RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with rate and time in the same unit (here per month).

1. $P=1200$, $R=3\%$ per month, $T=5$ months.
2. $\text{SI}=\frac{1200\times 3\times 5}{100}=12\times 15=180$

Trick: 3% per month for 5 months = 15% in total; 15% of 1200 = 180.

Answer: (D) 180

Concept: In a "buy X, get Y free" offer, $\text{Discount}\%=\frac{\text{free value}}{\text{total value received}}\times 100$.

1. Customer pays ₹6000 and receives goods worth $6000+1000=7000$.
2. Discount = ₹1000 on goods of ₹7000.
3. $\text{Discount}\%=\frac{1000}{7000}\times 100=\frac{100}{7}\approx 14.29\%$

Trick: Divide by the total goods received, not by the amount paid: $\frac{1}{7}=14.29\%$.

Answer: (D) $14.29\%$

Concept: Successive increases multiply: $\text{Final}=\text{Initial}\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})$.

1. Let sales two years ago be $x$.
2. $x\times 1.04\times 1.25=1.3x$
3. $1.3x=135200\Rightarrow x=\frac{135200}{1.3}=104000$

Trick: $4\%$ and $25\%$ together $=4+25+\frac{4\times 25}{100}=30\%$, so divide by 1.3.

Answer: (B) ₹1,04,000

Concept: With a faulty weight, the trader sells the price of 1 unit but gives less goods: $1+\text{Profit}=\frac{1+\text{claimed profit}}{1-\text{weight error}}$.

1. Let CP of 1 kg = ₹100. He charges $100\times 1.68=168$ for "1 kg".
2. He actually gives only $0.88$ kg, whose CP $=88$.
3. $\text{Profit}\%=\frac{168-88}{88}\times 100=\frac{80}{88}\times 100\approx 90.9\%\approx 91\%$

Trick: $\frac{1.68}{0.88}=\frac{21}{11}=1.909$ ⇒ about $91\%$ profit.

Answer: (A) $91\%$

Concept: Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ with ${\sin }^{2}A+{\cos }^{2}A=1$, and complementary angles: $\sin ({90}^{\circ }-\theta )=\cos \theta$.

1. ${\sin }^{3}A-{\cos }^{3}A=(\sin A-\cos A)(1+\sin A\cos A)$
2. So $\frac{{\sin }^{3}A-{\cos }^{3}A}{1+\sin A\cos A}=\sin A-\cos A$
3. $\sin {54}^{\circ }=\cos {36}^{\circ }$, so $2{\sin }^2{54}^{\circ }{\sec }^2{36}^{\circ }=2{\cos }^2{36}^{\circ }\times \frac{1}{{\cos }^2{36}^{\circ }}=2$
4. Total $=\sin A-\cos A+2$

Trick: Angles adding to ${90}^{\circ }$ (${54}^{\circ }+{36}^{\circ }$): $\sin$ of one $=\cos$ of the other, so ${\sin }^2{54}^{\circ }{\sec }^2{36}^{\circ }=1$.

Answer: (C) $\sin A-\cos A+2$

Concept: Take total work = LCM of the given days; then work per day of each person = total ÷ days.

1. LCM$(20,12,9)=180$ units. Shifa $=\frac{180}{20}=9$ units/day.
2. Shifa + Afiya $=\frac{180}{12}=15\Rightarrow$ Afiya $=6$.
3. All three $=\frac{180}{9}=20\Rightarrow$ Mariyam $=20-15=5$.
4. Shifa + Mariyam $=9+5=14$ units/day ⇒ time $=\frac{180}{14}=\frac{90}{7}$ days.

Trick: Directly: $\frac{1}{9}-\frac{1}{12}=\frac{1}{36}$ (Mariyam), then $\frac{1}{20}+\frac{1}{36}=\frac{7}{90}$.

Answer: (C) $\frac{90}{7}$

Concept: For the same distance, time is inversely proportional to speed: $\text{Distance}=\text{Speed}\times \text{Time}$.

1. Distance Delhi–Kanpur $=132\times 5=660\text{km}$.
2. Mira's time $=\frac{660}{60}=11$ hours.

Trick: $T_2=T_1\times \frac{S_1}{S_2}=5\times \frac{132}{60}=11$.

Answer: (C) 11 hours

Concept: Write the condition with present ages $F$ (father) and $S$ (son).

1. Six years ago: $F-6=2(S-6)+40\Rightarrow F=2S+34$.
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$.
3. $x=(2S+34)-2S=34$ years.

Trick: When father = twice son, $x=F-2S$ – here it is just the constant 34.

Answer: (D) 34

Concept: Share $=\frac{\text{its ratio term}}{\text{sum of terms}}\times \text{total}$.

1. Sum of ratio terms $=16+14+8=38$.
2. One part $=\frac{456}{38}=12$.
3. Difference $=(16-8)\times 12=96$.

Trick: Find one part first, then multiply the difference of ratio terms.

Answer: (C) 96

Concept: Increasing a number by $100\%$ doubles it: $x\times (1+\frac{100}{100})=2x$.

1. $2x=2690$
2. $x=\frac{2690}{2}=1345$

Trick: +100% = ×2, so just halve 2690.

Answer: (C) 1345

Concept: One day's work of C = combined one day's work − (A's + B's) one day's work.

1. A's 1-day work $=\frac{1}{19}$, B's $=\frac{1}{15}$, (A+B+C) $=\frac{1}{3}$.
2. C's 1-day work $=\frac{1}{3}-\frac{1}{19}-\frac{1}{15}=\frac{95-15-19}{285}=\frac{61}{285}$.
3. C alone takes $\frac{285}{61}=4\frac{41}{61}$ days.

Trick: LCM method: total work $=285$ units; A = 15, B = 19, all three = 95 units/day, so C $=95-34=61$ units/day.

Answer: (C) $4\frac{41}{61}$ days

Concept: Under compound interest, the multiplying factor for equal time periods multiplies: if money becomes $k$ times in $t$ years, it becomes $k^n$ times in $nt$ years.

1. In 3 years: amount $=3P$.
2. $27P=3^{3}P$, i.e. three such periods.
3. Time $=3\times 3=9$ years.

Trick: Power of the factor × time: $27=3^3\Rightarrow 3\times 3=9$ years.

Answer: (A) 9 years

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ (per pen).

1. CP of 1 pen $=\frac{150}{12}=$ ₹12.5.
2. SP of 1 pen = ₹17, profit = ₹4.5.
3. $\text{Profit}\%=\frac{4.5}{12.5}\times 100=36\%$.

Trick: Per dozen: SP $=17\times 12=204$, profit $=54$ on 150 $\Rightarrow \frac{54}{150}\times 100=36\%$.

Answer: (B) 36

Concept: Third proportional of $a$ and $b$ is $x$ where $a:b=b:x$, so $x=\frac{b^2}{a}$.

1. $8:56=56:x$.
2. $x=\frac{{56}^2}{8}=\frac{3136}{8}=392$.

Trick: $\frac{56}{8}=7$, so $x=56\times 7=392$.

Answer: (A) 392

Concept: Sum of the remaining numbers = total sum − known parts; then divide by their count.

1. Sum of first two $=2\times 80=160$.
2. Remaining two $=655-160-103=392$.
3. Average $=\frac{392}{2}=196$.

Trick: Always convert averages to sums first, then subtract.

Answer: (B) 196

Concept: Square root of a decimal: take the root of the digits and halve the number of decimal places.

1. $\sqrt{196}=14$.
2. $\sqrt{0.0324}=0.18$ (since ${18}^2=324$, 4 decimal places → 2).
3. $\sqrt{5.76}=2.4$ (since ${24}^2=576$).
4. $14+0.18-2.4=11.78$.

Yaad rakhein: Decimal places in the root = half of those under the root sign.

Answer: (B) 11.78

Concept: Check options: the pair must add to 81 and have LCM 420.

1. $420=2^2\times 3\times 5\times 7$; both numbers must divide 420.
2. 14 + 67, 26 + 55, 19 + 62: 67, 55, 26, 19, 62 do not divide 420.
3. $21=3\times 7$, $60=2^2\times 3\times 5$; LCM $=2^2\times 3\times 5\times 7=420$ and $21+60=81$.

Trick: Each number must be a factor of the LCM — eliminate options instantly.

Answer: (D) 21, 60

Concept: Profit share ∝ capital × time.

1. M's share $=\frac{3895+205}{2}=2050$, N's $=1845$; ratio $2050:1845=10:9$.
2. Let N's capital $=x$: $\frac{5(x+66000)}{7x}=\frac{10}{9}$.
3. $45x+2970000=70x\Rightarrow x=118800$.
4. M's capital $=118800+66000=$ ₹1,84,800.

Trick: Divide both shares by their difference (205) to get the ratio fast: $10:9$.

Answer: (C) ₹1,84,800

Concept: Both numbers must be factors of the LCM and add up to 115.

1. $600=2^3\times 3\times 5^2$.
2. 77, 38, 33, 82, 70, 45 are not factors of 600 → options A, B, C fail.
3. $40=2^3\times 5$, $75=3\times 5^2$; LCM $=2^3\times 3\times 5^2=600$ and $40+75=115$.

Trick: Use the factor test on the LCM to eliminate options.

Answer: (D) 40, 75

Concept: Evaluate powers and roots first (BODMAS), then add/subtract.

1. $7^2=49$, $\sqrt{{10}^2}=10$, $2\sqrt{64}=2\times 8=16$.
2. $49+10-16-6=37$.

Yaad rakhein: $\sqrt{a^2}=a$ for positive $a$.

Answer: (C) 37