RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Add the per-hour work of filling taps and subtract that of the emptying tap.

1. Net part filled per hour $=\frac{1}{4}+\frac{1}{30}-\frac{1}{30}=\frac{1}{4}$.
2. Full cistern takes 4 hours, so one-fourth takes $4\times \frac{1}{4}=1$ hour.

Trick: The 30-hour filler and 30-hour emptier cancel exactly; only the 4-hour tap matters.

Answer: (B) 1

Concept: Average $=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Sum $=72+68+74+70+76=360$.
2. Average $=\frac{360}{5}=72$.

Trick: Take 72 as base: deviations $0,-4,+2,-2,+4$ add to 0, so the average is exactly 72.

Answer: (D) 72

Concept: If two runners run in the same direction with speeds in the simplest ratio $a:b$, the number of distinct meeting points on the track is $|a-b|$ (for opposite directions it is $a+b$).

1. Speeds: Shahrukh $=\frac{1530}{85}=18\text{m/s}$, Amir $=\frac{1530}{90}=17\text{m/s}$.
2. Ratio $18:17$ is already in lowest terms.
3. Same direction → meeting points $=18-17=1$ (only at the starting point).

Trick: Ratio of speeds = inverse ratio of lap times $=90:85=18:17$; difference 1 → one meeting point.

Answer: (A) 1

Concept: For ungrouped frequency data, the mode is the observation with the highest frequency.

1. Number of plants: 14 cm → 1, 27 cm → 3, 23 cm → 6, 21 cm → 12, 19 cm → 7, 17 cm → 5.
2. Highest frequency is 12, for height 21 cm.

Trick: Mode = "most repeated"; do not average or pick the largest height (27 cm is a distractor).

Answer: (A) 21 cm

Concept: For equal distances at speeds $a$ and $b$, average speed $=\frac{2ab}{a+b}$.

1. Let morning speed $=v$, afternoon $=v-2$. Then $\frac{2v(v-2)}{2v-2}=4.8\Rightarrow \frac{v(v-2)}{v-1}=4.8$.
2. $v^2-2v=4.8v-4.8\Rightarrow v^2-6.8v+4.8=0\Rightarrow (v-6)(v-0.8)=0$.
3. $v=6$ (since $v>2$); speeds 6 and 4 km/h, sum $=10$ km/h.

Trick: Test options: sum 10 → speeds 6 and 4 → $\frac{2\times 6\times 4}{10}=4.8$ ✓ in one line.

Answer: (B) 10

Concept: Total = Average × Number. When a new member raises the average, his weight = new average + (old count × increase).

1. Total weight of 10 students $=10\times 50=500$ kg.
2. With the teacher, 11 persons have average 51 kg → total $=11\times 51=561$ kg.
3. Teacher's weight $=561-500=61$ kg.

Trick: New member = new average + (old number × rise) $=51+10\times 1=61$.

Answer: (B) 61 kg

Concept: For 2 years, difference between CI and SI $=P{\left(\frac{r}{100}\right)}^2$.

1. $P=4000,r=6\%$ → difference $=4000\times {\left(\frac{6}{100}\right)}^2=4000\times \frac{36}{10000}$.
2. $=\frac{144000}{10000}=14.40$.
3. Check: SI $=480$; CI $=4000({1.06}^2-1)=4000\times 0.1236=494.40$; difference $=14.40$ ✓.

Trick: 2-year difference = SI of one year's interest for one year: $240\times 6\%=14.40$.

Answer: (D) ₹14.40

Concept: Profit % is based on cost price; the number of articles does not matter if CP and SP per unit are compared.

1. CP of one article $=\frac{390}{12}=32.5$; SP of one article $=78$.
2. Profit per article $=78-32.5=45.5$.
3. Profit % $=\frac{45.5}{32.5}\times 100=\frac{91}{65}\times 100=1.4\times 100=140\%$.

Trick: Compare per dozen: SP of a dozen $=78\times 12=936$; $\frac{936}{390}=2.4$ → 140% profit. The "19 dozens" is a distractor.

Answer: (B) 140%

Concept: A number is divisible by 3 when the sum of its digits is a multiple of 3.

1. Digit sum of 54572A0 $=5+4+5+7+2+A+0=23+A$.
2. $23+A$ must be a multiple of 3 → $A=1$ (24), $4$ (27) or $7$ (30).
3. Minimum value $=1$.

Trick: $23÷3$ leaves remainder 2, so add $3-2=1$ → smallest A = 1.

Answer: (B) 1

Concept: Take CP = 100. Marked price = CP + markup; discount % is calculated on the marked price.

1. CP $=100$ → MP $=100+100\%=200$.
2. Required SP for 26% gain $=126$.
3. Discount $=200-126=74$ → Discount % $=\frac{74}{200}\times 100=37\%$.

Trick: $\text{SP}=\text{MP}(1-d)$ → $126=200(1-d)$ → $d=0.37$.

Answer: (D) 37%

Concept: A tangent is perpendicular to the radius at the point of contact, and a perpendicular from the centre to a chord bisects the chord.

1. AB touches the inner circle $C_2$ at P, so OP (radius of $C_2$) is perpendicular to AB.
2. For the outer circle $C_1$, AB is a chord and OP ⊥ AB passes through the centre O → P is the midpoint of AB.
3. Hence $AP=PB$. OP depends on the radius of $C_2$ and need not equal AP, 2PB or AB.

Yaad rakhein: In concentric circles, a chord of the bigger circle touching the smaller one is bisected at the point of contact.

Answer: (A) AP = PB

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

1. $4\times \frac{22}{7}\times r^2=154$.
2. $r^2=\frac{154\times 7}{4\times 22}=\frac{1078}{88}=12.25$.
3. $r=\sqrt{12.25}=3.5$ cm.

Trick: $154=22\times 7$, so $r^2=\frac{7\times 7}{4}$ → $r=\frac{7}{2}=3.5$.

Answer: (A) 3.5 cm

Concept: Solve the equation that has no unknown, then substitute each of its roots in the other equation.

1. $x^2+2x-3=0$ → $(x+3)(x-1)=0$ → $x=1$ or $x=-3$.
2. If $x=1$ is common: $1+1+p=0$ → $p=-2$.
3. If $x=-3$ is common: $9-3+p=0$ → $p=-6$ (not in options). So $p=-2$.

Trick: With a common root, plug the known roots directly – no need for the general common-root formula.

Answer: (B) −2

Concept: Perimeter of a sector $=2r+l$ (two radii + arc) and area $=\frac{1}{2}lr$.

1. $2\times 8+l=24$ → $l=8$ cm.
2. Area $=\frac{1}{2}\times l\times r=\frac{1}{2}\times 8\times 8=32$ cm².

Trick: Sector area $=\frac{1}{2}(\text{arc})(\text{radius})$ – works without knowing the angle, just like a triangle with base = arc and height = radius.

Answer: (A) 32 cm²

Concept: Weighted mean $=\frac{\sum w_i{x}_i}{\sum w_i}$. Here each number is its own weight, so $w_i=x_i$.

1. $\sum w_i{x}_i=1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55$.
2. $\sum w_i=1+2+3+4+5=15$.
3. Weighted mean $=\frac{55}{15}=\frac{11}{3}$.

Trick: For the first n natural numbers weighted by themselves, mean $=\frac{\sum n^2}{\sum n}=\frac{2n+1}{3}$ → $\frac{11}{3}$ for n = 5.

Answer: (C) $\frac{11}{3}$

Concept: First find the actual selling price in terms of CP from the loss condition, then apply the new fraction.

1. Let CP $=100$. Selling at $\frac{4}{11}$ of SP gives 20% loss → $\frac{4}{11}\text{SP}=80$ → SP $=220$.
2. New price $=71\%$ of $220=156.2$.
3. Profit $=156.2-100=56.2$ → profit % $=56.2\%$.

Trick: SP = 2.2 × CP, so any fraction $k$ of SP gives profit % $=220k-100$ → $220\times 0.71-100=56.2$.

Answer: (B) 56.2%

Concept: Bells ringing at different intervals ring together again after the LCM of the intervals.

1. $120=2^3\times 3\times 5$, $180=2^2\times 3^2\times 5$, $240=2^4\times 3\times 5$.
2. LCM $=2^4\times 3^2\times 5=720$ minutes $=12$ hours.
3. 9:00 a.m. + 12 h = 9:00 p.m.

Trick: In hours the intervals are 2, 3, 4 → LCM = 12 h, a much quicker calculation.

Answer: (A) 9 p.m.

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

1. $T=\frac{360\times 100}{7200\times 4}=\frac{36000}{28800}$.
2. $=\frac{5}{4}=1.25$ years.

Trick: One year's interest $=4\%$ of $7200=288$; $\frac{360}{288}=1.25$ years.

Answer: (C) 1.25

Concept: When three expressions are equal, subtract them pairwise; the differences multiply to a neat result.

1. From $p+\frac{1}{m}=m+\frac{1}{n}$: $p-m=\frac{1}{n}-\frac{1}{m}=\frac{m-n}{mn}$.
2. Similarly $m-n=\frac{n-p}{np}$ and $n-p=\frac{p-m}{pm}$.
3. Multiply all three: $(p-m)(m-n)(n-p)=\frac{(m-n)(n-p)(p-m)}{(pmn{)}^2}$.
4. Since $p\ne m\ne n$, cancel the non-zero product: $(pmn{)}^2=1\Rightarrow pmn=\pm 1$.

Yaad rakhein: $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$ (all distinct) $\Rightarrow abc=\pm 1$. Standard result – just recall it in the exam.

Answer: (C) $\pm 1$

Concept: Take the younger brother's age as $x$ and write every age in terms of $x$; "3 years ago" subtracts 3 from each person.

1. Younger brother $=x$, Vanraj $=2x$, other brother $=x-3$.
2. 3 years ago the ages were $x-3,2x-3,x-6$; sum $=4x-12$.
3. $4x-12=20\Rightarrow x=8$.
4. Vanraj's present age $=2x=16$ years.

Trick: Present sum $=20+3\times 3=29$, so $x+2x+(x-3)=29\Rightarrow x=8$ – add $3$ per person instead of subtracting thrice.

Answer: (A) 16