RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Read 'sells at 75% of the profit' as a selling price fixed at 75% profit, then reduced by the 14% discount; the second pen is sold at 48% profit.

1. Pen 1: $444\times 1.75=777$; after discount $777\times 0.86=668.22$; profit $=224.22$.
2. Pen 2: profit $=356\times 0.48=170.88$.
3. Total profit $=224.22+170.88=395.10$.

Trick: Pen 1 multiplier $1.75\times 0.86=1.505$ → profit $=444\times 0.505=224.22$.

Answer: (D) ₹395.10

Concept: Mean proportional between $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{0.06\times 6}=\sqrt{0.36}$
2. $=0.6$

Trick: Check: $0.06:0.6=0.6:6$ (both ratios are $1:10$).

Answer: (B) $0.6$

Concept: BODMAS: Bracket → Multiplication → Addition/Subtraction.

1. Bracket: $30+3=33$
2. $4\times 33=132$
3. $66-132+40=-26$

Trick: Add the positives first: $66+40=106$, then $106-132=-26$.

Answer: (A) $-26$

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

1. Wrong sum $=10\times 40=400$
2. Correct sum $=400-34+43=409$
3. Correct average $=\frac{409}{10}=40.9$

Trick: Average changes by $\frac{43-34}{10}=0.9$, so $40+0.9=40.9$.

Answer: (D) $40.9$

Concept: Expenditure = Price × Quantity, so multiply the two ratios term by term.

1. Gram : Pea expenditure $=(2\times 4):(5\times 3)$
2. $=8:15$

Trick: Compound ratio: multiply corresponding terms.

Answer: (C) $8:15$

Concept: Whole numbers start from 0, so the first 13 whole numbers are 0 to 12.

1. Sum $=0+1+\dots +12=\frac{12\times 13}{2}=78$
2. Average $=\frac{78}{13}=6$

Trick: Consecutive numbers: average = $\frac{\text{first}+\text{last}}{2}=\frac{0+12}{2}=6$. (Taking 1 to 13 wrongly gives 7.)

Answer: (A) $6$

Concept: If amounts are equal, $P_1(1+r{)}^{n_1}=P_2(1+r{)}^{n_2}$.

1. $A(1.1{)}^{11}=M(1.1{)}^{12}\Rightarrow A=1.1M$
2. $A:M=11:10$
3. Mahesh’s share $=\frac{10}{21}\times 1218=580$

Trick: Shares are in ratio $(1+r{)}^{\text{difference in years}}:1=11:10$.

Answer: (B) ₹$580$

Concept: Net filling rate = filling rate − emptying rate.

1. Net work per minute $=\frac{1}{9}-\frac{1}{90}=\frac{10-1}{90}=\frac{1}{10}$
2. Full tank takes $10$ minutes.
3. Half tank $=\frac{10}{2}=5$ minutes.

Trick: LCM method: capacity 90 units; A = $+10$, B = $-1$ → net 9 units/min; half = 45 units → 5 min.

Answer: (A) $5$

Concept: For 100% gain, SP = $2\times$ CP.

1. CP of 1 lemon $=₹\frac{1}{6}$
2. SP of 1 lemon $=2\times \frac{1}{6}=₹\frac{1}{3}$
3. For ₹1 he must sell $3$ lemons.

Trick: 100% gain → sell half as many for the same money: $\frac{6}{2}=3$.

Answer: (A) $3$

Concept: Form two equations from the two conditions.

1. $F+S=4S+18\Rightarrow F=3S+18$
2. After 5 years: $4(F+5)=14(S+5)-8\Rightarrow 4F=14S+42$
3. $4(3S+18)=14S+42\Rightarrow 12S+72=14S+42\Rightarrow S=15$
4. $F=63$, difference $=63-15=48$

Trick: Difference $F-S=2S+18$; with $S=15$ it is 48.

Answer: (C) $48$

Concept: Volume $=\pi r^{2}h$; total surface area $=2\pi r(r+h)$.

1. $\frac{22}{7}\times r^2\times 38=5852\Rightarrow r^2=\frac{5852\times 7}{22\times 38}=49\Rightarrow r=7$
2. TSA $=2\times \frac{22}{7}\times 7\times (7+38)$
3. $=44\times 45=1980{\text{cm}}^2$

Trick: $5852÷38=154=\pi r^2$ → $r^2=49$ directly.

Answer: (B) $1980{\text{cm}}^2$

Concept: Use $(a\pm b{)}^2=a^2\pm 2ab+b^2$ and $(a+b)(a-b)=a^2-b^2$.

1. $(5x-2y{)}^2=25x^2-20xy+4y^2$
2. $(2x+5y{)}^2=4x^2+20xy+25y^2$
3. $(5x+2y)(5x-2y)=25x^2-4y^2$
4. Sum $=54x^2+25y^2$

Trick: The $xy$ terms cancel ($-20xy+20xy$); add only $x^2$ and $y^2$ coefficients: $25+4+25=54$, $4+25-4=25$.

Answer: (C) $54x^2+25y^2$

Concept: Recognise perfect powers: $4^6=4096$, ${15}^4=50625$, ${28}^3=21952$, ${58}^2=3364$.

1. $\sqrt[6]{4096}=4$, $\sqrt[4]{50625}=15$
2. $\sqrt[3]{21952}=28$, $\sqrt{3364}=58$
3. Sum $=4+15+28+58=105$

Trick: Unit digits help: a 4th power ending in 5 → 15 (${15}^2=225$, ${225}^2=50625$); a cube ending in 2 → root ends in 8.

Answer: (B) $105$

Concept: $(a-b{)}^2=a^2+b^2-2ab$ and $(a+b{)}^2=a^2+b^2+2ab$.

1. $(a-b{)}^2=111-54=57$
2. $(a+b{)}^2=111+54=165$
3. $\frac{a-b}{a+b}=\sqrt{\frac{57}{165}}$ (positive since $a>b$)

Trick: Ratio of squares first, then take the square root – options without the root are the trap.

Answer: (B) $\sqrt{\frac{57}{165}}$

Concept: Largest container that measures both quantities exactly = HCF.

1. $102=2\times 3\times 17$
2. $224=2^5\times 7$
3. HCF $=2$ litres

Trick: Both even, $\frac{102}{2}=51$ is odd and shares no factor with $\frac{224}{2}=112$ → HCF 2.

Answer: (D) $2$

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=\frac{112}{2}=56\text{cm}$
2. $4\times \frac{22}{7}\times 56\times 56=88\times 448$
3. $=39424{\text{cm}}^2$

Trick: Using diameter: $S=\pi d^2=\frac{22}{7}\times {112}^2=22\times 16\times 112=39424$.

Answer: (D) $39,424{\text{cm}}^2$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. SI $=2220-1200=1020$
2. $1020=\frac{1200\times 20\times T}{100}=240T$
3. $T=\frac{1020}{240}=4.25$ years

Trick: Interest per year $=20\%$ of 1200 $=240$; $1020÷240=4.25$.

Answer: (C) $4.25$

Concept: $\text{MP}=\text{CP}\times \frac{125}{100}$ and $\text{SP}=\text{MP}\times \frac{84}{100}$.

1. $\text{CP}=1600\times \frac{100}{125}=1280$ (₹).
2. $\text{SP}=1600\times \frac{84}{100}=1344$ (₹).
3. $\text{Profit}\%=\frac{1344-1280}{1280}\times 100=\frac{64}{1280}\times 100=5\%$.

Trick: Net effect: $1.25\times 0.84=1.05$ → 5% profit directly.

Answer: (A) 5%

Concept: ∠BRP and ∠DSP are corresponding angles, so they are equal; ∠CSP and ∠DSP form a linear pair.

1. $2x+13=3x-22\Rightarrow x=35$.
2. $\angle DSP=3(35)-22={83}^{\circ }$.
3. $\angle CSP={180}^{\circ }-{83}^{\circ }={97}^{\circ }$.

Trick: Corresponding angles equal; angles on a straight line add to ${180}^{\circ }$.

Answer: (C) ${97}^{\circ }$

Concept: $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$.

1. Distance covered $=\frac{2}{3}\times 753=502\text{km}$ in $\frac{2}{3}\times 12=8$ hours.
2. Remaining distance $=753-502=251\text{km}$; time left $=12-8=4$ hours.
3. Required speed $=\frac{251}{4}=62.75\text{km/h}$.

Trick: Remaining $\frac{1}{3}$ distance in $\frac{1}{3}$ time → speed $=\frac{753}{12}=62.75$ (same average).

Answer: (D) 62.75