Q&A: practice questions with answers

Concept: Expand each bracket, then combine like terms.

1. $3(\frac{5}{3}{x}^2-28x+15)=5x^2-84x+45$
2. $-5(x^2+6x-15)=-5x^2-30x+75$
3. Add: $(5x^2-5x^2)+(-84x-30x)+(45+75)=-114x+120$

Trick: The $x^2$ terms cancel; only track the $x$ terms ($-84-30$) and constants ($45+75$).

Answer: (D) $-114x+120$

Concept: Total $=$ average $\times$ number; then share in ratio.

1. Total price $=3\times 5310=15930$.
2. Ratio sum $=3+5+7=15$.
3. Cheapest $=\frac{3}{15}\times 15930=3186$.

Trick: Average of the parts $=5$ parts $=5310$ → 1 part $=1062$ → 3 parts $=3186$.

Answer: (B) $3,186$

Concept: One-day work of C $=$ (A+B+C) work $-$ A's work $-$ B's work.

1. $\frac{1}{C}=\frac{1}{2}-\frac{1}{22}-\frac{1}{10}$
2. LCM 110: $\frac{55-5-11}{110}=\frac{39}{110}$
3. C alone $=\frac{110}{39}=2\frac{32}{39}$ days.

Trick: All three finish in 2 days, so C alone must take more than 2 days but fewer than A or B — only $2\frac{32}{39}$ fits.

Answer: (D) $2\frac{32}{39}$ days

Concept: Successive discounts $a\%$ and $b\%$ equal a single discount of $a+b-\frac{ab}{100}$.

1. Equivalent discount $=10+15-\frac{150}{100}=23.5\%$.
2. Difference in discount $=25\%-23.5\%=1.5\%$.
3. Difference in SP $=1.5\%$ of $38800=582$.

Trick: No need to find both SPs — just take the gap between the two equivalent discounts on the MP.

Answer: (D) ₹582

Concept: Let Ram $=R$, Ramya $=r$ and form two equations.

1. $2R=9r+25$ and $5r=R-4\Rightarrow R=5r+4$.
2. $2(5r+4)=9r+25\Rightarrow 10r+8=9r+25\Rightarrow r=17$.
3. $R=5(17)+4=89$; difference $=89-17=72$.

Trick: Substitute $R=5r+4$ straight away — one variable remains.

Answer: (A) $72$

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$.

1. Discount $=10\%$ of $400=40$.
2. SP $=400-40=360$.

Trick: 10% off → multiply MP by $0.9$: $400\times 0.9=360$.

Answer: (A) ₹360

Concept: Count days excluding the deposit day and including the withdrawal day; $\text{SI}=\frac{P\times R\times T}{100}$.

1. 2024 is a leap year: February $29-21=8$ days, March 31, April 22 → $61$ days.
2. $T=\frac{61}{366}$ year.
3. $\text{SI}=\frac{4000\times 6.75\times 61}{100\times 366}=\frac{16470}{366}=45$

Yaad rakhein: Count only one of the two end dates; with 365 days the SI is $45.12\approx 45$, the same option.

Answer: (A) ₹45

Concept: If the bases are equal, the exponents are equal.

1. Both $x$ and $y$ equal $7^{0.07}$, so $x=y$.
2. $x^z=y^3\Rightarrow (7^{0.07}{)}^z=(7^{0.07}{)}^3$.
3. $0.07z=0.21\Rightarrow z=3$.

Trick: $x=y$ (and $x\ne 1$), so $x^z=x^3$ gives $z=3$ at once.

Answer: (B) $3$

Concept: Distance $=$ speed $\times$ time (same units). The car's own length is negligible.

1. $3.6$ min $=\frac{3.6}{60}=0.06$ h.
2. Length $=79\times 0.06=4.74\text{km}$.

Trick: Minutes → hours: divide by 60; $3.6÷60=0.06$.

Answer: (C) $4.74\text{km}$

Concept: Product of two numbers $=$ HCF $\times$ LCM.

1. $55\times x=5\times 495=2475$
2. $x=\frac{2475}{55}=45$
3. Check: HCF(55, 45) $=5$ ✔

Trick: The other number must be a multiple of the HCF 5 — rules out 64, 8, 9 instantly.

Answer: (A) $45$

Concept: Check every number: the element just before AND just after it must both be symbols.

1. Numbers in order: 6, 5, 4, 9, 6.
2. First 6: before # (symbol), after 5 (number) ✘. 5: before 6 ✘. 4: before >, after 9 ✘. 9: before 4 ✘.
3. Last 6: before \ (symbol), after ^ (symbol) ✔.
4. Only one such number.

Trick: Numbers that sit in pairs (6 5, 4 9) can never qualify — only isolated numbers need checking.

Answer: (A) One

Concept: Track net east–west and north–south displacement with A as origin.

1. East–west: $+18-21+3=0$ → P is directly in line with A.
2. North–south: $+4-9=-5$ → P is 5 km south of A.
3. To return, drive 5 km towards north.

Trick: Four left turns from east: E → N → W → S → E; just add same-axis moves.

Answer: (B) 5 km towards north

Concept: Total $=$ rank from top $+$ rank from bottom $-1$.

1. Total $=69+13-1$
2. $=81$

Yaad rakhein: Subtract 1 because Tony is counted in both ranks.

Answer: (D) $81$

Concept: Fix the definite positions first (number from bottom $=1$ to top $=7$).

1. Only E above B → E $=7$, B $=6$.
2. Only two above F → F $=5$; G just below F → G $=4$.
3. A just above C in positions 3, 2, 1; A is not 3rd → A $=2$, C $=1$, D $=3$.
4. C is at the bottom, so 6 boxes are above it.

Trick: "Only X above Y" fixes the top two boxes at once.

Answer: (C) Six

Concept: Compare first-to-first and second-to-second letters by their positions.

1. TM - OO: T(20) $-5$ = O(15); M(13) $+2$ = O(15).
2. IY - DA: I(9) $-5$ = D(4); Y(25) $+2$ = A(1, wrap). NH - IJ: N(14) $-5$ = I(9); H(8) $+2$ = J(10).
3. CP - XS: C $-5$ = X ✔ but P(16) $+2$ = R, not S ✘.

Trick: Rule $(-5,+2)$; CP - XS uses $+3$ in the second letter.

Answer: (B) CP - XS

Concept: Follow each turn, then find net displacement from Y.

1. Directions: N 4, W 2, S 6, E 4, S 4 (right of east), W 2 (right of south).
2. East–west: $-2+4-2=0$.
3. North–south: $+4-6-4=-6$ → Z is 6 km south of Y.
4. So she must drive 6 km towards north.

Trick: Facing east, a right turn means south; facing south, a right turn means west.

Answer: (A) 6 km towards north

Concept: Write position values and check the gaps between consecutive letters.

1. EYTQ: 5, 25, 20, 17 → $+20,-5,-3$.
2. AUPM: 1, 21, 16, 13 → $+20,-5,-3$. BVQN: 2, 22, 17, 14 → $+20,-5,-3$.
3. FZVR: 6, 26, 22, 18 → $+20,-4,-4$ ✘.

Trick: The first two gaps often match; check the third letter early — FZV breaks the $-5$ step.

Answer: (B) FZVR

Concept: A word common to two sentences takes the code common to both.

1. Common word: 'is'.
2. Common code: 're'.
3. So 'is' → 're'.

Trick: Match common word with common code — no need to decode the rest.

Answer: (B) re

Concept: Swap the signs first, then apply BODMAS.

1. After swapping: $32-2+72÷9\times 5$
2. $72÷9=8$, $8\times 5=40$
3. $32-2+40=70$

Trick: Rewrite the whole expression with new signs before calculating anything.

Answer: (D) $70$

Concept: "All A are B" + "All B are C" ⇒ "All A are C"; but the reverse "All B are A" is not certain.

1. Fathers lie inside Brothers, and Brothers lie inside Sons → all fathers are sons: I follows.
2. Brothers is a bigger circle than Fathers; all brothers need not be fathers: II does not follow.

Trick: "All" can never be reversed directly.

Answer: (B) Only conclusion I follows.