Key idea: Find the area once, then each altitude is \( \frac{2 \times \text{Area}}{\text{base}} \).
1. In an isosceles triangle the altitude from A bisects BC, so BD = 40 cm and the altitude from A is \( \sqrt{50^2 - 40^2} = 30 \) cm.
2. Area \( = \frac{1}{2} \times 80 \times 30 = 1200 \) sq. cm.
3. The altitude to AB (and, by symmetry, to AC) is \( \frac{2 \times 1200}{50} = 48 \) cm.
4. Sum of altitudes \( = 30 + 48 + 48 = 126 \) cm.
Tip: The 50-40-30 split is the 3-4-5 triple scaled by 10; spotting it saves a square root.
Hence, the answer is 126.
Q2
Algebra - Progressions TYPE IN THE ANSWER
In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is
ABCD is a trapezium in which AB is parallel to \( \mathrm { DC } , \mathrm { AD } \) is perpendicular to AB , and \( \mathrm { AB } = 3 \mathrm { DC } \). If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm , then the area, in sq. cm , of the trapezium is
Key idea: A quadrilateral with an incircle has equal sums of opposite sides, and here the perpendicular side AD equals the diameter.
1. AD is perpendicular to both parallel sides, so AD = height = \( 2r = 6 \) cm.
2. Let DC = \( x \), so AB = \( 3x \). The slant side BC spans a horizontal gap of \( 3x - x = 2x \) and a vertical gap of 6, so \( BC = \sqrt{4x^2 + 36} \).
3. Tangential condition: \( AB + DC = AD + BC \Rightarrow 4x = 6 + \sqrt{4x^2 + 36} \).
5. So DC = 4, AB = 12, and the area \( = \frac{1}{2}(12 + 4) \times 6 = 48 \) sq. cm.
Check: BC = \( \sqrt{64 + 36} = 10 \), and 12 + 4 = 6 + 10.
Hence, the answer is (C) 48.
Q4
Algebra - Simple Equations MULTIPLE CHOICE
If \( \left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) = 25 \) and \( x > 0 \), then the value of \( \left( x ^ { 7 } + \frac { 1 } { x ^ { 7 } } \right) \) is
Tip: All options differ only in the last digit, so compute exactly: \( 72 \times 623 = 44856 \), then subtract 3.
Hence, the answer is (C) \( 44853\sqrt{3} \).
Q5
Arithmetic - Profit & Loss MULTIPLE CHOICE
The monthly sales of a product from January to April were 120, 135, 150 and 165 units, respectively. The cost price of the product was Rs. 240 per unit, and a fixed marked price was used for the product in all the four months. Discounts of 20%, 10% and 5% were given on the marked price per unit in January, February and March, respectively, while no discounts were given in April. If the total profit from January to April was Rs. 138825, then the marked price per unit, in rupees, was
2. Units sold \( = 120 + 135 + 150 + 165 = 570 \), so total cost \( = 570 \times 240 = 136800 \).
3. Revenue \( = 136800 + 138825 = 275625 \).
4. \( M = \frac{275625}{525} = 525 \).
Shortcut: Revenue must be 525 times an option; \( 525 \times 525 = 275625 \) matches at once.
Hence, the answer is (A) 525.
Q6
Arithmetic - Time, Speed & Distance MULTIPLE CHOICE
Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was
Key idea: After the stop, the time lost must be made up on the remaining distance \( D \) by the extra speed.
1. Let the initial speed be \( v \). With a 20-minute stop and speed \( v + 3 \):
\[ \frac{D}{v} - \frac{D}{v+3} = \frac{1}{3} \Rightarrow \frac{3D}{v(v+3)} = \frac{1}{3} \Rightarrow D = \frac{v(v+3)}{9}. \]
2. With a 30-minute stop (same point) and speed \( v + 5 \):
\[ \frac{5D}{v(v+5)} = \frac{1}{2} \Rightarrow D = \frac{v(v+5)}{10}. \]
3. Equating: \( 10(v + 3) = 9(v + 5) \Rightarrow v = 15 \) km/h.
Check: \( D = \frac{15 \times 18}{9} = 30 \) km, normally 2 h; at 18 km/h it takes 1 h 40 min, saving exactly 20 min. At 20 km/h it takes 1.5 h, saving 30 min.
Hence, the answer is (D) 15.
Q7
Algebra - Logarithms MULTIPLE CHOICE
The sum of all possible real values of \( x \) for which \( \log _ { x - 3 } \left( x ^ { 2 } - 9 \right) = \log _ { x - 3 } ( x + 1 ) + 2 \), is
4. \( x = \frac{3 \pm \sqrt{33}}{2} \). Since \( \sqrt{33} \approx 5.74 \), only \( \frac{3 + \sqrt{33}}{2} \approx 4.37 \) satisfies \( x > 3, x \ne 4 \); the other root is negative.
So the sum of all valid values is \( \frac{3 + \sqrt{33}}{2} \).
Tip: Option (D) 3 is the trap of adding both roots without checking the domain.
Hence, the answer is (C) \( \frac{3+\sqrt{33}}{2} \).
Q8
Arithmetic - Ratio, Proportion & Variation TYPE IN THE ANSWER
The ratio of the number of coins in boxes A and B was 17:7. After 108 coins were shifted from box A to box B, this ratio became 37:20. The number of coins that needs to be shifted further from A to B, to make this ratio 1:1, is
1. Let A = \( 17k \), B = \( 7k \). After shifting 108:
\[ \frac{17k - 108}{7k + 108} = \frac{37}{20} \Rightarrow 340k - 2160 = 259k + 3996 \Rightarrow 81k = 6156 \Rightarrow k = 76. \]
2. Initially A = 1292, B = 532, total = 1824. After the shift, A = 1184, B = 640.
3. For a 1:1 ratio each box needs \( 1824 / 2 = 912 \), so A must give \( 1184 - 912 = 272 \) more.
Tip: Whenever items only move between two containers, fix the total first. Every later ratio is just a split of that total.
Hence, the answer is 272.
Q9
Algebra - Surds & Indices TYPE IN THE ANSWER
If \( 12 ^ { 12 x } \times 4 ^ { 24 x + 12 } \times 5 ^ { 2 y } = 8 ^ { 4 z } \times 20 ^ { 12 x } \times 243 ^ { 3 x - 6 } \), where \( x , y \) and \( z \) are natural numbers, then \( x + y + z \) equals
Teams A, B, and C consist of five, eight, and ten members, respectively, such that every member within a team is equally productive. Working separately, teams A, B, and C can complete a certain job in 40 hours, 50 hours, and 4 hours, respectively. Two members from team A, three members from team B, and one member from team C together start the job, and the member from team C leaves after 23 hours. The number of additional member(s) from team B, that would be required to replace the member from team C, to finish the job in the next one hour, is
Key idea: Work out the rate of one member of each team, measured in "jobs per hour".
1. One A member: \( \frac{1}{40 \times 5} = \frac{1}{200} = \frac{2}{400} \). One B member: \( \frac{1}{50 \times 8} = \frac{1}{400} \). One C member: \( \frac{1}{4 \times 10} = \frac{1}{40} = \frac{10}{400} \).
2. Starting group rate: \( 2 \times \frac{2}{400} + 3 \times \frac{1}{400} + \frac{10}{400} = \frac{17}{400} \) per hour.
3. In 23 hours: \( \frac{391}{400} \) done, leaving \( \frac{9}{400} \) to finish in 1 hour.
4. Without the C member the group does \( \frac{7}{400} \) per hour, short by \( \frac{2}{400} \), which is the work of 2 B members.
Tip: Taking total work = 400 units makes every rate a whole number: A = 2, B = 1, C = 10 units per member per hour.
Hence, the answer is (A) 2.
Q11
Algebra - Quadratic Equations MULTIPLE CHOICE
If \( f ( x ) = \left( x ^ { 2 } + 3 x \right) \left( x ^ { 2 } + 3 x + 2 \right) \), then the sum of all real roots of the equation \( \sqrt { f ( x ) + 1 } = 9701 \), is
2. Case \( y + 1 = 9701 \): \( x^2 + 3x - 9700 = 0 \). Its discriminant is positive, so it has two real roots with sum \( -3 \).
3. Case \( y + 1 = -9701 \): \( x^2 + 3x + 9702 = 0 \). Discriminant \( 9 - 4 \times 9702 < 0 \), so no real roots.
So the sum of all real roots is \( -3 \).
Tip: \( x^2 + 3x \ge -\frac{9}{4} \) for every real \( x \), so \( y = -9702 \) is impossible without computing anything.
Hence, the answer is (B) \( -3 \).
Q12
Arithmetic - Average MULTIPLE CHOICE
The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
2. A 5% rise in the average means the total rises by \( 0.05 \times 18{,}00{,}000 = 90{,}000 \).
3. This equals 20% of the managers' total, so the managers' total \( = \frac{90000}{0.2} = 4{,}50{,}000 \).
4. Engineers' total \( = 18{,}00{,}000 - 4{,}50{,}000 = 13{,}50{,}000 \), so their average \( = \frac{1350000}{25} = 54{,}000 \).
Hence, the answer is (A) 54000.
Q13
Algebra - Simple Equations TYPE IN THE ANSWER
In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is
Key idea: Eliminate commerce using the head-count equation, then apply the constraint science ≤ arts.
1. Let the counts be \( s, a, c \): \( s + a + c = 1500 \) and \( 1100s + 1000a + 800c = 1550000 \).
2. Subtract 800 times the first equation: \( 300s + 200a = 350000 \Rightarrow 3s + 2a = 3500 \).
3. Since \( a \ge s \): \( 3500 = 3s + 2a \ge 5s \Rightarrow s \le 700 \).
4. \( s = 700 \) gives \( a = 700 \) and \( c = 100 \), which is valid. Fee check: \( 770000 + 700000 + 80000 = 1550000 \).
Hence, the answer is 700.
Q14
Modern Math - Sets MULTIPLE CHOICE
In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is half the number of students who chose both physics and mathematics. The maximum possible number of students who chose physics but not mathematics, is
Key idea: Use the union formula to link the pairwise overlaps with the triple overlap, then minimise the P∩M overlap.
1. Let \( n(P \cap C) = n(C \cap M) = x \) and \( n(P \cap M) = 2x \), each including the \( t \) students who chose all three.
2. \( 150 = 75 + 111 + 40 - 4x + t \Rightarrow t = 4x - 76 \).
3. Physics but not maths \( = 75 - 2x \), so we want the smallest \( x \).
4. \( t \ge 1 \Rightarrow 4x \ge 77 \Rightarrow x \ge 20 \) (and \( t \le x \) gives \( x \le 25 \)). Take \( x = 20, t = 4 \).
5. Check that every region is non-negative: P∩M only = 36, P∩C only = 16, C∩M only = 16, all three = 4, only P = 19, only M = 55, only C = 4. Total = 150. ✔
So the maximum is \( 75 - 40 = 35 \).
Hence, the answer is (B) 35.
Q15
Algebra - Functions & Graphs MULTIPLE CHOICE
For real values of \( x \), the range of the function \( f ( x ) = \frac { 2 x - 3 } { 2 x ^ { 2 } + 4 x - 6 } \) is
3. \( (8y - 1)(2y - 1) \ge 0 \Rightarrow y \le \frac{1}{8} \) or \( y \ge \frac{1}{2} \).
4. Checks: \( y = 0 \) occurs at \( x = \frac{3}{2} \); \( y = \frac{1}{8} \) at \( x = 3 \); \( y = \frac{1}{2} \) at \( x = 0 \). The excluded points \( x = 1, -3 \) never satisfy the quadratic (substituting them gives 1 = 0 and 9 = 0), so nothing is lost.
So the range is \( \left(-\infty, \frac{1}{8}\right] \cup \left[\frac{1}{2}, \infty\right) \).
Shortcut: \( f(0) = \frac{-3}{-6} = \frac{1}{2} \) already rules out (A) and (D).
Hence, the answer is (B).
Q16
Arithmetic - Time, Speed & Distance TYPE IN THE ANSWER
Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is
Check: 221 leaves remainder 5 on division by 9, and so does the number: \( 1 + 1 - 123 \equiv -121 \equiv 5 \pmod 9 \). Option (C) 212 also leaves 5, so the full count is still needed.
Hence, the answer is (B) 221.
Q18
Geometry - Triangles MULTIPLE CHOICE
In \( \triangle A B C , A B = A C = 12 \mathrm {~cm} \) and \( D \) is a point on side \( B C \) such that \( A D = 8 \mathrm {~cm} \). If \( A D \) is extended to a point \( E \) such that \( \angle A C B = \angle A E B \), then the length, in cm , of \( A E \) is
Key idea: Since AB = AC, \( \angle ABC = \angle ACB = \angle AEB \), which makes triangles ABD and AEB similar.
1. In triangles ABD and AEB: \( \angle BAD = \angle EAB \) (common angle, as D lies on AE) and \( \angle ABD = \angle ABC = \angle ACB = \angle AEB \).
2. So \( \triangle ABD \sim \triangle AEB \) (AA), giving \( \frac{AB}{AE} = \frac{AD}{AB} \).
Tip: Equivalently, \( \angle AEB = \angle ACB \) means A, B, E, C lie on one circle, and the result \( AD \cdot AE = AB^2 \) is a standard fact for an isosceles triangle's circumcircle.
Hence, the answer is (C) 18.
Q19
Arithmetic - Time & Work MULTIPLE CHOICE
The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to
Key idea: Measure the tank in units of pipe A's flow.
1. Let the flow rates be 4, 9 and 36 units per hour. A alone takes 15 h, so the tank holds \( 4 \times 15 = 60 \) units.
2. All three together: \( 4 + 9 + 36 = 49 \) units per hour.
3. Time \( = \frac{60}{49} \) h \( = \frac{3600}{49} \) min \( \approx 73.47 \) min.
The nearest option is 73 minutes.
Hence, the answer is (A) 73.
Q20
Algebra - Number Theory MULTIPLE CHOICE
For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is
Key idea: Subtracting the two sum equations links the thousands digit and the units digit.
1. Let the number be \( \overline{abcd} \). Then \( a + b + c = 15 \) and \( b + c + d = 16 \), so \( d = a + 1 \).
2. Also \( c = d + 6 \), so \( d \le 3 \). Since \( a \ge 1 \), \( d \ge 2 \).
3. \( d = 2 \): \( a = 1, c = 8, b = 15 - 1 - 8 = 6 \), giving 1682.
4. \( d = 3 \): \( a = 2, c = 9, b = 15 - 2 - 9 = 4 \), giving 2493.
5. Difference \( = 2493 - 1682 = 811 \).
Hence, the answer is (C) 811.
Q21
Algebra - Inequalities & Modulus TYPE IN THE ANSWER
Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose value lies between 150 and 500. If p is not less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is
Key idea: \( p + q = 900 - r \). The largest \( p \) comes from the smallest \( r \) with \( p = 0.7q \), and the smallest \( p \) from the largest \( r \) with \( p = 0.3q \).
1. Perfect squares between 150 and 500: \( 13^2 = 169 \) up to \( 22^2 = 484 \).
2. Maximum: \( r = 169 \Rightarrow p + q = 731 \). \( p \le 0.7(731 - p) \Rightarrow 1.7p \le 511.7 \Rightarrow p \le 301 \). With \( p = 301 \), \( q = 430 \) and \( 0.7 \times 430 = 301 \). ✔
3. Minimum: \( r = 484 \Rightarrow p + q = 416 \). \( p \ge 0.3(416 - p) \Rightarrow 1.3p \ge 124.8 \Rightarrow p \ge 96 \). With \( p = 96 \), \( q = 320 \) and \( 0.3 \times 320 = 96 \). ✔
4. Sum \( = 301 + 96 = 397 \).
Tip: \( p \le 0.7q \) is the same as \( p \le \frac{7}{17}(p+q) \), and \( p \ge 0.3q \) is \( p \ge \frac{3}{13}(p+q) \). Here 731 = 17 × 43 and 416 = 13 × 32, so both bounds are whole numbers.
Hence, the answer is 397.
Q22
Arithmetic - Mixture, Alligation, Removal & Replacement TYPE IN THE ANSWER
Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is
Check: water in A \( = \frac{960}{76} \) and alcohol \( = \frac{3600}{76} \), and \( 3600 : 960 = 15 : 4 \). ✔
Shortcut: For any one-round swap of \( v \) litres between equal volumes \( V \) of pure liquids, the final ratio in A is \( V : v \). Here \( 60 : v = 15 : 4 \) gives \( v = 16 \) directly.
