RRB NTPC Graduate CBT 1 2024 PYQs

Concept: If c is the third proportional of a and b, then $a:b=b:c$, i.e. $b^2=ac$.

1. $40:Z=Z:10$
2. $Z^2=40\times 10=400$
3. $Z=\sqrt{400}=20$ (positive value).

Trick: Third proportional of a and b $=\frac{b^2}{a}$; check: $\frac{{20}^2}{40}=10$ ✓.

Answer: (C) $20$

Concept: $\text{Profit}=\text{SP}-\text{CP}$; the cost price stays the same when SP changes.

1. Let CP = C, SP = S. Initial profit $=S-C$.
2. New: $2S-C=8(S-C)\Rightarrow 2S-C=8S-8C\Rightarrow 7C=6S$.
3. $S=\frac{7C}{6}$, profit $=\frac{C}{6}$.
4. Profit% $=\frac{1}{6}\times 100=16.67\%$.

Trick: Take CP = 6, SP = 7 (profit 1): new SP 14 → profit 8 = 8 times ✓ → $\frac{1}{6}=16.67\%$.

Answer: (D) $16.67$

Concept: Solve innermost brackets first; inside a bracket do ÷ and × from left to right.

1. $26÷\frac{4}{2}=26÷2=13$
2. $8÷2\times \frac{10}{5}=4\times 2=8$
3. $4\times 4÷3=\frac{16}{3}$
4. $13\times 8÷\frac{16}{3}=104\times \frac{3}{16}=19.5$

Trick: Dividing by a fraction = multiplying by its reciprocal: $÷\frac{16}{3}=\times \frac{3}{16}$.

Answer: (B) $19.5$

Concept: Average of groups $=\frac{\sum (\text{share}\times \text{average})}{\text{total share}}$.

1. Children $=100-20-30=50\%$.
2. Average $=0.2\times 50+0.3\times 40+0.5\times 30$
3. $=10+12+15=37\text{kg}$.

Trick: Percentages already add to 100 → just multiply each by its average and add.

Answer: (A) $37$

Concept: $44=4\times 11$ (co-prime), so the number must be divisible by both 4 and 11.

1. By 4: last two digits divisible by 4 → only 3344 (44) qualifies; 79, 67, 50 fail.
2. By 11 for 3344: $(3+4)-(3+4)=0$ ✓.
3. Check: $3344÷44=76$.

Trick: Divisible by a composite number → test its co-prime factors.

Answer: (D) 3344

Concept: Work is fixed: number of pipes × time = constant (inverse proportion).

1. $39\times 19=741$ pipe-minutes.
2. Time with 23 pipes $=\frac{741}{23}=32.217\dots$
3. $\approx 32.22$ minutes.

Trick: Fewer pipes → more time, so the answer must exceed 19; $M_1{T}_1=M_2{T}_2$.

Answer: (B) $32.22$

Concept: Same exam → same maximum marks; find it from Kalpana's data.

1. Maximum marks $=\frac{656}{0.32}=2050$.
2. Sudha's % $=\frac{697}{2050}\times 100=34\%$.

Trick: Ratio method: $\frac{697}{656}\times 32=\frac{17}{16}\times 32=34\%$ (since $656=41\times 16$, $697=41\times 17$).

Answer: (C) $34\%$

Concept: Area $=\sqrt{s(s-a)(s-b)(s-c)}$ and altitude $h=\frac{2\times \text{Area}}{\text{base}}$.

1. $s=\frac{109+171+100}{2}=190$.
2. Area $=\sqrt{190\times 81\times 19\times 90}=\sqrt{26316900}=5130{\text{cm}}^2$.
3. $h=\frac{2\times 5130}{171}=60\text{cm}$.

Trick: Check with Pythagoras: foot divides 171 into 91 and 80; ${91}^2+{60}^2={109}^2$ and ${80}^2+{60}^2={100}^2$ ✓.

Answer: (C) $60$

Concept: Median = middle value after arranging data in order; for n odd it is the ${\left(\frac{n+1}{2}\right)}^{th}$ value.

1. Ascending: 15, 25, 33, 34, 36, 36, 37, 54, 70, 74, 78.
2. $n=11$ → position $\frac{11+1}{2}=6$.
3. 6th value = 36.

Trick: Always sort first — the unsorted middle value (36) can mislead; here it coincides.

Answer: (B) 36

Concept: Multiples of 13 form an AP; average of first n multiples of k $=k\times \frac{n+1}{2}$.

1. First 17 multiples: $13\times 1,13\times 2,\dots ,13\times 17$.
2. Average $=13\times \frac{1+2+\dots +17}{17}=13\times \frac{17+1}{2}$.
3. $=13\times 9=117$.

Trick: Odd count → average = middle term: 9th multiple $=13\times 9=117$.

Answer: (C) 117

Concept: The value of $a+b$ is given directly, so only squaring is needed.

1. $a+b=50$
2. $(a+b{)}^2={50}^2=2500$.
3. ($a-b=21$ is extra data; it would be needed for $ab$ or $a^2+b^2$.)

Trick: Read what is asked — sometimes extra data is a distraction.

Answer: (B) $2500$

Concept: Greatest number leaving remainders $r_1,r_2$ $=\text{HCF}(a-r_1,b-r_2)$.

1. $137-7=130$ and $198-3=195$.
2. $130=2\times 5\times 13$, $195=3\times 5\times 13$.
3. HCF $=5\times 13=65$. (Divisor 65 is larger than both remainders ✓.)

Trick: Subtract the remainders first, then take the HCF.

Answer: (C) 65

Concept: $\text{SP}=\text{MP}\times (1-\text{discount})$; $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. SP = 50% of ₹1,032 = ₹516.
2. Profit = ₹516 − ₹300 = ₹216.
3. Profit% $=\frac{216}{300}\times 100=72\%$.

Trick: 50% discount → SP is simply half the MP.

Answer: (D) $72\%$

Concept: If a, b, c, d are in proportion, $a:b=c:d$, i.e. $ad=bc$ (product of extremes = product of means).

1. $14:7=12:y$
2. $14y=7\times 12=84$
3. $y=\frac{84}{14}=6$.

Trick: $14:7=2:1$, so y is half of 12 = 6.

Answer: (B) $6$

Concept: Area of rectangle = length × breadth; use $(x+y)(x-y)=x^2-y^2$.

1. Length $=\frac{2a+2\sqrt{5}}{4}=\frac{a+\sqrt{5}}{2}$.
2. Area $=\frac{a+\sqrt{5}}{2}\times \frac{a-\sqrt{5}}{2}=\frac{a^2-(\sqrt{5}{)}^2}{4}$.
3. $=\frac{a^2-5}{4}{\text{m}}^2$.

Trick: Spot the conjugate pair $(a+\sqrt{5})(a-\sqrt{5})$ — the roots vanish.

Answer: (B) $\left(\frac{a^2-5}{4}\right){\text{m}}^2$

Concept: Write the numbers as $10k,15k,13k$ and find $k$ from the given condition.

1. $10\%$ of first $=66\Rightarrow$ first $=660$, so $10k=660$, $k=66$.
2. Second $=15\times 66=990$, third $=13\times 66=858$.
3. Difference $=990-858=132$; $50\%$ of $132=66$.

Trick: Difference of 3rd and 2nd $=(15-13)k=2k$; half of it $=k=66$ directly.

Answer: (B) 66

Concept: In $△ABC\cong △PQR$, corresponding parts are equal: $\angle B=\angle Q$, $\angle C=\angle R$, $AC=PR$.

1. $\angle PQR=\angle ABC$: $3x-4=77\Rightarrow x=27$.
2. $PR=AC$: $x+3y=48\Rightarrow 27+3y=48\Rightarrow y=7$.
3. $\angle QRP=\angle BCA=(x-y{)}^{\circ }=(27-7{)}^{\circ }={20}^{\circ }$.

Trick: Match letters in order: A↔P, B↔Q, C↔R, so $\angle QRP$ pairs with $\angle BCA$.

Answer: (D) ${20}^{\circ }$

Concept: Median $=l+\frac{\frac{N}{2}-cf}{f}\times h$, where the median class contains the $\frac{N}{2}$-th value.

1. $N=11+18+22+30+40+16+33=170$, $\frac{N}{2}=85$.
2. Cumulative frequencies: $11,29,51,81,121,\dots$ → median class $55$–$65$.
3. $l=55,cf=81,f=40,h=10$.
4. Median $=55+\frac{85-81}{40}\times 10=55+1=56$.

Trick: Stop the cumulative total as soon as it crosses $\frac{N}{2}$ – that class is the median class.

Answer: (C) 56 years

Concept: For 2 years: $CI-SI=P{\left(\frac{r}{100}\right)}^2$ and $S{I}_{1\text{ yr}}=\frac{Pr}{100}$.

1. SI for 2 years $=40\Rightarrow$ SI for 1 year $=20$.
2. $CI-SI=41-40=1$ = interest on first year’s interest: $20\times \frac{r}{100}=1\Rightarrow r=5\%$.
3. $P=\frac{20\times 100}{5}=400$.

Trick: $P=\frac{(S{I}_1{)}^2}{CI-SI}=\frac{{20}^2}{1}=400$.

Answer: (A) 400

Concept: $SI=\frac{P\times R\times T}{100}$ with $T$ in years; count days excluding the first day and including the last.

1. Days: February $28-15=13$, March $31$, April $29$ → $73$ days $=\frac{73}{365}=\frac{1}{5}$ year.
2. $SI=\frac{2000\times 5.75\times 1}{100\times 5}=23$.

Trick: $73$ days $=\frac{1}{5}$ year – a very common exam figure.

Answer: (D) 23