Total $=\text{difference}\times \frac{40}{6}$.

Answer: (B) 1,96,980

Concept: Apply the percentages one after another on the remaining amount.

1. After debt: $70\%$ of salary remains.
2. Spends $20\%$ of this, saves $80\%$: $0.7\times 0.8S=0.56S$
3. $0.56S=602\Rightarrow S=\frac{602}{0.56}=1075$

Trick: Savings fraction $=\frac{7}{10}\times \frac{4}{5}=\frac{14}{25}$.

Answer: (A) 1,075

Concept: Profit is shared in the ratio capital × time.

1. $M+N=1494$, $M-N=498$ ⇒ $M=996$, $N=498$, ratio $2:1$
2. $\frac{3(N+48000)}{4N}=2\Rightarrow 3N+144000=8N$
3. $N=28800$, $M$'s capital $=28800+48000=76800$

Trick: Check: $3\times 76800=230400$, $4\times 28800=115200$ → $2:1$ ✓

Answer: (B) ₹76,800