Let the digits of the 3-digit number be p, q, &amp; r.2 &lt; p × q × r &lt; 7Therefore, p × q × r can take the values 3, 4, 5, or 6. Let's start with prime numbers 3 &amp; 5.Since they are prime, they can't be splitted, and hence if one of p,q or r is 3, the remaining two should be 1. So, the possible combinations are1, 1, 31, 3, 13, 1, 11, 1, 51, 5, 15, 1, 1 4 can be splitted as 2 × 2. Therefore, the possible combinations of p, q, r are1, 1, 41, 4, 14, 1, 11, 2, 22, 1, 22, 2, 16 can be split as 3 × 2. Therefore, the possible combinations of p, q, r are1, 1, 61, 6, 16, 1, 1 1, 2, 3 will also yield a product of 6. We can 3! = 6 combinations of p, q, r with 1, 2, 31, 2, 31, 3, 22, 1, 32, 3, 13, 1, 23, 2, 1 Therefore, the total number of possibilities are 3 + 3 + 3 + 3 + 3 + 6 = 21Hence, the answer is, "21"

Given f (5 + x) = f (5 – x)When x = 1, f(6) = f(4)When x = 2, f(7) = f(3)Imagine a graph and while joining (6,4) and (7,3) it looks symmetrical about 5Given 4 distinct real roots. Assume 12 as one of the roots.f(12) = 0 --&gt; Then f(5 + 7) = 0 which is same as f(5 – 7) = f(-2) = 0So the roots are in the form (5 + k) and (5 – k) {This is one pair}Sum of the roots = 5 + k + 5 – k = 10We have 4 roots, So two pairs =10 + 10 = 20Hence, the answer is, "20"

Let assume its equal after N yearsVeeru’s investment after N years = 10000 + 10000 × $$\frac{5}{100}$$ × NJoy’s investment after N years = 8000 + 8000 ×&nbsp;$$\frac{10}{100}$$ × (N – 2)Equating both10000 + 10000 × $$\frac{5}{100}$$ × N = 8000 + 8000 × $$\frac{10}{100}$$ ×(N – 2)10000 + 500N = 8000 + 800N – 16003600 = 300NN = 12The answer is, " 12"

Let’s take speed of the train to be x and the time taken be tSince speed is reduced to $$\frac{1}{3}$$ rd,New speed = $$\frac{x}{3}$$Since the speed is one-third, time taken will be tripled. T = 3tThis 3t is after the scheduled time, So extra 2t = 30 mintuest = 15 minutesTrain travels at x km/hr takes 15 minutes andTrain travels at $$\frac{x}{3}$$ km/hr takes 45 minutes So, the train usually takes 15 minutes to cover the distance.It travels 5 minutes at the usual speed. That is, it travels $$\frac{1}{3}$$ rd of the time at the usual speed. So it covers $$\frac{1}{3}$$ rd of the distance in 5 minutes. To reach it's destination in the on time, the train has to travel the remaining $$\frac{2}{3}$$ rds of the distance in 10 minutes. Since the train halts for 4 minutes, it should now cover the $$\frac{2}{3}$$ rds of the distance in 6 (10 - 4) minutes.In other words, the train has to cover the same distance in $$\frac{6}{10}$$ th of the usual time.In order to do so, the speed must be&nbsp;$$\frac{10}{6}$$ ths of the usual speed. Or the increased speed will be $$\frac{4}{6}$$ ths or $$\frac{2}{3}$$ rds of the usual speed. Which is an increase of 66.66% or nearly 67%.The answer is, "67"

log45 = log4y × log6√5log45 ÷ log4y = logy5 = log6√5 = k (Let’s assume this as k)6k&nbsp;= √5 ---&gt; On squaring this {6k}2&nbsp;= 5 log45 = log4y × log6√5logy5 = log6√5logy5 = kyk&nbsp;= 5 {6k}2&nbsp;= yk&nbsp;= 5{62}k&nbsp;= yk36k&nbsp;= yky = 36.Hence, the answer is, " 36"

2x&nbsp;+ 2-x&nbsp;is of the form y + $$\frac{1}{y}$$Minimum value the expression can take is ≥ 2So 2x&nbsp;+ 2-x&nbsp;≥ 2Now, 2 – (x – 2)2&nbsp;----&gt; (x – 2)2&nbsp;is ≥ 0So 2 – (x – 2)2 ≤ 2 (From 2 we are subtracting a non-negative number)Maximum value the expression can have is 2The only possibility is both sides are = 2If value = 2Then x = 020&nbsp;+ 20&nbsp;= 2 – (0 – 2)22 ≠ – 2There is no case where LHS = RHS = 2Hence 0 real-valued solutionsHence, the answer is, "0"

Let the meeting point be D. Let’s assume C1 takes t minutes to reach the meeting point. Then C2 takes same t minutes to reach the meeting point.<img src="https://online.2iim.com/CAT-question-paper/CAT-2020-Question-Paper-Slot-1-Quant/figa/q7_meeting_points.png">After reaching the destinations...C1 takes t minutes to cover AD, while C2 takes 20 minutesC1 takes 45 minutes to cover DB, while C2 takes t minutes the ratio of speeds of C1 and C2 on the stretch AD will be equal to the ratio of their speed on the stretch DB.When the distances are the same, the ratio of speeds is the inverse ratio of time taken.ratio of speeds of C1 and C2 on the stretch AD = $$\frac{20}{t}$$ratio of speeds of C1 and C2 on the stretch DB = $$\frac{t}{45}$$Therefore, $$\frac{20}{t}$$ = $$\frac{t}{45}$$t × t = 45 × 20t&nbsp;2&nbsp;= 900t = 30.$$\frac{Speed of C1}{Speed of C2}$$ = $$\frac{20}{t}$$$$\frac{60}{Speed of C2}$$ = $$\frac{20}{30}$$Speed of C2 = 90 kmphThe answer is, "90"

a + $$\frac{b + c}{2}$$ = 5 ---&gt; 2a + b + c = 10b +&nbsp;$$\frac{a + c}{2}$$ = 7 ---&gt; 2b + a + c = 14Solving these two, we get b – a = 4b = a + 4Substituting this in the first equation2a + a + 4 + c = 103a + c = 6Given all three as positive integers, maximum value we can get for a = 1Ex: When substituting a = 2, c = 0 (Not possible)So a = 1, then b = 5Sum of a + b = 6Hence, the answer is, "6"

642&nbsp;= 4096x = (4096)7 + 4√3x = (642)7 + 4√3√x = (64)7 + 4√3√x( 1 ÷ (7 + 4√3))&nbsp;= 641 ÷ (7 + 4√3) = $$\frac{1}{7 + 4√3}$$ = $$\frac{1 × (7 - 4√3)}{(7 + 4√3) × (7 - 4√3)}$$ = $$\frac{7 - 4√3}{49 - 48}$$ = 7 - 4√3 √x( 1 ÷ (7 + 4√3))&nbsp;= 64√x(7 - 4√3)&nbsp;= 64x(7/2 - (4√3)/2)&nbsp;= 64x(7/2 - 2√3)&nbsp;= 64$$\frac{x^{7/2}}{x^{2√3}}$$Hence, the answer is, "$$\frac{x^{7/2}}{x^{2√3}}$$"

Even numbers so b = 01100 1122 1144 1166 11882200 2222 2244 … …...9900 … … … …By adding$$\frac{5(1100 + ... + 9900) + 9(22 + 44 + 66 + 88)}{45}$$ {Since the term is common 5 times and 9 times}$$\frac{(5 × 100 × 11 × 45) + 9 × 22(1 + 2 + 3 + 4)}{45}$$= 5544Hence, the answer is, "5544"

x +&nbsp;$$\frac{1}{x}$$&nbsp;= yy2&nbsp;– 3y + 2 = 0y ≥ 2 &amp; y ≤ - 2(y – 1) (y – 2) = 0y = 1(Not possible) or y = 2Now, x +$$\frac{1}{x}$$&nbsp;= 2This has only one option x = 1. So, only one real rootHence, the answer is, " 1"

D = Price of Desktop and L = Price of Laptop0.2 D – 0.1 L = 2 % of 500000.2 D – 0.1 L = 10002D – L = 10000D + L = 50000Solving D = 20000The answer is, " 20000"

x0&nbsp;= 100 (If all 100 had their birthdays on January)If x0&nbsp;= 3 (Not possible, because all x1&nbsp;,.....x12&nbsp;together will be only 36)So x0&nbsp;will be lesser when the numbers are as close as possible$$\frac{100}{12}$$ =8.5So if we add 8 (12 times) = 96So, need to have 8,8,8,8,8,8,8,8,9,9,9,9 (Adding these 12 we will get 100)x0&nbsp;= 9The answer is, "9"

Two person to be A and BA = 2km/hr and B = 4 km/hrSpeed of the train = t km/hrGiven, Train crosses A in 90 seconds (Length of the train is the distance)t – 2 in 90 seconds and t – 4 in 100 secondsRatio of speed is opposite to time(t – 2) 90 = (t - 4) 100Solving t = 22km/hrFor A, Speed = t – 2 = 20 km/hr in 90 secondsSo travelling at 22 km/hr ----&gt; $$\frac{90 × 20}{22}$$ &nbsp;≅ 82The answer is, "82"

(x2-7x+11)($$x^{2}$$-13x+42)&nbsp;= 1Only possible thing is x^2-13x+42 = 0Or x2-7x+11 = 1Solving these two x2-13x+42 = 0(x – 6) (x – 7) = 0x = 6 or x = 7x2-7x+10 = 0(x – 2) (x – 5) = 0x = 2 or x = 5At this moment we have 4 values possible, But there is one more way we can arrive this(-1)Even&nbsp;= 1So, x2-7x+11 = -1(x – 3) (x – 4) = 0x = 3 or x = 4So, 4 + 2 = 6 valuesHence, the answer is, "6"

|x| – y = 1y = |x| – 1<img src="https://online.2iim.com/CAT-question-paper/CAT-2020-Question-Paper-Slot-1-Quant/figa/q16_modulus_graphs.png">We got a trapezium with h = 1And parallel sides = 2 and 4=&nbsp;$$\frac{1}{2}(2+4)×1$$= 3 square unitsHence, the answer is, " 3"

When we consider top part of the triangleHeight is = $$\frac{1}{3}$$ rd of original height<img src="https://online.2iim.com/CAT-question-paper/CAT-2020-Question-Paper-Slot-1-Quant/figa/q17_right_circular_cone.png" alt="Right Circular Cone">If height becomes&nbsp;$$\frac{1}{3}$$rd then radius also becomes&nbsp;$$\frac{1}{3}$$rdIf both height and radius become&nbsp;$$\frac{1}{3}$$rd, Volume will be $$\frac{1}{27}$$thSo, Top part’s volume = VRemaining part = 26VTotally = 27VGiven 26 V – V = 225V = 927 V = 243The answer is, "243"

Diagonals of a rhombus are perpendicular bisectors of each otherIf diagonals are 12 and 16Half of diagonals are 6 and 8 third side will become 10 = side of rhombusRadius of the circle will be perpendicular to the side of the rhombusBecause side of the rhombus is tangent to the circleThe triangle is taken by half a diagonal (6) and (8) from the original figure<img src="https://online.2iim.com/CAT-question-paper/CAT-2020-Question-Paper-Slot-1-Quant/figa/q18_triangle.png" alt="triangle"><img src="https://online.2iim.com/CAT-question-paper/CAT-2020-Question-Paper-Slot-1-Quant/figa/q18_rhombus.png" alt="rhombus">Just equating the areas$$\frac{1}{2}$$ × 8 × 6 = $$\frac{1}{2}$$ × 10 × hh =&nbsp;$$\frac{24}{5}$$ = same as radius of the circle (r)Now, area of circle : area of rhombusπ × $$\frac{24}{5}$$ × $$\frac{24}{5}$$ : $$\frac{1}{2}$$ × 12 × 166π : 25Hence, the answer is, "$$\frac{6π}{25}$$"

Speed are 8 km/hr and 15 km/hrSo it takes 15t times and 8t timesDifference of times = 15t – 8t = 7t7t = 35 minutes (10:15 min – 9:40 min)t = 5 minutesHe takes 75 minutes by travelling at 8 km/hr and 40 minutes if travelled at 15 km/hrHe starts at 9 am.So, 9:10 to 10:00 = 50 minutes15 km/hr = 40 minAnd at what speed he should travel, so as to reach there within 50 minutes\$$\frac{15 × 40}{50}$$ =12 km/hrThe answer is, "12"

Since the product is involved, we will keep the numbers as close as possiblebc = 96 and c &lt; 9b c12 × 8 = 9616 × 6 = 9624 × 4 = 96ab = 432a b12 × 36 = 43216 × 27 = 43224 × 18 = 432 So possible values of a = 36, b = 12, c = 8 Sum = 58a = 27, b = 16, c = 6 Sum = 49a = 18, b = 24, c = 4 Sum = 46Least possible value = 46Hence, the answer is, "46"

Taking log on both sidesy2log35 log 2 = log23 log 5Choosing the base to be 3y2&nbsp;log35 log32 = log23 log35y2&nbsp;log32 = log23 (Cancelling log35)y2&nbsp;= log23/log32y2&nbsp;= (log23)2y = - log23 (Given y is a negative number)y = log2$$\frac{1}{2}$$Hence, the answer is, "log2&nbsp;(1/3)"

Total area = 5 parts<img src="https://online.2iim.com/CAT-question-paper/CAT-2020-Question-Paper-Slot-1-Quant/figa/q22_circle_area.png">So, painted area = 135 ×&nbsp;$$\frac{3}{5}$$&nbsp;= π r2Solving r =&nbsp;$$\frac{9}{√π}$$Substituting r in l =&nbsp;$$\frac{135}{2r}$$l =&nbsp;$$\frac{15√π}{2}$$Perimeter of rectangle = 2(l + b)= 2($$\frac{15√π}{2}$$&nbsp;+&nbsp;$$\frac{18}{√π}$$)= 3√π (5 +&nbsp;$$\frac{12}{π}$$)The answer is, "3√π (5 +&nbsp;$$\frac{12}{π}$$)"

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