RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Use $(1-\sin A)(1+\sin A)={\cos }^{2}A$ and ${\sec }^{2}A-1={\tan }^{2}A$.

1. $\frac{2\cos A}{1-\sin A}+\frac{2\cos A}{1+\sin A}=\frac{2\cos A(1+\sin A+1-\sin A)}{1-{\sin }^{2}A}=\frac{4\cos A}{{\cos }^{2}A}=4\sec A$
2. $3\sqrt{{\sec }^{2}A-1}=3\sqrt{{\tan }^{2}A}=3\tan A$
3. Product $=4\sec A\times 3\tan A=12\sec A\tan A$

Trick: Put $A=0^{\circ }$: bracket $=4$, root part $=0$ → value 0; only $12\sec A\tan A$ (and $12\tan A$) give 0, and at $A={45}^{\circ }$ the value is $12\sqrt{2}$, which fits $12\sec A\tan A$.

Answer: (A) $12\sec A\tan A$

Concept: Work backwards: $\text{CP}=\frac{\text{SP}\times 100}{100\pm \text{P/L}\%}$. Anand's total cost includes the repair.

1. Bharath's CP $=\frac{4788\times 100}{56}=8550$ (44% loss).
2. This is Anand's SP at 52% profit: Anand's total cost $=\frac{8550\times 100}{152}=5625$
3. Original cost $=5625-127=5498$

Trick: Chain: $4788÷0.56÷1.52-127$; check $5498+127=5625$, $5625\times 1.52=8550$, $8550\times 0.56=4788$ ✔

Answer: (A) ₹$5498$

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

1. Expenditure $=100\%-46\%=54\%$ of salary
2. $=\frac{54}{100}\times 6150=3321$

Trick: $6150\times 0.5=3075$, $6150\times 0.04=246$; $3075+246=3321$.

Answer: (A) ₹$3321$

Concept: Rotating a right triangle about one leg gives a cone: that leg = height, other leg = radius, hypotenuse = slant height. $V=\frac{1}{3}\pi r^{2}h$.

1. Rotated about 9 cm side: $h=9$, $r=12$, $l=15$
2. $V=\frac{1}{3}\pi \times {12}^2\times 9=\frac{1}{3}\pi \times 144\times 9$
3. $=432\pi {\text{cm}}^3$

Trick: Axis side = height; the other leg (not the hypotenuse 15) is the radius.

Answer: (D) $432\pi$

Concept: Overall average $=\frac{\text{total of all scores}}{\text{number of students}}$.

1. First 12: $12\times 70=840$; remaining 18: $18\times 64=1152$
2. Total $=1992$
3. Average $=\frac{1992}{30}=66.4$

Trick: Deviation method: $64+\frac{12\times 6}{30}=64+2.4=66.4$.

Answer: (A) $66.4$

Concept: If A is $r\%$ more than B, then B is less than A by $\frac{r}{100+r}\times 100\%$.

1. $r=17$
2. $\frac{17}{117}\times 100=14.53\%$

Trick: Take B = 100, A = 117; difference 17 out of 117.

Answer: (C) $14.53\%$

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{gain}\%}{100}$ and $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. $\text{SP}=2107\times \frac{189}{100}=3982.23$
2. $\text{MP}\times \frac{21}{100}=3982.23$
3. $\text{MP}=\frac{3982.23\times 100}{21}=18963$

Trick: $\text{MP}=\text{CP}\times \frac{189}{21}=2107\times 9=18963$.

Answer: (D) ₹$18963$

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where the modal class has the highest frequency.

1. Highest frequency 34 → modal class 25–35: $l=25$, $f_1=34$, $f_0=31$, $f_2=13$, $h=10$
2. $\text{Mode}=25+\frac{34-31}{68-31-13}\times 10=25+\frac{3}{24}\times 10$
3. $=25+1.25=26.25$

Trick: Denominator $=(f_1-f_0)+(f_1-f_2)=3+21=24$.

Answer: (D) $26.25$

Concept: For $a{x}^3+b{x}^2+cx+d=0$: $\alpha +\beta +\gamma =-\frac{b}{a}$, $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$, $\alpha \beta \gamma =-\frac{d}{a}$.

1. Here $a=1$, $b=-3$, $c=4$, $d=-12$
2. Sum of products two at a time $=\frac{c}{a}=\frac{4}{1}=4$

Trick: Check: $x^3-3x^2+4x-12=(x-3)(x^2+4)$, roots $3,\pm 2i$: $3(2i)+3(-2i)+(2i)(-2i)=4$ ✔

Answer: (C) $4$

Concept: Sum $=\text{average}\times \text{count}$; express all numbers in terms of the smallest.

1. Let third $=x$, second $=2x$, first $=4x$
2. $x+2x+4x=3\times 154=462$ ⇒ $7x=462$ ⇒ $x=66$
3. First $=4x=264$

Trick: Ratio $4:2:1$ (total 7 parts); first $=\frac{4}{7}\times 462=264$.

Answer: (D) $264$

Concept: A triangle needs any 3 vertices; no three vertices of a polygon are collinear, so count $={}^n{C}_3$.

1. ${}^{10}{C}_3=\frac{10\times 9\times 8}{3\times 2\times 1}$
2. $=120$

Yaad rakhein: Diagonals of an $n$-gon $={}^n{C}_2-n$; triangles $={}^n{C}_3$.

Answer: (C) $120$

Concept: Downstream speed $=B+C$, upstream speed $=B-C$, so $C=\frac{\text{down}-\text{up}}{2}$.

1. $B+C=47$, $B-C=19$
2. $C=\frac{47-19}{2}=14\text{km/h}$ (boat speed $B=33\text{km/h}$)

Trick: Half the sum = boat speed, half the difference = stream speed.

Answer: (A) $14\text{km/h}$

Concept: $\text{CSA}=2\pi rh$, $\text{TSA}=2\pi r(r+h)$.

1. $2\pi rh=\frac{1}{2}\times 2\pi r(r+h)$
2. $2h=r+h$ ⇒ $r=h$
3. $r=53\text{cm}$

Yaad rakhein: CSA = half of TSA always means radius = height.

Answer: (A) $53$

Concept: $x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$ and $x^4+\frac{1}{x^4}={(x^2+\frac{1}{x^2})}^2-2$.

1. $x^2+\frac{1}{x^2}=(\sqrt{5}{)}^2-2=3$
2. $x^4+\frac{1}{x^4}=3^2-2=7$

Trick: Square and subtract 2 at each step: $\sqrt{5}\to 3\to 7$.

Answer: (A) $7$

Concept: Mode is the value that occurs most often.

1. 20 appears 3 times.
2. 31, 33 and 35 appear twice each; all others once.
3. So the mode is 20.

Trick: Tally quickly: the first three values 20, 20, 20 already give the highest count.

Answer: (A) $20$

Concept: Leak rate $=$ (filling rate without leak) $-$ (filling rate with leak).

1. X + Y: $\frac{1}{12}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}$ → 8 hours
2. With leak: $8+0.5=8.5=\frac{17}{2}$ hours → rate $\frac{2}{17}$
3. Leak $=\frac{1}{8}-\frac{2}{17}=\frac{17-16}{136}=\frac{1}{136}$
4. Leak empties the full tank in $136$ hours.

Trick: Shortcut: leak time $=\frac{T_1{T}_2}{T_2-T_1}=\frac{8\times 8.5}{0.5}=136$.

Answer: (D) $136$ hours

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

1. $59-5=54$ and $133-7=126$.
2. $54=2\times 3^3$, $126=2\times 3^2\times 7$.
3. $\text{HCF}=2\times 3^2=18$ (and $18>7$, so remainders 5 and 7 are valid).

Trick: Check options: $59÷18$ leaves 5, $133÷18$ leaves 7 ✓.

Answer: (A) 18

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

1. Per hour: $\frac{1}{12}-\frac{1}{21}=\frac{7-4}{84}=\frac{3}{84}=\frac{1}{28}$ of the tank.
2. Full tank takes 28 hours.
3. Half tank takes $\frac{28}{2}=14$ hours.

Trick: Net time $=\frac{12\times 21}{21-12}=\frac{252}{9}=28$ h; half of it $=14$ h.

Answer: (B) 14

Concept: Let walking one way take $w$ and driving one way take $d$.

1. $w+d=8\text{h}15\text{min}$.
2. Driving both ways saves 5 h: $2d=8\text{h}15\text{min}-5\text{h}=3\text{h}15\text{min}$, i.e. $w-d=5\text{h}$.
3. Walking both ways $=2w=(w+d)+(w-d)=8\text{h}15\text{min}+5\text{h}=13\text{h}15\text{min}$.

Trick: Walk both ways = given total + time gained by driving.

Answer: (D) 13 hours 15 minutes

Concept: Form an equation for the number $x$.

1. $\frac{x}{3}+10=65$.
2. $\frac{x}{3}=55\Rightarrow x=165$.
3. Sum of digits $=1+6+5=12$.

Trick: Reverse the steps: $(65-10)\times 3=165$.

Answer: (D) 12