25 February 2025 · Shift 1

Percentage to ratio Both salaries are compared with C's salary, so make C's salary the base. Let C's salary be 100 units. A is 40% more than C, so A =100+40=140 units. B is 70% more than C, so B =100+70=170 units. B:A=170:140=17:14 (dividing both terms by 10). Hence the correct option is (a).

Loss stated in units Let the cost price of one ball be x (in ₹). Cost price of 15 balls =15x and selling price of 15 balls =1220. Loss = CP - SP =15x-1220, and the question says this loss equals the CP of 5 balls =5x. 15x-1220=5x 10x=1220 $x=\frac{1220}{10}=122$ Hence the cost price of one ball is ₹122, option (b).

Basic percentage '2% of x' means $\frac{2}{100}\times x$. $\frac{2}{100}\times x=360$ $x=360\times \frac{100}{2}$ $x=360\times 50=18000$ Hence the correct option is (c).

Work with alternating helpers Take the total work as the LCM of 30, 33, 36 and 44, i.e. 1980 units. One day's work: $A=\frac{1980}{30}=66$, $B=\frac{1980}{33}=60$, $C=\frac{1980}{36}=55$, $D=\frac{1980}{44}=45$ units. A and B together do 66+60=126 units a day, and the helper changes every two days. One full cycle of 4 days does 362+342=704 units, so 8 days (two cycles) do 1408 units. Days 9 and 10 are C days: 1408+362=1770 units done, so 1980-1770=210 units are left. Day 11 is a D day and gives 171 units, leaving 210-171=39 units. Day 12 is also a D day, so the last bit needs $\frac{39}{171}=\frac{13}{57}$ of a day. Total time $=11+\frac{13}{57}=11\frac{13}{57}$ days. Hence the correct option is (c).

BODMAS with brackets Open the brackets from the inside out: round, then curly, then square. Round bracket: 6-5=1, so $6\left(6-5\right)\times 2=6\times 1\times 2=12$. Curly bracket: $\left\{12-12\right\}=0$. Now $3\times 0÷20=0$, so the square bracket becomes [2+0]=2. $79-10\times 2=79-20=59$ Hence the correct option is (a).

Marked price, discount, profit Take the cost price as 100 units, because both percentages are easy on 100. Marked price =100+50=150 units. The 12% discount is on the marked price, not on the cost price, so SP $=150\times \frac{88}{100}=132$ units. Profit =132-100=32 units. Profit $\%=\frac{32}{100}\times 100=32\%$ Hence the correct option is (b).

Simplification of brackets Innermost round bracket first: 6-4+6=8. Curly bracket: $\left\{6-8\right\}=-2$. Square bracket: [3-(-2)]=3+2=5. 2-5=-3 Hence the correct option is (b).

Melting and recasting a cube Melting changes the shape but not the volume, so the big cube's volume equals the total volume of the 8 small cubes. Volume of the big cube $=9^3=729$ cm³. Let the side of a small cube be a, so $8a^3=729$. $a^3=\frac{729}{8}$ $a=\sqrt[3]{\frac{729}{8}}=\frac{9}{2}=4.5$ Hence each small cube has a side of 4.5 cm, option (b).

Order of operations Innermost bracket first: 6-5=1. Curly bracket: $\frac{56}{8}+\frac{14}{1}\times 1=7+14=21$. Round bracket: $36÷9=4$. $\left(4\times 21\right)=84$ Hence the correct option is (a).

Average and a new member Total age of the 28 students $=28\times 40=1120$ years. With the teacher the group has 29 people, so their total age $=29\times 41=1189$ years. Age of the teacher =1189-1120=69 years. Shortcut: the teacher's age = new average + (number of students $\times$ rise in average) $=41+28\times 1=69$. Hence the correct option is (d).

Speed for the remaining distance Total time =2 hours 24 minutes $=2+\frac{24}{60}=2.4$ hours. One-third of 24 km =8 km, and this part is covered at 8 km/hr. Time for the first part $=\frac{8}{8}=1$ hour. Remaining distance =24-8=16 km and remaining time =2.4-1=1.4 hours. Required speed $=\frac{16}{1.4}=\frac{160}{14}=\frac{80}{7}=11\frac{3}{7}$ km/hr. Hence the correct option is (a).

BODMAS left to right Division and multiplication rank equally, so they are done left to right, before the subtraction. $32÷8\times 4=4\times 4=16$ $2\times 3=6$ 16-6=10 Hence the correct option is (d).

Ratio after new admissions Let the sections hold 5x, 4x and 7x students. After the admissions they hold 5x+15, 4x+16 and 7x+9, and these are in the ratio 10:9:11. Two of the three parts are enough: $\frac{5x+15}{4x+16}=\frac{10}{9}$. 9(5x+15)=10(4x+16) 45x+135=40x+160 5x=25, so x=5. The third part checks out too, since 40:36:44=10:9:11. Total =40+36+44=120 Hence the correct option is (b).

Compound interest to simple interest The two interests share the same principal, rate and time, so first find the principal from the compound interest. For 2 years, CI $=P\left({\left(1+\frac{R}{100}\right)}^2-1\right)$. $229.5=P\left({\left(1.04\right)}^2-1\right)=P\times 0.0816$ $P=\frac{229.5}{0.0816}=2812.5$ SI $=\frac{P\times R\times T}{100}=\frac{2812.5\times 4\times 2}{100}=225$ Check: for 2 years CI - SI = 4% of the first year's interest $=0.04\times 112.5=4.5$, and 225+4.5=229.5. Hence the simple interest is ₹225, option (b).

Gain given as selling price Let the selling price of one chair be s and its cost price be c. Gain on 150 chairs =150s-150c, and this equals the SP of 75 chairs =75s. 150s-150c=75s 75s=150c, so s=2c. Profit on one chair =2c-c=c, so profit $\%=\frac{c}{c}\times 100=100\%$. Hence the correct option is (c).

Average cost from a total Cost of 6 tables $=6\times 1500=9000$. So the 12 chairs together cost 12600-9000=3600. Average cost of one chair $=\frac{3600}{12}=300$. Hence the average cost of a chair is ₹300, option (d).

Literacy percentage of females Number of females =327000-224000=103000. Literate people $=65\%$ of 327000=212550. Literate males $=65\%$ of 224000=145600. Literate females =212550-145600=66950. Required percentage $=\frac{66950}{103000}\times 100=65\%$. The male rate is already equal to the overall rate, so the female rate has to match it as well - a quick way to spot the answer. Hence the correct option is (c).

Trains crossing in the same direction For two trains moving in the same direction the relative speed is the difference: 90-60=30 km/h. 30 km/h $=30\times \frac{5}{18}=\frac{25}{3}$ m/s. To cross completely, the faster train has to cover the sum of the two lengths =210+440=650 m. Time $=\frac{650}{\frac{25}{3}}=650\times \frac{3}{25}=78$ seconds. 78 seconds $=\frac{78}{60}=1.3$ minutes. Hence the correct option is (a).

Compounding over 10-month periods Interest is added every 10 months, so one conversion period is 10 months, not one year. Rate for one period $=12\%\times \frac{10}{12}=10\%$. $2\frac{1}{2}$ years =30 months, which is $\frac{30}{10}=3$ periods. $A=x{\left(1+\frac{10}{100}\right)}^3$ $75867=x\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}=x\times \frac{1331}{1000}$ $x=\frac{75867\times 1000}{1331}=57000$ Hence the correct option is (d).

Simple interest from a difference In simple interest, $SI=\frac{P\times R\times T}{100}$, and here R and T are the same for both men. So the extra interest comes only from the extra principal: 9400-5400=4000. $\frac{4000\times R\times 6}{100}=720$ 240R=720 $R=\frac{720}{240}=3$ Hence the rate of interest is 3% per annum, option (c).