7 February 2025 · Shift 3

Discount on marked price A discount of 26% leaves the selling price at 74% of the marked price. Formula: $SP=MP\times \left(1-\frac{d}{100}\right)$ Substituting: $2886=MP\times \frac{74}{100}$ $MP=\frac{2886\times 100}{74}=3900$ Discount = 3900-2886 = 1014 Hence the amount of discount is ₹1,014, so the correct option is (c).

Average of weights Total weight = average $\times$ number of persons, and let the weight of Kiran be K. Ashish + Kiran + Mahesh $=46\times 3=138$ kg Ashish + Kiran $=40\times 2=80$ kg Kiran + Mahesh $=43\times 2=86$ kg Adding the last two: 80+86 = 166 kg, which counts Kiran twice. So 138+K = 166 K = 166-138 = 28 kg Hence the weight of Kiran is 28 kg, so the correct option is (b).

Average and alligation Let the number of women be w, so the number of men is 96-w. Total age of all workers $=32\times 96=3072$ years 41(96-w)+25w = 3072 3936-41w+25w = 3072 16w = 3936-3072 = 864 $w=\frac{864}{16}=54$ Hence there are 54 women in the factory, so the correct option is (a).

Mean proportional The mean proportional of a and b is $\sqrt{ab}$. Given: $z=\sqrt{xy}$, so $z^2=xy$. Required value $=\sqrt{x^{2}z\times y^{2}z}$ $=\sqrt{x^2{y}^2{z}^2}$ $=xyz=\left(xy\right)\times z$ $=z^2\times z=z^3$ Hence the mean proportional is $z^3$, so the correct option is (c).

BODMAS simplification Work outwards from the innermost bracket (BODMAS). (4-2) = 2 and $21÷7=3$ $\frac{63}{9}=7$ and $\frac{14}{2}=7$ Inside the braces: $7+7\times 2=7+14=21$ Whole expression $=3\times 21=63$ Hence the value is 63, so the correct option is (b).

Mean proportional Formula: mean proportional of a and b $=\sqrt{ab}$ Substituting: $=\sqrt{9\times 49}$ $=\sqrt{441}$ = 21 Hence the mean proportional is 21, so the correct option is (d).

Percentage change in expenditure Expenditure = price $\times$ consumption. Take price = 100 and consumption = 100, so expenditure = 10000. New price $=100\times \frac{160}{100}=160$ New expenditure $=10000\times \frac{144}{100}=14400$ New consumption $=\frac{14400}{160}=90$ Fall in consumption = 100-90 = 10 out of 100, that is 10% Hence the consumption must be reduced by 10%, so the correct option is (c).

Trains crossing each other While crossing each other the trains together cover the sum of their lengths. Total length = 250+250 = 500 m Relative speed $=\frac{500}{25}=20$ m/s In km/hr: $20\times \frac{18}{5}=72$ km/hr In opposite directions the speeds add, so 45+x = 72 x = 72-45 = 27 Hence the other train runs at 27 km/hr, so the correct option is (c).

Profit and selling price Cost price of 400 pears $=\frac{1200}{100}\times 400=4800$ Selling price of 400 pears = 4800+1000 = 5800 Selling price of 1 pear $=\frac{5800}{400}=14.5$ Selling price of 1 dozen $=14.5\times 12=174$ Hence one dozen pears sell for ₹174, so the correct option is (a).

Simple and compound interest Simple interest of one year $=\frac{40}{2}=20$ Difference = 44-40 = 4, and for 2 years this is only the interest earned on the interest of the first year, ₹20. Rate $=\frac{4}{20}\times 100=20$, that is 20% per annum. Formula: $SI=\frac{P\times R\times T}{100}$ Substituting: $40=\frac{P\times 20\times 2}{100}$ $P=\frac{40\times 100}{40}=100$ Hence the principal is ₹100, so the correct option is (a).

HCF of two numbers A container that measures both amounts exactly must hold their HCF. $765=3\times 3\times 5\times 17$ and $833=7\times 7\times 17$ The only common factor is 17, so HCF = 17 Hence the maximum capacity is 17 litres, so the correct option is (d).

Train crossing a tunnel To cross a tunnel a train covers its own length plus the length of the tunnel. Total distance = 496+664 = 1160 m Speed $=18\times \frac{5}{18}=5$ m/s Time = distance $÷$ speed $=\frac{1160}{5}$ = 232 seconds Hence the train needs 232 seconds, so the correct option is (a).

Work done on alternate days A takes 12 days, B takes half of that, $\frac{12}{2}=6$ days, and C takes half of that of B, $\frac{6}{2}=3$ days. Total work = LCM of 12, 6 and 3 = 12 units One day of work: A $=\frac{12}{12}=1$, B $=\frac{12}{6}=2$, C $=\frac{12}{3}=4$ units The cycle B, A, C does 2+1+4 = 7 units in 3 days. Day 4 (B): 7+2 = 9 units, and day 5 (A): 9+1 = 10 units Remaining work = 12-10 = 2 units, and day 6 is the turn of C. Time C needs $=\frac{2}{4}=0.5$ day Total time = 5+0.5 = 5.5 days Hence the work is finished in 5.5 days, so the correct option is (a).

Compound interest and shares Mahesh is 15, so his money grows for 22-15 = 7 years. David is 16, so his money grows for 22-16 = 6 years. Formula: $A=P{\left(1+\frac{R}{100}\right)}^T$ Equal amounts: $M{\left(\frac{11}{10}\right)}^7=D{\left(\frac{11}{10}\right)}^6$ $\frac{M}{D}=\frac{10}{11}$, so their shares are in the ratio 10:11. 10+11 = 21 parts = ₹25,200, so 1 part $=\frac{25200}{21}=1200$ The share of David $=11\times 1200=13200$ Hence Ketan gave David ₹13,200, so the correct option is (d).

LCM by prime factors Break every number into its prime factors. The LCM takes the highest power of each prime: $2^2\times 3^2\times 7\times 31$ $=4\times 9\times 7\times 31$ $=252\times 31=7812$ Hence the LCM is 7812, so the correct option is (d).

Operations on integers By BODMAS, divide first, then multiply, then add and subtract. $\left(-60\right)÷\left(-15\right)=4$ $\left(-2\right)\times 6=-12$ The expression becomes -9-4+(-12) = -13-12 = -25 Hence the value is -25, so the correct option is (a).

Loss to profit selling price At an 8% loss the selling price is 92% of the cost price. Substituting: $20240=CP\times \frac{92}{100}$ $CP=\frac{20240\times 100}{92}=22000$ For a 12% profit the selling price is 112% of the cost price. $SP=22000\times \frac{112}{100}=24640$ Hence he should sell it for ₹24,640, so the correct option is (c).

Curved surface area of cylinder Formula: curved surface area of a cylinder $=2\pi rh$ Substituting: $=2\times \frac{22}{7}\times 7\times 9.5$ $=2\times 22\times 9.5$ $=44\times 9.5=418$ Hence the curved surface area is 418 cm², so the correct option is (d).

Percentage change in a fraction Let the original fraction be $\frac{n}{d}$. New fraction $=\frac{n\times \frac{110}{100}}{d\times \frac{90}{100}}=\frac{11n}{9d}$ Given: $\frac{11n}{9d}=\frac{11}{45}$ $\frac{n}{d}=\frac{11}{45}\times \frac{9}{11}=\frac{9}{45}$ $=\frac{1}{5}$ Hence the original fraction is $\frac{1}{5}$, so the correct option is (b).

Simple interest principal Formula: $SI=\frac{P\times R\times T}{100}$ Substituting: $360=\frac{P\times 2\times 4}{100}$ $360=\frac{8P}{100}$ $P=\frac{360\times 100}{8}$ = 4500 Hence the principal is ₹4,500, so the correct option is (d).