18 February 2025 · Shift 3

Area of a rhombus The diagonals of a rhombus bisect each other at right angles, so the rhombus splits into 4 equal right triangles. Formula: area of a rhombus $=\frac{1}{2}\times d_1\times d_2$ . Substituting $d_1=15$ and $d_2=24$ : area $=\frac{1}{2}\times 15\times 24$ $=\frac{360}{2}=180$ So the area is $180{cm}^2$ and the correct option is (b).

Time and work in cycles Take the total work as 4 times the LCM of 6, 8, 12 and 16, i.e. $4\times 48=192$ units, so that a quarter-day of work is a whole number. One day's work: Avi $=\frac{192}{6}=32$ , Avni $=\frac{192}{8}=24$ , Aarti $=\frac{192}{12}=16$ , Reeta $=\frac{192}{16}=12$ units. Each of them works only $\frac{1}{4}$ day per turn, so in one turn they do 8, 6, 4 and 3 units respectively. One complete cycle (Avi, Avni, Aarti, Reeta) =8+6+4+3=21 units. $192÷21$ gives 9 complete cycles, work done $=9\times 21=189$ units. Work still left =192-189=3 units. The next turn is Avi's, and he does 8 units in his quarter-day, which is more than the 3 units left, so the work ends inside his turn. So Avi was working when the work finished and the correct option is (c).

Mean proportional A is the mean proportional of 192 and 48, so cross-multiply $\frac{A}{192}=\frac{48}{A}$ . $A\times A=192\times 48$ $A^2=9216$ $A=\sqrt{9216}=96$ So A=96 and the correct option is (d).

Percentage of literates Number of females =385000-168000=217000 . Total literates $=52\%$ of $385000=\frac{52}{100}\times 385000=200200$ . Literate males $=21\%$ of $168000=\frac{21}{100}\times 168000=35280$ . Literate females =200200-35280=164920 . The percentage asked for is out of the female population, not the whole district: $\frac{164920}{217000}\times 100$ $=76\%$ , so the correct option is (c).

Speed, time and distance The distance is fixed, so speed and time are inversely proportional: speed $\times$ time is constant. The new speed is $\frac{1}{8}$ of the normal speed, so speed ratio (normal : new) =8:1 and time ratio =1:8 . Let 1 unit of time be t minutes; the delay is 8t-t=7t . 7t=21 t=3 So at his normal speed Hari would take 3 minutes and the correct option is (b).

LCM and HCF relation Rule: for two numbers, LCM $\times$ HCF = first number $\times$ second number. $198\times 9=99\times x$ 1782=99x $x=\frac{1782}{99}=18$ So the other number is 18 and the correct option is (c).

Cost price from gain percent Formula: selling price = cost price $\times \left(1+\frac{20}{100}\right)$ , because the gain is $20\%$ of the cost price. $222=\mathrm{CP}\times \frac{6}{5}$ $\mathrm{CP}=222\times \frac{5}{6}$ $\mathrm{CP}=185$ So the cost price is ₹185 and the correct option is (c); dividing by $\frac{6}{5}$ is the step, not taking $20\%$ off ₹222.

Average speed Average speed is total distance divided by total time; it is never the plain average of the four speeds. Total distance =2+2+2+2=8 km. Total time $=\frac{2}{10}+\frac{2}{20}+\frac{2}{30}+\frac{2}{40}=2\times \frac{12+6+4+3}{120}=\frac{50}{120}=\frac{5}{12}$ hours. Average speed $=8÷\frac{5}{12}=8\times \frac{12}{5}=\frac{96}{5}$ =19.2 km/hr, so the correct option is (c); the plain average of 10, 20, 30 and 40 would be 25, which is not asked.

Average age Total age of the 48 students $=48\times 10=480$ years. With the teacher the group has 49 members, so the new total $=49\times 11=539$ years. Teacher's age =539-480 =59 years, so the correct option is (c).

Weighted average The three quantities are unequal, so a weighted average is needed, with the number of barrels as the weights. Weighted total $=200\times 85+100\times 90+100\times 95$ =17000+9000+9500=35500 Total barrels =200+100+100=400 Average octane $=\frac{35500}{400}=88.75\%$ So the correct option is (a); the plain average of 85, 90 and 95 would be 90, which ignores the extra Regular barrels.

Third proportional If c is the third proportional to a and b , then a:b=b:c , which gives $c=\frac{b^2}{a}$ . $c=\frac{{30}^2}{18}=\frac{900}{18}$ c=50 So the third proportional is 50 and the correct option is (d).

Compound interest sharing Mandar is 13, so his money grows for 23-13=10 years; Ketan is 14, so his grows for 23-14=9 years. Formula: amount $=P{\left(1+\frac{5}{100}\right)}^n=P{\left(\frac{21}{20}\right)}^n$ . Equal amounts at age 23 give $M{\left(\frac{21}{20}\right)}^{10}=K{\left(\frac{21}{20}\right)}^9$ . Dividing both sides by ${\left(\frac{21}{20}\right)}^9$ : $M\times \frac{21}{20}=K$ , so M:K=20:21 . 20+21=41 units correspond to ₹24600, so 1 unit $=\frac{24600}{41}=600$ . Ketan's share $=21\times 600=12600$ , i.e. ₹12600, so the correct option is (c).

Percentage equation $4\%$ of x=36 means $\frac{4}{100}\times x=36$ . $x=36\times \frac{100}{4}$ $x=36\times 25$ x=900 , so the correct option is (a).

Rate of simple interest Formula: simple interest $=\frac{P\times R\times T}{100}$ ; here R is the same for both and T=4 for both. So the extra ₹720 of interest comes only from the extra principal =9400-5400=4000 . $\frac{4000\times R\times 4}{100}=720$ 160R=720 $R=\frac{720}{160}=4.5$ The rate is $4.5\%$ per annum, so the correct option is (b).

LCM of several numbers The smallest number divisible by all of them is their LCM. Prime factors: $12=2^2\times 3$ , $28=2^2\times 7$ , $36=2^2\times 3^2$ , $27=3^3$ . The LCM takes the highest power of each prime: $2^2\times 3^3\times 7$ . $=4\times 27\times 7$ =756 , so the correct option is (c); 756 is also the only option divisible by 27.

Successive discounts Successive discounts act one after another, so their multipliers are multiplied; they are never simply added. Selling price $=60000\times \left(1-\frac{10}{100}\right)\times \left(1-\frac{15}{100}\right)\times \left(1-\frac{20}{100}\right)$ $=60000\times \frac{9}{10}\times \frac{17}{20}\times \frac{4}{5}$ $=54000\times \frac{17}{20}=45900$ , and then $45900\times \frac{4}{5}=36720$ So the selling price is ₹36,720 and the correct option is (a).

Difference of CI and SI For 2 years the simple interest is the same in each year, so one year's interest $=\frac{40}{2}=20$ . For 2 years, compound interest minus simple interest is exactly the interest earned on the first year's interest. $45-40=20\times \frac{R}{100}$ $5=\frac{20R}{100}$ , so $R=25\%$ . Now use simple interest $=\frac{P\times R\times T}{100}$ : $40=\frac{P\times 25\times 2}{100}$ . $40=\frac{P}{2}$ P=80 , so the principal is ₹80 and the correct option is (a).

BODMAS simplification By BODMAS, clear the innermost bracket first: 8-5=3 . Inside the braces, multiplication comes before addition: $\frac{28}{3}\times 3=28$ . So the braces give $\frac{63}{7}+28=9+28=37$ . The first bracket gives $90÷3=30$ . $30\times 37=1110$ So the value is 1110 and the correct option is (d).

Profit by false weight Take the marked weight as 1000 g. It is $32\%$ light, so the customer actually receives 1000-320=680 g. The shopkeeper pays for 680 g but charges the price of 1000 g, so his cost is 680 and his gain is 1000-680=320 . Profit percentage $=\frac{320}{680}\times 100$ $=\frac{3200}{68}=\frac{800}{17}$ $=47\frac{1}{17}\%$ , so the correct option is (c). The gain must be divided by 680, his own cost, and not by 1000; dividing by 1000 would wrongly give $32\%$ .

HCF, LCM and ratio Let the numbers be x and 3x . As 1:3 is already in lowest terms, their HCF is x and their LCM is 3x . Rule: HCF $\times$ LCM = product of the two numbers. $x\times 3x=1200$ $3x^2=1200$ , so $x^2=400$ and x=20 . So the numbers are 20 and $3\times 20=60$ , which is option (a).