RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Let a pen cost ₹$p$ and a pencil ₹$c$; form two equations.

1. $6p+5c=259$ … (i)
2. $19(p-9)+2(c+4)=90\Rightarrow 19p+2c=253$ … (ii)
3. (i)$\times 2$: $12p+10c=518$; (ii)$\times 5$: $95p+10c=1265$ → $83p=747$, $p=9$
4. $5c=259-54=205\Rightarrow c=41$
5. $13p+2c=117+82=199$

Trick: Eliminate the variable whose coefficients are easy to equalise ($5c$ and $2c$ → $10c$).

Answer: (B) ₹$199$

Concept: A root of an equation satisfies it: substitute the common root.

1. Check: ${11}^2-9(11)-22=121-99-22=0$ ✓
2. Put $x=11$ in $2x^2-3x+c=0$: $2(121)-33+c=0$
3. $242-33+c=0\Rightarrow c=-209$

Trick: Only the equation containing c needs the substitution; the second one only confirms 11 is a root.

Answer: (D) $-209$

Concept: Solve brackets from inside out: ( ) → { } → [ ].

1. $(27÷9)=3$
2. $\{63-3\}=60$, then $60÷6=10$
3. $[7+10]=17$
4. $49+17=66$

Trick: Innermost bracket first, then division before addition.

Answer: (D) $66$

Concept: $\text{Sum}=\text{Average}\times \text{Count}$

1. First two numbers: $77\times 2=154$; third $=129$
2. Remaining two $=655-154-129=372$
3. Their average $=\frac{372}{2}=186$

Trick: Subtract known parts from the total, then divide by how many are left.

Answer: (B) $186$

Concept: Mode is the value that occurs most often.

1. Count: 9 → 4 times; 2 → 3 times; 4 → 3 times; 5 → 3 times
2. 1, 6, 8 → once each
3. Highest frequency belongs to 9.

Trick: Tally only the repeated values – the one with the longest tally is the mode.

Answer: (D) $9$

Concept: Convert all fractions to the common denominator 16.

1. $\frac{2}{8}=\frac{4}{16}$, so $\frac{4}{16}+\frac{4}{16}+\frac{1}{16}+\frac{4}{16}=\frac{13}{16}$
2. $1=\frac{16}{16}$
3. $\frac{13}{16}-\frac{16}{16}=-\frac{3}{16}$

Trick: $3\times \frac{2}{8}=\frac{3}{4}$; $\frac{3}{4}+\frac{1}{16}$ is less than 1, so the answer must be negative.

Answer: (C) $-\frac{3}{16}$

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

1. Mode 165 lies at the end of 160–165, so take modal class 160–165: $l=160$, $h=5$, $f_1=37$, $f_0=26$, $f_2=x$
2. $165=160+\frac{37-26}{74-26-x}\times 5$
3. $5=\frac{55}{48-x}\Rightarrow 48-x=11$
4. $x=37$

Trick: Plug the options in: only $x=37$ makes $\frac{11}{48-x}\times 5=5$.

Answer: (C) $37$

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. Interior sum $={360}^{\circ }+100\%\text{ of }{360}^{\circ }={720}^{\circ }$
2. $(n-2)\times 180=720\Rightarrow n=6$
3. Each interior angle $=\frac{720}{6}={120}^{\circ }$

Trick: Interior sum $=k\times {360}^{\circ }$ → $n=2k+2$. Here $k=2$, so $n=6$.

Answer: (D) $120$

Concept: Successive discounts multiply the remaining fractions: $\text{Remaining}=(1-d_1)(1-d_2)(1-d_3)$.

1. $0.94\times 0.72=0.6768$
2. $0.6768\times 0.85=0.57528$
3. Single discount $=1-0.57528=0.42472=42.47\%$

Trick: Two at a time: $6\%,28\%\to 6+28-\frac{6\times 28}{100}=32.32\%$; then with $15\%$: $32.32+15-\frac{32.32\times 15}{100}=42.47\%$.

Answer: (D) $42.47\%$

Concept: Convert each ratio to a decimal (or compare by cross-multiplication).

1. $\frac{16}{57}\approx 0.281$, $\frac{17}{53}\approx 0.321$
2. $\frac{24}{54}=\frac{4}{9}\approx 0.444$
3. $\frac{28}{60}=\frac{7}{15}\approx 0.467$
4. Check: $28\times 54=1512>24\times 60=1440$, so $28:60$ is the greatest.

Trick: Simplify first: $\frac{4}{9}$ vs $\frac{7}{15}$ → cross-multiply $60$ vs $63$.

Answer: (D) $28:60$

Concept: Average of the first $n$ natural numbers $=\frac{n+1}{2}$.

1. Here $n=17$.
2. Average $=\frac{17+1}{2}=\frac{18}{2}=9$.
3. Check: sum $=\frac{17\times 18}{2}=153$, and $153÷17=9$.

Trick: For consecutive numbers, average = middle term; the middle of 1 to 17 is 9.

Answer: (A) 9

Concept: Take CP $=100$; then MP and SP follow from the given percentages.

1. MP is 28% above CP: MP $=128$.
2. Profit 3%: SP $=103$.
3. Discount $=128-103=25$.
4. Discount% $=\frac{25}{128}\times 100=19.53\%$.

Trick: Discount% $=\frac{\text{markup}-\text{profit}}{100+\text{markup}}\times 100=\frac{28-3}{128}\times 100$. The MP ₹397 is not even needed.

Answer: (D) $19.53\%$

Concept: For $n$ (odd) observations in ascending order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ term.

1. Ascending: 11, 13, 21, 34, 58, 64, 65, 70, 73, 78, 88.
2. $n=11$, so median $=$ the $\frac{11+1}{2}=6^{\text{th}}$ term.
3. 6th term $=64$.

Trick: Always sort first – the 6th number in the given (unsorted) list is 73, a trap.

Answer: (A) 64

Concept: Use $\sec \theta =\frac{1}{\cos \theta }$, $\text{cosec}\theta =\frac{1}{\sin \theta }$ and ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $\sec \theta -\cos \theta =\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }$.
2. $\text{cosec}\theta -\sin \theta =\frac{{\cos }^2\theta }{\sin \theta }$.
3. $\sin \theta -\text{cosec}\theta =-\frac{{\cos }^2\theta }{\sin \theta }$.
4. Product $=\frac{{\sin }^2\theta }{\cos \theta }\times \frac{{\cos }^2\theta }{\sin \theta }\times (-\frac{{\cos }^2\theta }{\sin \theta })=-{\cos }^3\theta$.

Trick: The last bracket is negative (since $\sin \theta <\text{cosec}\theta$), so the answer must be negative; checking with $\theta ={30}^{\circ }$ gives $-{\cos }^3\theta$.

Answer: (D) $-{\cos }^3\theta$

Concept: Equal final amounts: $P_1(1+r{)}^{n_1}=P_2(1+r{)}^{n_2}$.

1. $P\times (1.1{)}^{11}=M\times (1.1{)}^{12}$ ⇒ $P=1.1M$, i.e. $P:M=11:10$.
2. $11+10=21$ parts $=$ ₹1617 ⇒ 1 part $=77$.
3. Piyush $=11\times 77=$ ₹847.

Trick: Shares are in ratio $(1+\frac{r}{100}{)}^{\text{gap}}:1=11:10$ for a 1-year gap at 10%.

Answer: (B) ₹847

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

1. $(a-b{)}^2=168-2\times 27=114$.
2. $(a+b{)}^2=168+2\times 27=222$.
3. $\frac{a-b}{a+b}=\sqrt{\frac{114}{222}}=\sqrt{\frac{57}{111}}$ (positive since $a>b$).

Trick: ${\left(\frac{a-b}{a+b}\right)}^2=\frac{S-2P}{S+2P}$ directly.

Answer: (A) $\sqrt{\frac{57}{111}}$

Concept: If $a:b::b:x$, then $b$ is the mean proportional and $x=\frac{b^2}{a}$.

1. $x=\frac{96.9\times 96.9}{5.1}$.
2. $\frac{96.9}{5.1}=19$.
3. $x=19\times 96.9=1841.1$.

Trick: Spot that $96.9=19\times 5.1$; the ratio is $1:19$, so $x=96.9\times 19$.

Answer: (D) 1841.1

Concept: Take total work = LCM of the days and find each person's daily work.

1. Total work $=\text{LCM}(11,33)=33$ units; A $=3$ units/day, B $=1$ unit/day.
2. In 3 days together: $3\times (3+1)=12$ units.
3. Remaining $=33-12=21$ units.
4. A alone: $\frac{21}{3}=7$ days.

Trick: Unit method avoids fractions completely.

Answer: (A) 7

Concept: $\sigma =\sqrt{\frac{\sum (x-\overline{x}{)}^2}{n}}$.

1. Sum $=552$, $n=8$, mean $\overline{x}=69$.
2. Deviations: $-4,-22,-10,6,13,22,-14,9$.
3. Squares: $16+484+100+36+169+484+196+81=1566$.
4. $\sigma =\sqrt{\frac{1566}{8}}=\sqrt{195.75}\approx 13.99$.

Trick: ${14}^2=196$, and $195.75$ is just below it, so $\sigma$ is just below 14.

Answer: (B) 13.99

Concept: $x^2+2xy+y^2=(x+y{)}^2$.

1. $100a^2=(10a{)}^2$ and $225b^2=(15b{)}^2$.
2. Middle term check: $2\times 10a\times 15b=300ab$ ✔ (positive).
3. So the expression $=(10a+15b{)}^2$.

Trick: All terms positive → '+' inside; $a^2$ has coefficient 100, so $10a$ comes with $a$.

Answer: (C) $(10a+15b{)}^2$