RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Apply successive percentage changes as multipliers.

1. After 30% discount: $8000\times 0.70=5600$
2. Add 12% tax: $5600\times 1.12=6272$

Trick: Net factor $0.7\times 1.12=0.784$ → $8000\times 0.784=6272$.

Answer: (D) ₹6,272

Concept: $1-2{\sin }^{2}A=\cos 2A=2{\cos }^{2}A-1$ and $\sqrt{{\csc }^{2}A-1}=\cot A$.

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

Trick: Put $A={45}^{\circ }$: first part undefined, so try $A={30}^{\circ }$: $\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{4}{\sqrt{3}}$ = $\sec {30}^{\circ }\csc {30}^{\circ }=\frac{2}{\sqrt{3}}\times 2$ ✓.

Answer: (A) $\sec A\csc A$

Concept: Square: perimeter $=4a$, area $=a^2$.

1. $a=\frac{248}{4}=62$ m
2. Area $={62}^2=3844{\text{m}}^2$

Trick: ${62}^2={60}^2+2\times 60\times 2+2^2=3600+240+4=3844$; a square of a number ending in 2 must end in 4.

Answer: (D) 3844

Concept: For equal time, profit is shared in the ratio of investments.

1. $1020:1440:1230=102:144:123$, sum $=369$.
2. Rahul’s share $=1260\times \frac{123}{369}=1260\times \frac{1}{3}=420$

Trick: $1230$ is exactly one-third of the total $3690$ → share $=\frac{1260}{3}$.

Answer: (A) ₹420

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. Sum $=40+49+56+73+39=257$
2. Average $=\frac{257}{5}=51.4$ kg

Trick: Deviations from 50: $-10,-1,+6,+23,-11$ → total $+7$ → $50+\frac{7}{5}=51.4$.

Answer: (A) 51.4

Concept: Increasing by 100% means the number becomes double.

1. $x\times \frac{200}{100}=2700$
2. $x=\frac{2700}{2}=1350$

Trick: +100% = ×2, so just halve the result.

Answer: (A) 1350

Concept: If SP of x articles = CP of y articles, then $\text{loss}\%=\frac{x-y}{x}\times 100$ (when $x>y$).

1. SP of 50 = CP of 5; let CP of 1 book = ₹1 → CP of 50 = ₹50, SP of 50 = ₹5.
2. Loss $=50-5=45$ → $\text{loss}\%=\frac{45}{50}\times 100=90\%$

Trick: $\frac{50-5}{50}\times 100=90\%$; SP is smaller than CP, so it is a loss.

Answer: (A) 90% loss

Concept: Eliminate using quick rules, then check the remaining divisor.

1. Ends in 1 → not divisible by 25; odd → not by 28 or 18.
2. $159916901÷19=8416679$ exactly ($19\times 8416679=159916901$).

Trick: Odd number → all even divisors out; 25 needs the last two digits 00, 25, 50 or 75. Only 19 remains.

Answer: (B) 19

Concept: Profit ratio = (capital × time) ratio.

1. $M+N=4896$, $M-N=288$ → M’s share $=2592$, N’s share $=2304$; ratio $=9:8$.
2. Let N’s capital $=x$: $\frac{3(x+70000)}{5x}=\frac{9}{8}$
3. $24x+1680000=45x\Rightarrow x=80000$
4. M’s capital $=80000+70000=150000$

Trick: Check option (A): $3\times 150000:5\times 80000=450000:400000=9:8$ ✓.

Answer: (A) ₹1,50,000

Concept: LCM takes the highest power of each factor appearing in the numbers.

1. Highest power of 3: $3^3$; of 8: $8^3$.
2. Highest power of 14: ${14}^2$; of 19: ${19}^2$.
3. LCM $=3^3\times 8^3\times {14}^2\times {19}^2$ (in primes $2^{11}\times 3^3\times 7^2\times {19}^2$ – the same value).

Trick: LCM = highest powers, HCF = lowest powers – only option D has every highest power.

Answer: (D) $3^3\times 8^3\times {14}^2\times {19}^2$

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

1. $2p+19c=118$ ...(i)
2. New prices: $5(p-5)+8(c+4)=144\Rightarrow 5p-25+8c+32=144\Rightarrow 5p+8c=137$ ...(ii)
3. (i)$\times 5$ − (ii)$\times 2$: $95c-16c=590-274\Rightarrow 79c=316\Rightarrow c=4$, so $p=\frac{118-76}{2}=21$
4. Required cost $=14\times 21+6\times 4=294+24=318$

Trick: Since $118-19c$ must be even, $c$ is even; try $c=4$: $p=21$ and $5(16)+8(8)=144$ ✔.

Answer: (A) ₹318

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $182\times x=91\times 546$
2. $x=\frac{91\times 546}{182}=\frac{546}{2}=273$
3. Check: $182=2\times 7\times 13$, $273=3\times 7\times 13$ → HCF $=91$, LCM $=546$ ✔

Trick: Both numbers are multiples of the HCF: $182=91\times 2$, so other $=91\times 3=273$ (since $\frac{546}{91}=6=2\times 3$).

Answer: (A) 273

Concept: If a number $x$ is increased by $r\%$, new value $=x\times \frac{100+r}{100}$.

1. $x\times \frac{160}{100}=3550$
2. $x=\frac{3550\times 100}{160}=\frac{3550\times 5}{8}=\frac{17750}{8}=2218.75$

Trick: $60\%$ increase = multiply by $\frac{8}{5}$; reverse by multiplying with $\frac{5}{8}$.

Answer: (A) 2218.75

Concept: A dishonest seller gains twice: on the marked-up price and on the short weight. $\text{Gain}\%=(\frac{\text{price received}}{\text{cost of goods given}}-1)\times 100$.

1. Let cost of 1 unit = ₹1. He claims 100 units and charges $100\times 1.39=139$.
2. He actually gives $100-17=83$ units, costing ₹83.
3. $\text{Gain}\%=\frac{139-83}{83}\times 100=\frac{5600}{83}\approx 67.47\%\approx 67\%$

Trick: $\frac{1+\frac{39}{100}}{1-\frac{17}{100}}=\frac{139}{83}\approx 1.675$ → about $67\%$.

Answer: (C) $67\%$

Concept: Take the square root of each term separately, then add/subtract.

1. $\sqrt{400}=20$
2. $\sqrt{0.0225}=\sqrt{\frac{225}{10000}}=\frac{15}{100}=0.15$
3. $\sqrt{9}=3$
4. $20+0.15-3=17.15$

Trick: Decimal places in a square root = half of those in the number: $0.0225$ has 4 places → root $0.15$ has 2.

Answer: (D) 17.15

Concept: If a number $N$ leaves remainders $r_1,r_2$ on dividing $a,b$, then $N=\text{HCF}(a-r_1,b-r_2)$.

1. $527-7=520$ and $632-8=624$
2. $520=2^3\times 5\times 13$, $624=2^4\times 3\times 13$
3. $\text{HCF}=2^3\times 13=104$
4. Check: $527=5\times 104+7$, $632=6\times 104+8$ ✔ (and $104>8$)

Trick: $624-520=104$; the HCF must divide this difference, and 104 itself divides both → answer 104.

Answer: (B) 104

Concept: If $9A=3B=7C=k$, then $A:B:C=\frac{1}{9}:\frac{1}{3}:\frac{1}{7}$; multiply by LCM of $9,3,7=63$.

1. $A:B:C=\frac{63}{9}:\frac{63}{3}:\frac{63}{7}=7:21:9$
2. Total parts $=7+21+9=37$
3. 1 part $=\frac{74000}{37}=2000$
4. $A=7\times 2000=14000$

Trick: The share is inversely proportional to the multiplier: smallest multiplier 3 (B) gets the largest share.

Answer: (C) ₹14,000

Concept: Expand each bracket, then combine like terms.

1. $x(6x-3)=6x^2-3x$
2. $5(x^2-4)=5x^2-20$
3. Sum: $6x^2-3x+5x^2-20+18=11x^2-3x-2$

Trick: Put $x=1$: expression $=3+(-15)+18=6$; option (B) gives $11-3-2=6$ ✔.

Answer: (B) $11x^2-3x-2$

Concept: To test $n$ for primality, check divisibility by primes up to $\sqrt{n}$. Here $\sqrt{476}<22$, so test 2, 3, 5, 7, 11, 13, 17, 19.

1. Even numbers 466, 468, 470, 472, 474, 476 are out; 475 ends in 5.
2. $469=7\times 67$, $471=3\times 157$, $473=11\times 43$
3. 467 is not divisible by 2, 3, 5, 7, 11, 13, 17 or 19 → prime.
4. So only one prime: 467.

Trick: Between consecutive odd numbers, quickly test 3 (digit sum), 7 and 11 (alternating sum) first – they remove most candidates.

Answer: (A) 1

Concept: $\text{Profit}=\text{SP of plates sold}-\text{CP of all plates}$; broken plates add to cost but give no revenue.

1. CP of 79 plates = ₹79
2. Plates left $=79-13=66$
3. SP $=66\times 5.90=389.40$
4. Profit $=389.40-79=310.40$

Trick: $66\times 5.9=66\times 6-66\times 0.1=396-6.6=389.4$.

Answer: (D) ₹310.40