RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Profit is shared in the ratio of investments (same time).

1. Total parts $=6+59+41=106$
2. 1 part $=\frac{3180}{106}=30$
3. $Y-Z=(59-41)\times 30=18\times 30=540$

Trick: Find the value of 1 part, then multiply by the difference of parts.

Answer: (D) ₹540

Concept: Work done $=\text{rate}\times \text{time}$; A’s rate $=\frac{1}{25}$, B’s rate $=\frac{1}{18}$ per day.

1. A in 4 days: $\frac{4}{25}$
2. Remaining: $1-\frac{4}{25}=\frac{21}{25}$
3. B’s time $=\frac{21}{25}\times 18=\frac{378}{25}=15\frac{3}{25}$ days

Trick: Remaining work × B’s total days = B’s time.

Answer: (A) $15\frac{3}{25}$

Concept: Correct sum = wrong sum − wrong value + right value.

1. Wrong sum $=10\times 40=400$
2. Correct sum $=400-33+50=417$
3. Correct average $=\frac{417}{10}=41.7$

Trick: Change in average $=\frac{50-33}{10}=1.7$ → $40+1.7$.

Answer: (D) 41.7

Concept: Net change $=a+b+\frac{ab}{100}$.

1. $a=-16$, $b=+10$
2. $-16+10+\frac{(-16)(10)}{100}=-6-1.6=-7.6\%$
3. So the final salary is 7.6% lower.

Trick: Check: $100\times 0.84\times 1.1=92.4$.

Answer: (D) 7.6%

Concept: For 2 years: $CI-SI=P{\left(\frac{R}{100}\right)}^2$.

1. $=17700\times {\left(\frac{4}{100}\right)}^2$
2. $=17700\times \frac{16}{10000}=28.32$

Trick: 2-year difference = interest on first year’s interest.

Answer: (B) ₹28.32

Concept: Stream speed $=\frac{\text{Downstream}-\text{Upstream}}{2}$.

1. Upstream $=\frac{184}{48}=\frac{23}{6}\text{km/h}$
2. Downstream $=\frac{184}{12}=\frac{46}{3}\text{km/h}$
3. Stream $=\frac{1}{2}(\frac{92}{6}-\frac{23}{6})=\frac{69}{12}=5.75\text{km/h}$

Trick: Same distance: stream $=\frac{d}{2}(\frac{1}{t_1}-\frac{1}{t_2})=92(\frac{1}{12}-\frac{1}{48})$.

Answer: (D) 5.75 km/hr

Concept: For a fixed selling amount, $\text{SP of }n\text{ items}=1.4\times \text{CP of }n\text{ items}$ when gain is $40\%$.

1. CP of 1 lemon = ₹$\frac{1}{7}$.
2. For 40% gain, SP of 1 lemon $=\frac{1}{7}\times \frac{140}{100}$ = ₹$\frac{1}{5}$.
3. So for ₹1 he must sell $1÷\frac{1}{5}=5$ lemons.

Trick: Lemons to sell $=\frac{7\times 100}{140}=5$.

Answer: (C) 5

Concept: Angle sum of a triangle is ${180}^{\circ }$; alternate interior angles between parallel lines are equal; angles opposite equal sides are equal.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{69}^{\circ }-{23}^{\circ }={88}^{\circ }$.
2. $BC\parallel AD$, so $\angle CAD=\angle BCA={88}^{\circ }$ (alternate angles).
3. $AC=CD$, so $\angle CDA=\angle CAD={88}^{\circ }$.
4. In $△ACD$: $\angle ACD={180}^{\circ }-{88}^{\circ }-{88}^{\circ }=4^{\circ }$.

Trick: $\angle ACD={180}^{\circ }-2({180}^{\circ }-{69}^{\circ }-{23}^{\circ })={180}^{\circ }-{176}^{\circ }=4^{\circ }$.

Answer: (A) $4^{\circ }$

Concept: $\text{Average}=\frac{\text{Sum}}{\text{Count}}$.

1. First 10 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
2. Sum $=2+3+5+7+11+13+17+19+23+29=129$.
3. Average $=\frac{129}{10}=12.9$.

Yaad rakhein: Sum of first 10 primes = 129; of first 25 primes (up to 100) = 1060.

Answer: (D) 12.9

Concept: The largest cone cut from a cube of edge $a$ has $r=\frac{a}{2}$ and $h=a$; $V=\frac{1}{3}\pi r^{2}h$.

1. $a=8$, so $r=4\text{cm}$, $h=8\text{cm}$.
2. $V=\frac{1}{3}\times \frac{22}{7}\times 4^2\times 8=\frac{22\times 128}{21}=\frac{2816}{21}$.
3. $\frac{2816}{21}=134\frac{2}{21}{\text{cm}}^3$.

Trick: Largest cone in cube $=\frac{\pi a^3}{12}=\frac{22\times 512}{84}=134\frac{2}{21}$.

Answer: (B) $134\frac{2}{21}$

Concept: Follow BODMAS – innermost bracket first.

1. $(1-5+4)=0$.
2. $7\times 0=0$, and $0÷91=0$.
3. $127-0=127$.

Trick: Spot the zero bracket – the whole term vanishes.

Answer: (C) 127

Concept: Net rate = filling rate − emptying rate.

1. Net rate per minute $=\frac{1}{21}-\frac{1}{24}=\frac{8-7}{168}=\frac{1}{168}$ of the tank.
2. Full tank takes 168 minutes together.
3. One-fourth tank takes $\frac{168}{4}=42$ minutes.

Trick: LCM method: tank = 168 units; filler 8, emptier 7 units per minute → net 1 unit; $\frac{1}{4}$ tank = 42 units = 42 min.

Answer: (A) 42

Concept: A product stays the same if the total shift of decimal points is zero.

1. Option (C): $6893.5\to 68.935$ (divided by ${10}^2$), $603.3\to 6033$ (×10), $0.3019\to 3.019$ (×10).
2. Net factor $=\frac{10\times 10}{100}=1$ → product unchanged ✓.
3. (A) and (B) give a net factor $\frac{1}{100}$; (D) gives 100.

Trick: Count decimal places: original 1 + 1 + 4 = 6; in (C) 3 + 0 + 3 = 6 – same total.

Answer: (C) $68.935\times 6033\times 3.019$

Concept: $\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=28+31+40+63+57+37+34+70+99=459$.
2. Number of observations = 9.
3. Mean $=\frac{459}{9}=51$.

Trick: Pair to round numbers: (31 + 99) + (63 + 37) + (70 + 40) + (28 + 57 + 34) = 130 + 100 + 110 + 119 = 459.

Answer: (D) 51

Concept: If $a:b::c:d$ then $ad=bc$; mean proportional $y$ of $p$ and $q$ satisfies $y^2=pq$.

1. $\frac{22+x}{26+x}=\frac{19+x}{21+x}\Rightarrow (22+x)(21+x)=(19+x)(26+x)$.
2. $462+43x=494+45x\Rightarrow 2x=-32\Rightarrow x=-16$ (check: 6, 10, 3, 5 are in proportion).
3. $2x=-32$, $4x-8=-72$; $y^2=(-32)(-72)=2304$.
4. $y>0\Rightarrow y=48$.

Trick: $y=\sqrt{32\times 72}=\sqrt{2304}=48$.

Answer: (A) 48

Concept: $x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$.

1. Here $x-\frac{1}{x}=4$.
2. $x^3-\frac{1}{x^3}=4^3+3\times 4=64+12$.
3. $=76$.

Trick: If $x-\frac{1}{x}=k$, then $x^3-\frac{1}{x^3}=k^3+3k$.

Answer: (A) 76

Concept: The largest container that exactly measures both quantities = HCF of the quantities.

1. $200=2^3\times 5^2$.
2. $274=2\times 137$ (137 is prime).
3. HCF $=2$ → maximum volume of each container = 2 litres.

Trick: $274-200=74=2\times 37$; HCF must divide 74 and 200 → only 2 works.

Answer: (A) 2

Concept: Simplify, then factorise by splitting the middle term.

1. $4m^2+6m+2=0\Rightarrow 2m^2+3m+1=0$.
2. $2m^2+2m+m+1=0\Rightarrow (2m+1)(m+1)=0$.
3. $m=-\frac{1}{2}$ or $m=-1$.

Trick: All coefficients positive → both roots negative; sum $=-\frac{6}{4}=-\frac{3}{2}$ and product $=\frac{2}{4}=\frac{1}{2}$ fit $-\frac{1}{2},-1$.

Answer: (C) $-\frac{1}{2}$ and $-1$

Concept: $\text{Savings}\%=100\%-\text{Expenditure}\%$.

1. Spends 40% → saves 60% of income.
2. $60\%$ of income = ₹24,000.
3. Income $=\frac{24000\times 100}{60}$ = ₹40,000.

Trick: 60% = 24,000 → 10% = 4,000 → 100% = 40,000.

Answer: (B) 40,000

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

1. Let CP $=C$, initial profit $=p$, so SP $=C+p$.
2. New SP $=2(C+p)$, new profit $=11p$: $2(C+p)-C=11p$.
3. $C+2p=11p\Rightarrow C=9p$.
4. Initial profit $\%=\frac{p}{9p}\times 100=\frac{100}{9}=11\frac{1}{9}\%$.

Trick: Doubling SP adds one more SP $=(C+p)$ to profit: $C+p=10p\Rightarrow C=9p$.

Answer: (C) $11\frac{1}{9}\%$