RRB NTPC Graduate CBT 1 2024 PYQs

Concept: If a number increases by $r\%$, new value $=\text{number}\times \frac{100+r}{100}$.

1. $57\frac{1}{2}\%=\frac{115}{2}\%$, so new value $=x\times \frac{315}{200}$.
2. $x\times \frac{315}{200}=378$
3. $x=\frac{378\times 200}{315}=240$

Trick: Check: $240+57.5\%\text{ of }240=240+138=378$ ✔

Answer: (A) $240$

Concept: The altitude to the base of an isosceles triangle bisects the base; $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Half base $=\frac{18}{2}=9\text{cm}$
2. $h=\sqrt{{41}^2-9^2}=\sqrt{1681-81}=\sqrt{1600}=40\text{cm}$
3. $\text{Area}=\frac{1}{2}\times 18\times 40=360{\text{cm}}^2$

Trick: (9, 40, 41) is a Pythagorean triplet. Or use $\frac{b}{4}\sqrt{4a^2-b^2}=\frac{18}{4}\times 80=360$.

Answer: (C) $360$

Concept: Least number to be added $=\text{divisor}-\text{remainder}$.

1. $22938÷7$: $7\times 3276=22932$, remainder $=6$
2. Number to add $=7-6=1$
3. $22938+1=22939=7\times 3277$ ✔

Trick: Divide by 7, find remainder; add (7 − remainder).

Answer: (C) $1$

Concept: Volume of shell $=\frac{4}{3}\pi (R^3-r^3)$ and $\text{Mass}=\text{Density}\times \text{Volume}$.

1. With $r=0$, $R=3$: $V=\frac{4}{3}\times \frac{22}{7}\times 27=\frac{792}{7}{\text{cm}}^3$
2. $\text{Mass}=7\times \frac{792}{7}=792\text{g}$
3. $792\text{g}=0.792\text{kg}$

Trick: The density 7 cancels the 7 of $\pi =\frac{22}{7}$: mass $=\frac{4}{3}\times 22\times 27=792$ g.

Answer: (C) $0.792$

Concept: A quadratic $a{x}^2+bx+c=0$ has equal roots when the discriminant $D=b^2-4ac=0$.

1. Here $a=1$, $b=4$, $c=K$.
2. $4^2-4\times 1\times K=0$
3. $16-4K=0\Rightarrow K=4$

Trick: Equal roots → perfect square: $x^2+4x+4=(x+2{)}^2$, so $K={\left(\frac{b}{2}\right)}^2=4$.

Answer: (C) $4$

Concept: $\text{Overall average}=\frac{\text{Total sum}}{\text{Total count}}$.

1. Sum of 15 boys $=15\times 160=2400$
2. Remaining $50-15=35$ boys: $35\times 168=5880$
3. Average $=\frac{2400+5880}{50}=\frac{8280}{50}=165.6\text{cm}$

Trick: Deviation method: $160+\frac{35\times 8}{50}=160+5.6=165.6$.

Answer: (B) $165.6$

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

1. Let B = 100, then A = 112.
2. B is less than A by 12.
3. $\frac{12}{112}\times 100=10.71\%$

Trick: $\frac{r}{100+r}$: $\frac{12}{112}=\frac{3}{28}\approx 10.71\%$.

Answer: (B) $10.71\%$

Concept: Arrange the data in order; for $n$ odd, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 15, 33, 34, 54, 60, 65, 66, 68, 70, 71, 77
2. $n=11$, median $={\left(\frac{11+1}{2}\right)}^{\text{th}}=6^{\text{th}}$ value
3. $6^{\text{th}}$ value $=65$

Trick: 5 values below and 5 above: 15, 33, 34, 54, 60 | 65 | 66, 68, 70, 71, 77.

Answer: (D) $65$

Concept: At compound interest, if money becomes 2 times in $n$ years, it becomes $2^k$ times in $kn$ years.

1. $P(1+r{)}^{11}=2P$
2. $4P=2^{2}P=P{[(1+r{)}^{11}]}^2=P(1+r{)}^{22}$
3. So time $=2\times 11=22$ years.

Trick: Doubling time multiplies: 2× in 11 → 4× in 22 → 8× in 33.

Answer: (D) $22$

Concept: Write initial money and spending in ratio form; equal remaining amounts give an equation.

1. Money: $17x$ and $14x$; spending ratio $12:8=3:2$, i.e. $3k$ and $2k$.
2. $17x-3k=3392$ and $14x-2k=3392$
3. From the second: $k=7x-1696$; then $17x-21x+5088=3392\Rightarrow x=424$
4. Sohit $=17\times 424=₹7208$

Trick: Check: $k=2968-1696=1272$; Sohit spends $3816$, $7208-3816=3392$ ✔

Answer: (B) ₹7,208

Concept: Downstream speed $=u+v$, upstream speed $=u-v$; so current speed $v=\frac{\text{down}-\text{up}}{2}$.

1. Downstream $=46\text{km/h}$, upstream $=36\text{km/h}$.
2. $v=\frac{46-36}{2}=\frac{10}{2}=5\text{km/h}$.
3. (Boat speed in still water $=\frac{46+36}{2}=41\text{km/h}$.)

Trick: Current = half the difference; boat = half the sum.

Answer: (A) $5\text{km/h}$

Concept: Work together: combined rate = sum of individual rates. Take LCM of 26 and 36 as total capacity.

1. LCM$(26,36)=468$ units. A fills $\frac{468}{26}=18$ units/h, B fills $\frac{468}{36}=13$ units/h.
2. Together $=18+13=31$ units/h.
3. Half cistern $=234$ units → time $=\frac{234}{31}=7\frac{17}{31}$ hours.

Trick: Full tank together $=\frac{ab}{a+b}=\frac{26\times 36}{62}=\frac{468}{31}$; halve it.

Answer: (C) $7\frac{17}{31}$ hours

Concept: Mode is the value that occurs most often in the data.

1. Count: 49 occurs 4 times.
2. 43, 44, 46, 52, 53, 54 occur 2 times each; the rest once.
3. Highest frequency is for 49.

Trick: Tally only the repeating values – one with the most ticks is the mode.

Answer: (A) 49

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

1. $ab=(4+2\sqrt{3})(16-8\sqrt{3})$.
2. $=64-32\sqrt{3}+32\sqrt{3}-16\times 3=64-48=16$.
3. $\sqrt{16}=4$.

Trick: $16-8\sqrt{3}=4(4-2\sqrt{3})$, so product $=4(16-12)=16$ quickly.

Answer: (A) 4

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

1. $\text{SP}=626\times \frac{135}{100}=845.10$.
2. $\text{MP}=845.10\times \frac{100}{90}=939$.

Trick: $\text{MP}=\text{CP}\times \frac{135}{90}=626\times 1.5=939$.

Answer: (A) ₹939

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$: $D>0$ real distinct, $D=0$ equal, $D<0$ imaginary.

1. $a=2,b=1,c=3$.
2. $D=1-4\times 2\times 3=1-24=-23$.
3. $D<0$ → roots are imaginary (not real).

Trick: If $a$ and $c$ have the same sign and $b^2<4ac$, roots are imaginary.

Answer: (C) Imaginary roots

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Total distance $=5+5=10\text{km}$.
2. Time $=1+1\frac{15}{60}=2\frac{1}{4}=\frac{9}{4}$ h.
3. Average speed $=10÷\frac{9}{4}=\frac{40}{9}\text{km/h}$.

Trick: Never average the two speeds directly; use total distance over total time.

Answer: (B) $\frac{40}{9}$

Concept: Sum of ages = average × number; each person ages by the same years.

1. 2 years ago A + B + C $=3\times 66=198$ → now $=198+6=204$.
2. 2 years later A + C $=2\times 89=178$ → now $=178-4=174$.
3. B now $=204-174=30$; after 4 years $=34$.

Trick: Bring every sum to “now” first, then subtract.

Answer: (C) 34

Concept: Missing value = total of all − total of the rest.

1. Total of 7 subjects $=7\times 76=532$.
2. Total of 6 subjects $=6\times 70=420$.
3. English $=532-420=112$.

Trick: English $=70+7\times (76-70)=70+42=112$.

Answer: (C) 112

Concept: Use the angle sum in a triangle, alternate angles for parallel lines and equal base angles of an isosceles triangle.

1. In $△ABC$: $\angle ACB={180}^{\circ }-{13}^{\circ }-{164}^{\circ }=3^{\circ }$.
2. $BC\parallel AD$ → $\angle CAD=\angle ACB=3^{\circ }$ (alternate angles).
3. $AC=CD$ → $\angle CDA=\angle CAD=3^{\circ }$.
4. $\angle ACD={180}^{\circ }-3^{\circ }-3^{\circ }={174}^{\circ }$.

Trick: In isosceles $△ACD$, apex angle $={180}^{\circ }-2\times$ base angle.

Answer: (D) ${174}^{\circ }$