RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Expand each bracket, then collect like terms ($x^2$, $x$, constants).

1. $4(\frac{6}{4}{x}^2-25x+12)=6x^2-100x+48$
2. $-6(x^2+5x-13)=-6x^2-30x+78$
3. Add: $(6x^2-6x^2)+(-100x-30x)+(48+78)=-130x+126$

Trick: The $x^2$ terms cancel ($6-6=0$), so only check the $x$-coefficient: $-100-30=-130$.

Answer: (A) $-130x+126$

Concept: A number is divisible by 5 if its last digit is 0 or 5; by 3 if its digit sum is divisible by 3; by 15 if by both 3 and 5; by 14 only if it is even and divisible by 7.

1. Last digit of 386937355 is 5 ⇒ divisible by 5.
2. Digit sum $=3+8+6+9+3+7+3+5+5=49$, not a multiple of 3 ⇒ not divisible by 3, hence not by 15.
3. The number is odd ⇒ not divisible by 14.

Trick: Odd number ending in 5 → rule out 14 at once; then a digit sum test settles 3 and 15.

Answer: (C) 5

Concept: In a partnership, profit is shared in the ratio of (capital $\times$ time).

1. $M+N=6705$ and $M-N=745$ ⇒ $M=3725$, $N=2980$; ratio $=3725:2980=5:4$.
2. Let N’s capital $=x$; M’s $=x+79000$. Then $\frac{7(x+79000)}{6x}=\frac{5}{4}$.
3. $28x+2212000=30x$ ⇒ $x=1106000$.
4. M’s capital $=1106000+79000=1185000$, i.e. ₹11,85,000.

Trick: Shares from sum & difference: $\frac{6705+745}{2}=3725$. Check option D: $\frac{1185000\times 7}{1106000\times 6}=\frac{5}{4}$ ✓.

Answer: (D) ₹11,85,000

Concept: Follow BODMAS: innermost bracket first, then multiplication, addition, and finally division.

1. $(5-4)=1$
2. $2\times 1=2$, so $\{20+2\}=22$
3. $682÷22=31$

Yaad rakhein: Remove brackets from inside out: ( ) → { } → [ ].

Answer: (A) 31

Concept: With a free item, the seller receives money for 15 books but hands over 16. Equate selling price to cost price $\times 1.35$.

1. Let CP of one book $=c$, MP $=m$. SP of one book after 4% discount $=0.96m$.
2. Money received for 15 books $=15\times 0.96m=14.4m$; cost of 16 books given $=16c$.
3. $14.4m=1.35\times 16c=21.6c$ ⇒ $\frac{m}{c}=\frac{21.6}{14.4}=1.5$.
4. So MP is $50\%$ above CP.

Trick: $\frac{m}{c}=\frac{16\times 135}{15\times 96}=\frac{3}{2}$ – just multiply the "books given ÷ books paid" with $\frac{100+\text{gain}}{100-\text{discount}}$.

Answer: (D) $50\%$

Concept: Profit $=$ total selling price of the mirrors actually sold $-$ total cost price of all mirrors bought.

1. Total CP $=$ ₹83.
2. Mirrors left $=83-22=61$.
3. SP $=61\times 3.60=$ ₹219.60.
4. Profit $=219.60-83=$ ₹136.60.

Trick: $61\times 3.6=61\times 3+61\times 0.6=183+36.6=219.6$.

Answer: (B) ₹136.60

Concept: Total literates $=$ literate males $+$ literate females; find literate females, then divide by the female population.

1. Total literate $=47\%\times 362000=170140$.
2. Literate males $=35\%\times 181000=63350$.
3. Females $=362000-181000=181000$; literate females $=170140-63350=106790$.
4. Female literacy $=\frac{106790}{181000}\times 100=59\%$.

Trick: Males = females (equal halves), so overall % is their average: $\frac{35+f}{2}=47$ ⇒ $f=59$.

Answer: (D) $59\%$

Concept: $\text{Average}=\frac{\text{Total runs}}{\text{Number of matches}}$; combine totals, not averages.

1. Runs in 23 matches $=23\times 40=920$.
2. Runs in next 10 matches $=10\times 16=160$.
3. Total $=1080$ in 33 matches; average $=\frac{1080}{33}=32.727\dots \approx 32.73$.

Trick: Weighted shortcut: $40-\frac{10\times (40-16)}{33}=40-7.27=32.73$.

Answer: (D) 32.73

Concept: Total $=$ average $\times$ number of items; then split the total in the given ratio.

1. Total price $=3\times 5325=15975$.
2. Ratio parts $=3+5+7=15$ ⇒ 1 part $=\frac{15975}{15}=1065$.
3. Cheapest item $=3\times 1065=3195$.

Trick: Average of $3:5:7$ is the middle term 5, so 5 parts $=5325$ ⇒ 1 part $=1065$ ⇒ 3 parts $=3195$.

Answer: (C) 3,195

Concept: Use ${\sec }^2\theta -{\tan }^2\theta =1$, i.e. $(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$.

1. $\sec \theta +\tan \theta =s$ ⇒ $\sec \theta -\tan \theta =\frac{1}{s}$.
2. Adding: $2\sec \theta =s+\frac{1}{s}=\frac{s^2+1}{s}$; subtracting: $2\tan \theta =\frac{s^2-1}{s}$.
3. $\frac{s^2-1}{s^2+1}=\frac{2\tan \theta }{2\sec \theta }=\frac{\tan \theta }{\sec \theta }=\frac{\sin \theta }{\cos \theta }\times \cos \theta =\sin \theta$.

Trick: Put $\theta =0^{\circ }$: $s=1$ gives $\frac{0}{2}=0=\sin 0^{\circ }$, while $\sec 0^{\circ }=1$ and $\cot 0^{\circ }$ is undefined.

Answer: (C) $\sin \theta$

Concept: In a trapezium, the segment joining the midpoints of the two diagonals is parallel to the parallel sides and equals half their difference.

1. $AB\parallel DC$ with $AB=12$, $DC=30$.
2. $EF=\frac{DC-AB}{2}=\frac{30-12}{2}$
3. $EF=\frac{18}{2}=9\text{cm}$ (BC and AD are not needed).

Trick: Midpoints of diagonals → half the difference; midpoints of the legs (median) → half the sum.

Answer: (A) 9

Concept: If $a^2+b^2=c^2$, the triangle is right-angled and Area $=\frac{1}{2}\times a\times b$.

1. ${28}^2+{45}^2=784+2025=2809$
2. ${53}^2=2809$ → right angle between the sides 28 and 45.
3. Area $=\frac{1}{2}\times 28\times 45=630{\text{cm}}^2$

Yaad rakhein: (28, 45, 53) is a Pythagorean triplet – no need for Heron's formula.

Answer: (B) 630

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$

1. Times: $\frac{198}{66}=3\text{h}$, $\frac{584}{73}=8\text{h}$, $\frac{656}{82}=8\text{h}$ → total $19\text{h}$.
2. Total distance $=198+584+656=1438\text{km}$
3. Average speed $=\frac{1438}{19}\approx 75.68\text{km/h}$

Trick: Never average the three speeds; each leg here divides exactly, so find the times first.

Answer: (B) 75.68

Concept: Make the decimal places equal, find the LCM of the whole numbers, then put the decimal back.

1. $6.3=\frac{630}{100}$, $0.07=\frac{7}{100}$
2. $\text{LCM}(630,7)=630$ since $630=7\times 90$.
3. LCM $=\frac{630}{100}=6.3$

Trick: 6.3 is itself a multiple of 0.07 ($0.07\times 90=6.3$), so the larger number is the LCM.

Answer: (C) 6.3

Concept: For two numbers, $\text{HCF}\times \text{LCM}=\text{first}\times \text{second}$.

1. $11\times 330=110\times x$
2. $x=\frac{3630}{110}=33$
3. Check: $\text{HCF}(110,33)=11$ and $\text{LCM}=330$ ✓

Trick: The other number must be a multiple of the HCF 11 – only 33 among the options is.

Answer: (A) 33

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

1. ${(1+\frac{r}{100})}^2=\frac{12321}{10000}$
2. $1+\frac{r}{100}=\sqrt{\frac{12321}{10000}}=\frac{111}{100}$
3. $r=11\%$

Trick: ${111}^2=12321$ – an amount of the form $(100+r{)}^2$ gives the rate directly.

Answer: (C) $11\%$

Concept: Find each value separately and subtract.

1. $65\%\text{ of }75=\frac{65\times 75}{100}=48.75$
2. $\frac{2}{5}\text{ of }45=18$
3. Difference $=48.75-18=30.75$

Trick: $65\%$ of 75 = $75\%$ of 65 $=\frac{3}{4}\times 65=48.75$.

Answer: (B) 30.75

Concept: Form two equations in the pen price $p$ and pencil price $c$ and solve.

1. $6p+19c=275$
2. $15(p-7)+4(c+7)=132\Rightarrow 15p+4c=209$
3. Eliminate $c$: $285p+76c=3971$ and $24p+76c=1100$ → $261p=2871$, $p=11$, so $c=\frac{275-66}{19}=11$.
4. $6p+2c=66+22=88$

Trick: Try $p=c$: $25p=275$ gives $p=11$, and $15\times 11+4\times 11=209$ fits the second equation too.

Answer: (A) ₹88

Concept: For a number ending in 5, $(10n+5{)}^2=n(n+1)\times 100+25$.

1. Here $n=14$: $14\times 15=210$
2. Write 25 after it: ${145}^2=21025$

Trick: The square of any number ending in 5 ends in 25 – this alone rules out 20975 and 21075; 21005 does not end in 25 either.

Answer: (B) 21025

Concept: Let mother = $M$, daughter = $D$; turn both statements into equations.

1. $2M=4D+26$
2. $3D=M-7\Rightarrow M=3D+7$
3. $2(3D+7)=4D+26\Rightarrow 2D=12$, $D=6$, $M=25$
4. Difference $=25-6=19$ years

Trick: Check: $2\times 25=50=4\times 6+26$ ✓ and $3\times 6=18=25-7$ ✓

Answer: (C) 19